136 free, high-quality past-paper-style questions on Gravitational fields for IB Physics SL, each with a full mark scheme. Covers: Newton's law of gravitation; Kepler's laws; Gravitational field lines; and more. For when you have finished the past papers, or want to practise one topic at a time.D.1 at Higher Level →Read the D.1 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.659 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1D-1A-01Newton's law of gravitationPaper 1AEasy1 mark
Two small spheres, each of mass 5.0 kg, have their centres 0.20 m apart. G = 6.67 × 10−11 N m² kg−2.
Which statement about the planets of the Solar System agrees with Kepler's laws of orbital motion?
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Step 1Kepler's first law: each orbit is an ellipse with the Sun at one focus. (The second law implies the speed varies; the third law, T² ∝ r³, gives longer periods farther out.)
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Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: this is Kepler's first law.
BThe Sun is at a focus, not at the centre. Only for a circle do the focus and the centre coincide.
CEqual areas in equal times means the planet moves faster when it is nearer the Sun, so its speed is not constant on an ellipse.
DThis inverts Kepler's third law: T² ∝ r³, so the period increases with distance from the Sun.
3D-1A-05Gravitational field linesPaper 1AEasy1 mark
Which diagram best represents the gravitational field lines around an isolated, uniform spherical planet?
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Step 1The force on a small test mass points straight towards the centre of a uniform sphere, so the lines are radial and point inwards; being radial, they crowd together near the surface where the field is strongest.
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Answer C
Answer: C · 1 stage of work, one mark
Every option, and why
AThe arrows point outwards. Gravity is always attractive, so the field points towards the planet.
BClosed loops are a feature of magnetic fields. Gravitational field lines end on the mass.
CCorrect: radial lines pointing inwards, closer together near the surface where the field is strongest.
DThe lines curve round the planet. For a uniform sphere the force on a test mass points straight at the centre, so the lines must be radial.
7D-1A-61Gravitational field linesPaper 1AEasy1 mark
The diagram shows some of the gravitational field lines around two asteroids X and Y. A small rock is momentarily at rest at point P, which lies on one of the field lines.
Which arrow shows the direction of the acceleration of the rock at P?
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Step 1The acceleration of the rock is in the direction of the gravitational force on it, which is the direction of the resultant field g at P.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2The direction of the resultant field at a point is along the tangent to the field line through that point, in the direction of the arrows on the line: arrow B.
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Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AArrow A points straight at the centre of X. That would be the direction of the field of X alone; the field of Y also acts at P, so the resultant is along the field line, not towards X.
BCorrect: the resultant field, and so the acceleration, is along the tangent to the field line at P, in the direction shown by the arrows on the line.
CArrow C points straight at the centre of Y. The field of Y alone would point this way, but the field of X also acts at P, and X is the more massive body (more field lines end on it).
DArrow D is perpendicular to the field line. A gravitational force acts along the field line, never across it.
8D-1A-67Newton's law of gravitationPaper 1AEasy1 mark
Two small satellites fly in formation with their centres 2.0 m apart. The gravitational force between them is F. One satellite is replaced by another of twice the mass, and the separation of the centres is increased to 6.0 m.
What is the new gravitational force between the satellites?
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Step 1F ∝ m1m2/r2: the product of the masses doubles and the separation is multiplied by 3.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2New force = F × 2/32 = 2F/9.
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: the force is multiplied by 2 for the mass and divided by 32 = 9 for the separation, giving 2F/9.
BThis doubles both masses, multiplying the product by 4. Only one of the satellites was replaced.
CThe separation has not been squared. Newton’s law is an inverse-square law, so tripling the separation divides the force by 9.
DThis multiplies by the square of the distance ratio instead of dividing by it. A larger separation must give a smaller force.
9IBP2S-028-03-01Gravitational field linesPaper 1AEasy1 mark
A groundskeeper models the gravitational field over a football stadium, from pitch level up to the roof 40 m above. Which description of the field lines in this region is most appropriate?
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A 40 m region is tiny compared with Earth's radius (about 6.4×106 m). The radial lines are therefore almost parallel, and their spacing barely changes. The field is effectively uniform: parallel, equally spaced lines pointing towards Earth (downwards). Field lines show the direction of the force on a mass, so they cannot point upwards. Converging lines would describe a point mass at pitch level, not Earth.
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Answer B
Answer: B
Exam tip: Over regions much smaller than a planet's radius, draw parallel, equally spaced lines. Equal spacing means equal field strength.
10IBP2S-028-03-02Variation of g with distancePaper 1AEasy1 mark
A proposed space habitat is a thin, uniform spherical shell of mass M and radius R. What is the magnitude of the gravitational field strength due to the shell at a point a distance r from its centre, where r > R?
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Outside the shell, a spherically symmetric mass distribution produces the same field as a point mass at its centre. So g = GM/r2, with r measured from the centre. Being hollow only matters for points inside the shell. Option B measures the distance from the surface, and option D wrongly treats the field as uniform.
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Answer A
Answer: A
Common error: In Newton's law, distances are always measured between centres, never from a surface.
11IBP2S-028-03-08Extended bodies as point massesPaper 1AEasy1 mark
A survey probe skims 10 km above the surface of a uniform spherical moon of radius 1700 km. The field strength at the probe is calculated using g = GM/r2, where M is the moon's mass. Which value of r should be used?
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Outside a spherically symmetric body, its field is the same as that of a point mass at its centre. So r is the distance from the centre: 1700 km + 10 km = 1710 km. Using 10 km measures from the surface. Using 1700 km ignores the height, and using 1690 km subtracts the height instead of adding it.
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Answer D
Answer: D
Common error: Even when the height is small, the point-mass model needs the distance from the centre of the body.
12IBP2S-028-03-09Gravitational field linesPaper 1AEasy1 mark
An asteroid has an irregular, elongated shape. What happens to its gravitational field lines at distances very much larger than its length?
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At distances much larger than the body's size, the details of its shape stop mattering. Any finite mass then behaves like a point mass at its centre of mass. The field lines become radial and point inwards, with a 1/r2 field. Parallel lines describe a uniform field near a large surface. The field weakens with distance but never becomes exactly zero.
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Answer A
Answer: A
Exam tip: Two situations justify the point-mass model: a spherically symmetric body, or a distance much larger than the body's size.
13IBP2S-028-01-01Kepler's second lawPaper 1AEasy1 mark
A space probe moves in an elliptical orbit around Jupiter. Its closest distance from the centre of Jupiter is 5×108 m and its farthest distance is 5×109 m. What is the ratio of the probe's speed at closest approach to its speed at the farthest point?
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By Kepler's second law the line from Jupiter to the probe sweeps equal areas in equal times. At the two extreme points the velocity is perpendicular to the radius, so the area swept in a short time Δt is ½rvΔt. This makes rv the same at both points. The speed ratio is therefore 5×109 / 5×108 = 10.
Why the other options are wrong: • \(\sqrt{10}\) uses the circular-orbit result v ∝ 1/√r, which does not apply here. • 100 wrongly treats speed like an inverse-square quantity. • 0.1 inverts the ratio.
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Answer B
Answer: B
Common error: Do not use the circular-orbit relation v ∝ 1/√r for a single elliptical orbit. At perihelion and aphelion, use r × v = constant.
14IBP2S-028-01-02Geostationary orbitsPaper 1AEasy1 mark
An engineering team is planning to place a weather satellite in a geostationary orbit around Earth. Which of the following is not a requirement for the orbit to be geostationary?
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A geostationary satellite must do three things: • orbit above the equator, • move west to east, • have a period equal to one rotation of Earth.
For a circular orbit, GMm/r² = mω²r. The satellite mass m cancels, so the orbital radius and period depend only on Earth's mass. A satellite of any mass can therefore be geostationary.
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Answer D
Answer: D
Exam tip: Whenever you set gravitational force equal to centripetal force, the orbiting mass cancels. Orbits depend on the central mass only.
15IBP2S-028-01-06Geostationary orbitsPaper 1AEasy1 mark
A telecommunications satellite is in geostationary orbit directly above an office building on the equator. Which of the following correctly compares the satellite with the building?
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Both the satellite and the building complete one revolution in one sidereal day, so they have the same angular speed ω.
Linear speed is v = ωr. The satellite's orbital radius (about 4.2×107 m) is much larger than Earth's radius, so the satellite has a greater linear speed.
Why the other options are wrong: • Option A wrongly assumes that 'staying above the same point' means equal speed. • Options C and D confuse angular speed with linear speed.
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Answer B
Answer: B
Common error: Being 'stationary relative to the ground' means equal angular speed, not equal linear speed.
16IBP2S-028-00-01Newton's law of gravitationPaper 1AEasy1 mark
Two uncrewed spacecraft drift near each other in deep space. Each behaves as a point mass, and the gravitational force between them is F. One spacecraft then takes on cargo, so its mass doubles, and the separation between them becomes three times larger. What is the new gravitational force between them?
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F = Gm1m2/r2. Doubling one mass multiplies F by 2. Tripling the separation multiplies F by 1/32 = 1/9. The new force is 2/9 F. Answer 2/3 F treats the force as inversely proportional to r, not r2. 18F multiplies by r2 instead of dividing. 4/9 F wrongly squares the mass factor as well.
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Answer A
Answer: A
Common error: Only the separation is squared in Newton's law. Each mass appears to the first power.
17IBP2S-028-00-02Combining two gravitational fieldsPaper 1AEasy1 mark
A space telescope orbits at a height above Earth's surface equal to one Earth radius. The gravitational field strength at Earth's surface is 9.8 N kg−1. What is the gravitational field strength at the telescope's position?
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Distance is measured from Earth's centre, so r = R + R = 2R. Since g ∝ 1/r2, g = 9.8/4 ≈ 2.5 N kg−1. Answer 9.8 comes from using the height (R) as the distance. Answer 4.9 uses a 1/r relationship. Answer 0 confuses the apparent weightlessness of orbit with zero field.
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Answer C
Answer: C
Exam tip: Always convert a height above the surface into a distance from the centre before using g = GM/r².
18IBP2S-028-00-05Gravitational field linesPaper 1AEasy1 mark
Which statement defines the gravitational field strength at a point?
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Gravitational field strength is g = F/m: the force per unit mass on a small test mass placed at the point. The test mass must be small so that it does not disturb the field. The first option defines gravitational potential (apart from its sign). The third option omits the 'per unit mass'. The fourth option refers to the source's own motion, not the field it creates.
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Answer B
Answer: B
Exam tip: A full definition includes 'per unit mass' and 'small test mass'. Leaving either out can lose the mark.
19IBP2S-028-00-10Gravitational field strengthPaper 1AEasy1 mark
A planetary rover of mass 1000 kg is sent to a planet with 0.10 times Earth's mass and 0.50 times Earth's radius. The gravitational field strength at Earth's surface is 9.8 N kg−1. What is the weight of the rover on the planet's surface?
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gplanet = 9.8 × 0.10/0.502 = 3.92 N kg−1, so W = 1000 × 3.92 ≈ 3.9 kN. Answer 2.0 kN divides by the radius ratio without squaring it. Answer 1.0 kN ignores the change in radius. Answer 9.8 kN is the rover's weight on Earth; the rover's mass is unchanged, but its weight is not.
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Answer D
Answer: D
Common error: Mass stays the same on every planet. Weight changes with the local value of g.
20IBP2S-028-02-01Variation of g with distancePaper 1AEasy1 mark
A mission team is planning a lander for an exoplanet. The exoplanet has twice the mass of Earth and twice the radius of Earth. The gravitational field strength at Earth's surface is 9.8 N kg−1. What is the gravitational field strength at the surface of the exoplanet?
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Surface field strength is g = GM/R2. Doubling M doubles g. Doubling R divides g by 22 = 4. The overall factor is 2/4 = 1/2, so g = 9.8/2 = 4.9 N kg−1. Option C assumes the two changes cancel, which ignores the square on R. Option D treats g as proportional to R. Option A divides by 2 twice.
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Answer B
Answer: B
Common error: Field strength depends on 1/R², not 1/R. Always square the radius factor before combining it with the mass factor.
21IBP2S-028-02-02Kepler's second lawPaper 1AEasy1 mark
A space probe tracks a comet as it moves along a highly elliptical orbit around the Sun. Which statement about the comet's motion is correct?
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Kepler's second law says that the line from the Sun to the comet sweeps out equal areas in equal time intervals. Near the Sun this line is short. To sweep the same area in the same time, the comet must cover a longer arc, so it moves fastest at perihelion. Option A confuses equal areas with equal speeds. Option B gets the trend backwards. Option D is wrong because in an ellipse the force is generally not perpendicular to the velocity, so the speed changes around the orbit.
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Answer C
Answer: C
Exam tip: Equal areas in equal times means the orbiting body moves fast when close to the central body and slow when far away.
22IBP2S-028-02-10Newton's law of gravitationPaper 1AEasy1 mark
Two small asteroids attract each other with a gravitational force F. Each asteroid gains mass through collisions until its mass is doubled, and their centre-to-centre separation also doubles. What is the new gravitational force between them?
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Newton's law of gravitation gives F = Gm1m2/r2. Doubling both masses multiplies F by 2 × 2 = 4. Doubling r divides F by 22 = 4. The overall factor is 4/4 = 1, so the force is unchanged. Option C doubles only one mass. Option D ignores the change in separation. Option A ignores the change in mass.
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Answer B
Answer: B
Exam tip: Deal with each change as a separate multiplying factor, then combine the factors at the end.
23IBP2S-028-04-01Weightlessness & free fallPaper 1AEasy1 mark
An astronaut on board a space station in a circular orbit 400 km above the Earth floats freely inside the cabin. Which statement best explains why the astronaut experiences apparent weightlessness?
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At 400 km the field strength is still about 8.7 N kg−1, roughly 90% of its surface value, so gravity is far from negligible. Gravity is the only significant force on both the astronaut and the station. It provides the centripetal force, so both fall towards the Earth with the same acceleration. Because there is no relative acceleration, the floor exerts no normal force on the astronaut, and so there is no sensation of weight. The centripetal force is not an extra outward force, and the Moon's field at that height is tiny.
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Answer C
Answer: C
Common error: The centripetal force is not an additional force. In orbit it is the gravitational force itself, directed towards the centre.
24IBP2S-028-04-02Geostationary orbitsPaper 1AEasy1 mark
A telecommunications company plans to launch a geostationary satellite. Which of the following is not a requirement for a geostationary orbit?
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For a circular orbit, GMm/r2 = mv2/r. The satellite mass m cancels, so the orbital radius and speed for a given period do not depend on the satellite's mass. Staying above one point on the ground does require three things: an equatorial orbit, a period equal to one Earth rotation, and motion in the same (west-to-east) direction as the Earth's rotation.
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Answer D
Answer: D
Exam tip: Whenever orbital speed, period or radius is involved, check whether the satellite's mass cancels. For circular orbits it always does.
25IBP2S-028-04-03Variation of g with distancePaper 1AEasy1 mark
A lander descends towards Mars, which has radius R. At what height above the Martian surface is the gravitational field strength one-ninth of its value at the surface?
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g ∝ 1/r2, where r is measured from the centre of Mars. For g to fall to one-ninth, r must be three times larger, so r = 3R. The height above the surface is therefore 3R − R = 2R. Option C gives the distance from the centre rather than the height. Option D assumes r = 9R, which comes from forgetting that the inverse-square law involves the square of r.
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Answer B
Answer: B
Common error: In Newton's law, r is measured from the centre of the planet, not from its surface. Subtract the radius to obtain the height.
26D-1A-02Gravitational field strengthPaper 1AMedium1 mark
Which statement about the gravitational field strength g at a point is correct?
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Step 1g = F/m is defined as force per unit mass on a small point mass. Doubling the test mass doubles F, so F/m is unchanged: the field belongs to the point, not to the test mass.
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Answer C
Answer: C · 1 stage of work, one mark
Every option, and why
AThis confuses the field strength with the force. The force doubles with the mass, but the force per unit mass does not.
BThe acceleration equals g only in free fall. If other forces act, for example a normal force, the acceleration is different.
CCorrect: g = F/m is the force per unit mass on a small test mass, so the test mass cancels.
DThe unit is right, but g is a vector. It has the direction of the force on a test mass, towards the attracting body.
32D-1A-30Weightlessness & free fallPaper 1AMedium1 mark
An astronaut in a space station 400 km above the Earth's surface floats freely inside the station.
Which statement is correct?
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Step 1At 400 km the field is still about 8.7 N kg−1, so the astronaut has weight.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2The astronaut and the station fall towards the Earth with the same acceleration, so no contact force acts between them: this is apparent weightlessness.
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Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThe field there is about 8.7 N kg−1, only about 11 % less than at the surface.
BThere is no outward force on the astronaut; the unbalanced gravitational force provides the centripetal acceleration.
CCorrect: gravity gives the astronaut and the floor the same acceleration, so no contact force acts between them.
DWeight is the gravitational force, which does not depend on whether there is air.
The planets of a star of mass M move in circular orbits. For each planet, T² = Kr³, where T is the orbital period, r is the orbital radius and K is a constant.
The moons of one of these planets also move in circular orbits and obey T² = K′r³. The mass of the planet is much smaller than M.
Which row gives K and compares K′ with K?
KComparison
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Step 1For a circular orbit the gravitational force provides the centripetal force: GMm/r² = m(4π²r/T²), so T² = (4π²/GM)r³ and K = 4π²/GM.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2For the moons the central mass is the planet's mass, which is smaller than M, so K′ = 4π²/(G × planet mass) is larger than K.
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: K = 4π²/GM, and because the constant is inversely proportional to the central mass, the smaller planet gives the larger constant K′.
BThe expression is right, but Kepler's constant is not universal: it depends on the mass of the central body.
CThis is r³/T², the reciprocal of the constant asked for; with that expression the comparison would indeed be K′ < K.
DThis inverts the expression and also assumes that the same constant applies to every orbiting system.
35D-1A-47Gravitational field linesPaper 1AMedium1 mark
Two stars of masses 4M and M are a fixed distance apart. The stars are drawn as dots of the same size.
Which diagram best represents the gravitational field lines in the region around the stars?
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Step 1Gravitational field lines point towards each mass, and the number of lines ending on a mass is proportional to the mass, so four times as many lines end on 4M as on M.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2At the null point G(4M)/x² = GM/(d − x)², so x = 2(d − x): the point where no line passes is one third of the separation from the smaller star. Only diagram C shows all three features.
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Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThe pattern is right, but the arrows point away from the stars; gravitational fields are always attractive, so the lines point towards the masses.
BThis is the pattern for two equal masses: equal numbers of lines and a null point midway between the stars.
CCorrect: inward lines, four times as many on 4M, and the null point nearer the smaller star.
DThe labels are the wrong way round for the pattern: the null point is shown nearer the larger star, and the larger star has fewer lines.
36D-1A-62Extended bodies as point massesPaper 1AMedium1 mark
Newton’s law of gravitation is used to calculate the gravitational force that a body exerts on a small object, treating the body as a point mass at its centre.
For which body is this treatment NOT a good approximation?
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Step 1A body acts as a point mass at its centre if its mass is distributed with spherical symmetry (at any distance outside it), OR if the distance to the object is very large compared with the size of the body.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2The space station is not spherical and the astronaut is only 5.0 m away from a body 100 m long, so neither condition is met.
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Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThe asteroid is a uniform sphere, so outside it the field is exactly that of a point mass at its centre, even though the spacecraft is close. Choosing A assumes the distance must always be large.
BThe cube is not spherical, but 50 km is about 25 000 times its width, so it behaves as a point mass. Choosing B assumes that only spheres can be treated as point masses.
CThe Earth is very nearly spherically symmetric, so the satellite can be close to it. Choosing C assumes that 400 km is too close, but this condition is not needed for a spherical body.
DCorrect: the station is neither spherical nor small compared with the 5.0 m distance, so the parts of it at different distances cannot be replaced by a single point at its centre.
A comet moves around the Sun in an elliptical orbit. Its greatest distance from the Sun is six times its smallest distance. At both of these points the velocity of the comet is perpendicular to the line joining it to the Sun.
Which row gives the ratio of the comet’s speed, and the ratio of the magnitude of its acceleration, at the closest point to those at the farthest point?
Diagram not to scale
speed at closest/speed at farthestacceleration at closest/acceleration at farthest
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Step 1Kepler’s second law: in a short time Δt the area swept out is ½rvΔt at both points, so rPvP = rAvA and vP/vA = rA/rP = 6.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2The acceleration is the gravitational field strength of the Sun, a = GM/r2, so aP/aA = 62 = 36.
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Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
A√6 comes from using v2 ∝ 1/r, which is true only when comparing circular orbits. On one elliptical orbit the speeds at the two ends follow from equal areas: v ∝ 1/r.
BThe speed ratio is right, but the acceleration has been taken as proportional to 1/r. The gravitational force, and so the acceleration, follows an inverse-square law.
CCorrect: equal areas give vP/vA = 6, and the inverse-square law gives aP/aA = 62 = 36.
DThe acceleration ratio is right, but the inverse-square law has also been applied to the speed. Equal areas in equal times make the speed inversely proportional to r at these points.
Planet P moves in a circular orbit of radius r around a star of mass M. Its orbital period is T. Planet Q moves in a circular orbit of radius 2r around a different star of mass 2M.
For a circular orbit, T2 = 4π2r3/(GM). What is the orbital period of Q?
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Step 1T2 ∝ r3/M, so TQ2/T2 = 23/2 = 4.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2TQ = √4 × T = 2T.
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Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThis inverts the relationship, taking T2 ∝ M/r3. A larger orbit takes longer, so TQ must be greater than T here.
BCorrect: TQ2/T2 = (23)/2 = 4, so TQ = 2T.
CThis uses T2 ∝ r3 as if both planets orbited the same star. The mass of the star appears in the constant, and doubling it reduces the period.
DThis uses the right ratio, 23/2 = 4, but it is the ratio of T2, not of T. The square root has not been taken.
Two satellites, X and Y, are both in geostationary orbits around the Earth. The mass of Y is three times the mass of X.
Which row compares the orbital speed of Y and the gravitational force on Y with those of X?
Orbital speed of YGravitational force on Y
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Step 1Both satellites have the period of the Earth’s rotation, so (from GMm/r2 = 4π2mr/T2) both have the same orbital radius and so the same speed v = 2πr/T: the mass m cancels.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2At the same radius the gravitational force GMm/r2 is proportional to m, so the force on Y is three times that on X.
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: the satellite’s mass cancels when GMm/r2 = mv2/r, so the speed is the same, but the force GMm/r2 is three times larger for Y.
BThe speed is right, but this confuses the force with the gravitational field strength. g is the same at both satellites; the force mg is proportional to the mass.
CThe force is right, but the speed has been found from F = mv2/r using the mass of X for both. The extra mass of Y needs exactly the extra force, so the speed is unchanged.
DThe speed is right, but g = F/m has been rearranged wrongly as F = g/m. The force is mg, so it is larger for the more massive satellite.
40D-1A-69Gravitational field strengthPaper 1AMedium1 mark
A probe on the surface of Callisto, a moon of Jupiter, releases a small sphere from rest. The graph shows how the velocity v of the sphere varies with time t as it falls. Callisto has no atmosphere and may be treated as a uniform sphere of radius 2.41 × 106 m.
What is the mass of Callisto?
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Step 1The gradient of the v–t graph is the acceleration of free fall: g = 3.0/2.4 = 1.25 m s−2 (= 1.25 N kg−1).
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2g = GM/R2, so M = gR2/G = 1.25 × (2.41 × 106)2/6.67 × 10−11 = 1.09 × 1023 kg ≈ 1.1 × 1023 kg.
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Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThe radius has not been squared: M = gR/G gives 4.5 × 1016 kg. The field strength depends on 1/R2.
BThe gradient has been calculated upside down, as Δt/Δv = 0.80, instead of Δv/Δt = 1.25 m s−2.
CCorrect: g = 1.25 N kg−1 from the gradient and M = gR2/G = 1.09 × 1023 kg.
DThis uses the diameter, 4.82 × 106 m, as the distance from the centre. The field at the surface depends on the radius.
41IBP2S-028-03-03Variation of g with distancePaper 1AMedium1 mark
Two rocky exoplanets, X and Y, are uniform spheres of the same density. The radius of X is twice the radius of Y. What is the ratio of the surface gravitational field strengths?
(field strength at surface of X) / (field strength at surface of Y)
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Mass is density × volume, so M ∝ ρR3. Then g = GM/R2 ∝ ρR3/R2 = ρR. With equal densities, doubling R doubles g, so the ratio is 2. Option D (8) uses the mass ratio but ignores the larger radius. Option A (1/2) applies the inverse square law with the mass unchanged.
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Answer B
Answer: B
Exam tip: For equal densities, write M in terms of ρ and R first. Then g ∝ ρR follows in one line.
42IBP2S-028-03-04Gravitational field linesPaper 1AMedium1 mark
The field lines around an isolated spherical moon of radius R are radial. A communications relay is placed at a height of 2R above the moon's surface. What is the density of field lines at the relay, as a fraction of the density at the surface?
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The same number of field lines crosses every sphere centred on the moon. Line density is therefore proportional to 1/r2, the same dependence as the field strength. The relay is at r = R + 2R = 3R from the centre, so the density is (1/3)2 = 1/9 of the surface value. Option B (1/4) uses the height 2R instead of the distance from the centre. Option A (1/3) treats the dependence as 1/r.
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Answer C
Answer: C
Common error: A height above the surface is not the distance from the centre. Add the radius first.
43IBP2S-028-03-05Gravitational field linesPaper 1AMedium1 mark
Which statement about gravitational field lines is correct?
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A field line is drawn so that its tangent at any point shows the direction of the force on a small test mass. Gravity is always attractive, so lines point towards masses and end on them. Lines never cross, because the field has only one direction at each point. Midway between equal masses the field is zero, and no line passes through that point. Closer lines mean a stronger field, not a weaker one.
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Answer C
Answer: C
Exam tip: Learn two rules together: direction = tangent to the line, and strength = how close the lines are.
44IBP2S-028-03-06Combining two gravitational fieldsPaper 1AMedium1 mark
A binary asteroid system has two spherical bodies of masses M and 4M. Their centres are a distance d apart. A small probe is placed on the line joining the centres, at the point where the resultant gravitational field is zero. How far is the probe from the centre of the body of mass M?
Diagram not to scale
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Let x be the distance from M. The fields cancel when GM/x2 = G(4M)/(d − x)2. This gives (d − x)/x = 2, so x = d/3. The null point lies closer to the smaller mass. Option A (d/5) uses the mass ratio 4 without taking the square root. Option D (2d/3) is the distance from the larger body.
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Answer B
Answer: B
Common error: Field strength depends on 1/r². The distance ratio is the square root of the mass ratio, not the mass ratio itself.
45IBP2S-028-01-03Kepler's third lawPaper 1AMedium1 mark
Two moons, X and Y, orbit the same exoplanet in circular orbits. The orbital radius of X is four times that of Y. What is \(\frac{T_X}{T_Y}\), the ratio of their orbital periods?
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Kepler's third law gives T² ∝ r³, so T ∝ r3/2.
The ratio is therefore 43/2 = (√4)³ = 2³ = 8.
Why the other options are wrong: • 2 takes the square root only. • 4 assumes T ∝ r. • 16 assumes T ∝ r².
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Answer C
Answer: C
Exam tip: To evaluate x³/² without a calculator, take the square root first and then cube it.
An exoplanet orbits a star whose mass is four times the mass of the Sun. Its circular orbit has the same radius as Earth's orbit around the Sun. What is the orbital period of the exoplanet?
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The period of a circular orbit is T = 2π√(r³/GM). At fixed r, this gives T ∝ 1/√M.
With four times the mass, the period is 1 year / √4 = 0.5 years.
Why the other options are wrong: • 1 year assumes the period does not depend on the central mass. In fact only the orbiting mass cancels. • 0.25 years assumes T ∝ 1/M. • 2 years inverts the dependence.
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Answer B
Answer: B
Common error: The period does not depend on the mass of the orbiting body, but it does depend on the mass of the central body.
47IBP2S-028-01-07Kepler's third lawPaper 1AMedium1 mark
A planet has the same mass as Earth but rotates once every 12 hours. The radius of Earth's geostationary orbit is rG. What is the approximate radius of the orbit in which a satellite stays above a fixed point on this planet's equator?
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The central mass is the same, so Kepler's third law gives r ∝ T2/3.
Halving the period gives r = (0.5)2/3 rG ≈ 0.63 rG.
Why the other options are wrong: • 0.50 assumes r ∝ T. • 0.71 uses r ∝ T1/2. • 0.25 uses r ∝ T².
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Answer C
Answer: C
Exam tip: Rearrange T² ∝ r³ into r ∝ T²/³ before you substitute a ratio.
48IBP2S-028-01-09Kepler's second lawPaper 1AMedium1 mark
Some communication satellites serving high-latitude regions use highly elongated elliptical orbits, with the farthest point of the orbit above the region being served. Why do such satellites spend most of each orbit near their farthest point from Earth?
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By Kepler's second law, the radius line sweeps area at a constant rate. When the satellite is far from Earth, a given area corresponds to only a short arc of the orbit. The satellite therefore moves slowly there and spends a large fraction of each orbit near that point.
Why the other options are wrong: • The gravitational force is smaller near the farthest point, but it is never zero. • The period is a property of the whole orbit, not of one position along it. • The speed is smallest, not greatest, at the farthest point.
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Answer D
Answer: D
Exam tip: Link 'equal areas in equal times' directly to speed: large r means low speed, and small r means high speed.
49IBP2S-028-00-03Density, mass and gPaper 1AMedium1 mark
An exoplanet survey identifies a rocky planet with the same mean density as Earth but twice Earth's radius. The gravitational field strength at Earth's surface is g. What is the gravitational field strength at the surface of the exoplanet?
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With M = ρ(4/3)πR3, g = GM/R2 = (4/3)πGρR. At constant density g ∝ R, so doubling R gives 2g. Answer 8g accounts for the eight-fold mass but forgets the larger radius. Answer g assumes equal density means equal g. Answer g/2 keeps the mass fixed while increasing the radius.
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Answer B
Answer: B
Exam tip: For planets of equal density, g = (4/3)πGρR shows that surface field strength is directly proportional to radius.
50IBP2S-028-00-04Variation of g with distancePaper 1AMedium1 mark
A planet has four times the mass of Earth. Its surface gravitational field strength is the same as Earth's. Earth's mean density is ρE. What is the mean density of the planet?
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Equal g means M/R2 is the same for both planets. Four times the mass therefore requires R2 four times larger, so R = 2RE. Density ∝ M/R3 = 4/23 = 1/2, giving ρE/2. Answer ρE assumes equal g implies equal density. Answer ρE/8 uses the larger radius but forgets the larger mass. Answer 4ρE ignores the change in radius.
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Answer D
Answer: D
Common error: Equal surface g does not mean equal density. g depends on the product ρR.
51IBP2S-028-00-06Combining two gravitational fieldsPaper 1AMedium1 mark
A lunar probe travels along the line joining the centres of Earth and the Moon, which are a distance d apart. Earth's mass is 81 times the Moon's mass. At what distance from Earth's centre is the resultant gravitational field strength due to Earth and the Moon zero?
Diagram not to scale
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At the null point GME/x2 = GMM/(d − x)2. This gives x/(d − x) = √81 = 9, so x = 0.90d. Answer 0.99d (≈ 81d/82) comes from forgetting the square root. Answer 0.50d ignores the different masses. Answer 0.10d gives the distance from the Moon instead of from Earth.
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Answer D
Answer: D
Common error: The field depends on 1/r², so the ratio of distances equals the square root of the ratio of masses.
52IBP2S-028-00-07Newton's law of gravitationPaper 1AMedium1 mark
A skydiver of mass 80 kg falls towards Earth before opening her parachute. Which statement about the gravitational interaction between the skydiver and Earth is correct?
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Newton's law of gravitation gives one force magnitude, GMm/r2, acting on each body (a Newton's third law pair). Earth's acceleration F/M is tiny because M is enormous, while the skydiver's acceleration is about 9.8 m s−2. Equal forces do not mean equal accelerations, since a = F/m.
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Answer C
Answer: C
Common error: A larger mass does not exert a larger gravitational force. Both bodies feel forces of the same magnitude.
53IBP2S-028-00-08Variation of g with distancePaper 1AMedium1 mark
The gravitational field strength at the surface of a planet of radius R is 1.21 times the field strength at the orbit of a space station. What is the height of the space station above the planet's surface?
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g ∝ 1/r2, so (r/R)2 = 1.21 and r = 1.1R. The height above the surface is r − R = 0.10R. Answer 0.21R omits the square root. Answer 1.10R is the distance from the centre, not the height. Answer 1.21R makes both mistakes.
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Answer A
Answer: A
Exam tip: Finish by checking whether the question asks for distance from the centre or height above the surface.
54IBP2S-028-02-03Kepler's second lawPaper 1AMedium1 mark
A weather satellite and a communications relay satellite both move in circular orbits around Earth. The orbital radius of the relay satellite is four times that of the weather satellite. What is the ratio of their orbital periods, (period of relay satellite)/(period of weather satellite)?
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By Kepler's third law, T2 ∝ r3, so T ∝ r3/2. The period ratio is 43/2 = (√4)3 = 23 = 8. Option A uses T ∝ √r. Option B assumes T ∝ r. Option D uses T ∝ r2.
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Answer C
Answer: C
Exam tip: Rewrite Kepler's third law as T ∝ r³/² before substituting a ratio. Choosing ratios that are perfect squares makes the arithmetic quick.
55IBP2S-028-02-04Combining two gravitational fieldsPaper 1AMedium1 mark
A spacecraft travels in a straight line from Earth towards the Moon. The mass of the Moon is 1/81 of the mass of Earth. The centre-to-centre distance between Earth and the Moon is D. At what distance from the centre of Earth is the resultant gravitational field strength due to Earth and the Moon zero?
Diagram not to scale
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At the null point, GME/x2 = GMM/(D − x)2. This gives x/(D − x) = √81 = 9, so x = 9D/10 = 0.90D. Option D uses the mass ratio 81 directly instead of its square root. Option A ignores the difference in mass. Option B wrongly takes the mass fraction as the distance fraction.
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Answer C
Answer: C
Common error: Because field strength depends on 1/r², the distances to the null point are in the ratio of the square roots of the masses.
56IBP2S-028-02-05Weightlessness & free fallPaper 1AMedium1 mark
An astronaut aboard a space station orbiting 400 km above Earth's surface floats freely inside the cabin. Which statement best explains why the astronaut feels weightless?
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At 400 km altitude, g is still about 8.7 N kg−1, roughly 89% of the surface value. The astronaut and the station are both in free fall: each has a centripetal acceleration equal to the local g. The floor therefore does not push on the astronaut, and the absence of a normal contact force is what 'weightlessness' means. Options A and D wrongly assume gravity becomes negligible. Option B invents an outward force that would leave no resultant force to keep the station in orbit.
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Answer C
Answer: C
Common error: Orbiting objects are not beyond gravity. Gravity supplies the centripetal force, and it is the lack of a contact force that makes them feel weightless.
57IBP2S-028-02-06Extended bodies as point massesPaper 1AMedium1 mark
A space agency wants a navigation beacon to have a gravitational field strength due to Earth equal to one-ninth of the value at Earth's surface. Earth's radius is R. At what altitude above Earth's surface must the beacon be placed?
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Since g ∝ 1/r2, reducing g to 1/9 of the surface value requires r = 3R measured from Earth's centre. The altitude above the surface is therefore 3R − R = 2R. Option C gives the distance from the centre, not the altitude. Option D assumes g ∝ 1/r, giving r = 9R and an altitude of 8R. Option A has no valid basis.
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Answer B
Answer: B
Exam tip: In g = GM/r², r is always measured from the centre of the planet. Subtract the planet's radius to find the altitude.
58IBP2S-028-02-07Kepler's third lawPaper 1AMedium1 mark
For each of the large moons of Jupiter, a student calculates T2/r3, where T is the orbital period and r is the orbital radius. She finds that the value is the same for every moon. Which quantity does this value depend on?
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For a circular orbit, GMm/r2 = m4π2r/T2, so T2/r3 = 4π2/(GM). Here M is the mass of the central body, Jupiter. The moon's mass m cancels, so options A and D are wrong. Jupiter's radius does not appear in the expression, so option C is wrong.
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Answer B
Answer: B
Exam tip: Measuring the period and radius of any satellite gives the mass of the body it orbits. This is how planetary masses are found.
59IBP2S-028-02-09Geostationary orbitsPaper 1AMedium1 mark
A space-debris removal tug moves from one circular orbit around Earth to a higher circular orbit with four times the orbital radius. What is the ratio (new orbital speed)/(original orbital speed)?
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Gravity provides the centripetal force: GMm/r2 = mv2/r, so v = √(GM/r). Speed is therefore proportional to r−1/2, and 4−1/2 = 1/2. Option B assumes v ∝ 1/r. Option A confuses speed with the inverse of the period ratio, 43/2 = 8. Option D wrongly assumes that a larger orbit requires a higher speed.
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Answer C
Answer: C
Common error: Higher orbits are slower. Orbital speed falls as r⁻¹/², even though the circumference of the orbit increases.
60IBP2S-028-04-04Gravitational field linesPaper 1AMedium1 mark
A space probe drifts along the line joining the centres of a dwarf planet and its moon. The centres are a distance D apart, and the dwarf planet's mass is 64 times that of the moon. How far from the centre of the dwarf planet is the point where the net gravitational field strength is zero?
Diagram not to scale
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At the null point, GMp/x2 = GMm/(D − x)2. This gives x/(D − x) = √64 = 8, so x = 8D/9. The null point lies closer to the less massive body. Option D comes from omitting the square root (x/(D − x) = 64). Option A is the distance measured from the moon instead of from the dwarf planet. Option B ignores the difference in mass.
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Answer C
Answer: C
Exam tip: The ratio of the distances to the null point is the square root of the mass ratio. The null point is always nearer the smaller mass.
61IBP2S-028-04-05Gravitational field linesPaper 1AMedium1 mark
Two uniform spherical asteroids of masses M and 4M have their centres a distance 2r apart. What is the net gravitational field strength at the midpoint between their centres?
Diagram not to scale
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The midpoint is a distance r from each centre. The field due to 4M is 4GM/r2, and the field due to M is GM/r2. These fields point in opposite directions, each towards its own mass. The net field is therefore 3GM/r2 towards the larger mass. Option B adds the two fields as if they were in the same direction. Option D uses 2r instead of r as the distance.
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Answer A
Answer: A
Common error: Field strength is a vector. Between two masses the individual fields point in opposite directions, so you subtract their magnitudes.
62IBP2S-028-04-06Kepler's third lawPaper 1AMedium1 mark
Astronomers find that exoplanet B orbits its star with a period 27 times that of exoplanet A, which orbits the same star. Both orbits are circular. What is the ratio of orbital radii, (radius of B)/(radius of A)?
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Kepler's third law gives T2 ∝ r3, so r ∝ T2/3. The ratio is 272/3 = (271/3)2 = 32 = 9. Option A takes only the cube root. Option C assumes r ∝ T. Option D inverts the law to r ∝ T3/2.
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Answer B
Answer: B
Exam tip: Write the law as r³ ∝ T² before rearranging. Raising both sides to the power 1/3 gives r ∝ T²/³.
63IBP2S-028-04-07Orbits and satellitesPaper 1AMedium1 mark
Atmospheric drag gradually causes a small satellite to move into a lower circular orbit. How have the satellite's orbital speed and orbital period changed?
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For a circular orbit, v = √(GM/r), so a smaller radius means a larger orbital speed. The period T = 2πr/v depends on a smaller distance travelled at a greater speed, so T decreases (consistent with T2 ∝ r3). Drag removes total energy, and the satellite spirals inwards. As it does so, it loses more gravitational potential energy than the energy dissipated, so its kinetic energy increases.
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Answer C
Answer: C
Common error: Drag does not leave the satellite moving more slowly in its new orbit. A lower circular orbit always has a higher orbital speed.
64D-1A-08Combining two gravitational fieldsPaper 1AHard1 mark
Two spheres of mass M and 9M have their centres a distance d apart. At point P on the line joining their centres, between the spheres, the resultant gravitational field strength is zero.
What is the distance of P from the centre of the sphere of mass M?
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Step 1At P the two fields are equal and opposite: GM/x² = G(9M)/(d − x)².
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Taking square roots: d − x = 3x.
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Step 3x = d/4 (closer to the smaller mass).
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Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis sets d − x = 9x, using the mass ratio instead of its square root.
BCorrect: (d − x)/x = √9 = 3, so x = d/4.
CThe midpoint is the null point only for equal masses.
DThis is the distance of P from the larger sphere, not from the sphere of mass M.
66D-1A-28Combining two gravitational fieldsPaper 1AHard1 mark
Two spheres of mass M and 4M have their centres a distance d apart. Point P lies on the line through both centres, a distance d from the centre of the sphere of mass M, on the side away from the sphere of mass 4M.
What is the magnitude of the resultant gravitational field strength at P?
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Step 1Field of M at P: GM/d². Field of 4M at P (distance 2d): 4GM/(2d)² = GM/d².
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2P is outside the pair, so both fields point from P towards the spheres and add: 2GM/d².
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Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThe two contributions are equal, but at P they point the same way; subtracting them to get zero treats them as opposite.
BThis subtracts the fields and uses GM/(2d)² for the larger sphere, leaving out its factor 4.
CCorrect: GM/d² + 4GM/(2d)² = 2GM/d², both directed from P towards the spheres.
DThis places the sphere of mass 4M a distance d from P instead of 2d.
A communications satellite moves around the Earth in a highly elliptical orbit with the centre of the Earth at one focus. At time t = 0 the satellite is at its closest point to the Earth. The period of the orbit is T.
Which graph shows how the distance r of the satellite from the centre of the Earth varies with t over one orbit?
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Step 1Kepler’s second law: the line from the Earth to the satellite sweeps out equal areas in equal times, so the satellite moves fastest when r is smallest and slowest when r is largest.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2So r changes rapidly near t = 0 and t = T, and the satellite spends most of the period far from the Earth, with a broad maximum of r at T/2: graph A.
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: r rises steeply from its minimum, stays near its maximum for most of the orbit (the satellite moves slowly there) and falls steeply back to the minimum.
BGraph B is the result of assuming the satellite moves at constant speed around the ellipse. Kepler’s second law requires it to move much faster near the Earth.
CGraph C is the result of assuming the satellite turns through equal angles about the Earth in equal times. That makes it spend most of the time close to the Earth, the opposite of the real motion.
DGraph D shows the speed of the satellite, which is greatest at the closest point, not its distance. It starts at a maximum, but r starts at its minimum.
70D-1A-66Gravitational field linesPaper 1AHard1 mark
The gravitational field of an isolated uniform spherical planet is represented by radial field lines. The field strength at a point is represented by the number of field lines per unit area there, measured on a surface perpendicular to the lines.
Which statement explains why this representation agrees with g = GM/r2?
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Step 1Every line that ends on the planet passes through every sphere of radius r centred on it, so the number N crossing each sphere is the same.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Lines per unit area = N/(4πr2) ∝ 1/r2, the same dependence on r as GM/r2.
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Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AStraight lines show only the direction of the field. The lines spread out along their length, so the field strength changes along each line.
BThis uses the spacing of lines in a flat drawing. The field is three-dimensional, so the strength depends on lines per unit AREA, which falls as 1/r2, not as 1/r.
CThe lines do spread out, but a linear decrease would not match GM/r2. The line density falls as 1/r2, approaching zero only gradually.
DCorrect: a fixed number of lines crosses a spherical area 4πr2, so the number per unit area, and hence the field strength represented, is proportional to 1/r2.
71IBP2S-028-03-07Variation of g with distancePaper 1AHard1 mark
A planet of radius R is a uniform sphere. A weather balloon is at a height h where the gravitational field strength is 64% of its value at the surface. What is h?
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The ratio of field strengths is g/g0 = R2/(R + h)2 = 0.64. Taking the square root, R/(R + h) = 0.8, so R + h = 1.25R and h = 0.25R. Option A (0.20R) comes from 1 − 0.8. Option C (0.36R) comes from 1 − 0.64. Option D (0.56R) comes from 1/0.64 − 1, which forgets the square root.
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Answer B
Answer: B
Exam tip: Write the ratio as (R/r)², take the square root, then find r. Subtract R only at the end.
72IBP2S-028-03-10Newton's law of gravitationPaper 1AHard1 mark
In a laboratory demonstration, two identical uniform lead spheres are placed in contact. They are then replaced by two lead spheres, also in contact, each with twice the radius. By what factor does the gravitational force between the spheres increase?
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Each sphere's mass is proportional to r3, so each mass increases by a factor of 8. The product of the masses therefore increases by 64. The spheres are in contact, so their centres are 2r apart, and this distance doubles. The inverse square law divides the force by 4. The overall factor is 64/4 = 16. Option D (64) forgets that the separation of the centres also increases.
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Answer C
Answer: C
Common error: When spheres touch, the separation of their centres depends on their radii. Changing the size changes both the masses and the distance.
A comet orbits the Sun in an elliptical orbit. Its closest distance to the Sun is 1 AU and its farthest distance is 7 AU. Earth orbits at 1 AU with a period of 1 year. What is the orbital period of the comet?
Diagram not to scale
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For an elliptical orbit, Kepler's third law uses the semi-major axis a. This is half the sum of the closest and farthest distances: a = (1 + 7)/2 = 4 AU.
Then T = a3/2 = 43/2 = 8 years.
Why the other options are wrong: • 19 years uses the farthest distance: 73/2 ≈ 18.5. • 23 years uses the full major axis: 83/2 ≈ 22.6. • 64 years uses T ∝ a³.
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Answer A
Answer: A
Common error: For an ellipse, r in Kepler's third law is the semi-major axis: the average of the closest and farthest distances. It is not the farthest distance.
74IBP2S-028-01-10Kepler's third lawPaper 1AHard1 mark
The Moon orbits Earth with a period of 27.3 days at an orbital radius rM. A geostationary satellite has a period of 1.00 day and an orbital radius rG. Which value is closest to \(\frac{r_M}{r_G}\)?
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Both bodies orbit Earth, so Kepler's third law gives r ∝ T2/3.
The ratio is rM/rG = 27.32/3. Since 272/3 = 9, the value is about 9.1. This is consistent with rM ≈ 3.8×108 m and rG ≈ 4.2×107 m.
Why the other options are wrong: • 5.2 is 27.31/2. • 27 assumes r ∝ T. • 143 is 27.33/2, which applies the law the wrong way round.
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Answer B
Answer: B
Exam tip: Look for a nearby perfect cube to estimate powers quickly: 27²/³ = (27¹/³)² = 3² = 9.
75IBP2S-028-00-09Newton's law of gravitationPaper 1AHard1 mark
Two identical uniform steel ball bearings rest in contact with each other. The gravitational force between them is F. They are replaced by two ball bearings of the same steel, each of twice the radius, also in contact. What is the new gravitational force?
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Each mass is m = ρ(4/3)πr3 and the centre-to-centre separation is 2r. So F = Gm2/(2r)2 ∝ r6/r2 = r4. Doubling r multiplies the force by 24 = 16. Answer 64F accounts for the masses but not the larger separation. Answer F/4 accounts only for the separation. Answer 4F wrongly treats each mass as proportional to r.
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Answer B
Answer: B
Exam tip: When the size changes at fixed density, express mass in terms of radius before applying Newton's law.
76IBP2S-028-02-08Density, mass and gPaper 1AHard1 mark
Two spherical planets, A and B, have the same gravitational field strength at their surfaces. The radius of A is twice the radius of B. What is the ratio (mean density of A)/(mean density of B)?
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Write the mass as M = (4/3)πR3ρ. Then g = GM/R2 = (4/3)πGρR, so g ∝ ρR. Equal g with RA = 2RB requires ρA = ρB/2. Option B is the ratio of 1/R2 (and it is the inverse of the mass ratio, which is 4). Option A divides the mass ratio by the volume ratio incorrectly. Option D inverts the correct ratio.
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Answer C
Answer: C
Exam tip: Substituting M = (4/3)πR³ρ into g = GM/R² gives the useful result g ∝ ρR.
A student uses a computer simulation to investigate how the period T of a planet in a circular orbit depends on the mass M of its star. The orbital radius is kept at 1.50 AU. For each value of M, the student uses the simulation clock to measure the time t for five complete orbits. Judging when an orbit is complete gives an uncertainty of ±5 days in t.
The student tests the hypothesis T ∝ Mn by plotting lg (T/day) against lg (M/M☉), where M☉ is the mass of the Sun. Four of the points have been plotted.
M / M☉
t / day
T / day
lg (M / M☉)
lg (T / day)
0.40
5308
−0.398
3.026
0.70
4006
801.2
1.00
3357
671.4
0.000
2.827
1.50
2738
547.6
0.176
2.738
2.00
2376
475.2
3.00
1934
386.8
0.477
2.587
Graph drawn to scale
(a)
Calculate T, with its absolute uncertainty, for M = 0.40 M☉.
(2)
(b)
Complete the table for M = 0.70 M☉ and M = 2.00 M☉.
(1)
(c)
Plot the two missing points on the graph and draw the line of best fit for the data.
(2)
(d)
Determine the gradient of the line.
(2)
(e)
For a circular orbit, Newton's law of gravitation predicts T ∝ M−1/2 at a fixed orbital radius. Outline whether the data support this prediction.
(1)
(f)
Suggest why the student timed five orbits rather than one.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
T = 5308/5 = 1062 day
✓ 1
ΔT = 5/5 = ±1 day, so T = (1062 ± 1) day
✓ 1
Accept 1061.6 ± 1.0 day. The uncertainty must be divided by 5 as well.
Part (b)
M = 0.70: −0.155 and 2.904; M = 2.00: 0.301 and 2.677
✓ 1
All four values required, to at least 3 decimal places.
Part (c)
Points plotted at (−0.155, 2.904) and (0.301, 2.677) to within half a small square
✓ 1
Allow ECF from (b).
Single straight line passing through (or very close to) all six points
✓ 1
Do not accept a line joining the dots.
Part (d)
Gradient calculated from a triangle using at least half the length of the line
✓ 1
gradient = −0.50
✓ 1
Accept −0.48 to −0.52. The negative sign is required.
Part (e)
Yes: the gradient is (very close to) −0.5 = n, and the points lie on a straight line
✓ 1
Allow ECF from (d) if the conclusion is consistent with the candidate's gradient.
Part (f)
The judgement uncertainty (±5 days) is the same whatever the time measured, so timing five orbits makes the percentage uncertainty in T five times smaller
✓ 1
OWTTE
Answers: (a) T = (1062 ± 1) day · (b) −0.155, 2.904; 0.301, 2.677 · (d) −0.50 (the remaining parts are explanations — see the table above)
78D-1B-28Gravitational field strengthPaper 1BEasy10 marks
A student measures the gravitational field strength g in the school laboratory. A force sensor connected to a data logger is clamped above the bench. Slotted masses m are hung from the sensor and the force F that the sensor records is noted. The masses are known to a much smaller percentage uncertainty than the forces. The force readings fluctuate, so the uncertainty in each value of F is ±0.05 N.
The graph shows the data. The error bars are too small to be seen on this scale.
m / kg
0.100
0.200
0.300
0.400
0.500
0.600
F / N
1.12
2.05
3.09
4.03
5.04
5.99
Graph drawn to scale
(a)
State the resolution of the force sensor.
(1)
(b)
Draw the line of best fit for the data.
(1)
(c)
Determine the gradient of your line, including its unit.
(2)
(d)
State what the gradient of the line represents.
(1)
(e)
The line does not pass through the origin. Identify the type of error that this shows and state its effect on your answer to (c).
(2)
(f)
The student proposes repeating the experiment in a mountain hut 2.5 km higher than the school, to show that g decreases with height. The radius of the Earth is 6.37 × 106 m.
(i)
Show that g in the hut is expected to be about 0.08 % smaller than in the school.
(1)
(ii)
Determine whether the student’s method could detect this change.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
0.01 N
✓ 1
Do not accept 0.05 N (that is the uncertainty, not the resolution).
Part (b)
single straight line passing through or very close to all the points, with points on both sides, not forced through the origin
✓ 1
Do not accept a line joining the points.
Part (c)
gradient calculated from a triangle using at least half the length of the line
✓ 1
gradient = 9.8 N kg−1
✓ 1
Accept 9.6–10.0. Accept m s−2. Unit required for this mark.
Part (d)
the gravitational field strength g «force per unit mass»
✓ 1
Accept «g».
Part (e)
systematic error «zero error / offset of the sensor»
✓ 1
Do not accept «random error».
it adds the same amount to every reading, so it shifts the line without changing the gradient: the value of g is unaffected
✓ 1
OWTTE
Part (f)(i)
ghut/g = (6.37 × 106/6.3725 × 106)2 = 0.99922, so g is 0.078 % «≈ 0.08 %» smaller
✓ 1
Must see full substitution OR answer to at least 2 s.f. Accept the approximation 2h/R.
Part (f)(ii)
percentage uncertainty in a single reading of F is at least 0.05/5.99 ≈ 0.8 % OR percentage uncertainty in the gradient ≈ 2 %
✓ 1
Any sensible estimate of the percentage uncertainty of the method. Allow ECF from (c).
this is about ten (or more) times larger than 0.08 %, so the change could NOT be detected
✓ 1
MP2 only scores if a comparison with 0.08 % is made.
Answers: (c) 9.8 N kg−1 · (f)(i) 0.078 % (the remaining parts are explanations — see the table above)
79D-1B-01Gravitational field strengthPaper 1BMedium10 marks
A student uses a published satellite catalogue to test how the gravitational field strength g of the Earth varies with the distance r from the Earth's centre. For five satellites in near-circular orbits, the catalogue lists the mean altitude h above the Earth's surface and the orbital period T.
For a satellite in a circular orbit, g at the orbit is equal to the centripetal acceleration a = 4π²r/T². The radius of the Earth is 6.37 × 106 m.
The graph shows a against 1/r² for four of the satellites.
Satellite
h / km
T / min
r / 106 m
a / m s−2
1/r² / 10−15 m−2
A
410
92.6
6.78
8.67
21.8
B
1336
112.2
7.71
6.71
16.8
C
8062
287.6
D
20180
717.6
26.6
0.565
1.42
E
35786
1435.7
42.2
0.224
0.563
Graph drawn to scale
(a)
Show that a for satellite B is about 6.7 m s−2.
(1)
(b)
Complete the table for satellite C.
(2)
(c)
Plot the data point for satellite C on the graph and draw the line of best fit for the data.
(2)
(d)
Determine the gradient of the line. State its unit.
(2)
(e)
Determine the mass of the Earth. Compare your answer with the accepted value of 5.97 × 1024 kg.
(2)
(f)
Suggest one advantage of plotting lg a against lg r for these data.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
r = 6.37 × 106 + 1.336 × 106 = 7.706 × 106 m AND a = 4π² × 7.706 × 106/(112.2 × 60)² = 6.71 m s−2
✓ 1
Must see full substitution with T in seconds OR answer to at least 3 s.f.
Part (b)
r = 14.4 × 106 m AND 1/r² = 4.80 × 10−15 m−2
✓ 1
Accept 1/r² in the range 4.79–4.81.
a = 4π² × 14.43 × 106/(287.6 × 60)² = 1.91 m s−2
✓ 1
Must be given to 3 s.f., consistent with the rest of the table.
Part (c)
Point plotted at (4.80, 1.91) to within half a small square
✓ 1
Single straight line passing through (or very close to) all five points and the origin
✓ 1
Do not accept a line joining the dots or a curve.
Part (d)
Gradient calculated from a triangle using at least half the length of the line
✓ 1
gradient = 3.99 × 1014 m³ s−2
✓ 1
Accept 3.90 × 1014 to 4.10 × 1014. Accept N m² kg−1. Unit required for this mark.
Part (e)
M = gradient/G = 3.99 × 1014/6.67 × 10−11 = 5.98 × 1024 kg
✓ 1
Allow ECF from (d). Accept 5.85 × 1024 to 6.15 × 1024 kg.
Differs from the accepted value by about 0.1 %, so the two agree / the data support g = GM/r²
✓ 1
The comparison must be consistent with the candidate's value; a percentage difference or a clear statement of closeness is required. OWTTE
Part (f)
The points would be spread more evenly across the graph (here four of the five points lie near the ends of the range) OR the gradient gives the power of r directly, so the inverse-square relationship (gradient −2) is tested without being assumed
✓ 1
OWTTE
Answers: (a) a = 6.71 m s−2 · (b) r = 14.4 × 106 m; a = 1.91 m s−2; 1/r² = 4.80 × 10−15 m−2 · (d) 3.99 × 1014 m³ s−2 · (e) M = 5.98 × 1024 kg (the remaining parts are explanations — see the table above)
A student uses a published exoplanet database to test Kepler's third law for the seven planets that orbit the star TRAPPIST-1. The table gives the orbital radius a of each planet in astronomical units (AU) and its orbital period T in days. The orbits may be treated as circular.
The student tests the relationship T = kan, where k and n are constants, by plotting lg (T / day) against lg (a / AU). Five of the points have been plotted.
Planet
a / AU
T / day
lg (a / AU)
lg (T / day)
b
0.01154
1.511
−1.938
0.179
c
0.01580
2.422
d
0.02227
4.049
−1.652
0.607
e
0.02925
6.101
−1.534
0.785
f
0.03849
9.208
−1.415
0.964
g
0.04683
12.35
h
0.06189
18.77
−1.208
1.273
Graph drawn to scale
(a)
Complete the table for planets c and g.
(2)
(b)
Plot the points for planets c and g on the graph and draw the line of best fit for the data.
(2)
(c)
Determine the gradient of the line.
(2)
(d)
Kepler's third law predicts that T² is proportional to a³. Outline whether the data support this prediction.
(1)
(e)
The relationship T² = 4π²a³/(GM) applies, where M is the mass of the star. Determine, using the data for planet e, the mass of TRAPPIST-1 in kg. (1 AU = 1.50 × 1011 m)
(2)
(f)
The database gives a for planet e as (0.02925 ± 0.00025) AU; the uncertainty in T is negligible. Determine the absolute uncertainty in your answer to (e) and discuss whether your value agrees with the published mass of 0.0898 solar masses. (Mass of the Sun = 1.99 × 1030 kg)
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
lg (a / AU): c = −1.801; g = −1.329
✓ 1
Accept 3 or 4 decimal places; do not accept 2 d.p.
lg (T / day): c = 0.384; g = 1.092
✓ 1
Accept 3 or 4 decimal places; do not accept 2 d.p.
Part (b)
Both points plotted correctly to within half a small square
✓ 1
Allow ECF from (a).
Single straight line of best fit through (or very close to) all seven points
✓ 1
Part (c)
Gradient calculated from a large triangle (at least half the length of the line)
✓ 1
gradient = 1.50
✓ 1
Accept 1.46 to 1.54. No unit.
Part (d)
Gradient ≈ 1.5 = 3/2, so T ∝ a3/2; squaring gives T² ∝ a³, which supports the prediction
✓ 1
Allow ECF from (c) if the candidate's gradient is within ±3 % of 1.5.
Part (e)
a = 0.02925 × 1.50 × 1011 = 4.388 × 109 m AND T = 6.101 × 86 400 = 5.271 × 105 s
✓ 1
Both conversions needed.
M = 4π²a³/(GT²) = 1.80 × 1029 kg
✓ 1
Accept 1.79–1.81 × 1029 kg.
Part (f)
percentage uncertainty in M = 3 × (0.00025/0.02925) × 100 = 2.6 %
✓ 1
ΔM = 0.05 × 1029 kg, so M = (1.80 ± 0.05) × 1029 kg
✓ 1
MP2 is for matching precision of value and uncertainty. Allow ECF from (e).
Published mass = 0.0898 × 1.99 × 1030 = 1.79 × 1029 kg, which lies within the range 1.75–1.85 × 1029 kg, so the values agree
✓ 1
Accept comparison in solar masses: 0.0904 ± 0.0023 M☉.
Answers: (a) c: −1.801, 0.384; g: −1.329, 1.092 · (c) 1.50 · (e) M = 1.80 × 1029 kg · (f) M = (1.80 ± 0.05) × 1029 kg (the remaining parts are explanations — see the table above)
A student tests Kepler's second law using a published ephemeris (a table of predicted planetary positions) for Mars. For five 10-day intervals, each starting a given number of days after Mars passes perihelion (its closest point to the Sun), the student records the angle Δθ through which Mars moves around the Sun during the interval and the distance r of Mars from the Sun at the middle of the interval. The ephemeris gives angles to the nearest 0.01°.
For a short interval, the area swept out by the line joining Mars to the Sun is approximately ½r²Δθ, with Δθ in radians.
Start of interval / days after perihelion
r / AU
Δθ / °
1/r² / AU−2
0
1.3815
6.35
0.524
95
1.4459
5.79
181
1.5549
5.01
0.414
267
1.6411
4.50
343
1.6659
4.36
0.360
Graph drawn to scale
(a)
Complete the table.
(1)
(b)
Draw the line of best fit for the data.
(1)
(c)
Explain how the graph supports Kepler's second law.
(2)
(d)
Calculate, in m², the area swept out in 10 days starting at perihelion. (1 AU = 1.50 × 1011 m)
(2)
(e)
Explain why the student used the angle moved in 10 days rather than the angle moved in 1 day or in 100 days.
(2)
(f)
Determine the percentage difference between the values of r²Δθ at perihelion (0 days) and at 343 days, and comment on your answer.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
0.478 AND 0.371 AU−2
✓ 1
Both needed, to 3 s.f. (3 d.p.).
Part (b)
Single straight line through (or very close to) all the points, passing through (or very close to) the origin
✓ 1
Part (c)
The graph is a straight line through the origin, so Δθ ∝ 1/r² / r²Δθ is constant
✓ 1
The area swept out in each 10-day interval, ½r²Δθ, is therefore the same: equal areas are swept out in equal times
✓ 1
OWTTE
Part (d)
r = 1.3815 × 1.50 × 1011 = 2.072 × 1011 m AND Δθ = 6.35 × π/180 = 0.1108 rad
✓ 1
area = ½ × (2.072 × 1011)² × 0.1108 = 2.38 × 1021 m²
✓ 1
Accept 2.37–2.39 × 1021 m².
Part (e)
In 1 day Δθ is only about 0.4–0.6°, so the ±0.005° rounding gives a percentage uncertainty about ten times larger
✓ 1
OWTTE
In 100 days r changes considerably, so ½r²Δθ with a single value of r is no longer a good approximation to the area
✓ 1
OWTTE
Part (f)
r²Δθ = 12.12 and 12.10 AU² degrees, a difference of about 0.16 %, which is within the rounding uncertainty of the angles (0.005/6.35 + 0.005/4.36 ≈ 0.19 %), so r²Δθ is constant: consistent with Kepler's second law
✓ 1
Accept any comparison showing the difference is very small / within uncertainty.
Answers: (a) 0.478, 0.371 AU−2 · (d) 2.38 × 1021 m² · (f) about 0.16 % (the remaining parts are explanations — see the table above)
82D-1B-21Density, mass and gPaper 1BMedium12 marks
A student uses a published database of Solar System bodies to test the hypothesis that all large solid bodies in the Solar System have the same mean density. The table gives the mean radius R and the gravitational field strength g at the surface for eight bodies.
The graph shows g against R for seven of the bodies.
Body
R / 106 m
g / m s−2
Mercury
2.44
3.70
Venus
6.05
8.87
Earth
6.37
9.81
Mars
3.39
3.73
Moon
1.74
1.62
Io
1.82
1.80
Ganymede
2.63
1.43
Titan
2.57
1.36
Graph drawn to scale
(a)
Show that, for a uniform sphere of mean density ρ, g = 4πGρR/3.
(1)
(b)
Plot the point for Mars on the graph.
(1)
(c)
Draw the straight line through the origin that best fits the points for Mercury, Venus and the Earth.
(1)
(d)
Determine, using the gradient of your line, the mean density of these three planets.
(3)
(e)
Discuss whether the data support the hypothesis that all eight bodies have the same mean density.
(2)
(f)
Tracking of a spacecraft orbiting the asteroid Vesta gives g = (0.250 ± 0.005) m s−2 at a mean radius R = (263 ± 5) km. Determine the mean density of Vesta and its absolute uncertainty.
(3)
(g)
Vesta is noticeably flattened and irregular in shape. Outline why this makes the value in (f) less reliable.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M = ρ × 4πR³/3 substituted into g = GM/R² to give g = 4πGρR/3
✓ 1
Must see the substitution of the volume of a sphere.
Part (b)
Point at (3.39, 3.73) to within half a small square
✓ 1
Part (c)
Single straight line from the origin passing between/through the three points, close to all of them
✓ 1
Line must start at the origin.
Part (d)
Gradient from a large triangle ≈ 1.51 m s−2 per 106 m = 1.51 × 10−6 s−2
✓ 1
Accept 1.46–1.55 m s−2 per 106 m. Power of ten may be shown in MP2.
ρ = 3 × gradient/(4πG)
✓ 1
Allow ECF from MP1.
ρ = 5.4 × 103 kg m−3
✓ 1
Accept 5.2 × 103–5.6 × 103 kg m−3. Allow ECF from MP1.
Part (e)
No: the points for Mars, the Moon, Io, Ganymede and Titan lie well below the line, by far more than any plausible uncertainty in the data
✓ 1
OWTTE
so these bodies have smaller mean densities; Ganymede and Titan, about one third of the planets' value, the least dense
✓ 1
Accept a calculated density for any one of these bodies as evidence. OWTTE
MP3 is for matching the precision of value and uncertainty. Allow ECF from MP1 and MP2.
Part (g)
The model (point mass at the centre / g = GM/R² with a single radius) needs a spherically symmetric body; for a flattened body g varies over the surface and the "mean radius" is not well defined
✓ 1
OWTTE
Answers: (d) gradient ≈ 1.51 m s−2 per 106 m; ρ ≈ 5.4 × 103 kg m−3 · (f) ρ = (3.4 ± 0.1) × 103 kg m−3(the remaining parts are explanations — see the table above)
A student photographs Jupiter and its moon Callisto through a small telescope on 17 clear nights over 36 days. On each image the student measures the separation x of Callisto from the centre of Jupiter, in units of the radius RJ of Jupiter measured on the same image. x is positive when Callisto is east of Jupiter. The uncertainty in each value of x is ±0.8 RJ. The Earth lies very close to the plane of Callisto’s orbit, which may be treated as a circle.
The graph shows the results.
For a circular orbit of radius r and period T around a body of mass M, T2 = 4π2r3/(GM).
Graph drawn to scale
(a)
Draw a smooth curve of best fit for the data.
(1)
(b)
Determine, using the graph, the orbital period of Callisto in days, and its absolute uncertainty.
(2)
(c)
Outline why the greatest value of x is equal to the radius of Callisto’s orbit.
(1)
(d)
RJ = 7.15 × 107 m. Show, using the graph, that the radius of Callisto’s orbit is about 1.9 × 109 m.
(1)
(e)
Determine the mass of Jupiter.
(2)
(f)
The greatest value of x is (26.3 ± 1.0) RJ. Determine the absolute uncertainty in your answer to (e), and discuss whether your result agrees with the accepted value of 1.90 × 1027 kg.
(3)
(g)
Suggest one change to the observations that would reduce the uncertainty in the mass.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
smooth sinusoidal curve through (almost) all the error bars, with equal positive and negative maxima of about 26 RJ
✓ 1
Do not accept straight segments joining the points.
Part (b)
time for one period read between alternate zero crossings (e.g. ≈ 2.3 d and ≈ 19.0 d, or ≈ 10.6 d and ≈ 27.3 d) OR between two maxima / minima
✓ 1
Any valid method using a full cycle (or more) of the curve.
T = 16.7 ± 0.3 d
✓ 1
Accept 16.3–17.1 d. Accept an uncertainty from 0.1 to 0.5 d.
Part (c)
the orbit is seen edge-on, so x is greatest when Callisto is at 90° to the line of sight from the Earth, and its distance from Jupiter is then all across the line of sight «x = r»
✓ 1
OWTTE. Accept a sketch showing the circle seen edge-on.
Part (d)
r = 26 × 7.15 × 107 ≈ 1.86 × 109 m «≈ 1.9 × 109 m»
✓ 1
Accept a greatest value of x from 25.5 to 27. Must see full substitution OR answer to 3 s.f.
Part (e)
M = 4π2r3/(GT2) with T in seconds «16.7 × 86 400 = 1.44 × 106 s»
✓ 1
M = 1.9 × 1027 kg
✓ 1
Accept 1.8–2.0 × 1027 kg. Allow ECF from (b) and (d). Award [2] for CNA.
Allow ECF from the candidate’s uncertainty in T. Award [0] for MP1 if the powers 3 and 2 are not used.
M = (1.9 ± 0.3) × 1027 kg
✓ 1
MP2 is for matching the precision of value and uncertainty.
the accepted value lies within this range, so the result agrees with it
✓ 1
Allow ECF. The comparison must be consistent with the candidate’s range.
Part (g)
observe over more orbits / a longer time so that the percentage uncertainty in T is smaller OR take more images near the greatest separations / use a higher magnification so the percentage uncertainty in the amplitude «which is cubed» is smaller
✓ 1
Accept any specific, relevant change. Do not accept «take more readings» without detail.
Answers: (b) 16.7 ± 0.3 d · (e) 1.9 × 1027 kg · (f) ±0.3 × 1027 kg (the remaining parts are explanations — see the table above)
84D-1B-30Combining two gravitational fieldsPaper 1BMedium11 marks
A student uses a computer simulation of a planet A and its moon B, whose centres are fixed a distance d = 4.00 × 108 m apart. The student sets the ratio k = (mass of A)/(mass of B) and then moves a probe along the line joining the centres until the resultant gravitational field strength at the probe is zero. The distance x of this point from the centre of A is read from a scale on the screen with an uncertainty of ±0.05 × 108 m.
The student assumes that the field strength of each body is proportional to 1/rn, where r is the distance from its centre, and tests whether n = 2.
k
x / 108 m
lg k
lg((d − x)/x)
2
2.32
0.301
−0.140
5
2.79
0.699
−0.363
10
3.01
1.000
−0.483
20
3.30
50
3.47
1.699
−0.816
100
3.66
2.000
−1.032
Graph drawn to scale
(a)
Show that lg((d − x)/x) = −(1/n) lg k.
(2)
(b)
Complete the table for k = 20.
(2)
(c)
Explain why the error bars become larger as k increases.
(1)
(d)
Plot the point for k = 20 and draw the line of best fit for the data.
(2)
(e)
Determine n and its uncertainty, using lines of maximum and minimum gradient. State whether the data are consistent with an inverse-square law.
(3)
(f)
Suggest why the student chose values of k that increase by roughly the same factor each time rather than by the same amount.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
at the zero point GMA/xn = GMB/(d − x)n OR ((d − x)/x)n = MB/MA = 1/k
✓ 1
Must see full substitution of the two field expressions.
taking logarithms: n lg((d − x)/x) = −lg k «and rearranging»
for large k the point is close to B, so d − x is small and the same ±0.05 × 108 m is a much larger fractional uncertainty in d − x «and so in the logarithm»
✓ 1
OWTTE
Part (d)
point plotted at (1.30, −0.67) within half a small square
✓ 1
Allow ECF from (b).
single straight line passing through all the error bars
✓ 1
The line need not be forced through the origin.
Part (e)
best gradient ≈ −0.51 AND n = −1/gradient ≈ 2.0
✓ 1
Accept gradient −0.54 to −0.48.
max and min gradient lines through all the error bars give gradients of about −0.53 and −0.47, so n from about 1.9 to 2.1
✓ 1
Accept any pair of lines that pass through all the error bars. Accept Δn from ±0.1 to ±0.3.
n = 2.0 ± 0.1; the range includes 2, so the data are consistent with an inverse-square law
✓ 1
Allow ECF. The conclusion must agree with the candidate’s range.
Part (f)
the points are spread evenly along the lg k axis «a range from 2 to 100 in equal steps would crowd the points at large lg k»
✓ 1
OWTTE
Answers: (b) 1.301, −0.673 · (e) n = 2.0 ± 0.1 (the remaining parts are explanations — see the table above)
A student uses a mission-planning database of near-circular orbits around Mars that have been used or proposed for spacecraft. For seven orbits, labelled A to G, the database gives the mean altitude h of the spacecraft above the surface of Mars and the orbital period T. Because the orbits are not exactly circular and Mars is not exactly spherical, each value of h has an uncertainty of ±50 km; the uncertainty in T is negligible.
Mars may be treated as a uniform sphere of radius R and mass M. For a circular orbit of radius r around a uniform sphere of mass M, T2 = 4π2r3/(GM).
The student plots a graph of h against T2/3. Six of the points have been plotted. The error bars are too small to be shown.
Orbit
h / km ± 50 km
T / min
T2/3 / min2/3
A
350
114.3
23.6
B
830
140.7
27.1
C
1470
170.4
30.7
D
2570
234.6
E
3940
314.9
46.3
F
5180
402.8
54.5
G
6380
486.4
61.8
Graph drawn to scale. The point for orbit D has not been plotted.
(a)
Outline why the graph is expected to be a straight line that meets the h-axis at h = −R.
(2)
(b)
Complete the table for orbit D.
(1)
(c)
Plot the point for orbit D on the graph and draw the line of best fit. Extend your line to meet the h-axis.
(2)
(d)
Determine, using your graph, the radius of Mars.
(2)
(e)
The student estimates that the uncertainty in the value of R found from the graph is ±0.10 × 106 m. Determine, using your answer to (d) and the data for orbit G, the mass of Mars and its absolute uncertainty. Ignore all other uncertainties.
(3)
(f)
The accepted value of the mass of Mars is 6.42 × 1023 kg. Comment on your answer to (e).
(1)
(g)
Suggest, with a reason, one change to the set of orbits used that would reduce the uncertainty in the value of R found from the graph.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Mars acts as a point mass at its centre «uniform sphere», so r is measured from the centre: r = R + h
✓ 1
OWTTE. Must refer to the centre of Mars or to r = R + h.
r = (GM/4π2)1/3T2/3, so h = (GM/4π2)1/3T2/3 − R: linear in T2/3 with intercept −R where T2/3 = 0
✓ 1
Must see the rearranged equation OR the gradient (GM/4π2)1/3 and intercept −R identified.
Part (b)
T2/3 = 234.62/3 = 38.0 «min2/3»
✓ 1
Accept 38.03. Do not accept 38.
Part (c)
Point plotted at (38.0, 2.57) to within half a small square
✓ 1
Allow ECF from (b).
Single straight line through or very close to all seven points, extended to meet the h-axis «at about −3.4»
✓ 1
The line must reach T2/3 = 0.
Part (d)
Intercept read from the h-axis OR calculated from the gradient and a point on the line, e.g. gradient = (6.07 − 1.33)/(60 − 30) = 0.158 «103 km min−2/3» and R = 0.158 × 60 − 6.07 ≈ 3.4 «103 km»
✓ 1
Allow ECF from (c).
R = 3.40 × 106 m
✓ 1
Accept 3.33 × 106 to 3.47 × 106 m. Unit and power of ten required. Award [2] for CNA.
Part (e)
r = 3.40 × 106 + 6.38 × 106 = 9.78 × 106 m AND T = 486.4 × 60 = 2.918 × 104 s
✓ 1
Award [0] for MP1 if h is used as the orbital radius «gives 1.8 × 1023 kg».
M = 4π2r3/(GT2) = 4π2 × (9.78 × 106)3/(6.67 × 10−11 × (2.918 × 104)2) = 6.50 × 1023 kg
✓ 1
Allow ECF from (d): accept 6.36 × 1023 to 6.64 × 1023 kg for R from 3.33 × 106 to 3.47 × 106 m.
ΔM/M = 3 × 0.10/9.78 = 0.031, so M = (6.5 ± 0.2) × 1023 kg
✓ 1
Factor of 3 needed. Uncertainty in r is the uncertainty in R. MP3 is for matching the precision of value and uncertainty: accept (6.50 ± 0.20) × 1023 kg. Allow ECF from MP2.
Part (f)
The accepted value lies within the range «6.3 × 1023 to 6.7 × 1023 kg», so the result agrees with it
✓ 1
Allow ECF from (e). A comparison with the range is needed, not just 'close'.
Part (g)
Include orbits at lower altitudes «closer to the surface» / spread over a wider range, so the line is extrapolated a shorter distance and small changes in its gradient change the intercept less
✓ 1
Accept: use many more orbits so that the scatter has less effect on the gradient. A reason is required. Do not accept 'repeat the readings'.
Answers: (b) 38.0 min2/3 · (d) R = 3.40 × 106 m · (e) M = (6.5 ± 0.2) × 1023 kg (the remaining parts are explanations — see the table above)
86D-1B-02Variation of g with heightPaper 1BHard11 marks
A student investigates how the gravitational field strength g varies with height h above the ground. A portable gravimeter is carried by lift to different floors of a tall office building. On each floor it measures the change Δg in g relative to a reference point at street level. Each value of Δg is the mean of several readings and has an uncertainty of ±10 μm s−2; the uncertainty in h is negligible.
For h much smaller than the radius R of the Earth, g = GM/(R + h)² leads to Δg = −(2g0/R)h, where g0 = 9.81 m s−2 is the field strength at street level.
The graph shows the data with error bars.
h / m
Δg / μm s−2
40
−111
80
−236
120
−345
160
−467
200
−576
240
−701
Graph drawn to scale
(a)
The gravimeter displays readings to the nearest 0.1 μm s−2. Suggest why the uncertainty in Δg is much larger than this.
(1)
(b)
Draw the line of best fit for the data.
(1)
(c)
Determine the gradient of the line and its absolute uncertainty. Draw lines of maximum and minimum gradient on the graph.
(3)
(d)
Determine, from your answer to (c), the radius R of the Earth and its absolute uncertainty.
(2)
(e)
The accepted value of R is 6.37 × 106 m. Comment on the student's result.
(1)
(f)
The building has a very large mass. Explain how the gravitational attraction of the building affects the measured values of Δg and hence the value of R obtained.
(2)
(g)
Suggest one improvement to the investigation.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Repeated readings vary/fluctuate (e.g. because of vibration or sway of the building, lifts, traffic, wind), so the spread of the readings, not the resolution, sets the uncertainty
✓ 1
OWTTE
Part (b)
Single straight line passing through all the error bars with the points scattered on both sides
✓ 1
Line does not need to pass through the origin.
Part (c)
Best gradient = −2.92 × 10−6 s−2
✓ 1
Accept −2.87 to −2.97 × 10−6 s−2 (μm s−2 per m). Sign not required.
Max and min gradient lines drawn through all the error bars, with gradients of about −3.00 × 10−6 and −2.85 × 10−6 s−2
✓ 1
Uncertainty = ½(max − min) ≈ ±0.07 × 10−6 s−2
✓ 1
Accept ±0.05 to ±0.12 × 10−6 s−2. Allow the larger of (max − best) and (best − min).
Part (d)
R = 2g0/|gradient| = 2 × 9.81/2.92 × 10−6 = 6.7 × 106 m
✓ 1
Allow ECF from (c).
ΔR = R × (Δgradient/gradient) ≈ 0.2 × 106 m, so R = (6.7 ± 0.2) × 106 m
✓ 1
MP2 is for the uncertainty with matching precision of value and uncertainty. Accept an uncertainty range 6.54–6.88 × 106 m from the max/min gradients.
Part (e)
The accepted value lies (just) outside the student's range, so the result is not consistent with it: there is a systematic error / the result is too large
✓ 1
Allow ECF from (d).
Part (f)
On the upper floors most of the building's mass is below the gravimeter and pulls downwards (at street level it is above and pulls upwards), so Δg is less negative than for the Earth alone
✓ 1
OWTTE
The magnitude of the gradient is too small, so R = 2g0/|gradient| is too large — consistent with (e)
✓ 1
Award MP2 only if the direction of the effect on the gradient is correct.
Part (g)
Repeat the measurements on a structure of much smaller mass (e.g. an open lattice mast) OR correct the readings for the building's own attraction OR extend the range of heights to reduce the percentage uncertainty in the gradient
✓ 1
Accept any specific, relevant improvement. Do not accept “repeat the readings” without further detail.
Answers: (c) −2.92 × 10−6 s−2 ± 0.07 × 10−6 s−2 · (d) R = (6.7 ± 0.2) × 106 m (the remaining parts are explanations — see the table above)
87D-1B-11Newton's law of gravitationPaper 1BHard11 marks
A student uses a school torsion balance to investigate the gravitational force between two spheres. A small lead sphere of mass 15.0 g is fixed to each end of a light beam hanging from a fine wire. A large lead sphere of mass 1.50 kg is placed near each small sphere, and the gravitational attraction twists the wire until the beam comes to rest. The balance has been calibrated so that the force F on each small sphere is found from the deflection of a laser spot.
The centre-to-centre separation r of each pair of spheres is varied. The uncertainty in r is ±0.5 mm and the uncertainty in F is ±0.20 × 10−10 N. The graph shows F against 1/r² with error bars.
r / m
1/r² / m−2
F / 10−10 N
0.046
473
6.93
0.052
5.26
0.060
278
4.07
0.070
204
2.90
0.085
138
2.07
0.100
100
1.40
Graph drawn to scale
(a)
Calculate 1/r² for r = 0.052 m and its absolute uncertainty.
(2)
(b)
Draw the line of best fit for the data.
(1)
(c)
Determine, using lines of maximum and minimum gradient, a value for the gravitational constant G and its absolute uncertainty.
(4)
(d)
The accepted value of G is 6.67 × 10−11 N m² kg−2. Comment on the student's result.
(1)
(e)
Each small sphere is also attracted by the large sphere on the far side of the beam. Explain the effect of this on the value of G obtained.
(2)
(f)
Suggest one change to the apparatus that would reduce the percentage uncertainty in F.
Accept ±7 m−2 (±7.1). Award [0] for MP2 if the fractional uncertainty is not doubled.
Part (b)
Single straight line passing through all the error bars, close to the origin
✓ 1
Part (c)
Best gradient ≈ 1.46 × 10−12 N m²
✓ 1
Accept 1.40–1.52 × 10−12 N m².
G = gradient/(Mm) = 1.46 × 10−12/(1.50 × 0.0150) = 6.5 × 10−11 N m² kg−2
✓ 1
Allow ECF from the candidate's gradient.
Max and min gradient lines drawn through all error bars, gradients ≈ 1.61 × 10−12 and 1.34 × 10−12 N m²
✓ 1
Accept any pair of lines that pass through all the error bars.
ΔG = ½(7.17 − 5.94) × 10−11 ≈ 0.6 × 10−11, so G = (6.5 ± 0.6) × 10−11 N m² kg−2
✓ 1
Accept ±0.4 to ±0.8 × 10−11. MP4 is for matching the precision of value and uncertainty.
Part (d)
The accepted value lies within the student's range of values, so the result is consistent with it (although the student's value is lower)
✓ 1
Allow ECF from (c).
Part (e)
The far large sphere pulls each small sphere in the opposite direction to the near sphere, so the net force measured is smaller than GMm/r²
✓ 1
OWTTE
The measured F values, the gradient and hence G are all too small / G is underestimated
✓ 1
Award MP2 only if the direction of the effect is correct.
Part (f)
Use large spheres of greater mass (or denser material) so the forces are larger while the absolute uncertainty in reading the deflection stays the same
✓ 1
OWTTE. Accept a longer light path to the scale so that the laser-spot deflection for a given force is larger.
Answers: (a) 1/r² = (370 ± 7) m−2 · (c) G = (6.5 ± 0.6) × 10−11 N m² kg−2(the remaining parts are explanations — see the table above)
An astronomy club combines observations of a comet to find its distance r from the Sun at different positions on its orbit. θ is the angle, measured at the Sun, between the line from the Sun to the comet and the line from the Sun to the comet’s closest point. Each value of r has an uncertainty of ±2 %.
If the orbit is an ellipse with the Sun at one focus, then 1/r = (1 + e cos θ)/p, where e (the eccentricity) and p are constants for the orbit.
θ / °
r / AU
cos θ
(1/r) / AU−1
0
0.733
1.000
1.364 ± 0.027
25
0.778
0.906
1.285 ± 0.026
50
0.851
0.643
1.175 ± 0.024
75
1.045
0.259
0.957 ± 0.019
100
1.353
125
1.840
−0.574
0.543 ± 0.011
150
2.615
−0.866
0.382 ± 0.008
170
3.060
−0.985
0.327 ± 0.007
Graph drawn to scale
(a)
State Kepler’s first law.
(1)
(b)
Complete the table for θ = 100°, including the absolute uncertainty in 1/r.
(2)
(c)
Plot the point for θ = 100° and draw the line of best fit for the data.
(2)
(d)
Determine the gradient of your line and its intercept on the 1/r axis.
(2)
(e)
Determine e and the greatest distance of the comet from the Sun.
(2)
(f)
Outline how the graph supports Kepler’s first law for this comet.
(1)
(g)
A database gives e = 0.618 for this comet. Calculate the percentage difference between your value of e and this value.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
each planet / orbiting body moves in an ellipse with the Sun «central body» at one focus
the data fit a straight line, so they fit the equation of an ellipse whose focus is at the Sun «the point from which r and θ are measured»
✓ 1
OWTTE
Part (g)
(0.620 − 0.618)/0.618 × 100 ≈ 0.3 %
✓ 1
Allow ECF from (e): any percentage consistent with the candidate’s e.
Answers: (b) −0.174; 0.739 ± 0.015 AU−1 · (d) gradient 0.52 AU−1, intercept 0.83 AU−1 · (e) e = 0.62; 3.2 AU (the remaining parts are explanations — see the table above)
89D-2-19Gravitational field strengthPaper 2Easy10 marks
Rhea is one of the moons of Saturn. It may be treated as a uniform sphere of mass 2.31 × 1021 kg and radius 7.64 × 105 m, and it has no atmosphere. A robotic lander of mass 450 kg is designed to explore its surface.
G = 6.67 × 10−11 N m² kg−2; gravitational field strength at the surface of the Earth = 9.81 N kg−1.
(a)
Show that the gravitational field strength at the surface of Rhea is about 0.26 N kg−1.
(1)
(b)
The engine of the lander can provide a maximum upward thrust of 300 N.
(i)
Determine whether this engine could hold the lander at rest just above the surface of Rhea, and whether it could do so just above the surface of the Earth.
(2)
(ii)
Calculate the initial acceleration of the lander when it lifts off from Rhea with the engine at maximum thrust.
(2)
(c)
The lander releases a small sensor from rest at a height of 12.0 m above the surface.
(i)
Calculate the time taken for the sensor to reach the surface.
(2)
(ii)
Outline why the gravitational field strength may be treated as constant during this fall.
(1)
(d)
A rock sample collected by the lander has a weight of 0.50 N on Rhea.
weight on Rhea = 450 × 0.264 = 119 N, which is less than 300 N, so the lander can hover above Rhea
✓ 1
Allow ECF from (a).
weight on the Earth = 450 × 9.81 = 4.41 × 103 N, which is greater than 300 N, so it cannot hover above the Earth
✓ 1
A conclusion without a calculated weight scores [0] for that mark.
Part (b)(ii)
resultant «upward» force = 300 − 119 = 181 N
✓ 1
Allow ECF from (b)(i).
a = 181 / 450 = 0.40 m s−2
✓ 1
Accept 0.40–0.41 m s−2. Award [2] for CNA.
Part (c)(i)
s = ½gt², so t = √(2 × 12.0 / 0.264)
✓ 1
Allow ECF from (a).
t = 9.5 s
✓ 1
Accept 9.5–9.6 s. Award [2] for CNA. An answer of 1.6 s «uses g = 9.81 N kg−1» scores [0].
Part (c)(ii)
12.0 m is negligible compared with the radius «7.64 × 105 m», so the distance from the centre and hence g change by a negligible fraction «≈ 0.003 %»
✓ 1
OWTTE. Accept “the field near the surface is «approximately» uniform over such a small height”.
Part (d)
mass of sample = 0.50 / 0.264 = 1.89 kg «the mass is the same on both bodies»
✓ 1
Allow ECF from (a).
weight on the Earth = 1.89 × 9.81 = 18.6 N
✓ 1
Accept 18.6–18.9 N. Award [2] for CNA.
Answers: (a) 0.264 N kg−1 · (b)(i) 119 N on Rhea (yes); 4.41 × 103 N on the Earth (no) · (b)(ii) 0.40 m s−2 · (c)(i) 9.5 s · (d) 18.6 N (the remaining parts are explanations — see the table above)
Mars has two small moons, Phobos and Deimos. Both move in almost circular orbits in the plane of the equator of Mars and in the same direction as Mars rotates. The table gives data for the two moons.
Mars rotates once every 24.6 h. Mass of Mars = 6.42 × 1023 kg; mass of Phobos = 1.07 × 1016 kg; G = 6.67 × 10−11 N m² kg−2.
Moon
orbital radius r / m
orbital period T / h
Phobos
9.38 × 106
7.65
Deimos
2.35 × 107
30.3
(a)
State Kepler's third law.
(1)
(b)
Show that the data in the table are consistent with Kepler's third law.
(2)
(c)
A communications satellite is to be placed in a circular equatorial orbit in which it stays above the same point on the surface of Mars.
Determine whether the radius of this orbit lies between the orbits of Phobos and Deimos.
(3)
(d)
For an observer on Mars, the stars and Deimos rise in the east.
Explain why Phobos rises in the west.
(2)
(e)
Calculate the magnitude of the gravitational force between Mars and Phobos.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
the square of the orbital period is proportional to the cube of the orbital radius «for bodies orbiting the same central body»
✓ 1
Accept T² ∝ r³ with symbols defined. Accept «mean distance» or «semi-major axis» for radius.
Part (b)
T²/r³ for Phobos = 7.65² / (9.38 × 106)³ = 7.09 × 10−20 «h² m−3» AND for Deimos = 30.3² / (2.35 × 107)³ = 7.07 × 10−20 «h² m−3»
✓ 1
Accept the ratio method: (30.3/7.65)² = 15.7 and (2.35 × 107/9.38 × 106)³ = 15.7.
the two values are equal «to 2 s.f., within the precision of the data», so the law holds
✓ 1
Part (c)
the period of the satellite must equal the rotation period of Mars, 24.6 h
✓ 1
r = 9.38 × 106 × (24.6/7.65)2/3 = 2.04 × 107 m
✓ 1
Accept use of Deimos data «same answer». Accept 2.0–2.1 × 107 m. Accept a derivation of r from GM if the mass of Mars is used.
9.38 × 106 m < 2.04 × 107 m < 2.35 × 107 m, so yes, the orbit lies between them
✓ 1
MP3 only scores with a value of r in the range.
Part (d)
the period of Phobos «7.65 h» is shorter than the rotation period of Mars «24.6 h» «so Phobos moves round faster than the surface turns»
✓ 1
Phobos overtakes the observer, moving eastwards across the sky relative to the surface, so it appears in the west and sets in the east
Answers: (b) T²/r³ ≈ 7.1 × 10−20 h² m−3 for both · (c) r = 2.04 × 107 m — yes · (e) 5.21 × 1015 N (the remaining parts are explanations — see the table above)
91D-2-51Variation of g with heightPaper 2Easy11 marks
Io, a moon of Jupiter, may be treated as a uniform sphere of mass 8.93 × 1022 kg and radius 1.82 × 106 m. Io has no atmosphere. Volcanoes on Io throw material up to a height of 330 km above its surface.
Axes for (b)(ii)
(a)
Show that the gravitational field strength at the surface of Io is about 1.8 N kg−1.
(1)
(b)
(i)
Assume that the gravitational field strength is uniform. Calculate the vertical speed with which material must leave a vent to reach a height of 330 km.
(2)
(ii)
Calculate the gravitational field strength at a height of 330 km. Hence sketch, on the axes, how the gravitational field strength varies with height h from 0 to 330 km.
(3)
(iii)
State and explain whether your answer to (b)(i) is an overestimate or an underestimate of the speed actually needed.
(2)
(c)
A fragment leaves a vent at 1.0 km s−1 at 45° to the horizontal.
(i)
Assuming a uniform field of 1.8 N kg−1 and a flat surface, calculate the time taken for the fragment to land.
(2)
(ii)
The fragment lands about 560 km from the vent. Suggest why the assumptions in (c)(i) are not reliable for this flight.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
g = 6.67 × 10−11 × 8.93 × 1022/(1.82 × 106)2 = 1.798 N kg−1
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (b)(i)
v2 = 2gh «0 = v2 − 2gh» with h = 3.30 × 105 m
✓ 1
v = √(2 × 1.80 × 3.30 × 105) = 1090 m s−1 «≈ 1.1 km s−1»
✓ 1
Award [2] for CNA. Award [1 max] if h is not converted to metres.
Part (b)(ii)
g = 6.67 × 10−11 × 8.93 × 1022/(2.15 × 106)2 = 1.29 N kg−1
✓ 1
Do not accept 3.30 × 105 m used as the distance from the centre.
curve starting at 1.8 N kg−1 at h = 0 and ending at about 1.3 N kg−1 at 330 km
✓ 1
Allow ECF from the calculated value.
curve decreasing with a gradient whose magnitude decreases «concave upwards», not a straight line and not reaching zero
✓ 1
Part (b)(iii)
the field «deceleration» is less than 1.8 N kg−1 at greater heights
✓ 1
so less speed is needed to reach 330 km: (b)(i) is an overestimate
✓ 1
MP2 only scores if MP1 scores.
Part (c)(i)
vertical component = 1000 sin 45° = 707 m s−1 AND t = 2vy/g
✓ 1
t = 2 × 707/1.8 = 786 s
✓ 1
Accept 780–790 s. Award [2] for CNA.
Part (c)(ii)
560 km is a large fraction of Io’s radius, so the surface curves away beneath the fragment AND the field lines are radial, so over this distance the field changes direction (and decreases with height) — it is not uniform
✓ 1
Either idea for the mark. OWTTE
Answers: (a) 1.80 N kg−1 · (b)(i) 1090 m s−1 · (b)(ii) 1.29 N kg−1 · (c)(i) 786 s (the remaining parts are explanations — see the table above)
92D-2-55Newton's law of gravitationPaper 2Easy11 marks
During a ship-to-ship transfer, oil is pumped from tanker A to tanker B while the two ships lie side by side. The diagram is a view from above. The centres of mass of the tankers, marked +, are 60 m apart.
Mass of A = 1.6 × 108 kg; mass of B = 8.0 × 107 kg.
Diagram NOT to scale
(a)
State Newton’s universal law of gravitation.
(2)
(b)
Calculate the gravitational force between the tankers, treating them as point masses.
(2)
(c)
Draw, on the diagram, arrows to represent the gravitational force on A and the gravitational force on B.
(1)
(d)
Calculate the acceleration of each tanker caused by this force alone.
(2)
(e)
The transfer takes 6 hours. Determine whether the gravitational attraction alone could close a gap of 1.0 m between the tankers in this time.
(2)
(f)
(i)
Outline why treating the tankers as point masses is not a good approximation here.
(1)
(ii)
Suggest why, in practice, the crews do not need to take the gravitational attraction into account.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
the force between two point masses is proportional to the product of their masses
✓ 1
and inversely proportional to the square of their separation «and is attractive»
✓ 1
Accept F = Gm1m2/r2 with all symbols defined for [2].
Part (b)
F = 6.67 × 10−11 × 1.6 × 108 × 8.0 × 107/602
✓ 1
F = 237 N
✓ 1
Award [2] for CNA. Accept 2.4 × 102 N.
Part (c)
arrow on A towards B and arrow on B towards A, along the line joining the centres, of EQUAL length
✓ 1
Both arrows needed.
Part (d)
aA = 237/1.6 × 108 = 1.5 × 10−6 m s−2
✓ 1
Allow ECF from (b).
aB = 237/8.0 × 107 = 3.0 × 10−6 m s−2
✓ 1
aB = 2aA.
Part (e)
relative acceleration = 4.4 × 10−6 m s−2 AND s = ½at2 gives t = √(2 × 1.0/4.4 × 10−6) = 671 s «≈ 11 min»
✓ 1
Allow ECF. Accept a calculation of the distance closed in 6 hours (≈ 1 km) instead.
this is much less than 6 hours, so yes, gravity alone could close the gap «if nothing else acted»
✓ 1
MP2 only scores if a comparison with 6 hours is made. Award [1 max] if only one tanker’s acceleration is used and the conclusion is consistent.
Part (f)(i)
the tankers are far from spherical and their size «hundreds of metres long» is not small compared with the 60 m separation
✓ 1
OWTTE
Part (f)(ii)
other forces, such as those from mooring lines, fenders, wind, waves and water resistance, are very much larger than about 240 N
✓ 1
OWTTE
Answers: (b) 237 N · (d) 1.5 × 10−6 m s−2 and 3.0 × 10−6 m s−2 · (e) 671 s (the remaining parts are explanations — see the table above)
93D-2-56Weightlessness & free fallPaper 2Easy11 marks
A research aircraft gives passengers periods of “weightlessness”. At a height of 7.5 km it is moving at 160 m s−1 at 45° above the horizontal. From then on, its engines are adjusted so that their thrust exactly balances air resistance, and the only unbalanced force on the aircraft is its weight. The aircraft follows a curved path until it is again moving at 45° below the horizontal at the same height.
Radius of the Earth = 6.37 × 106 m; gravitational field strength at sea level = 9.81 N kg−1.
Diagram for (d)(i)
(a)
Calculate the gravitational field strength at a height of 7.5 km, and the percentage by which it is less than at sea level.
(2)
(b)
Use g = 9.81 N kg−1 in (b).
(i)
Calculate the greatest height gained by the aircraft during the manoeuvre.
(2)
(ii)
Calculate the duration of the manoeuvre.
(2)
(c)
Explain why the passengers float freely inside the cabin even though a gravitational force acts on them.
(2)
(d)
On another flight the aircraft follows a flatter path, so that its acceleration is 1.62 m s−2 vertically downwards. This makes the passengers feel as if they were on the Moon.
(i)
Draw, on the diagram, the forces acting on a passenger standing on the floor. The lengths of your arrows should show the relative sizes of the forces.
(2)
(ii)
Calculate the normal force on a passenger of mass 75 kg.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
g = 9.81 × (6.37 × 106/6.3775 × 106)2 = 9.787 N kg−1
✓ 1
Accept GM/r2 with M = 5.97 × 1024 kg «9.77–9.79».
less by 0.24 % «about 0.2 %»
✓ 1
Accept 0.2–0.3 %. Allow ECF.
Part (b)(i)
vertical component of velocity = 160 sin 45° = 113 m s−1
✓ 1
h = 1132/(2 × 9.81) = 652 m
✓ 1
Award [2] for CNA. Accept 650–655 m.
Part (b)(ii)
vertical velocity changes from +113 m s−1 to −113 m s−1 «t = 2vy/g»
✓ 1
t = 2 × 113/9.81 = 23 s
✓ 1
Award [2] for CNA. Award [1 max] for 11.5 s «time to the top only».
Part (c)
the passengers and the aircraft have the same acceleration g «both are in free fall»
✓ 1
so no contact force from the floor or seat is needed to accelerate a passenger; there is no motion of a passenger relative to the cabin
✓ 1
OWTTE. Do not accept «there is no gravity».
Part (d)(i)
downward arrow labelled weight / gravitational force AND upward arrow labelled normal / reaction force, both on the passenger
✓ 1
Do not award if any other force is shown.
weight arrow clearly longer than the normal force arrow «normal force ≈ 0.83 × weight»
✓ 1
Part (d)(ii)
N = m(g − a) = 75 × (9.81 − 1.62) = 614 N
✓ 1
Answers: (a) 9.79 N kg−1, 0.24 % · (b)(i) 652 m · (b)(ii) 23 s · (d)(ii) 614 N (the remaining parts are explanations — see the table above)
94IBP2S-028-01-11Geostationary orbitsPaper 2Easy6 marks
Satellites in a global navigation system move in circular orbits around Earth. Each satellite has an orbital period of 11.97 hours. Mass of Earth M = 5.97×1024 kg; radius of Earth R = 6.37×106 m; G = 6.67×10−11 N m2 kg−2.
(a)
State Kepler's third law.
(1)
(b)
Show that, for a satellite in a circular orbit of radius r, \(T^2 = \frac{4\pi^2 r^3}{GM}\).
(2)
(c)
Calculate the altitude of a navigation satellite above Earth's surface.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The square of the orbital period is proportional to the cube of the orbital radius (or semi-major axis): T² ∝ r³
✓ 1
Part (b)
Gravitational force provides the centripetal force: GMm/r² = mv²/r (or mω²r)
✓ 1
Substitutes v = 2πr/T (or ω = 2π/T) and rearranges to T² = 4π²r³/GM
✓ 1
Part (c)
T = 11.97 × 3600 = 4.31×104 s
✓ 1
r = (GMT²/4π²)1/3 = 2.66×107 m
✓ 1
altitude = r − R = 2.02×107 m
✓ 1
Award full marks for a correct answer from a valid method.
—
Common error: The equation gives the orbital radius, measured from Earth's centre. Subtract Earth's radius to find the altitude.
95IBP2S-028-01-15Geostationary orbitsPaper 2Easy5 marks
A household satellite TV dish points permanently in one fixed direction. It receives signals from a satellite in geostationary orbit. The geostationary orbit is at an altitude of 3.58×107 m above Earth's surface. The speed of electromagnetic waves is 3.00×108 m s−1.
(a)
Explain why the dish does not need to move to track the satellite.
(2)
(b)
A ground station on the equator lies directly below the satellite. Calculate the minimum time for a signal to travel from the ground station up to the satellite and back down to the ground.
(2)
(c)
Outline why geostationary satellites are not used for detailed photography of Earth's surface.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The satellite's period equals Earth's rotation period, and it orbits above the equator in the same direction as Earth rotates
✓ 1
So the satellite stays in a fixed position relative to the dish
✓ 1
Part (b)
Total distance = 2 × 3.58×107 = 7.16×107 m
✓ 1
t = 7.16×107/3.00×108 = 0.239 s
✓ 1
Part (c)
They are very far from Earth (about 36 000 km), so the images resolve much less detail than images taken from a low orbit
✓ 1
Common error: Use the altitude, not the orbital radius, for the distance between a ground station and the satellite.
96IBP2S-028-00-11Newton's law of gravitationPaper 2Easy5 marks
At an athletics competition, two men's shots are placed next to each other on the ground. Each shot is a uniform metal sphere of mass 7.26 kg. Their centres are 0.30 m apart.
(a)
State Newton's universal law of gravitation.
(2)
(b)
Calculate the magnitude of the gravitational force between the two shots.
(2)
(c)
Outline why this force does not cause the shots to roll towards each other.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Every two point masses attract each other with a force proportional to the product of their masses . The force is inversely proportional to the square of the distance between them .
✓ 2
Part (b)
F = Gm1m2/r2 = 6.67×10−11 × 7.26 × 7.26/0.302 .
✓ 1
F = 3.9×10−8 N .
✓ 1
Part (c)
The gravitational force (about 4×10−8 N, roughly 10−9 of a shot's weight of about 71 N) is far smaller than the maximum resistive force or friction the ground can provide, so it is balanced and there is no resultant force .
✓ 1
Exam tip: For uniform spheres, use the centre-to-centre distance for r.
97IBP2S-028-02-11Gravitational field strengthPaper 2Easy4 marks
An engineering team is testing a robotic lander of mass 850 kg for a mission to Mars. Mass of Mars = 6.42×1023 kg Radius of Mars = 3.39×106 m
(a)
Define gravitational field strength at a point.
(1)
(b)
Calculate the gravitational field strength at the surface of Mars.
(2)
(c)
Calculate the weight of the lander on the surface of Mars.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The gravitational force per unit mass acting on a small (test) mass placed at that point
✓ 1
Part (b)
g = GM/R2 = (6.67×10−11 × 6.42×1023)/(3.39×106)2
✓ 1
g = 3.73 N kg−1
✓ 1
Part (c)
W = mg = 850 × 3.73 = 3.17×103 N (allow error carried forward from the field strength calculated in the previous part)
✓ 1
Common error: A definition that leaves out 'per unit mass' (or the idea of a small test mass) describes a force, not a field strength, and does not earn the mark.
98IBP2S-028-02-12Orbits and satellitesPaper 2Easy6 marks
A sports scientist imagines a high jumper competing in a pressurised dome on the Moon. On both Earth and the Moon, the athlete leaves the ground with a vertical take-off speed of 4.5 m s−1 for the centre of mass. Air resistance can be ignored. Mass of the Moon = 7.35×1022 kg Radius of the Moon = 1.74×106 m
(a)
Show that the gravitational field strength at the Moon's surface is about 1.6 N kg−1.
(2)
(b)
Calculate the maximum rise of the athlete's centre of mass on Earth.
(1)
(c)
Calculate the maximum rise of the athlete's centre of mass on the Moon.
(2)
(d)
Outline why the gravitational field strength can be treated as constant during the jump.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
g = GM/R2 = (6.67×10−11 × 7.35×1022)/(1.74×106)2
✓ 1
g = 1.62 N kg−1, given to at least 3 significant figures
✓ 1
Part (b)
h = v2/(2g) = 4.52/(2 × 9.8) = 1.0 m (accept 1.03 m)
✓ 1
Part (c)
h = 4.52/(2 × 1.62)
✓ 1
h = 6.3 m (accept any answer from 6.2 m to 6.3 m, including 6.25 m; allow a carried-forward value of g from the first part, and 1.6 N kg−1 gives 6.3 m)
✓ 1
Part (d)
The height of the jump (a few metres) is negligible compared with the Moon's radius (about 1.7×106 m), so the distance from the Moon's centre, and hence g, hardly changes
✓ 1
Exam tip: In a 'show that' question, give your answer to more significant figures than the value quoted, and show the substitution clearly.
99IBP2S-028-04-11Geostationary orbitsPaper 2Easy7 marks
Satellites in a global navigation system orbit the Earth in circular orbits with a period of 11.97 hours. The signals they transmit allow receivers in phones and cars to determine their position. Mass of Earth = 5.97×1024 kg Radius of Earth = 6.37×106 m G = 6.67×10−11 N m2 kg−2
(a)
Show that the orbital radius of one of these satellites is about 2.66×107 m.
(3)
(b)
Calculate the height of the satellite above the Earth's surface.
(1)
(c)
Calculate the orbital speed of the satellite.
(2)
(d)
A replacement satellite of twice the mass is placed in the same orbit. State and explain whether its orbital speed is different.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Equates the gravitational force to the centripetal force: GMm/r2 = 4π2mr/T2, so r3 = GMT2/(4π2)
✓ 1
Converts the period: T = 11.97 × 3600 = 4.31×104 s
✓ 1
r = (6.67×10−11 × 5.97×1024 × (4.31×104)2 / 4π2)1/3 = 2.656×107 m, quoted to more than 2 significant figures
✓ 1
Part (b)
h = 2.66×107 − 6.37×106 = 2.02×107 m
✓ 1
Part (c)
v = 2πr/T = 2π × 2.66×107 / 4.31×104
✓ 1
v = 3.87×103 m s−1 (accept 3.87–3.88×103 m s−1)
✓ 1
Part (d)
The speed is the same, because the satellite mass cancels in GMm/r2 = mv2/r, giving v = √(GM/r)
✓ 1
Exam tip: In a 'show that' question, give the final value to at least one more significant figure than the value stated, so that you have clearly calculated it.
100IBP2S-028-04-12Geostationary orbitsPaper 2Easy8 marks
A polar-orbiting weather satellite images the whole Earth's surface as the planet rotates beneath it. The satellite moves in a circular orbit at a height of 820 km above the Earth's surface. Mass of Earth = 5.97×1024 kg Radius of Earth = 6.37×106 m G = 6.67×10−11 N m2 kg−2
(a)
The satellite moves at constant speed. Explain why it is nevertheless accelerating.
(2)
(b)
Calculate the gravitational field strength at the satellite's orbit.
(2)
(c)
Calculate the orbital speed of the satellite.
(2)
(d)
Determine the number of complete orbits the satellite makes in 24 hours.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The direction of the velocity changes continuously, so the velocity changes
✓ 1
A resultant force (gravity) acts towards the centre of the Earth, perpendicular to the velocity
✓ 1
Part (b)
r = 6.37×106 + 8.20×105 = 7.19×106 m
✓ 1
g = GM/r2 = 7.70 N kg−1
✓ 1
Part (c)
v2/r = g, or v = √(GM/r)
✓ 1
v = √(7.70 × 7.19×106) = 7.44×103 m s−1; ECF from (b)
✓ 1
Part (d)
T = 2πr/v = 2π × 7.19×106 / 7.44×103 = 6.07×103 s; ECF from (c)
✓ 1
Number of orbits = 86400 / 6070 = 14.2, so 14 complete orbits
✓ 1
Common error: Using the height above the surface instead of the distance from the Earth's centre is the most frequent way to lose marks in orbit calculations.
101D-2-01Gravitational field strengthPaper 2Medium12 marks
A rover on the surface of Mars carries a simple pendulum of length 0.800 m. A camera on the rover records that the pendulum completes 20 small oscillations in 58.4 s.
The radius of Mars is 3.39 × 106 m.
Diagram NOT to scale
(a)
(i)
State what is meant by the gravitational field strength at a point.
(1)
(ii)
Show that the gravitational field strength at the surface of Mars is about 3.7 N kg−1.
(2)
(b)
Calculate the mass of Mars.
(2)
(c)
Diagram 1 shows Mars. Diagram 2 shows a small region of space just above the surface near the rover, greatly magnified.
(i)
Draw, on Diagram 1, the gravitational field lines of Mars.
(2)
(ii)
Draw, on Diagram 2, the gravitational field lines in this region. Outline why they differ from the lines you drew on Diagram 1.
(2)
(d)
A pendulum clock keeps correct time on the Earth, where the gravitational field strength is 9.81 N kg−1. The clock is taken to Mars. Determine whether the clock runs fast or slow on Mars, and by how many minutes it is wrong after one hour of real time.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
«gravitational» force per unit mass acting on a small «point» mass placed at that point
✓ 1
Accept g = F/m with the symbols defined. Do not accept «acceleration due to gravity» alone.
Part (a)(ii)
T = 58.4/20 = 2.92 s
✓ 1
g = 4π²l/T² = 4π² × 0.800/2.92² = 3.70 N kg−1
✓ 1
Must see full substitution OR answer to at least 3 s.f. Unrounded answer 3.704 N kg−1.
Part (b)
M = gR²/G = 3.70 × (3.39 × 106)²/6.67 × 10−11
✓ 1
Allow ECF from (a)(ii).
M = 6.38 × 1023 kg
✓ 1
Award [2] for CNA. Accept 6.3–6.4 × 1023 kg.
Part (c)(i)
straight radial lines, evenly spaced around the planet, meeting the surface at right angles
✓ 1
At least four lines. Lines end at the surface; they must not cross.
arrows on the lines pointing towards Mars «towards its centre»
✓ 1
Do not award if any arrow points away from the planet.
Part (c)(ii)
parallel, equally spaced, vertical lines with arrows pointing down towards the surface
✓ 1
the region is very small compared with the radius of Mars, so the radial lines are almost parallel there and the field strength is almost the same everywhere in it «uniform field»
✓ 1
OWTTE
Part (d)
T ∝ 1/√g, so TMars/TEarth = √(9.81/3.70) = 1.63
✓ 1
Allow ECF from (a)(ii).
the period is longer on Mars, so the clock ticks less often and runs slow «loses time»
✓ 1
MP2 only scores if the period on Mars is stated or shown to be longer.
in one hour the clock records 3600/1.63 = 2210 s, so it is about 23 minutes slow
✓ 1
Accept 22–24 minutes.
Answers: (a)(ii) g = 3.70 N kg−1 · (b) M = 6.38 × 1023 kg · (d) runs slow by about 23 min (the remaining parts are explanations — see the table above)
102D-2-02Variation of g with distancePaper 2Medium11 marks
Mercury may be treated as a uniform sphere of radius R = 2.44 × 106 m. The graph shows how the gravitational field strength g due to Mercury varies with the distance r from the centre of Mercury, for r ≥ R.
G = 6.67 × 10−11 N m² kg−2.
Graph drawn to scale
(a)
Determine, using the graph, the mass of Mercury.
(2)
(b)
Planet X, a uniform sphere, has the same mass as Mercury but twice its radius.
(i)
Sketch, on the graph, the variation of the gravitational field strength due to planet X with distance from its centre, from the surface of X outwards.
(2)
(ii)
Explain why the two graphs are the same at distances greater than 4.88 × 106 m.
(2)
(iii)
Deduce (mean density of planet X)/(mean density of Mercury).
(1)
(c)
A lander of mass 620 kg approaches Mercury.
(i)
Determine the magnitude of the gravitational force on the lander when it is at a height of 2R above the surface of Mercury.
(3)
(ii)
State the magnitude and direction of the gravitational force that the lander exerts on Mercury at this instant.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
a point read correctly from the curve, e.g. g = 1.38 N kg−1 at r = 4.0 × 106 m OR g = 3.70 N kg−1 at r = 2.44 × 106 m
✓ 1
Any correct reading; r must be in metres.
M = gr²/G = 1.38 × (4.0 × 106)²/6.67 × 10−11 = 3.3 × 1023 kg
✓ 1
Accept 3.2–3.4 × 1023 kg. Award [2] for CNA.
Part (b)(i)
line starts at r = 4.88 × 106 m «2R» with g ≈ 0.92 N kg−1 «one quarter of Mercury's surface value»
✓ 1
Accept a start between 0.85 and 1.0 N kg−1.
for r > 4.88 × 106 m the line lies on Mercury's curve; no line drawn for r < 4.88 × 106 m
✓ 1
Do not award MP2 for a separate curve that lies above or below Mercury's.
Part (b)(ii)
outside a uniform «spherically symmetric» body, the body acts as a point mass at its centre
✓ 1
OWTTE
g = GM/r² depends only on M and r, and the two masses are equal «so the radius of the body does not matter outside it»
✓ 1
Part (b)(iii)
1/8 «the volume is 2³ = 8 times larger for the same mass»
✓ 1
Accept 0.125.
Part (c)(i)
distance from the centre r = 3R = 7.32 × 106 m
✓ 1
Do not accept 2R as the distance.
g = 3.70/3² = 0.411 N kg−1
✓ 1
Accept a value read from the graph, 0.40–0.42 N kg−1, or GM/r² with ECF from (a).
F = 620 × 0.411 = 255 N
✓ 1
Accept 248–260 N. Award [1 max] for 574 N «r = 2R used».
Part (c)(ii)
255 N directed towards the lander «away from the centre of Mercury»
✓ 1
Allow ECF from (c)(i). Both magnitude and direction needed. Newton's third law.
Answers: (a) 3.3 × 1023 kg · (b)(iii) 1/8 · (c)(i) 255 N · (c)(ii) 255 N (the remaining parts are explanations — see the table above)
103D-2-15Orbits and satellitesPaper 2Medium12 marks
A lunar orbiter moves in a circular orbit at a height of 100 km above the surface of the Moon.
Mass of the Moon = 7.35 × 1022 kg. Radius of the Moon = 1.74 × 106 m.
(a)
(i)
Explain why the speed of the orbiter is constant even though a resultant force acts on it.
(2)
(ii)
Show that the speed of the orbiter is about 1.6 km s−1.
(2)
(iii)
Calculate the time for one orbit, in minutes.
(1)
(b)
Tracking shows that orbits this low are disturbed above some regions of the Moon, where there are large masses of unusually dense rock below the surface.
(i)
State the condition under which the Moon can be treated as a point mass at its centre.
(1)
(ii)
Suggest why these dense regions disturb a low orbit much more than an orbit far from the Moon.
(3)
(c)
The orbiter must get rid of 350 W of waste thermal energy. It uses a flat radiator panel of area 0.80 m² and emissivity 0.90, which emits from one face only.
(i)
Determine the temperature of the panel when it emits 350 W. Ignore any radiation absorbed by the panel.
(2)
(ii)
Suggest why the panel must operate at a higher temperature when its emitting face points towards the sunlit surface of the Moon.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
the gravitational force is always perpendicular to the velocity «directed towards the centre of the Moon»
✓ 1
so no work is done on the orbiter; its kinetic energy «speed» is unchanged and only the direction of its velocity changes
✓ 1
OWTTE
Part (a)(ii)
GMm/r² = mv²/r with r = 1.74 × 106 + 1.00 × 105 = 1.84 × 106 m
✓ 1
MP1 requires the orbit radius, not the height.
v = √(6.67 × 10−11 × 7.35 × 1022/1.84 × 106) = 1.63 × 103 m s−1
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (a)(iii)
T = 2π × 1.84 × 106/1630 = 7.1 × 103 s = 118 min
✓ 1
Allow ECF from (a)(ii). Accept 117–119 min.
Part (b)(i)
its mass must be distributed with spherical symmetry «density depends only on the distance from the centre»
✓ 1
Accept «uniform sphere». Accept «the distance from the Moon is very large compared with its radius». OWTTE
Part (b)(ii)
a low orbit passes close to the dense regions
✓ 1
the gravitational force decreases with the square of the distance, so at short range the extra pull of a dense region is relatively large «and is not directed towards the centre of the Moon»
✓ 1
far from the Moon the distance is large compared with the size of the Moon, so it behaves as a point mass and local differences in density have little effect
✓ 1
OWTTE
Part (c)(i)
350 = 0.90 × 5.67 × 10−8 × 0.80 × T⁴
✓ 1
T = 304 K
✓ 1
Award [2] for CNA. Accept 300–305 K.
Part (c)(ii)
the panel then absorbs infrared radiation emitted by the hot lunar surface «and reflected sunlight», so it must emit more power to remove the same 350 W
✓ 1
OWTTE
Answers: (a)(ii) v = 1.63 × 103 m s−1 · (a)(iii) 118 min · (c)(i) 304 K (the remaining parts are explanations — see the table above)
A spacecraft travels from the Earth to Mars on a transfer orbit around the Sun. The transfer orbit is half of an ellipse. Its closest point to the Sun, P, lies on the orbit of the Earth, 1.00 AU from the Sun. Its farthest point from the Sun, A, lies on the orbit of Mars, 1.52 AU from the Sun. The engines are switched off between P and A.
Treat the orbits of the Earth and Mars as circles. The period of the Earth's orbit is 1.00 year.
Diagram NOT to scale
(a)
(i)
State Kepler's first law as it applies to the transfer orbit.
(1)
(ii)
Draw and label arrows on the diagram to show the direction of the gravitational force on the spacecraft and the direction of its velocity when it is at X.
(2)
(iii)
Explain, with reference to your answer to (a)(ii), why the spacecraft slows down as it moves from P to A.
(2)
(b)
(i)
Calculate (gravitational field strength of the Sun at P)/(gravitational field strength of the Sun at A).
(1)
(ii)
State Kepler's second law.
(1)
(iii)
At P and at A the velocity of the spacecraft is perpendicular to the line joining it to the Sun. The speed of the spacecraft at P is 32.7 km s−1. Using Kepler's second law, determine its speed at A.
(3)
(c)
Kepler's third law also applies to an elliptical orbit if the orbital radius is replaced by half of the sum of the greatest and the smallest distances from the Sun.
(i)
Determine the time taken by the spacecraft to travel from P to A, in days.
(3)
(ii)
Suggest why the spacecraft must be launched from the Earth at a particular time.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
the transfer orbit is an ellipse with the Sun at one focus
✓ 1
Do not accept «the Sun at the centre».
Part (a)(ii)
force arrow from X pointing towards the Sun, labelled
✓ 1
velocity arrow from X along the tangent to the path, in the direction of motion «towards A», labelled
✓ 1
The velocity arrow must not be drawn perpendicular to the force arrow.
Part (a)(iii)
the angle between the force and the velocity is greater than 90°, so the force has a component opposite to the velocity
✓ 1
Allow ECF from (a)(ii).
so the gravitational force does negative work on the spacecraft and its kinetic energy decreases
✓ 1
Accept «the component of the force along the path decelerates it». OWTTE
Part (b)(i)
g ∝ 1/r², so the ratio is (1.52/1.00)² = 2.31
✓ 1
Part (b)(ii)
the line joining the Sun and an orbiting body sweeps out equal areas in equal times
✓ 1
Part (b)(iii)
in a short time Δt the area swept out is ½rvΔt, at P and at A
✓ 1
equal areas in equal times, so rPvP = rAvA
✓ 1
vA = 32.7 × 1.00/1.52 = 21.5 km s−1
✓ 1
Award [1 max] for a bare answer with no use of Kepler's second law. Do not accept 49.7 km s−1.
Part (c)(i)
half of the sum of the distances = (1.00 + 1.52)/2 = 1.26 AU
✓ 1
T² ∝ r³ compared with the Earth: T = 1.00 × 1.263/2 = 1.41 years
✓ 1
P to A is half an orbit: ½ × 1.414 × 365 = 258 days
✓ 1
Award [2 max] for a whole orbit «516 days». Accept 257–259 days.
Part (c)(ii)
Mars must reach A at the same time as the spacecraft, so the Earth and Mars must be in a particular relative position at launch
✓ 1
OWTTE
Answers: (b)(i) 2.31 · (b)(iii) 21.5 km s−1 · (c)(i) 258 days (the remaining parts are explanations — see the table above)
105D-2-18Geostationary orbitsPaper 2Medium13 marks
A communications satellite S is in a geostationary orbit around the Earth. The period of the orbit is 8.62 × 104 s.
Mass of the Earth = 5.97 × 1024 kg. Radius of the Earth = 6.37 × 106 m.
The equipment on S needs 4.0 kW of electrical power. In sunlight the power comes from solar panels; in the Earth's shadow it comes from batteries.
Diagram NOT to scale
(a)
(i)
Draw an arrow on the diagram to show the direction of the velocity of S.
(1)
(ii)
Outline why a geostationary orbit must lie in the plane of the equator.
(2)
(iii)
Show that the radius of the orbit is about 4.2 × 107 m.
(2)
(b)
(i)
Calculate the speed of S.
(1)
(ii)
For a few weeks each year S passes through the Earth's shadow once a day. Treat the shadow as a band whose width is equal to the diameter of the Earth. Estimate the longest time that S spends in the shadow.
(2)
(iii)
The fully charged batteries can deliver 1.8 × 107 J. Determine whether they can keep the equipment working for the whole of the time in (b)(ii).
(2)
(c)
(i)
Calculate the shortest time for a radio signal to travel from a ground station directly below S up to S and back to the ground.
(2)
(ii)
Suggest one reason why some internet services use satellites in low orbits rather than geostationary satellites.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
arrow at S along the tangent to the orbit, pointing anticlockwise «the same sense as the Earth's rotation»
✓ 1
Part (a)(ii)
the gravitational force on S is directed towards the centre of the Earth, so the centre of the orbit must be the centre of the Earth
✓ 1
S must stay above one point on the rotating Earth, which moves in a circle about the axis; only the equator's circle has the centre of the Earth as its centre
✓ 1
OWTTE
Part (a)(iii)
GMm/r² = 4π²mr/T², so r³ = GMT²/4π²
✓ 1
r = (6.67 × 10−11 × 5.97 × 1024 × (8.62 × 104)²/4π²)1/3 = 4.22 × 107 m
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (b)(i)
v = 2πr/T = 2π × 4.22 × 107/8.62 × 104 = 3.07 × 103 m s−1
✓ 1
Allow ECF from (a)(iii).
Part (b)(ii)
distance travelled in the shadow ≈ 2 × 6.37 × 106 = 1.27 × 107 m
✓ 1
t = 1.27 × 107/3.07 × 103 = 4.1 × 103 s «≈ 69 min»
✓ 1
Allow ECF from (b)(i). Accept 4.0–4.2 × 103 s.
Part (b)(iii)
energy needed = 4.0 × 103 × 4.1 × 103 = 1.6 × 107 J «1.7 × 107 J with the unrounded time»
✓ 1
Allow ECF from (b)(ii).
this is less than 1.8 × 107 J, so the batteries are «just» sufficient
✓ 1
The conclusion must be consistent with the candidate's energy.
Part (c)(i)
height = 4.22 × 107 − 6.37 × 106 = 3.58 × 107 m
✓ 1
t = 2 × 3.58 × 107/3.00 × 108 = 0.24 s
✓ 1
Award [1 max] for 0.28 s «orbit radius used instead of height».
Part (c)(ii)
the signal travels a much shorter distance, so the delay is much smaller
✓ 1
Accept another valid reason, e.g. less powerful transmitters are needed. OWTTE
Answers: (a)(iii) r = 4.22 × 107 m · (b)(i) v = 3.07 × 103 m s−1 · (b)(ii) 4.1 × 103 s · (b)(iii) 1.6–1.7 × 107 J «sufficient» · (c)(i) 0.24 s (the remaining parts are explanations — see the table above)
106D-2-22Variation of g with distancePaper 2Medium9 marks
A geophysicist searches for buried ore with a gravimeter, an instrument that can detect a change in the gravitational field strength of 5.0 × 10−8 N kg−1. An ore body may be modelled as a uniform sphere of radius 40 m whose centre is 150 m below the surface. Its density is 2500 kg m−3 greater than that of the surrounding rock.
G = 6.67 × 10−11 N m² kg−2; g at the surface = 9.81 N kg−1; radius of the Earth R = 6.37 × 106 m.
Diagram not to scale. A is the point on the surface above the centre of the ore body.
(a)
Show that the extra mass of the ore body, compared with the same volume of surrounding rock, is about 6.7 × 108 kg.
(1)
(b)
Calculate the increase in the gravitational field strength at the surface directly above the centre of the ore body.
(2)
(c)
Determine whether the gravimeter could detect the same ore body if its centre were 1.2 km below the surface.
(2)
(d)
The gravimeter contains a mass hanging from a spring, which is extended by 0.040 m.
Calculate the change in the extension of the spring caused by the ore body in (b).
(2)
(e)
For a small height h above the surface, the gravitational field strength of the Earth decreases by approximately 2gh/R.
(i)
Determine the change in height of the gravimeter that would produce the same change in gravitational field strength as the ore body in (b).
(1)
(ii)
Suggest, using your answer to (e)(i), one precaution the geophysicist must take during the survey.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Δm = 4/3 × π × 40³ × 2500 = 6.70 × 108 «kg»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (b)
Δg = GΔm/d² = 6.67 × 10−11 × 6.7 × 108 / 150² «the sphere acts as a point mass at its centre»
ALTERNATIVE: Δg falls by a factor (1200/150)² = 64, giving 3.1 × 10−8 N kg−1.
this is less than 5.0 × 10−8 N kg−1, so it could not be detected
✓ 1
Conclusion must follow from a calculation. Allow ECF from (b).
Part (d)
kx = mg, so x ∝ g and Δx = x Δg/g
✓ 1
Δx = 0.040 × 1.99 × 10−6 / 9.81 = 8.1 × 10−9 m
✓ 1
Accept 8.1–8.2 × 10−9 m. Allow ECF from (b). Award [2] for CNA.
Part (e)(i)
h = Δg R / 2g = 2.0 × 10−6 × 6.37 × 106 / (2 × 9.81) = 0.65 m
✓ 1
Accept 0.63–0.66 m. Allow ECF from (b).
Part (e)(ii)
measure the height of the gravimeter at every station to within a few centimetres and correct the readings for it
✓ 1
OWTTE. Accept any precaution clearly linked to height errors.
Answers: (a) 6.70 × 108 kg · (b) 2.0 × 10−6 N kg−1 · (c) 3.1 × 10−8 N kg−1; no · (d) 8.1 × 10−9 m · (e)(i) 0.65 m (the remaining parts are explanations — see the table above)
107D-2-40Extended bodies as point massesPaper 2Medium13 marks
A spacecraft flies past a comet nucleus that consists of two roughly spherical lobes, P and Q, resting in contact. Each lobe may be modelled as a uniform sphere. P has a radius of 8.0 km and a mass of 1.29 × 1015 kg; Q has a radius of 5.0 km and a mass of 3.14 × 1014 kg. The centres of the lobes are 13.0 km apart.
The diagram shows the comet nucleus and a point X on the line through both centres. G = 6.67 × 10−11 N m² kg−2.
Diagram drawn to scale
(a)
Outline two different conditions under which an extended body can be treated as a point mass when calculating its gravitational field.
(2)
(b)
Calculate the gravitational force that holds the two lobes together.
(2)
(c)
The navigation software of the spacecraft models the whole comet nucleus as a single point mass of 1.60 × 1015 kg at the centre of P.
(i)
Determine the percentage by which the navigation model underestimates the gravitational field strength at X.
(4)
(ii)
Determine whether the model gives the field strength to within 1 % at point Y, which lies on the same line, beyond Q, 1.00 × 103 km from the centre of P.
(3)
(d)
Draw, on the diagram, the gravitational field lines of the comet nucleus, both close to its surface and far from it.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
a uniform «or spherically symmetric» sphere acts as a point mass at its centre for all points outside it
✓ 1
Accept «spherical body of uniform density».
any body acts as a point mass when the distance to the point is very large compared with the size of the body
✓ 1
Part (b)
each lobe is a uniform sphere, so the centre-to-centre distance 13.0 km is used: F = 6.67 × 10−11 × 1.29 × 1015 × 3.14 × 1014 / (1.30 × 104)²
✓ 1
Do not award MP1 if a radius «8.0 km or 5.0 km» is used as the separation.
model: g = G × 1.60 × 1015 / (1.000 × 106 )² = 1.067 × 10−7 «N kg−1»
✓ 1
Values to at least 3 s.f. are needed to see the difference.
difference ≈ 0.8 %, which is less than 1 %, so yes «Y is far away compared with the size of the comet nucleus»
✓ 1
MP3 only scores if MP1 and MP2 score.
Part (d)
close to the comet nucleus: lines directed into the comet nucleus, ending on its surface at right angles to it and bent towards both lobes «denser near P»
✓ 1
Arrows required on at least two lines.
far from the comet nucleus: lines radial and evenly spaced, pointing towards a single central point
✓ 1
Answers: (b) 1.60 × 1011 N · (c)(i) 1.31 × 10−4 N kg−1 at X, about 25 % underestimate · (c)(ii) about 0.8 % — yes (the remaining parts are explanations — see the table above)
108D-2-41Newton's law of gravitationPaper 2Medium11 marks
A planet P orbits a star S of mass 0.90 M☉. The star and the planet both move in circles, with the same period of 6.80 days, about a fixed point O on the line joining them, as shown. The circle of the star is very small, so the distance between S and P may be taken as the orbital radius of the planet.
An astronomer on the Earth, which lies in the plane of the orbits, measures the Doppler shift of a line of wavelength 656.3 nm in the spectrum of the star. The largest shift is 1.20 × 10−4 nm.
M☉ = 1.99 × 1030 kg; G = 6.67 × 10−11 N m² kg−2; c = 3.00 × 108 m s−1.
Diagram NOT to scale
(a)
Draw and label, on the diagram, arrows to represent the gravitational force on the planet and the gravitational force on the star.
(2)
(b)
Show that the orbital radius of the planet is about 1.0 × 1010 m.
(2)
(c)
Calculate the speed of the star.
(2)
(d)
Determine the mass of the planet.
(4)
(e)
In most planetary systems the Earth does not lie in the plane of the orbits.
Suggest how this affects a mass found by the method in (d).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
arrow on P pointing towards S AND arrow on S pointing towards P, both along the line SP
✓ 1
the two arrows are of equal length «Newton's third law pair»
Must see full substitution OR answer to 3 s.f. Period must be in seconds «5.88 × 105 s».
Part (c)
Δλ/λ ≈ v/c ⇒ v = 3.00 × 108 × 1.20 × 10−4 / 656.3
✓ 1
v = 54.9 m s−1
✓ 1
Award [2] for CNA.
Part (d)
speed of the planet vP = 2πr/T = 2π × 1.01 × 1010 / 5.88 × 105 = 1.08 × 105 m s−1
✓ 1
Accept 1.07–1.09 × 105 m s−1, depending on the value of r used.
the forces on the star and on the planet are equal in magnitude «Newton's third law», so MaS = maP
✓ 1
both have the same period, so a = 2πv/T ∝ v, giving MvS = mvP
✓ 1
Accept M rS = m rP with r = vT/2π.
m = 1.79 × 1030 × 54.9 / 1.08 × 105 = 9.1 × 1026 kg
✓ 1
Allow ECF from (b) and (c). Accept 8.9–9.3 × 1026 kg.
Part (e)
only the component of the star's velocity along the line of sight is measured, so the speed of the star and hence the mass found are too small «the method gives a lower limit»
✓ 1
OWTTE
Answers: (b) 1.01 × 1010 m · (c) 54.9 m s−1 · (d) 9.1 × 1026 kg (the remaining parts are explanations — see the table above)
109D-2-52Gravitational field strengthPaper 2Medium13 marks
Two identical satellites, L (leading) and T (trailing), follow each other around the same circular orbit 490 km above the surface of the Earth, 220 km apart. A microwave link measures very small changes in their separation, from which changes in the Earth’s gravitational field are mapped.
Mass of the Earth = 5.97 × 1024 kg; radius of the Earth = 6.37 × 106 m.
Diagram NOT to scale
(a)
(i)
Calculate the gravitational field strength of the Earth at the orbit.
(1)
(ii)
Show that the orbital speed of the satellites is about 7.6 km s−1.
(1)
(iii)
Calculate the time interval between L and T passing over the same point.
(1)
(b)
The diagram shows the satellites when a region A, which has more mass than its surroundings, is directly below the midpoint between them.
(i)
Draw, on the diagram, arrows to show the direction of the gravitational force exerted by the extra mass in region A on L and on T.
(2)
(ii)
Explain why the separation of the satellites starts to decrease at this time.
(2)
(c)
The extra mass in region A is 2.0 × 1014 kg. Treating it as a point mass at the surface, calculate the gravitational field strength that it produces at a point on the orbit directly above it, as a fraction of the Earth’s field strength at the orbit.
(2)
(d)
Region A is an ice sheet about 300 km across.
(i)
Outline why treating the extra mass as a point mass gives an inaccurate field strength at the satellites.
(1)
(ii)
Over one year the measurements show that the ice sheet loses 2.0 × 1014 kg. All the meltwater flows into the oceans, of total area 3.6 × 1014 m2. Calculate the resulting rise in sea level. Density of water = 1000 kg m−3.
(2)
(e)
Suggest one reason why the satellites are placed in a low orbit.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
g = 6.67 × 10−11 × 5.97 × 1024/(6.86 × 106)2 = 8.46 N kg−1
✓ 1
Award [0] if 4.90 × 105 m is used as the distance.
Part (a)(ii)
mg = mv2/r so v = √(8.46 × 6.86 × 106) = 7619 m s−1
✓ 1
Must see full substitution OR answer to 3 s.f. Accept GMm/r2 = mv2/r.
Part (a)(iii)
t = 2.20 × 105/7.6 × 103 = 29 s
✓ 1
Accept 28–30 s.
Part (b)(i)
arrow from L pointing towards A «down and backwards» AND arrow from T pointing towards A «down and forwards»
✓ 1
Both arrows needed, each starting on its satellite.
arrows symmetrical about the vertical through A «equal angles»
✓ 1
Part (b)(ii)
the force on L has a component opposite to its velocity, so L slows down «slightly»
✓ 1
the force on T has a component along its velocity, so T speeds up and gains on L
✓ 1
OWTTE. Award [1 max] for «A attracts both satellites towards each other» without reference to the velocity.
Part (c)
gA = 6.67 × 10−11 × 2.0 × 1014/(4.90 × 105)2 = 5.6 × 10−8 N kg−1
✓ 1
The distance is the height of the orbit, 4.90 × 105 m.
fraction = 5.6 × 10−8/8.46 = 6.6 × 10−9
✓ 1
Allow ECF from (a)(i).
Part (d)(i)
the ice sheet is not spherical and its size «300 km» is not small compared with its distance «490 km» from the satellites
the field strength of a mass anomaly decreases as 1/r2, so closer to the surface the effect on the satellites «and the change in separation» is larger / easier to measure
✓ 1
OWTTE
Answers: (a)(i) 8.46 N kg−1 · (a)(iii) 29 s · (c) 5.6 × 10−8 N kg−1; 6.6 × 10−9 · (d)(ii) 5.6 × 10−4 m (the remaining parts are explanations — see the table above)
110D-2-54Orbits and satellitesPaper 2Medium12 marks
The asteroid Didymos has a small moon, Dimorphos. In a test of asteroid deflection, a spacecraft was crashed into Dimorphos. Before the impact, Dimorphos moved in a circular orbit of radius 1.19 km around the centre of Didymos.
Mass of Didymos = 5.4 × 1011 kg; radius of Didymos = 390 m; mass of Dimorphos = 4.3 × 109 kg.
Diagram not to scale. Radius of Didymos = 390 m.
(a)
(i)
Show, using Newton’s law of gravitation, that the orbital speed of Dimorphos was about 0.17 m s−1.
(2)
(ii)
Calculate the orbital period of Dimorphos, in hours.
(1)
(b)
The spacecraft had a mass of 580 kg and hit Dimorphos head-on at 6.1 km s−1, in the direction opposite to the velocity of Dimorphos. Ignore any material thrown out by the impact.
(i)
Calculate the change in the speed of Dimorphos.
(2)
(ii)
Explain why the reduction in speed makes Dimorphos move closer to Didymos.
(2)
(c)
After the impact, the orbit was again close to a circle and the orbital period was 4.8 % shorter. Determine the new orbital radius.
(2)
(d)
The measured change in the momentum of Dimorphos was about 3.6 times the momentum of the spacecraft. Suggest why.
(2)
(e)
Calculate the gravitational field strength at the surface of Didymos.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
GMm/r2 = mv2/r «so v2 = GM/r»
✓ 1
v = √(6.67 × 10−11 × 5.4 × 1011/1190) = 0.174 m s−1
✓ 1
Must see full substitution OR answer to 3 s.f. Do not accept 1.19 used in place of 1190 m.
Part (a)(ii)
T = 2π × 1190/0.174 = 4.3 × 104 s = 11.9 h
✓ 1
Allow ECF. Accept 11.8–12.0 h.
Part (b)(i)
momentum of spacecraft = 580 × 6.1 × 103 = 3.5 × 106 kg m s−1 «transferred to Dimorphos»
✓ 1
Δv = 3.5 × 106/4.3 × 109 = 8.2 × 10−4 m s−1
✓ 1
Award [2] for CNA. Ignore the very small mass of the spacecraft in the total.
Part (b)(ii)
at the original radius the gravitational force is now greater than the centripetal force mv2/r needed for the smaller speed
✓ 1
so the resultant force pulls Dimorphos inwards and its path curves towards Didymos
✓ 1
OWTTE. MP2 only scores if MP1 scores.
Part (c)
T2 ∝ r3 «Kepler’s third law, same central mass» so rnew = r(0.952)2/3
✓ 1
rnew = 1152 m «≈ 1.15 km»
✓ 1
Accept 1.14–1.16 km. Award [2] for CNA.
Part (d)
material «ejecta» was thrown out of Dimorphos backwards, towards the incoming spacecraft
✓ 1
by conservation of momentum «Newton’s third law» Dimorphos received extra momentum in the opposite direction to the ejecta
✓ 1
OWTTE
Part (e)
g = 6.67 × 10−11 × 5.4 × 1011/3902 = 2.4 × 10−4 N kg−1
✓ 1
Answers: (a)(ii) 11.9 h · (b)(i) 8.2 × 10−4 m s−1 · (c) 1152 m · (e) 2.4 × 10−4 N kg−1(the remaining parts are explanations — see the table above)
111IBP2S-028-03-13Extended bodies as point massesPaper 2Medium8 marks
A space probe measures the gravitational field strength g at several distances r from the centre of a planet. The planet is believed to be spherically symmetric.
The measurements are: r = 1.0×107 m, g = 3.80 m s−2 r = 2.0×107 m, g = 0.95 m s−2 r = 3.0×107 m, g = 0.42 m s−2 r = 4.0×107 m, g = 0.24 m s−2
G = 6.67×10−11 N m2 kg−2
Data from the question
(a)
Show that the data are consistent with an inverse square law.
(2)
(b)
Determine the mass of the planet.
(2)
(c)
The radius of the planet is 6.0×106 m. Calculate the gravitational field strength at its surface.
(2)
(d)
Explain why these measurements cannot show whether the planet has uniform density or a dense core surrounded by lighter rock.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Calculates gr2 for at least three data points: 3.80×1014, 3.80×1014, 3.78×1014, 3.84×1014 N m2 kg−1
✓ 1
States that gr2 is constant to within about 1%, so g ∝ 1/r2
✓ 1
Part (b)
Uses GM = gr2 ≈ 3.8×1014 N m2 kg−1
✓ 1
Obtains M = 3.8×1014 / 6.67×10−11 = 5.7×1024 kg
✓ 1
Part (c)
Uses g = GM/R2 = 3.8×1014 / (6.0×106)2; error carried forward from the mass found in the previous part is allowed
✓ 1
Obtains g = 11 m s−2 (10.6 m s−2)
✓ 1
Part (d)
States that all the measurements are taken outside the planet, where the field of a spherically symmetric mass is the same as that of a point mass at its centre
✓ 1
Concludes that g depends only on the total mass and on r, so the data reveal the total mass but not how it is distributed
✓ 1
Exam tip: To test whether y ∝ 1/x², show that the product yx² is constant. Use at least three data points.
112IBP2S-028-03-14Combining two gravitational fieldsPaper 2Medium7 marks
A mission is planned to a binary asteroid. The primary body has a mass of 5.2×1011 kg and the smaller secondary body has a mass of 4.0×109 kg. Both bodies may be treated as spheres, and their centres are 1.2 km apart.
Diagram not to scale
(a)
Outline what is meant by gravitational field strength at a point.
(1)
(b)
Explain why there is a point on the line joining the centres where the resultant gravitational field is zero.
(2)
(c)
Determine the distance of this point from the centre of the primary body.
(3)
(d)
State one feature of the field-line pattern at this point.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The gravitational force per unit mass acting on a small test mass placed at that point
✓ 1
Part (b)
Between the bodies, the two fields point in opposite directions, each towards its own body
✓ 1
Close to each body its own field is the larger, so somewhere between them the two magnitudes are equal and the fields cancel
✓ 1
Part (c)
GM1/x2 = GM2/(d − x)2, where x is measured from the primary
✓ 1
x/(d − x) = √(5.2×1011/4.0×109) = √130 = 11.4
✓ 1
x = 1200 × 11.4/12.4 = 1.10×103 m (about 1.1 km)
✓ 1
Part (d)
Any one of: no field line passes through the point; the field-line density there is zero; lines near the point curve away towards one body or the other
✓ 1
Common error: Placing the null point nearer the larger mass is a frequent slip. The point always lies closer to the smaller mass, because the weaker field has to be sampled at a shorter distance to match the stronger one.
113IBP2S-028-03-15Kepler's third lawPaper 2Medium7 marks
A communications satellite is to be placed in an 'areostationary' orbit around Mars. In this orbit its period equals the rotation period of Mars. Mass of Mars = 6.42×1023 kg Rotation period of Mars = 8.86×104 s G = 6.67×10−11 N m2 kg−2
(a)
State the assumption about Mars that allows its gravitational effect on the satellite to be calculated as if all its mass were at its centre.
(1)
(b)
Show that the orbital radius r satisfies r3 = GMT2/4π2.
(2)
(c)
Calculate the orbital radius.
(2)
(d)
Calculate the orbital speed of the satellite.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Mars is spherically symmetric (its density depends only on distance from the centre)
✓ 1
Part (b)
The gravitational force provides the centripetal force: GMm/r2 = mv2/r, with v = 2πr/T
✓ 1
Substitute and rearrange: GM/r2 = 4π2r/T2, so r3 = GMT2/4π2
114IBP2S-028-01-12Geostationary orbitsPaper 2Medium8 marks
A broadcasting company plans to launch a satellite into geostationary orbit. Earth rotates once, relative to the distant stars, in 86 164 s. Mass of Earth = 5.97×1024 kg; G = 6.67×10−11 N m2 kg−2.
(a)
State two features of a geostationary orbit, other than its period.
(2)
(b)
Show that the radius of a geostationary orbit is about 4.2×107 m.
(2)
(c)
Calculate the orbital speed of the satellite.
(2)
(d)
Explain why geostationary satellites cannot provide reliable coverage to research stations close to the North Pole.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
First valid feature
✓ 1
Second valid feature
✓ 1
Valid features (any two): the orbit lies in the plane of the equator / directly above the equator; the satellite moves in the same direction as Earth's rotation (west to east); the orbit is circular, at a fixed altitude.
Obtains r³ ≈ 7.49×1022 m³ and r ≈ 4.215×107 m, stated to more than 2 s.f.
✓ 1
Part (c)
Uses v = 2πr/T = 2π × 4.215×107 / 86164, or v = √(GM/r)
✓ 1
Obtains v ≈ 3.07×103 m s−1; accept 3.07×103 to 3.08×103 m s−1
✓ 1
Allow ECF from (b).
—
Part (d)
Geostationary satellites must orbit above the equator
✓ 1
From near the pole, the satellite is at or below the horizon, or only just above it, so there is no clear line of sight to it
✓ 1
Exam tip: In a 'show that' question, give your answer to more significant figures than the value quoted in the question. This shows you have calculated it rather than copied it.
116IBP2S-028-01-16Geostationary orbitsPaper 2Medium7 marks
A crewed space station orbits Earth in a circular orbit at an altitude of 420 km. Mass of Earth = 5.97×1024 kg; radius of Earth = 6.37×106 m; G = 6.67×10−11 N m2 kg−2.
(a)
Calculate the orbital period of the space station.
(3)
(b)
A cargo module with a mass much larger than that of the station, but still negligible compared with the mass of Earth, is placed in the same orbit. Explain why its orbital period is the same as that of the station.
(2)
(c)
A satellite is to be placed in a circular orbit with exactly twice the period of the space station. Determine the altitude of this orbit.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
r = 6.37×106 + 4.20×105 = 6.79×106 m
✓ 1
T = 2π√(r³/GM), from GMm/r² = 4π²mr/T²
✓ 1
T = 5.57×103 s (≈ 92.8 min)
✓ 1
Part (b)
The gravitational force provides the centripetal force: GMm/r² = 4π²mr/T², so both sides are proportional to the orbiting mass m
✓ 1
m cancels, so T = 2π√(r³/GM) depends only on Earth's mass and the orbital radius, not on the mass of the orbiting body
✓ 1
Part (c)
Using T² ∝ r³: r2 = r1 × 22/3 = 6.79×106 × 1.587 = 1.078×107 m
✓ 1
altitude = 1.078×107 − 6.37×106 = 4.41×106 m (accept 4.4×106 m)
✓ 1
A full method using T = 2π√(r³/GM) with T = 1.114×104 s is equally valid.
—
ECF from the radius in (a).
—
Exam tip: Using a ratio with Kepler's third law avoids recalculating with G and M. Keep unrounded intermediate values until the final step, especially before subtracting two similar numbers.
117IBP2S-028-00-12Density, mass and gPaper 2Medium9 marks
A space agency is planning a lander mission to one of three moons of a gas giant. Data for each moon are given below. Treat each moon as a uniform sphere.
Moon | radius R / 106 m | surface field strength g / N kg−1 X | 2.6 | 1.4 Y | 1.8 | 1.8 Z | 0.76 | 0.22
(a)
Show that the mean density ρ of a moon is given by \(\rho = \frac{3g}{4\pi G R}\).
(2)
(b)
Calculate the mean density of each moon.
(3)
(c)
Rock has a density of about 3×103 kg m−3 and water ice about 1×103 kg m−3. Identify which moon is most likely made mainly of ice and which is most likely made mainly of rock.
(2)
(d)
Determine the mass of moon Y.
(2)
Show mark scheme
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Notes
Part (a)
Uses g = GM/R2 together with M = ρ × (4/3)πR3 .
✓ 1
Substitutes to get g = (4/3)πGρR, then rearranges to ρ = 3g/(4πGR) .
Z is mainly ice, since its density is about 1.0×103 kg m−3, close to that of ice .
✓ 1
Y is mainly rock, since its density is about 3.6×103 kg m−3, the closest to that of rock . (X, at about 1.9×103 kg m−3, is probably a mixture of rock and ice.) ECF from (b).
✓ 1
Part (d)
M = gR2/G = 1.8 × (1.8×106)2/6.67×10−11 .
✓ 1
M = 8.7×1022 kg . Also accept M = ρ × (4/3)πR3, with ECF from (b).
✓ 1
Exam tip: Convert radii given in units of 10⁶ m into metres before substituting.
118IBP2S-028-00-13Variation of g with distancePaper 2Medium8 marks
A probe approaching a dwarf planet uses an on-board accelerometer to measure the gravitational field strength g at several distances r from the dwarf planet's centre. The dwarf planet is a uniform sphere of radius 1.2×106 m.
r / 106 m | g / N kg−1 1.5 | 0.61 2.0 | 0.34 3.0 | 0.152 4.0 | 0.086
Data from the question
(a)
Show that the data are consistent with g being inversely proportional to r2.
(2)
(b)
Determine the mass of the dwarf planet.
(2)
(c)
Calculate the gravitational field strength at the surface of the dwarf planet.
(2)
(d)
Determine the mean density of the dwarf planet.
(2)
Show mark scheme
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Notes
Part (a)
Calculates g r2 for at least three rows. The four values are 1.37, 1.36, 1.37 and 1.38 ×1012 N m2 kg−1 .
✓ 1
States that the product is constant within experimental uncertainty, so g ∝ 1/r2 .
✓ 1
Part (b)
Uses g = GM/r2, so GM = g r2 ≈ 1.37×1012 N m2 kg−1 .
✓ 1
M = 1.37×1012/6.67×10−11 ≈ 2.05×1022 kg. Accept 2.0–2.1×1022 kg .
✓ 1
Part (c)
Uses g = GM/R2 = 1.37×1012/(1.2×106)2 .
✓ 1
g ≈ 0.95 N kg−1. Accept 0.95–0.97 N kg−1 if M = 2.1×1022 kg is used. ECF from (b) .
✓ 1
Part (d)
V = (4/3)π(1.2×106)3 = 7.24×1018 m3 .
✓ 1
ρ = M/V = 2.05×1022/7.24×1018 ≈ 2.8×103 kg m−3. Accept 2.8–2.9×103 kg m−3. ECF from (b) .
✓ 1
Exam tip: To test a proportionality from a table, calculate the product or ratio for several rows and comment on whether it stays constant.
119IBP2S-028-00-16Newton's law of gravitationPaper 2Medium7 marks
A gold bar is hung from a precision electronic spring balance at sea level. The same bar and balance are then taken to a mountain observatory 5.0 km above sea level. Earth's radius is 6.37×106 m. Ignore the effects of Earth's rotation.
(a)
Calculate the ratio \(\frac{g_{\text{observatory}}}{g_{\text{sea level}}}\).
(2)
(b)
At sea level the balance reads 12.000 N. Determine its reading at the observatory.
(1)
(c)
Explain why an equal-arm beam balance, comparing the bar with standard masses, would balance at the same setting in both places.
(2)
(d)
Determine the height above sea level at which the spring balance reading would be 1.0% lower than at sea level.
(2)
Show mark scheme
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Notes
Part (a)
Ratio = (R/(R + h))2 = (6.37×106/6.375×106)2 .
✓ 1
Ratio = 0.9984 .
✓ 1
Part (b)
Reading = 12.000 × 0.9984 = 11.981 N . ECF from (a).
✓ 1
Part (c)
The bar and the standard masses are in the same field, so both weights change by the same factor .
✓ 1
The change in g cancels, so the beam balance compares masses, which do not change .
✓ 1
Part (d)
R/(R + h) = √0.99, so R + h = 6.37×106/0.99499 = 6.402×106 m .
✓ 1
h ≈ 3.2×104 m .
✓ 1
Exam tip: When an answer is very close to 1, keep at least four significant figures in intermediate steps so the small change is not lost to rounding.
121IBP2S-028-02-14Geostationary orbitsPaper 2Medium7 marks
A satellite in a global navigation system moves in a circular orbit around Earth with a period of 11 hours 58 minutes. Mass of Earth = 5.97×1024 kg Gravitational constant G = 6.67×10−11 N m2 kg−2
(a)
Show that the orbital radius of the satellite is about 2.7×107 m.
(3)
(b)
Calculate the gravitational field strength due to Earth at the position of the satellite.
(2)
(c)
The satellite moves at constant speed. Explain why it is nevertheless accelerating.
(2)
Show mark scheme
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Notes
Part (a)
Equating gravitational force to centripetal force: GMm/r2 = m4π2r/T2, so r3 = GMT2/(4π2)
✓ 1
Converting the period: T = 11 × 3600 + 58 × 60 = 43 080 s
✓ 1
r = (6.67×10−11 × 5.97×1024 × 43 0802/(4π2))1/3 = 2.66×107 m, which is about 2.7×107 m
✓ 1
Part (b)
g = GM/r2 = 6.67×10−11 × 5.97×1024/(2.66×107)2
✓ 1
g ≈ 0.565 N kg−1. Accept any answer from 0.55 to 0.57 N kg−1, depending on the rounding of r; using 2.7×107 m gives 0.55 N kg−1
✓ 1
Part (c)
Velocity is a vector. The direction of the satellite's velocity changes continuously, so its velocity changes
✓ 1
The gravitational force gives the satellite a centripetal acceleration directed towards Earth's centre. This acceleration is equal in size to the field strength calculated in the previous part
✓ 1
Exam tip: The field strength at a satellite's position equals its centripetal acceleration. As a quick check, 4π²r/T² should give the same value.
122IBP2S-028-04-13Combining two gravitational fieldsPaper 2Medium9 marks
A planetary-defence mission studies a binary asteroid system. A small moonlet orbits a larger primary asteroid. Treat both bodies as stationary uniform spheres, and ignore their orbital motion and rotation. Mass of primary = 5.4×1011 kg Mass of moonlet = 4.3×109 kg Radius of moonlet = 75 m Distance between centres = 1.19 km G = 6.67×10−11 N m2 kg−2 P is the point on the moonlet's surface nearest the primary. Q is the point on the moonlet's surface furthest from the primary. Both lie on the line joining the centres.
Diagram not to scale. Radius of moonlet = 75 m.
(a)
Show that the point between the bodies where the net gravitational field strength is zero is about 1.09 km from the centre of the primary.
(3)
(b)
Determine the magnitude and direction of the net gravitational field strength at P.
(3)
(c)
Determine the magnitude of the net gravitational field strength at Q.
(2)
(d)
Suggest why loose boulders are more easily lifted off the moonlet's surface at P than at Q.
(1)
Show mark scheme
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Notes
Part (a)
GM1/x2 = GM2/(d − x)2
✓ 1
x/(d − x) = √(5.4×1011 / 4.3×109) = 11.2
✓ 1
x = 1190 × 11.2/12.2 = 1.09×103 m
✓ 1
Part (b)
Field due to moonlet = 6.67×10−11 × 4.3×109 / 752 = 5.10×10−5 N kg−1
✓ 1
Field due to primary at 1115 m = 6.67×10−11 × 5.4×1011 / 11152 = 2.90×10−5 N kg−1
✓ 1
Net field = 2.2×10−5 N kg−1, directed towards the centre of the moonlet (away from the primary)
✓ 1
Part (c)
Field due to primary at 1265 m = 6.67×10−11 × 5.4×1011 / 12652 = 2.25×10−5 N kg−1, in the same direction as the moonlet's field at Q (both towards the primary)
✓ 1
Net field = 5.10×10−5 + 2.25×10−5 = 7.35×10−5 N kg−1. Accept 7.3×10−5 to 7.4×10−5. Allow follow-through from the moonlet's field calculated in the previous part
✓ 1
Part (d)
At P the two fields act in opposite directions and partly cancel. The net field, and so a boulder's weight, is therefore smaller than at Q, where the fields add
✓ 1
Exam tip: Before adding field strengths, decide the direction of each field at the point concerned. Each field always points towards the mass that produces it.
123IBP2S-028-04-14Kepler's third lawPaper 2Medium9 marks
Mission planners for a proposed Uranus orbiter use data for some of the planet's moons. All orbits may be taken as circular. Moon : orbital radius r / 108 m : orbital period T / days Miranda : 1.299 : 1.413 Ariel : 1.909 : 2.520 Titania : 4.363 : 8.706 Oberon : 5.835 : ? G = 6.67×10−11 N m2 kg−2; 1 day = 86400 s
(a)
Show, using the data for two of the moons, that the data are consistent with Kepler's third law.
(2)
(b)
Determine the mass of Uranus.
(3)
(c)
Calculate the orbital period of Oberon in days.
(2)
(d)
The orbiter is to be placed in a circular orbit of radius 1.0×108 m around Uranus. Calculate its orbital speed.
(2)
Show mark scheme
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Notes
Part (a)
Calculates T2/r3 for two moons, e.g. Miranda 0.911 and Titania 0.913, in units of days2 (108 m)−3
✓ 1
States that the values are equal to within about 0.2%, so T2 ∝ r3
✓ 1
Part (b)
M = 4π2r3/(GT2)
✓ 1
Uses SI units, e.g. Titania: r3 = 8.305×1025 m3 and T = 7.522×105 s
✓ 1
M = 8.7×1025 kg
✓ 1
Part (c)
T = 8.706 × (5.835/4.363)3/2
✓ 1
T = 13.5 days
✓ 1
Part (d)
v = √(GM/r) = √(6.67×10−11 × 8.7×1025 / 1.0×108)
✓ 1
v = 7.6×103 m s−1; ECF from (b)
✓ 1
Common error: Ratios such as T²/r³ can be compared in any units, but the mass calculation must use seconds and metres.
124D-2-04Combining two gravitational fieldsPaper 2Hard12 marks
Pluto and its largest moon, Charon, may be treated as uniform spheres. In this question treat both bodies as stationary and ignore the gravitational fields of all other bodies.
The distance between the centres of Pluto and Charon is 1.96 × 107 m.
mass / kg
radius / m
Pluto
1.30 × 1022
1.19 × 106
Charon
1.59 × 1021
6.06 × 105
Axes drawn to scale
(a)
Outline why there is a point on the line between the centres of Pluto and Charon at which the resultant gravitational field strength is zero.
(2)
(b)
(i)
Show that this point is about 1.45 × 107 m from the centre of Pluto.
(2)
(ii)
A probe rests on the surface of Charon at the point nearest to Pluto. Determine the magnitude and direction of the resultant gravitational field strength at the probe.
(3)
(c)
On the axes, sketch a graph to show how the resultant gravitational field strength varies along the line joining the centres, from the surface of Pluto to the surface of Charon. Take field strength directed towards Charon as positive.
(3)
(d)
A small probe is placed at rest at the point in (b)(i) and is then displaced slightly towards Charon. Explain why the probe does not return to that point.
(2)
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Notes
Part (a)
between the bodies the two fields act in opposite directions «each towards its own body»
✓ 1
each field decreases with distance from its own body «inverse square», so at some point the two magnitudes are equal
Must see full substitution OR answer to at least 3 s.f. Unrounded answer 1.452 × 107 m.
Part (b)(ii)
field of Charon at its surface = 6.67 × 10−11 × 1.59 × 1021/(6.06 × 105)² = 0.289 N kg−1
✓ 1
field of Pluto at the probe = 6.67 × 10−11 × 1.30 × 1022/(1.90 × 107)² = 2.40 × 10−3 N kg−1
✓ 1
The distance from the centre of Pluto must be 1.96 × 107 − 6.06 × 105 m.
resultant = 0.286 N kg−1 towards the centre of Charon
✓ 1
Direction required. Award [2 max] if the fields are added (0.291 N kg−1).
Part (c)
negative near the surface of Pluto «about −0.6 N kg−1» and positive near the surface of Charon «about +0.3 N kg−1», with the magnitude at Pluto clearly the larger
✓ 1
Values need not be exact.
curve crosses zero once, at about 1.45 × 107 m «closer to Charon»
✓ 1
Allow ECF from (b)(i).
curve steepest close to each surface and flatter in between «1/r² shape»
✓ 1
Do not award for straight-line segments.
Part (d)
closer to Charon, Charon's field is larger and Pluto's is smaller, so the resultant field points towards Charon
✓ 1
the resultant force accelerates the probe further away from the point «towards Charon»; the equilibrium is unstable
✓ 1
OWTTE
Answers: (b)(i) 1.45 × 107 m · (b)(ii) 0.286 N kg−1 towards the centre of Charon (the remaining parts are explanations — see the table above)
Many small asteroids are "rubble piles": collections of loose rocks held together only by their own gravity.
Asteroid R is modelled as a uniform sphere of radius 250 m and mean density 2.0 × 103 kg m−3. It rotates about its axis once every 3.6 hours.
Diagram NOT to scale
(a)
(i)
Show that the mass of R is about 1.3 × 1011 kg.
(1)
(ii)
Calculate the gravitational field strength at the surface of R.
(2)
(b)
A loose rock of mass 5.0 kg rests on the surface of R at its equator, as shown in the diagram.
(i)
Draw and label, on the diagram, the forces acting on the rock. The lengths of your arrows should show the relative sizes of the forces.
(2)
(ii)
Determine the normal force exerted on the rock by the surface.
(4)
(c)
(i)
Show that the shortest period of rotation that a rubble pile of mean density ρ can have without losing rocks from its equator is √(3π/(Gρ)).
(2)
(ii)
Calculate this shortest period for R, in hours.
(1)
(iii)
Surveys show that almost no asteroids wider than a few hundred metres rotate faster than once every 2.2 hours, but many smaller asteroids rotate once every few minutes. Suggest what this shows about these small, fast-rotating asteroids.
(1)
Show mark scheme
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Notes
Part (a)(i)
M = ρ × (4/3)πR³ = 2.0 × 103 × (4/3)π × 250³ = 1.31 × 1011 kg
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (a)(ii)
g = GM/R² = 6.67 × 10−11 × 1.31 × 1011/250²
✓ 1
g = 1.40 × 10−4 N kg−1
✓ 1
Award [2] for CNA.
Part (b)(i)
weight «gravitational force» towards the centre of R AND normal «reaction» force away from the centre, both labelled
✓ 1
Do not award if a «centripetal force» is added as a third force.
weight arrow longer than the normal force arrow, both starting on the rock
✓ 1
Part (b)(ii)
ω = 2π/(3.6 × 3600) = 4.85 × 10−4 rad s−1
✓ 1
ALTERNATIVE: v = 2πR/T = 0.121 m s−1, then a = v²/R.
centripetal acceleration a = ω²R = 5.88 × 10−5 m s−2
✓ 1
the resultant force is towards the centre: mg − N = ma, so N = m(g − a)
✓ 1
Allow ECF from (a)(ii).
N = 5.0 × (1.40 × 10−4 − 5.88 × 10−5) = 4.1 × 10−4 N «4.0 × 10−4 N with unrounded values»
✓ 1
Accept 4.0–4.1 × 10−4 N. Award [1 max] for N = mg = 7.0 × 10−4 N.
Part (c)(i)
at the shortest period N = 0, so gravity alone provides the centripetal force: GM/R² = ω²R
✓ 1
with M = (4/3)πρR³: ω² = 4πGρ/3, so T = 2π/ω = √(3π/(Gρ))
✓ 1
R must be seen to cancel.
Part (c)(ii)
T = √(3π/(6.67 × 10−11 × 2.0 × 103)) = 8.4 × 103 s = 2.3 h
✓ 1
Accept 2.3–2.4 h.
Part (c)(iii)
they cannot be held together by gravity alone, so they must be single solid bodies «held together by the strength of the rock»
✓ 1
OWTTE
Answers: (a)(ii) g = 1.40 × 10−4 N kg−1 · (b)(ii) N = 4.0–4.1 × 10−4 N · (c)(ii) 2.3 h (the remaining parts are explanations — see the table above)
A star has a mass M = 0.80 M☉ and a luminosity of 0.35 L☉. A planet moves around the star in a circular orbit of radius r with a period T of 150 days. For a circular orbit, T² = (4π²/GM) r³.
M☉ = 1.99 × 1030 kg; L☉ = 3.83 × 1026 W; G = 6.67 × 10−11 N m² kg−2; σ = 5.67 × 10−8 W m−2 K−4.
(a)
State Kepler's third law.
(1)
(b)
Calculate r.
(2)
(c)
Show that the intensity of the star's radiation at the distance of the planet is about 1.8 × 103 W m−2.
(1)
(d)
The planet has an albedo of 0.30 and its surface emits as a black body.
(i)
Determine the equilibrium temperature of the planet's surface, ignoring any effect of an atmosphere.
(3)
(ii)
Suggest why the actual mean surface temperature of the planet could be higher than your answer to (d)(i).
(2)
(e)
The luminosity of the star is known only to within ±10 %.
Estimate the resulting absolute uncertainty in your answer to (d)(i).
(2)
Show mark scheme
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Notes
Part (a)
the square of the orbital period is proportional to the cube of the «mean» orbital radius «for bodies orbiting the same central body»
✓ 1
Accept T² ∝ r³ with the symbols identified. Accept “semi-major axis” for radius.
Part (b)
T = 150 × 86 400 = 1.30 × 107 s AND M = 0.80 × 1.99 × 1030 = 1.59 × 1030 kg substituted into r³ = GMT²/4π²
✓ 1
r = 7.67 × 1010 m
✓ 1
Accept 7.6–7.7 × 1010 m «≈ 0.51 AU». Award [2] for CNA. Do not award MP2 if T is left in days.
127D-2-21Newton's law of gravitationPaper 2Hard11 marks
At new moon the Moon M lies on the straight line between the Sun S and the Earth E, as shown in the diagram.
Mass of the Sun = 1.99 × 1030 kg; mass of the Earth = 5.97 × 1024 kg; mass of the Moon = 7.35 × 1022 kg; distance between the centres of the Earth and the Moon = 3.84 × 108 m; distance between the centres of the Sun and the Moon = 1.50 × 1011 m; G = 6.67 × 10−11 N m² kg−2.
Diagram NOT to scale
(a)
(i)
Show that the gravitational force exerted by the Earth on the Moon is about 2.0 × 1020 N.
(1)
(ii)
Calculate the gravitational force exerted by the Sun on the Moon.
(1)
(b)
Draw and label, on the diagram, arrows to represent the two gravitational forces acting on the Moon. The lengths of your arrows should show the relative sizes of the forces.
(2)
(c)
Determine the magnitude and direction of the resultant gravitational field strength at the centre of the Moon.
(3)
(d)
Part (a) shows that the Sun pulls on the Moon more strongly than the Earth does.
(i)
Calculate the gravitational field strength due to the Sun at the centre of the Earth at the same instant.
(1)
(ii)
Explain, using your answers to (c) and (d)(i), why the Moon nevertheless continues to orbit the Earth.
one arrow from M towards S and one from M towards E, each labelled «e.g. force of Sun, force of Earth»
✓ 1
Arrows should start at, or touch, M.
arrow towards S about twice as long as arrow towards E «ratio 2.2»
✓ 1
Accept a length ratio between 1.8 and 2.6. Allow ECF from (a).
Part (c)
field of the Sun = 5.90 × 10−3 N kg−1 towards S
✓ 1
ALTERNATIVE: resultant force from (a) divided by the mass of the Moon, for MP1 and MP2.
field of the Earth = 2.70 × 10−3 N kg−1 towards E
✓ 1
resultant = 3.20 × 10−3 N kg−1 towards the Sun
✓ 1
Accept 3.1–3.3 × 10−3 N kg−1. Direction needed for MP3. Allow ECF from (a).
Part (d)(i)
g = 6.67 × 10−11 × 1.99 × 1030 / (1.504 × 1011)² = 5.87 × 10−3 N kg−1
✓ 1
Must see the distance 1.50 × 1011 + 3.84 × 108 = 1.504 × 1011 m used. Accept 5.86–5.88 × 10−3 N kg−1.
Part (d)(ii)
the Sun gives the Earth and the Moon almost the same acceleration «≈ 5.9 × 10−3 m s−2», so the Earth and the Moon move around the Sun together
✓ 1
OWTTE
the difference between the Sun's field at the Moon and at the Earth is only about 3 × 10−5 N kg−1
✓ 1
Allow ECF from (c) and (d)(i).
this is about 1 % of the Earth's field at the Moon «2.70 × 10−3 N kg−1», so relative to the Earth the Moon's motion is controlled by the Earth's pull
✓ 1
Award MP3 only if a comparison with the Earth's field at the Moon is made.
Answers: (a)(i) 1.98 × 1020 N · (a)(ii) 4.34 × 1020 N · (c) 3.20 × 10−3 N kg−1 towards the Sun · (d)(i) 5.87 × 10−3 N kg−1(the remaining parts are explanations — see the table above)
128D-2-38Combining two gravitational fieldsPaper 2Hard13 marks
An infrared space telescope is placed at a point L2 on the straight line through the Sun and the Earth, on the far side of the Earth from the Sun, 1.50 × 109 m from the centre of the Earth. In a simple model, the telescope moves in a circle around the Sun and stays on the Sun–Earth line, so that its period is the same as that of the Earth.
Treat the Sun, the Earth and the telescope as point masses, treat the orbit of the Earth as a circle, and ignore the Moon and the motion of the Sun.
Mass of the Sun = 1.99 × 1030 kg; mass of the Earth = 5.97 × 1024 kg; distance between the centres of the Sun and the Earth = 1.496 × 1011 m; orbital period of the Earth = 3.156 × 107 s; G = 6.67 × 10−11 N m² kg−2.
Diagram NOT to scale
(a)
Draw and label, on the diagram, arrows to represent the gravitational force exerted on the telescope by the Sun and by the Earth. The lengths of your arrows should show the relative sizes of the forces.
(2)
(b)
(i)
Calculate the gravitational field strength due to the Sun at L2.
(1)
(ii)
Determine the magnitude and direction of the resultant gravitational field strength at L2.
(3)
(c)
(i)
Explain, without calculation, why a body in a circular orbit around the Sun at the distance of L2 would have a period of more than one year if the Earth were absent.
(2)
(ii)
Determine whether the resultant field at L2 allows the telescope to move in a circle around the Sun with the same period as the Earth.
(3)
(d)
Radio signals travel between the telescope and the Earth at 3.00 × 108 m s−1.
(i)
Calculate the time taken by a radio signal to travel from the telescope to the Earth.
(1)
(ii)
Suggest why L2 is a good position for a telescope that detects faint infrared radiation from distant objects.
(1)
Show mark scheme
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Notes
Part (a)
arrow from the telescope towards the Sun AND arrow from the telescope towards the Earth, both pointing the same way «towards the Sun», each labelled
✓ 1
Arrows must start at, or touch, the telescope.
the arrow for the Sun is much longer than the arrow for the Earth «about 30 times»
✓ 1
Accept any clearly longer Sun arrow (at least 5 times).
comparison: this equals the resultant field strength found in (b)(ii) «to 3 s.f.»
✓ 1
Allow ECF from (b)(ii).
so yes: the resultant gravitational force provides exactly the centripetal force needed for a period of one year
✓ 1
Do not award MP3 if only the Sun's field is compared. MP3 only scores if MP2 scores.
Part (d)(i)
t = 1.50 × 109 / 3.00 × 108 = 5.0 s
✓ 1
Part (d)(ii)
the Sun and the Earth «and the Moon» are always in nearly the same direction as seen from L2, so a single shield can block their radiation and keep the telescope cold «its own infrared emission does not swamp faint sources»
✓ 1
OWTTE
Answers: (b)(i) 5.81 × 10−3 N kg−1 · (b)(ii) 5.99 × 10−3 N kg−1 towards the Sun · (c)(ii) required 5.99 × 10−3 m s−2 — yes · (d)(i) 5.0 s (the remaining parts are explanations — see the table above)
A binary star system consists of two stars, X and Y, that move in circular orbits about a fixed point C on the line joining them, as shown. Each star is acted on only by the gravitational force of the other. The distance between the centres of the stars is 18.0 AU, the radius of the orbit of X is 6.0 AU and the period of both orbits is 42.0 years.
For this system, T2 = 4π2d3/(G(MX + MY)), where d is the distance between the stars. 1 AU = 1.50 × 1011 m; 1 year = 3.16 × 107 s; mass of the Sun M☉ = 1.99 × 1030 kg; luminosity of the Sun L☉ = 3.83 × 1026 W; radius of the Sun = 6.96 × 108 m.
Diagram drawn to scale
(a)
The stars move anticlockwise. Draw, on the diagram, arrows to represent the velocity of X and the velocity of Y at the instant shown.
(2)
(b)
Explain why the mass of X is greater than the mass of Y.
(2)
(c)
(i)
Show that the total mass of the two stars is about 6.6 × 1030 kg.
(2)
(ii)
Calculate the mass of Y in solar masses.
(2)
(d)
Y has a luminosity of 0.0070 L☉ and a surface temperature of 1.8 × 104 K.
(i)
Determine the radius of Y.
(2)
(ii)
Identify the type of star that Y is. Give two reasons for your answer.
(2)
(e)
The separation of the stars is found from their angular separation as seen from the Earth. Outline what else must be measured to find the separation in metres.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
arrow on X pointing down the page AND arrow on Y pointing up the page, each along the tangent to its orbit «perpendicular to the line XY»
✓ 1
Arrows must start on the stars.
arrow on Y about twice as long as the arrow on X «same period, so v = 2πr/T ∝ r»
✓ 1
Accept a length ratio from 1.5 to 2.5.
Part (b)
the forces on the stars are equal in magnitude AND the stars have the same period «same ω»
✓ 1
MXω2rX = MYω2rY, so the star on the smaller circle «X» has the greater mass
✓ 1
OWTTE. Accept MX/MY = rY/rX = 2.
Part (c)(i)
d = 18.0 × 1.50 × 1011 = 2.70 × 1012 m AND T = 42.0 × 3.16 × 107 = 1.33 × 109 s
✓ 1
Both conversions needed.
MX + MY = 4π2(2.70 × 1012)3/(6.67 × 10−11 × (1.33 × 109)2) = 6.61 × 1030 kg
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (c)(ii)
MY = Mtotal × rX/d = (1/3) × 6.6 × 1030 = 2.2 × 1030 kg
✓ 1
Allow ECF from (b). Award [0] for MP1 if the masses are taken as equal.
= 1.1 M☉
✓ 1
Accept 1.1–1.2 M☉. Award [1 max] for 2.2 M☉ «mass of X».
radius less than 0.01 of the Sun’s «about the size of the Earth» although its surface temperature is high AND luminosity far below that of a main-sequence star of similar mass / lies below the main sequence on an HR diagram
✓ 1
Two distinct reasons needed for this mark. Allow ECF from (d)(i) only if the radius is much smaller than the Sun’s.
Part (e)
the distance from the Earth to the binary system «e.g. from stellar parallax»
✓ 1
Do not accept «the orbital period».
Answers: (c)(i) 6.6 × 1030 kg · (c)(ii) 1.1 M☉ · (d)(i) 6.0 × 106 m (the remaining parts are explanations — see the table above)
A proposed space elevator is a cable that rises vertically from a point on the equator and rotates with the Earth. The Earth rotates once every 8.62 × 104 s. A climber of mass 500 kg moves slowly up the cable, so at every height it moves in a circle about the Earth’s axis with the same angular velocity ω as the Earth.
The graph shows how the gravitational field strength g of the Earth varies with the distance r from the centre of the Earth. Mass of the Earth = 5.97 × 1024 kg.
Graph drawn to scale
(a)
Show that ω = 7.3 × 10−5 rad s−1.
(1)
(b)
(i)
Draw, on the graph, a line to show how the centripetal acceleration ω2r of the climber varies with r.
(2)
(ii)
Determine, using the graph, the value of r at which the cable exerts no force on the climber.
(1)
(iii)
State the significance of this value of r.
(1)
(c)
Determine the magnitude and direction of the force that the cable exerts on the climber when it is at r = 3.0 × 107 m.
(3)
(d)
At r = 5.0 × 107 m the climber lets go of the cable. State and explain how the climber moves immediately afterwards, relative to the Earth.
(2)
(e)
The cable is held taut by a large mass attached to its far end, beyond the value of r in (b)(ii). Explain why the mass must be placed beyond this value of r.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
ω = 2π/8.62 × 104 = 7.289 × 10−5 rad s−1
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (b)(i)
straight line with positive gradient that would pass through the origin
✓ 1
line passing through about (2.0, 0.11) and (5.0, 0.27)
✓ 1
Values within half a small square. ω2 = 5.3 × 10−9 s−2.
Part (b)(ii)
r where the line crosses the curve ≈ 4.2 × 107 m
✓ 1
Accept 4.1–4.3 × 107 m. Allow ECF from the candidate’s line.
Part (b)(iii)
it is the radius of a geostationary orbit: gravity alone provides exactly the centripetal force for a period equal to the Earth’s rotation
✓ 1
OWTTE
Part (c)
gravitational force = 500 × 0.442 = 221 N «g read from the graph ≈ 0.44 N kg−1»
✓ 1
Accept g from the graph 0.43–0.45 N kg−1.
force needed for circular motion = 500 × (7.29 × 10−5)2 × 3.0 × 107 = 80 N
✓ 1
cable force = 221 − 80 = 142 N, directed away from the Earth «upwards along the cable»
✓ 1
Accept 135–150 N. Direction required.
Part (d)
at this r the gravitational force is less than the force needed for circular motion at the climber’s speed «ω2r > g»
✓ 1
Allow ECF from (b)(i).
so the climber moves outwards, away from the Earth «it does not fall»
✓ 1
MP2 only scores if MP1 scores.
Part (e)
below this radius each part of the cable needs less force than gravity provides, so the cable below is pulled downwards / towards the Earth
✓ 1
beyond it, gravity is too small to provide the centripetal force, so the mass pulls outwards and keeps the cable in tension
✓ 1
OWTTE
Answers: (b)(ii) 4.2 × 107 m · (c) 142 N away from the Earth (the remaining parts are explanations — see the table above)
131D-2-57Newton's law of gravitationPaper 2Hard12 marks
A solar-sail spacecraft has a total mass of 16 kg, including a flat, perfectly reflecting sail of area 1600 m2. When the sail faces the Sun, sunlight of intensity I exerts a force F = 2IA/c on it, directed away from the Sun, where A is the area of the sail.
Luminosity of the Sun = 3.83 × 1026 W; mass of the Sun = 1.99 × 1030 kg; 1 AU = 1.50 × 1011 m.
Axes for (d)
(a)
Show that the intensity of sunlight at a distance of 1.0 AU from the Sun is about 1.4 kW m−2.
(1)
(b)
The spacecraft is 1.0 AU from the Sun with its sail facing the Sun.
(i)
Calculate the force exerted by sunlight on the sail.
(1)
(ii)
Calculate the gravitational force exerted by the Sun on the spacecraft.
(1)
(c)
Show that the ratio (force exerted by sunlight)/(gravitational force) does not depend on the distance from the Sun.
(2)
(d)
Sketch, on the axes, graphs to show how each of the two forces varies with distance d from the Sun between 0.5 AU and 3.0 AU. Label your graphs.
(2)
(e)
Determine the greatest total mass of a spacecraft with this sail that could stay at rest relative to the Sun at any distance.
(2)
(f)
The 16 kg spacecraft moves in a circular orbit around the Sun at 1.0 AU, with its sail always facing the Sun. Determine its orbital period, in days.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
I = 3.83 × 1026/(4π × (1.50 × 1011)2) = 1355 W m−2
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (b)(i)
F = 2 × 1355 × 1600/3.00 × 108 = 1.4 × 10−2 N
✓ 1
Accept 0.014–0.015 N.
Part (b)(ii)
F = 6.67 × 10−11 × 1.99 × 1030 × 16/(1.50 × 1011)2 = 0.094 N
✓ 1
Part (c)
sunlight force = 2LA/(4πd2c) ∝ 1/d2 AND gravitational force = GMm/d2 ∝ 1/d2
✓ 1
Must see full substitution of I = L/(4πd2).
so d2 cancels in the ratio «ratio = LA/(2πcGMm) ≈ 0.15»
✓ 1
Part (d)
two curves, each decreasing with d with the 1/d2 shape, gravity starting at about 0.38 N at 0.5 AU and sunlight at about 0.058 N
✓ 1
Allow ECF from (b).
sunlight curve about 0.15 of the gravity curve at every distance; the curves do not cross
✓ 1
Part (e)
the forces balance when m = LA/(2πcGM) OR mass = 16 × 0.153
✓ 1
Allow ECF from (b).
m = 2.4 kg
✓ 1
Accept 2.4–2.5 kg.
Part (f)
resultant force towards the Sun = 0.0944 − 0.0144 = 0.0799 N
✓ 1
Allow ECF from (b).
0.0799 = 16v2/1.50 × 1011 so v = 2.74 × 104 m s−1
✓ 1
T = 2πr/v = 3.44 × 107 s = 398 days
✓ 1
Accept 395–400 days. Award [3] for CNA. Award [2 max] for 366 days «force of sunlight ignored».
Answers: (a) 1355 W m−2 · (b)(i) 1.4 × 10−2 N · (b)(ii) 0.094 N · (e) 2.4 kg · (f) 398 days (the remaining parts are explanations — see the table above)
Mission planners send a probe from Earth to Mars along an elliptical transfer orbit around the Sun. The orbit's closest point to the Sun (perihelion) is at Earth's orbit, 1.00 AU from the Sun. Its farthest point (aphelion) is at Mars's orbit, 1.52 AU from the Sun. Assume Earth's and Mars's orbits are circular. Earth's orbital period is 365 days.
Diagram not to scale. Orbits of Earth and Mars shown as circles about the Sun.
(a)
Outline Kepler's second law.
(1)
(b)
Show that the semi-major axis of the transfer orbit is 1.26 AU.
(1)
(c)
Determine the time, in days, taken for the probe to travel from Earth to Mars.
(3)
(d)
Explain, with reference to energy, why the probe's speed decreases as it travels from perihelion to aphelion.
(2)
(e)
The speed of the probe at perihelion is 32.7 km s−1. Calculate its speed at aphelion.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The line joining a planet (or probe) to the Sun sweeps out equal areas in equal time intervals
✓ 1
Part (b)
a = (1.00 + 1.52)/2 = 1.26 AU
✓ 1
Part (c)
By Kepler's third law with Earth as reference: T = 1.263/2 × 365 days = 1.41 years (≈ 516 days)
✓ 1
The journey from perihelion to aphelion is half an orbit
✓ 1
Time ≈ 258 days (accept 257–259 days)
✓ 1
ECF from (b)
—
Part (d)
As the probe moves away from the Sun, its gravitational potential energy increases (the force has a component opposite to its velocity, so it does negative work)
✓ 1
Total energy is conserved, so kinetic energy, and therefore speed, decreases
✓ 1
Part (e)
At perihelion and aphelion the velocity is perpendicular to the radius. Equal areas swept per unit time gives rpvp = rava
✓ 1
va = 32.7 × 1.00/1.52 = 21.5 km s−1
✓ 1
Common error: A transfer from Earth to Mars covers only half of the ellipse. Many answers give the full period by mistake.
133IBP2S-028-00-14Variation of g with distancePaper 2Hard10 marks
A transit survey finds a rocky exoplanet with radius 1.6 times Earth's radius. Measurements of the motion of its star give the planet's mass as 6.0 times Earth's mass. Earth's mean density is 5.5×103 kg m−3 and the gravitational field strength at Earth's surface is 9.8 N kg−1.
(a)
Show that the surface gravitational field strength of a uniform spherical planet of density ρ and radius R is \(g = \frac{4}{3}\pi G\rho R\).
(2)
(b)
Determine the gravitational field strength at the exoplanet's surface.
(2)
(c)
Determine the mean density of the exoplanet.
(2)
(d)
Suggest what the answer to (c) indicates about the composition of the exoplanet.
(2)
(e)
On Earth, an astronaut in a spacesuit can raise her centre of mass by 0.50 m in a vertical jump. Estimate the height she could reach on the exoplanet, assuming the same take-off speed.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Uses g = GM/R2 with M = ρ(4/3)πR3 .
✓ 1
Cancels R2 to obtain g = (4/3)πGρR .
✓ 1
Part (b)
gp/gE = 6.0/1.62 = 2.34 .
✓ 1
gp = 2.34 × 9.8 ≈ 23 N kg−1 .
✓ 1
Part (c)
ρp/ρE = 6.0/1.63 = 1.46, or equivalently (gp/gE)/1.6 using (a) .
✓ 1
ρp = 1.46 × 5.5×103 ≈ 8.1×103 kg m−3 . ECF from (b).
✓ 1
Part (d)
Its density is far greater than that of ice or water (about 1×103 kg m−3) or gas, so the planet must be made mainly of rock and metal .
✓ 1
Its density is higher than Earth's. This may be because it has a larger proportion of iron (a larger core) or because its greater mass compresses its interior more strongly. Either explanation is credited if it is linked to the higher density .
✓ 1
Part (e)
h = v2/2g, so for the same take-off speed h ∝ 1/g .
✓ 1
h = 0.50/2.34 ≈ 0.21 m . ECF from (b).
✓ 1
Exam tip: Working in ratios to Earth avoids needing the values of G, ME and RE and reduces arithmetic slips.
134IBP2S-028-00-15Newton's law of gravitationPaper 2Hard10 marks
In a laboratory torsion-balance experiment, a small uniform lead sphere of mass 0.75 kg is attracted by a large uniform lead sphere of mass 160 kg. The distance between their centres is 0.23 m. The gravitational force between them is measured as 1.50×10−7 N.
(a)
Determine the value of the gravitational constant G obtained from this measurement.
(2)
(b)
Explain why the separation used in Newton's law is the distance between the centres of the spheres.
(2)
(c)
The centre-to-centre separation is measured to ±0.005 m. All other uncertainties are negligible. Determine the percentage uncertainty in G.
(2)
(d)
Using your value of G from (a), together with g = 9.8 N kg−1 and Earth's radius 6.37×106 m, determine the mass of Earth.
(2)
(e)
Determine the mean density of Earth.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
G = Fr2/(m1m2) = 1.50×10−7 × 0.232/(160 × 0.75) .
✓ 1
G = 6.6×10−11 N m2 kg−2 (6.61×10−11) .
✓ 1
Part (b)
Outside a uniform sphere, its gravitational field is the same as that of a point mass located at its centre .
✓ 1
Newton's law applies to point masses, so r must be measured between the centres .
✓ 1
Part (c)
Percentage uncertainty in r = 0.005/0.23 × 100 = 2.17% .
✓ 1
G ∝ r2, so the percentage uncertainty in G = 2 × 2.17 ≈ 4.3%. Accept answers from 4.3% to 4.4%, depending on rounding .
✓ 1
Part (d)
M = gR2/G = 9.8 × (6.37×106)2/(6.61×10−11) .
✓ 1
M = 6.0×1024 kg . Allow the answer carried forward from the value of G found in (a).
✓ 1
Part (e)
V = (4/3)π(6.37×106)3 = 1.08×1021 m3 .
✓ 1
ρ = M/V = 6.01×1024/1.08×1021 ≈ 5.6×103 kg m−3 . Allow the answer carried forward from the mass found in (d).
✓ 1
Common error: Students often use the percentage uncertainty in r as the percentage uncertainty in G. G depends on r², so double the percentage uncertainty in r. Keep unrounded intermediate values to avoid rounding drift.
135IBP2S-028-02-15Kepler's third lawPaper 2Hard8 marks
A space telescope detects an exoplanet by the dip in starlight each time the planet passes in front of its star. The mid-times of four consecutive transits are shown below. The star has mass 2.2×1030 kg. Assume the orbit is circular and the planet's mass is negligible compared with the star's.
Transit number | Mid-time / days 1 | 0.00 2 | 3.52 3 | 7.04 4 | 10.56
1 day = 86 400 s
(a)
Determine the orbital period of the planet, in seconds.
(1)
(b)
Determine the orbital radius of the planet.
(3)
(c)
Calculate the gravitational field strength due to the star at the planet's orbit.
(2)
(d)
A second planet is later found orbiting the same star. The star's gravitational field strength at its orbit is 1/16 of the value in (c). Determine the orbital period of the second planet, in days.
T ∝ r3/2, so T = 43/2 × 3.52 = 8 × 3.52 = 28 days (ECF from (a))
✓ 1
Exam tip: For the period, divide the total time span by the number of intervals between transits, not by the number of transits. This uses all of the data.
136IBP2S-028-02-16Variation of g with distancePaper 2Hard8 marks
Mining geologists use a sensitive gravimeter to search for a buried ore deposit. The deposit is modelled as a sphere of radius 200 m with its centre 600 m below a flat surface. The density of the ore is greater than the density of the surrounding rock by 2500 kg m−3. The gravimeter measures the extra vertical gravitational field strength, Δg, caused by this excess mass.
Diagram not to scale
(a)
Outline why the deposit's gravitational effect at the surface can be calculated as if its excess mass were concentrated at its centre.
(1)
(b)
Calculate the excess mass of the deposit.
(2)
(c)
Calculate Δg at the point on the surface directly above the centre of the deposit.
(2)
(d)
Determine Δg at a point on the surface 800 m horizontally from the point directly above the deposit.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Outside a uniform spherical mass distribution, the field is the same as that of a point mass at the centre, and the gravimeter is outside the sphere
✓ 1
Part (b)
Δm = Δρ × (4/3)πr3 = 2500 × (4/3)π × (200)3
✓ 1
Δm = 8.4×1010 kg
✓ 1
Part (c)
Δg = GΔm/d2 = 6.67×10−11 × 8.4×1010/6002
✓ 1
Δg = 1.6×10−5 N kg−1 (allow error carried forward from the excess mass)
✓ 1
Part (d)
Distance from the centre of the deposit = √(6002 + 8002) = 1000 m
✓ 1
Magnitude of the extra field = GΔm/10002 = 5.6×10−6 N kg−1
✓ 1
Vertical component = 5.6×10−6 × (600/1000) = 3.4×10−6 N kg−1 (allow error carried forward from the excess mass)
✓ 1
Common error: Away from the point directly above the deposit, the extra field is not vertical. The gravimeter detects only the vertical component, so resolve the field before giving the answer.
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