123 free, high-quality past-paper-style questions on Gravitational fields for IB Physics HL, each with a full mark scheme. Covers: Work from equipotentials; Newton's law of gravitation; Escape speed; and more. For when you have finished the past papers, or want to practise one topic at a time.D.1 at Standard 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.559 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1D-1A-29Work from equipotentialsPaper 1AEasy1 mark
The diagram shows equipotential surfaces around a planet. A spacecraft of mass 2000 kg moves from point A to point B.
What is the minimum work that must be done on the spacecraft?
Equipotential surfaces (dashed) around a planet. The values are the gravitational potential on each surface.Show mark scheme
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Step 1A is on the −60 MJ kg−1 surface and B on the −30 MJ kg−1 surface: ΔV = −30 − (−60) = +30 MJ kg−1.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2W = mΔV = 2000 × 30 × 106.
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Step 3W = 6.0 × 1010 J, independent of the path taken.
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Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis is the potential difference alone (30 MJ), not multiplied by the mass.
BCorrect: W = mΔV = 2000 × 3.0 × 107 J.
CThis uses the potential at A (−60) rather than the difference between A and B.
2D-1A-38Newton's law of gravitationPaper 1AEasy1 mark
A space probe is at a distance d from the centre of an asteroid and the gravitational force between them is F. The probe then releases a lander containing half of its mass, and moves away until it is at a distance 3d from the centre of the asteroid.
What is the gravitational force between the asteroid and the probe now?
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Step 1F ∝ m1m2/r²: the probe mass is multiplied by ½ and the distance by 3.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2New force = F × ½ × 1/3² = F/18.
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: ½ × 1/9 = 1/18.
BThis ignores the halving of the probe mass.
CThis halves the mass but does not square the distance: ½ × 1/3.
DThis doubles the mass instead of halving it: 2 × 1/9.
A planet moves around a star in an elliptical orbit. No other bodies affect its motion.
I. The star is at one focus of the ellipse.
II. The gravitational force on the planet is perpendicular to the velocity of the planet at every point of the orbit.
III. The kinetic energy of the planet is greatest at the point of the orbit closest to the star.
Which statements are correct?
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Step 1Kepler's first law: every planet moves in an ellipse with the star at one focus, so I is correct.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The gravitational force is directed towards the star. Except at the closest and furthest points, the velocity is not perpendicular to the line to the star, so the force has a component along (or against) the velocity and the speed changes. II is incorrect.
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Step 3The total energy is constant. Closest to the star the gravitational potential energy is least (most negative), so the kinetic energy is greatest. III is correct: I and III only.
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Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis accepts II, which is true only for a circular orbit (or at the two ends of the major axis), and misses that III follows from energy conservation.
BCorrect: the star is at a focus, and the kinetic energy is greatest where the potential energy is least.
CThis places the star at the centre of the ellipse instead of at a focus, and applies the circular-orbit result in II to an ellipse.
DThis treats all three statements as true, applying the circular-orbit property in II to an elliptical orbit.
5D-1A-61Uniform field near the surfacePaper 1AEasy1 mark
A small region just above the flat surface of a planet is considered. The equipotential surfaces in each diagram are drawn at equal intervals of gravitational potential.
Which diagram shows the gravitational field lines (solid) and the equipotential surfaces (dashed) in this region?
Field lines are drawn solid, with arrows; equipotential surfaces are dashed. Each diagram shows the same small region above the surface.Show mark scheme
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Step 1Close to the surface, over a small region, the field is uniform: the field lines are parallel, equally spaced and directed towards the planet (the direction of the force on a mass).
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Equipotential surfaces are perpendicular to the field lines, so they are horizontal.
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Step 3Since g = −ΔVg/Δr is constant, equal intervals of potential are equal distances apart: diagram D.
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Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThe spacing grows with height as it would far from a planet, where g decreases; over a small region near the surface g is constant, so the spacing is equal.
BThe arrows point away from the planet; the field direction is the direction of the force on a mass, which is towards the planet.
CThe equipotentials are drawn parallel to the field lines; they must be perpendicular, because no work is done moving a mass along an equipotential.
DCorrect: parallel, equally spaced field lines towards the surface, with perpendicular equipotentials equally spaced.
6IBP2H-029-00-01Kepler's second lawPaper 1AEasy1 mark
A comet moves in an elliptical orbit around the Sun. Its closest distance from the Sun is 0.50 AU and its greatest distance is 20 AU. What is the ratio \(\frac{\text{speed at closest approach}}{\text{speed at greatest distance}}\)?
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At the two extreme points the velocity is perpendicular to the radius. By Kepler's second law the area swept per unit time is constant, so ½rpvp = ½rava. This is equivalent to conservation of angular momentum. Therefore vp/va = ra/rp = 20/0.50 = 40. Option A inverts the ratio. Option B uses a square-root dependence on distance, which applies to circular orbits of different radii, not to one ellipse. Option D squares the ratio, treating speed like field strength.
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Answer C
Answer: C
Common error: The relation v ∝ 1/√r is for comparing different circular orbits. Within a single elliptical orbit, rv is constant at perihelion and aphelion.
7IBP2H-029-00-07Equipotential surfacesPaper 1AEasy1 mark
A rover on Mars climbs a slope where the gravitational field is uniform with strength 3.7 N kg−1. What is the vertical separation between equipotential surfaces that differ in gravitational potential by 10 J kg−1?
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In a uniform field ΔV = gΔh, so Δh = ΔV/g = 10/3.7 = 2.7 m. Option A uses Earth's field strength of 9.8 N kg−1. Option C multiplies ΔV by g. Option B calculates g/ΔV.
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Answer D
Answer: D
Common error: Equipotentials in a uniform field are equally spaced horizontal planes. A weaker field means a larger spacing for the same ΔV, not a smaller one.
8IBP2H-029-01-03Equipotential surfacesPaper 1AEasy1 mark
Equipotential surfaces are drawn around an isolated spherical planet. Adjacent surfaces differ by the same potential difference. Which statement is correct?
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Field strength equals the magnitude of the potential gradient, g = −ΔV/Δr. Further from the planet g is smaller, so a larger Δr is needed for the same ΔV. The surfaces therefore spread out. Equal spacing (A) would only occur in a uniform field. Option B reverses the trend. Field lines are always perpendicular to equipotentials (D is wrong).
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Answer C
Answer: C
Exam tip: Closely spaced equipotentials mean a strong field. Widely spaced equipotentials mean a weak field.
A planet has radius R. The gravitational field strength at its surface is g0. At what height above the surface is the field strength equal to g0/9?
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g ∝ 1/r2, so g = g0/9 requires r = 3R from the centre. The height above the surface is 3R − R = 2R. Option B is the distance from the centre, not the height. Option C assumes g ∝ 1/r and then subtracts R. Option D assumes g ∝ 1/r and measures from the centre.
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Answer A
Answer: A
Common error: In g = GM/r², r is measured from the centre of the planet. Height above the surface is r − R.
10IBP2H-029-01-07Work from equipotentialsPaper 1AEasy1 mark
Near a moon, two equipotential surfaces have potentials of −4.8 MJ kg−1 and −4.2 MJ kg−1. They are 250 km apart, measured radially. What is the average gravitational field strength between the surfaces?
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Magnitude: |g| = |ΔV|/Δr = (0.6×106 J kg−1)/(2.5×105 m) = 2.4 N kg−1. The field points towards lower (more negative) potential. The more negative surface is closer to the moon, so the field points towards the moon. Option B has the wrong direction. Option C comes from dividing 0.6 MJ kg−1 by 250 km without converting either prefix. Option D divides a single potential (4.8 MJ kg−1) by the separation instead of using the potential difference.
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Answer A
Answer: A
Common error: Field strength comes from the potential difference divided by the distance, not from the potential itself. Convert MJ to J and km to m before dividing.
11IBP2H-029-02-01Equipotential surfacesPaper 1AEasy1 mark
Two uniform spheres are held with their centres a distance d apart. The gravitational force between them is F. The mass of each sphere is doubled and the separation of their centres is increased to 3d. What is the new gravitational force?
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F = Gm1m2/r2. Doubling both masses multiplies the product m1m2 by 2 × 2 = 4. Tripling the separation divides the force by 32 = 9. The new force is therefore 4F/9. Option C comes from doubling only one mass. Option B treats the force as inversely proportional to r rather than r2. Option A makes both mistakes.
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Answer D
Answer: D
Common error: The product of the masses appears in Newton's law, so changing both masses changes the force by the product of the two factors.
12IBP2H-029-02-08Kepler's first lawPaper 1AEasy1 mark
A satellite of mass m moves in a circular orbit of radius r at speed v around a planet of mass M. What is the work done by the gravitational force on the satellite during one complete orbit?
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A circular orbit lies on an equipotential surface, so the potential does not change and W = mΔV = 0. Equivalently, the gravitational force is always perpendicular to the velocity. Option B multiplies the force by the distance travelled, which ignores the direction of the force. Option C is the size of the potential energy. Option D is the kinetic energy, which stays constant rather than being supplied by gravity.
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Answer A
Answer: A
Exam tip: No work is done moving along an equipotential surface. Look for this whenever the force is perpendicular to the motion.
13IBP3H-D1-20r32-01Kepler's second lawPaper 1AEasy1 mark
A satellite that relays data from tracking buoys on fishing fleets moves in a circular orbit of radius r around Earth with orbital speed v. A second satellite, in a circular orbit of radius 4r around the same planet, has orbital speed
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The gravitational force provides the centripetal force, so GMm/r² = mv²/r and v = √(GM/r), giving v ∝ 1/√r. Increasing r by a factor of 4 divides v by √4 = 2, so the speed is v/2. v/4 treats v as proportional to 1/r; v/16 treats it as proportional to 1/r²; 2v wrongly has speed increasing with orbital radius.
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Answer C
Answer: C
Common error: Orbital speed depends on the inverse square root of the radius, not on the inverse of the radius.
A robotic lander can visit two rocky planets with the same mean density. Planet B has twice the radius of planet A. The escape speed from the surface of planet A is v. The escape speed from the surface of planet B is
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Escape speed is v = √(2GM/R). With equal density, M ∝ R³, so v² ∝ R³/R = R² and v ∝ R. Doubling the radius doubles the escape speed to 2v. √2 v comes from taking M ∝ R; 4v comes from forgetting the square root; √8 v comes from using M ∝ R³ but forgetting to divide by R.
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Answer A
Answer: A
Exam tip: When two planets share the same density, write the mass as density × (4/3)πR³ before comparing escape speeds.
15IBP3H-D1-20r32-07Kepler's third lawPaper 1AEasy1 mark
A planet has two moons in circular orbits. The inner moon has an orbital radius of 2.0 × 105 km and a period of 3.0 days. The outer moon has an orbital radius of 8.0 × 105 km. The period of the outer moon is
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Kepler's third law gives T² ∝ r³. The radius increases by a factor of 4, so T increases by 43/2 = 8, giving 3.0 × 8 = 24 days. 12 days takes T ∝ r; 48 days takes T ∝ r² (factor 16); 6 days takes T ∝ √r.
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Answer A
Answer: A
Common error: Kepler's third law is T² ∝ r³, so the period scales as r³/², not as r.
A communications satellite moves in a circular orbit of radius r about a planet with speed v. It is moved to a new circular orbit of radius 4r about the same planet. What is its orbital speed in the new orbit?
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Gravitational force provides the centripetal force, so v = √(GM/r) and v ∝ 1/√r. Quadrupling r halves the speed, giving v/2. The value v/4 treats the speed as inversely proportional to r. The value v/16 treats it as inversely proportional to r2. The value 2v has the speed increasing, whereas a larger orbit needs a smaller speed.
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Answer C
Answer: C
Common error: Do not mix up the orbital speed, which varies as 1/√r, with the field strength, which varies as 1/r².
17IBP3H-D1-20r47-02Escape speed and orbital speedPaper 1AEasy1 mark
The gravitational field strength at the surface of Earth is 9.8 N kg−1. An astronaut of mass 60 kg stands on a platform held at rest at a height above the surface equal to the radius of Earth. What is the gravitational force on the astronaut at this height?
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At a height equal to Earth's radius the distance from Earth's centre is 2R, so the field strength falls to 9.8/4 = 2.45 N kg−1 and the force is 60 × 2.45 = 147 N. The value 588 N is the force at the surface. The value 294 N comes from halving the field for a doubled distance. The value 0 N confuses a weightless feeling in orbit with the absence of gravity, but the force is still large.
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Answer B
Answer: B
Exam tip: Measure distance in the inverse-square law from the centre of the planet, not from its surface.
18IBP3H-D1-20r47-06Resultant field of two bodiesPaper 1AEasy1 mark
Two identical spherical stars, each of mass M, have their centres a distance 2d apart. What is the magnitude of the resultant gravitational field strength at the point midway between their centres?
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Each star produces a field of magnitude GM/d2 at the midpoint, but the two fields point in opposite directions towards the two stars, so they cancel and the resultant is zero. The value GM/d2 counts only one star. The value 2GM/d2 adds the two magnitudes without considering direction. The value GM/(2d2) comes from a wrong distance and a single star.
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Answer A
Answer: A
Common error: Gravitational field strength is a vector: fields from two bodies on opposite sides of a point subtract rather than add.
Planet X moves in a circular orbit of radius r around a star of mass M with period T. Planet Y moves in a circular orbit of radius 2r around a different star of mass 4M.
What is the orbital period of Y?
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Step 1Gravity provides the centripetal force: GMm/r² = m(4π²/T²)r, so T² = 4π²r³/(GM).
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2For Y, r³ is multiplied by 8 and M by 4, so T² is multiplied by 8/4 = 2.
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Step 3Period of Y = √2 T.
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Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: T² ∝ r³/M gives a factor of √(8/4).
BThis uses T ∝ r³/M without taking the square root: 8/4 = 2.
CThis ignores the different mass of the star: T²/r³ is the same only for orbits around the same central body. √8 = 2√2.
DThis puts the mass in the numerator (T² ∝ Mr³): √(8 × 4) = 4√2.
Newton's universal law of gravitation, F = Gm1m2/r², is used to calculate the gravitational force between the bodies in each situation below, with r taken as the distance between their centres.
I. Two uniform solid lead spheres of radii 5.0 cm and 2.0 cm whose surfaces are touching, with r = 7.0 cm.
II. An irregularly shaped asteroid 3 km long and a planet whose centre is 2 × 108 m away.
III. Two identical steel rods, each 1.0 m long, lying parallel and side by side with their centres 5.0 cm apart.
In which situations does the law give the force correctly?
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Step 1A uniform sphere (spherically symmetric mass distribution) acts on bodies outside it as if all its mass were at its centre, at any separation — so I is valid even with the spheres touching.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Any body acts as a point mass when the separation is much larger than its size: 3 km ≪ 2 × 108 m, so II is valid.
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Step 3The rods are neither spherical nor far apart (5.0 cm ≪ 1.0 m), so different parts are at very different distances and the formula with the centre separation fails: III is not valid.
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Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis rejects II, assuming a non-spherical body can never be treated as a point mass — but it can when the separation is much larger than its size.
BThis rejects I, assuming spheres must be far apart compared with their radii — but uniform spheres act as point masses at any separation outside them.
CCorrect: I (uniform spheres) and II (separation ≫ size) only.
DThis assumes the law always works with the distance between centres; for long rods close together it does not.
24D-1A-32Equipotential surfacesPaper 1AMedium1 mark
Gravitational equipotential surfaces around an isolated spherical planet of radius R are drawn at equal intervals of potential. Close to the surface of the planet, adjacent surfaces are a distance s apart, where s is much smaller than R.
What is the distance between adjacent surfaces at a height 2R above the surface?
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Step 1g = −ΔVg/Δr, so for a fixed interval ΔVg the separation Δr = ΔVg/g is inversely proportional to g.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2g = GM/r², so the separation is proportional to r².
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Step 3At a height 2R the distance from the centre is 3R: separation = 3² × s = 9s.
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Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis treats the field as uniform, so that the surfaces are equally spaced everywhere. That is true only close to the surface.
BThis takes the separation as proportional to r, as if g were proportional to 1/r.
CThis uses the height 2R as the distance from the centre: 2² = 4.
The graph shows the variation of the gravitational potential Vg with distance r from the centre of a planet. The tangent to the curve at r = 8.0 × 106 m is drawn.
Which row gives the magnitude and the direction of the gravitational field strength at r = 8.0 × 106 m?
Variation of gravitational potential with distance from the centre of a planet. The dashed line is the tangent to the curve at r = 8.0 × 106 m; it passes through (0, −5.0) and (16, 0).
Magnitude / m s−2Direction
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Step 1Gradient of the tangent = ΔVg/Δr = (0 − (−5.0) MJ kg−1)/(16 × 106 m − 0) = 5.0 × 106/1.6 × 107 = 0.31 J kg−1 m−1.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2g = −ΔVg/Δr = −0.31 m s−2.
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Step 3The negative sign means the field points in the direction of decreasing r: towards the planet.
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Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis ignores the prefix M on the potential axis (5.0 J kg−1 instead of 5.0 × 106 J kg−1) while converting the distance: 5.0/1.6 × 107.
BThe magnitude is right but the sign of g = −ΔVg/Δr has been dropped: a positive potential gradient gives a field towards smaller r.
CCorrect: magnitude 0.31 m s−2, directed towards the planet.
DThis is the gradient of the chord between r = 4.0 and 12.0 × 106 m on the curve ((−1.67 − (−5.0))/8.0 = 0.42), not of the tangent at 8.0 × 106 m.
27D-1A-35Gravitational potential energyPaper 1AMedium1 mark
Two small moonlets, each of mass m, are at rest a distance a apart, far from any other mass.
What is the minimum work needed to increase their separation to 3a, leaving them at rest?
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Step 1The gravitational potential energy of the two-body system is Ep = −Gm²/r; the minimum work needed equals the increase in Ep (no kinetic energy is given to the moonlets).
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Initially Ep = −Gm²/a; finally Ep = −Gm²/(3a).
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Step 3W = −Gm²/(3a) − (−Gm²/a) = 2Gm²/(3a).
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Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis is the magnitude of the final potential energy only; the work done is the change in potential energy.
BCorrect: the potential energy rises from −Gm²/a to −Gm²/(3a), an increase of 2Gm²/(3a).
CThis treats the potential energy as varying as 1/r², like the force: (1 − 1/9)Gm²/a.
DThis counts the energy twice, once for each moonlet in the field of the other; the potential energy belongs to the pair, so there is only one term −Gm²/r.
Five moons move in circular orbits around the same planet. The graph shows the square of the orbital speed v² of each moon against the reciprocal of its orbital radius, 1/r, with the best-fit line.
What is the mass of the planet?
Square of orbital speed against the reciprocal of orbital radius for five moons of one planet.Show mark scheme
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Step 1vorbital = √(GM/r), so v² = GM × (1/r): the graph is a straight line through the origin with gradient GM.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Gradient of the line = 3.0 × 108 m² s−2 ÷ 8.0 × 10−9 m−1 = 3.75 × 1016 m³ s−2 (accept 3.7–3.8 × 1016).
29IBP2S-028-01-05Kepler's second lawPaper 1AMedium1 mark
An asteroid moves in an elliptical orbit around the Sun. Which quantity remains constant throughout the orbit?
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The gravitational force always acts along the line to the Sun. It therefore exerts no torque about the Sun, so the asteroid's angular momentum about the Sun is conserved. This is the physical origin of Kepler's second law: the rate at which area is swept out, ½rv⊥, is proportional to the angular momentum.
The other quantities all change: • the speed changes around an ellipse, • the kinetic energy therefore changes too, • the force magnitude changes because r changes.
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Answer C
Answer: C
Exam tip: Equal areas in equal times is equivalent to conservation of angular momentum for a central force.
Two exoplanets orbit the same star in circular orbits. The orbital period of planet Q is 8 times the orbital period of planet P. What is \(\frac{\text{orbital radius of Q}}{\text{orbital radius of P}}\)?
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Kepler's third law gives T2 ∝ r3, so r ∝ T2/3. The ratio is therefore 82/3 = (81/3)2 = 22 = 4. Option C takes only the cube root and forgets the square. Option B assumes r ∝ T. Option A uses r ∝ T3/2, which swaps the exponents.
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Answer D
Answer: D
Exam tip: Write T² ∝ r³ first, then make the unknown the subject before substituting. This avoids swapping the powers 2/3 and 3/2.
31IBP2H-029-00-03Variation of g with heightPaper 1AMedium1 mark
Planet X has gravitational field strength g at its surface. Planet Y has twice the mean density of X and half the radius of X. Both planets are uniform spheres. What is the gravitational field strength at the surface of Y?
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For a uniform sphere, M = (4/3)πR3ρ. Then g = GM/R2 = (4/3)πGρR, so g ∝ ρR. For Y, ρR = (2ρ)(R/2) = ρR, so the field strength is unchanged at g. Option A considers only the halved radius. Option C considers only the doubled density. Option D keeps the mass fixed and applies 1/R2, ignoring the fact that halving R reduces the mass.
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Answer B
Answer: B
Exam tip: When density and radius are given, express g in terms of them: g = (4/3)πGρR. This turns the comparison into a one-line ratio.
32IBP2H-029-00-04Resultant field of two bodiesPaper 1AMedium1 mark
The mass of the Earth is 81 times the mass of the Moon. A spacecraft travels along the line joining their centres. At point N the resultant gravitational field strength due to the Earth and the Moon is zero. What is \(\frac{\text{distance from N to the centre of the Earth}}{\text{distance from N to the centre of the Moon}}\)?
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At N the two fields are equal in magnitude: GME/x2 = GMM/y2. This gives x2/y2 = ME/MM = 81, so x/y = 9. N is therefore much closer to the Moon. Option D forgets the square root. Option B inverts the ratio. Option A takes a fourth root.
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Answer C
Answer: C
Common error: The field-free point is closer to the smaller mass. Check that your ratio places it on the Moon's side.
Two rocky planets A and B have the same mean density. The radius of B is twice the radius of A. The escape speed from the surface of A is v. What is the escape speed from the surface of B?
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vesc = √(2GM/R) and M = (4/3)πR3ρ. Therefore vesc = R√(8πGρ/3), so vesc ∝ R at fixed density. Doubling R gives 2v. Option C keeps M constant while R doubles. Option B assumes that equal density means equal escape speed. Option D uses M ∝ R3 but forgets to divide by R.
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Answer A
Answer: A
Exam tip: At constant density both g and escape speed are proportional to R. Writing M in terms of ρ and R first makes the scaling clear.
34IBP2H-029-01-01Launch and escape from orbitPaper 1AMedium1 mark
A satellite moves in a circular orbit around a planet. Its kinetic energy is Ek. Air resistance and other bodies are negligible. What is the minimum kinetic energy that must be added to the satellite so that it escapes from the planet?
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For a circular orbit, Ek = GMm/2r and Ep = −GMm/r, so the total energy is −GMm/2r = −Ek. To escape, the total energy must be raised to zero, so an extra Ek is needed. The kinetic energy doubles, so the speed rises by a factor of √2. Option A comes from taking the total energy as −½Ek. Option C confuses the energy added with the final kinetic energy. Option D applies the speed ratio to the energy.
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Answer B
Answer: B
Exam tip: For any circular orbit, Etotal = −Ek = ½Ep. Remembering this one result gives a quick route through most orbital energy problems.
35IBP2H-029-01-02Resultant field of two bodiesPaper 1AMedium1 mark
The mass of the Earth is 81 times the mass of the Moon. The distance between their centres is d. A spacecraft lies on the line joining the centres, at the point where the resultant gravitational field strength is zero. How far is the spacecraft from the centre of the Earth?
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At the zero-field point, GME/x2 = GMM/(d − x)2. This gives x/(d − x) = √81 = 9, so x = 9d/10 = 0.90d. Option A gives the distance from the Moon, not the Earth. Option B ignores the unequal masses. Option D uses the mass ratio without taking the square root (81/82 ≈ 0.99).
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Answer C
Answer: C
Common error: Field strength depends on 1/r². The distance ratio is therefore the square root of the mass ratio, not the mass ratio itself.
36IBP2H-029-01-05Gravitational potential energyPaper 1AMedium1 mark
Two identical stars, each of mass M, are a distance d apart. What are the gravitational field strength g and the gravitational potential V at the midpoint between them?
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The two field vectors are equal in size and opposite in direction, so they cancel and g = 0. Potential is a scalar, and each star contributes −GM/(d/2) = −2GM/d. The total is V = −4GM/d. Option A treats potential as if it cancels like a vector. Option C adds the fields as scalars and cancels the potentials. Option D uses distance d instead of d/2, or counts only one star.
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Answer B
Answer: B
Exam tip: Add field strengths as vectors. Add potentials as signed scalars. Zero field does not mean zero potential.
Planet X has the same radius as the Earth but twice the Earth's mean density. What is the ratio \(\frac{\text{escape speed from X}}{\text{escape speed from Earth}}\)?
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vesc = √(2GM/R) and M = (4/3)πR3ρ, so vesc = R√(8πGρ/3) ∝ R√ρ. With the same R and double ρ, the ratio is √2. Option A assumes only the radius matters. Option D forgets the square root. Option B inverts the dependence.
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Answer C
Answer: C
Exam tip: Replacing M with (4/3)πR³ρ often turns a comparison question into a simple proportionality.
38IBP2S-028-04-08Launch and escape from orbitPaper 1AMedium1 mark
A navigation satellite in a circular orbit around the Earth has a gravitational potential energy of −8.0 GJ. What are its kinetic energy and its total energy?
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For a circular orbit, Ek = GMm/(2r) and Ep = −GMm/r, so Ek = −½Ep = 4.0 GJ. The total energy is Ek + Ep = 4.0 − 8.0 = −4.0 GJ. The total is negative, which shows the satellite is bound. Option B assumes Ek = |Ep|, which would describe a body moving at escape speed. Option C subtracts the kinetic energy instead of adding it.
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Answer A
Answer: A
Exam tip: For any circular orbit, Ek : Ep : Etotal = 1 : −2 : −1.
39IBP2S-028-04-09Potential gradientPaper 1AMedium1 mark
A spacecraft travels in a straight line from the Earth towards the Moon. N is the point between them where the net gravitational field strength is zero. Which statement about the gravitational potential at N is correct, considering only points on the line between the surfaces of the Earth and the Moon?
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Field strength is the negative of the potential gradient, g = −ΔV/Δr. At N the field is zero, so the potential gradient is zero and V has a turning point there. Moving from N towards either body, the potential becomes more negative. N is therefore a maximum of V, although V is still negative there. Option A confuses zero field with zero potential. The potentials of the two bodies are both negative, so they add rather than cancel.
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Answer C
Answer: C
Common error: Zero field does not mean zero potential. The potentials of two masses always add to give a more negative value.
40IBP2H-029-02-02Launch and escape from orbitPaper 1AMedium1 mark
A satellite is in a circular orbit of radius r around a planet. Its kinetic energy is K. Its engines move it to a new circular orbit of radius 2r. Air resistance is negligible. What is the minimum work done by the engines?
Diagram not to scale
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For a circular orbit, Ek = GMm/2r = K, Ep = −GMm/r = −2K and the total energy is E = −K. At radius 2r the total energy is −K/2. The work done by the engines equals the increase in total energy: −K/2 − (−K) = K/2. Option C is the gain in potential energy alone, which ignores the fall in kinetic energy. Option D adds the size of the kinetic energy change to the potential energy change. Option A assumes that the kinetic energy lost exactly pays for the potential energy gained.
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Answer B
Answer: B
Exam tip: In a circular orbit Ek : Ep : Etotal = 1 : −2 : −1. Write all three in terms of one quantity before comparing two orbits.
41IBP2H-029-02-03Work from equipotentialsPaper 1AMedium1 mark
A 500 kg probe starts at rest at point P on the −40 MJ kg−1 equipotential surface of a moon. It is moved to point Q on the −25 MJ kg−1 surface and then to point R on the −30 MJ kg−1 surface, where it is again at rest. What is the net work done on the probe by the external agent from P to R?
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Gravity is a conservative field, so the work done depends only on the start and end potentials: W = mΔV = 500 × (−30 − (−40)) × 106 = 5.0 × 109 J. The path through Q has no effect. Option C uses only the first stage, P to Q. Option D adds the sizes of both stages (15 + 5 MJ kg−1). Option A uses only the second stage, Q to R.
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Answer B
Answer: B
Exam tip: In a gravitational field, net work between two points is m × (Vfinal − Vinitial), whatever route is taken.
42IBP2H-029-02-04Orbital energyPaper 1AMedium1 mark
A communications company needs a satellite that remains above the same point on the Earth's surface at all times. Which statement about this geostationary satellite is correct?
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The gravitational force points towards the Earth's centre, so every orbit is centred on that point. To stay fixed above one surface point, the orbit must lie in the equatorial plane. The satellite must also move in the same direction as the Earth rotates (west to east) with the same period. Option A is wrong because an inclined 24-hour orbit drifts north and south. Option B is wrong because the orbital radius does not depend on the satellite's mass. Option D confuses equal angular velocity with equal linear speed: the satellite is much further from the axis, so it moves much faster.
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Answer C
Answer: C
Common error: A geostationary satellite has the same angular velocity as the Earth's surface, not the same linear speed.
43IBP2H-029-02-06Launch and escape from orbitPaper 1AMedium1 mark
A spacecraft is in a circular orbit around a planet. Atmospheric effects are negligible. A short engine burn changes its speed, with no change of position. By what factor must its kinetic energy be multiplied so that it just escapes from the planet?
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In a circular orbit, Ek = GMm/2r and Ep = −GMm/r. To just escape, the total energy must be zero, so the new Ek must equal GMm/r. This is twice the orbital kinetic energy. Option A is the factor by which the speed increases. Option D squares 2 instead of √2. Option C adds kinetic energy equal to |Ep| on top of the kinetic energy the spacecraft already has.
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Answer B
Answer: B
Common error: Escape requires a total energy of zero. The kinetic energy the spacecraft already has counts towards this.
44IBP3H-D1-20r32-03Launch and escape from orbitPaper 1AMedium1 mark
A satellite carrying an advertising display for a fashion label is in a circular orbit around Earth. Its kinetic energy is K. Taking gravitational potential energy to be zero at infinite separation, the total energy of the satellite is
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For a circular orbit K = GMm/2r and the potential energy is −GMm/r = −2K. The total energy is K + (−2K) = −K. +K ignores the potential energy; −2K is the potential energy alone, not the total; −K/2 comes from halving the potential energy instead of adding the kinetic energy.
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Answer D
Answer: D
Common error: In a circular orbit the total energy equals minus the kinetic energy, and the potential energy is minus twice the kinetic energy.
45IBP3H-D1-20r32-04Potential gradientPaper 1AMedium1 mark
A probe near a planet measures a gravitational potential of −4.0 × 107 J kg−1 at a distance of 8.0 × 106 m from the centre of the planet. Assuming the planet is spherical and isolated, the magnitude of the potential gradient at this point, in J kg−1 m−1, is
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V = −GM/r, so dV/dr = GM/r² = −V/r. The magnitude is (4.0 × 107)/(8.0 × 106) = 5.0 J kg−1 m−1, which equals the field strength in N kg−1. 0.20 inverts the ratio; 2.5 divides by 2r; 3.2 × 1014 multiplies V by r.
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Answer B
Answer: B
Exam tip: Potential gradient has units J kg⁻¹ m⁻¹, which are the same as N kg⁻¹; check that your answer has these units.
46IBP3H-D1-20r32-06Resultant field of two bodiesPaper 1AMedium1 mark
A robotic rover is designed to hop across a small asteroid of radius 1.5 × 104 m. The gravitational field strength at the surface of the asteroid is 0.060 N kg−1. The minimum launch speed that would allow the rover to leave the asteroid for ever is closest to
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Using GM = gR², v = √(2gR) = √(2 × 0.060 × 1.5 × 104) = √1800 = 42 m s−1. 30 m s−1 is √(gR), the speed for a circular orbit at the surface; 21 m s−1 is √(gR/2); 1800 m s−1 forgets to take the square root.
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Answer D
Answer: D
Exam tip: Escape speed can be found from the surface field strength and radius without knowing the mass: v = √(2gR).
47IBP3H-D1-20r47-03Gravitational potentialPaper 1AMedium1 mark
Near the surface of a small moon, a probe climbs vertically through 300 m. The gravitational potential increases by 2.4 kJ kg−1 during the climb. What is the gravitational field strength in this region?
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The field strength is the magnitude of the potential gradient, so g = ΔV/Δr = 2400 J kg−1 / 300 m = 8.0 N kg−1. The value 0.125 comes from inverting the ratio. The value 7.2 × 105 comes from multiplying the two quantities. The value 8.0 × 103 forgets to convert kJ to J.
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Answer D
Answer: D
Common error: Convert kilojoules to joules before dividing; the potential gradient must be in J kg⁻¹ per metre.
48IBP3H-D1-20r47-04Escape speedPaper 1AMedium1 mark
The escape speed from the surface of Earth is 11.2 km s−1. A rocky planet has the same average density as Earth but twice the radius. What is the escape speed from the surface of this planet?
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With the same density, mass scales with volume, so the planet has 8 times the mass of Earth. Then v = √(2GM/R) increases by a factor √(8/2) = 2, giving 22.4 km s−1. The value 15.8 km s−1 forgets that the mass has also increased (factor √2 from the radius alone). The value 31.7 km s−1 uses the mass change only (√8). The value 44.8 km s−1 treats the speed as proportional to the square of the factor 2.
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Answer A
Answer: A
Exam tip: When a planet is scaled at constant density, write the mass as proportional to the cube of the radius before using any formula.
49IBP3H-D1-20r47-05Orbital speedPaper 1AMedium1 mark
Two moons, X and Y, orbit the same planet in circular orbits. The orbital radius of Y is 4 times that of X. The orbital period of X is T. What is the orbital period of Y?
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Kepler's third law gives T2 ∝ r3, so T ∝ r3/2. With r multiplied by 4, T is multiplied by 43/2 = 8. The value 2T uses T ∝ √r. The value 4T uses T ∝ r. The value 16T uses T ∝ r2.
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Answer D
Answer: D
Common error: Remember that it is the square of the period that is proportional to the cube of the radius.
50IBP3H-D1-20r47-08Launch and escape from orbitPaper 1AMedium1 mark
A satellite in a circular orbit close to the top of the atmosphere slowly loses energy through collisions with air molecules. It remains in an almost circular orbit. Which row describes the changes to its orbital radius and its speed?
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Losing total energy (which is −GMm/2r) makes the total energy more negative, so the orbital radius decreases. Since v = √(GM/r), a smaller radius means a larger speed. The rows with a decreasing speed ignore this relation, and the rows with an increasing radius contradict the loss of energy.
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Answer B
Answer: B
Common error: Do not assume that losing energy must slow the satellite down; in a lower orbit it moves faster.
51D-1A-02Escape speed and orbital speedPaper 1AHard1 mark
A small probe is launched vertically upwards from the surface of a non-rotating planet that has no atmosphere. The planet has mass M and radius R. The launch speed is equal to the orbital speed of a satellite in a circular orbit just above the surface, √(GM/R).
What is the maximum height reached by the probe above the surface of the planet?
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Step 1The field is not uniform over this range, so use energy conservation with Ep = −GMm/r: ½mv² − GMm/R = −GMm/r at the highest point, a distance r from the centre.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2With v² = GM/R: GMm/2R − GMm/R = −GMm/2R, so r = 2R.
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Step 3The height above the surface is r − R = R.
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Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis assumes a uniform field equal to the surface value g = GM/R² all the way up: h = v²/2g = R/2. The field weakens with height, so the probe rises further.
BCorrect: the probe reaches r = 2R, which is a height R above the surface.
CThis gives the distance from the centre of the planet (2R), not the height above the surface.
DThis omits the ½ in the kinetic energy: mv² = GMm/R is exactly the energy needed to escape, but the probe has only half of it (GMm/2R).
53D-1A-22Variation of g with heightPaper 1AHard1 mark
The graph shows how the gravitational field strength g varies with the distance r from the centre of a planet of radius 4.0 × 106 m, for points outside the planet.
What is the minimum energy needed to move a mass of 1.0 kg from the surface of the planet to a point where r = 8.0 × 106 m?
Variation of gravitational field strength with distance from the centre of the planet (outside the planet).Show mark scheme
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Step 1g = −ΔVg/Δr, so the change in gravitational potential between two points is the area under the g–r graph between them.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The area between r = 4.0 × 106 m and 8.0 × 106 m is about 16 squares of 1 N kg−1 × 106 m, i.e. 1.6 × 107 J kg−1 (accept 1.5–1.7 × 107). Check: Vg ∝ −1/r, so ΔVg = 8.0 × 4.0 × 106 × (1 − ½).
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Step 3Minimum energy for 1.0 kg = 1.6 × 107 J.
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Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis multiplies the distance moved by the smallest field strength on the interval, 2.0 N kg−1 at r = 8.0 × 106 m.
BCorrect: the area under the curve is 1.6 × 107 J kg−1.
CThis joins the end points with a straight line: ½(8.0 + 2.0) × 4.0 × 106. The curve lies below this chord, so the trapezium overestimates the area.
DThis assumes g keeps its surface value 8.0 N kg−1 all the way: mgΔr = 1.0 × 8.0 × 4.0 × 106 J.
54D-1A-36Launch and escape from orbitPaper 1AHard1 mark
A probe starts from rest on the surface of a non-rotating moon of radius R that has no atmosphere. The minimum energy needed to place the probe in a circular orbit of radius 2R is E.
What is the minimum additional energy the probe needs, once in this orbit, to escape completely from the moon's gravitational field?
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Step 1At rest on the surface the total energy is −GMm/R. In a circular orbit of radius 2R the total energy is −GMm/2(2R) = −GMm/4R.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2So E = GMm/R − GMm/4R = 3GMm/4R.
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Step 3To escape, the total energy must be raised to zero: additional energy = GMm/4R = E/3.
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Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: the orbiting probe needs GMm/4R more, which is one third of E.
BThis supplies the magnitude of the potential energy in orbit, GMm/2R, forgetting that the probe already has kinetic energy GMm/4R.
CThis assumes the escape speed is twice the orbital speed, so the kinetic energy must be quadrupled: an extra 3 × GMm/4R = E.
DThis is the escape energy from rest on the surface, GMm/R, ignoring the energy E already supplied.
The graph shows lg T against lg r for the moons of two planets X and Y, where T is the orbital period and r the radius of the circular orbit of a moon.
What is (mass of X)/(mass of Y)?
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Step 1T² = 4π²r³/(GM), so lg T = 1.5 lg r + lg(2π) − ½ lg(GM): both lines have gradient 1.5.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2At the same r, the line for X is 0.30 higher: lg TX − lg TY = −½ lg(MX/MY) = 0.30.
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Step 3lg(MX/MY) = −0.60, so MX/MY = 10−0.6 = 0.25.
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Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: a longer period at the same radius means a smaller mass, and T ∝ M−1/2.
BThis omits the factor ½, treating T as proportional to 1/M: 10−0.3 = 0.50.
CThis omits the factor ½ and takes the longer period to mean the larger mass: 100.3 = 2.0.
DThis takes the longer period to mean the larger mass: 100.6 = 4.0.
A small satellite is in a circular orbit around the Earth. Atmospheric drag acts on it and it slowly spirals into lower orbits that are always approximately circular. Which describes the changes in the kinetic energy of the satellite and in the total energy of the satellite–Earth system?
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For a circular orbit, Ek = GMm/2r and Etotal = −GMm/2r. Drag does negative work, so the total energy decreases. A decrease in −GMm/2r means that r decreases. Smaller r makes GMm/2r larger, so the kinetic energy increases. The gravitational potential energy (−GMm/r) falls by twice the amount that the kinetic energy gains. Option A assumes drag simply slows the satellite. Option C ignores the energy removed by drag. Option D ignores the change of radius.
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Answer B
Answer: B
Exam tip: For circular orbits remember Etotal = −Ek and Ep = −2Ek. Any loss of total energy appears as a gain in kinetic energy.
57IBP2H-029-00-08Kepler's second lawPaper 1AHard1 mark
Two planets orbit the same star in circular orbits. The orbital radius of the outer planet is 4 times the orbital radius of the inner planet. What is \(\frac{\text{area swept per unit time by the outer planet's radius line}}{\text{area swept per unit time by the inner planet's radius line}}\)?
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The area swept per unit time is πr2/T. By Kepler's third law T ∝ r3/2, so the area rate is proportional to r2/r3/2 = r1/2. The ratio is therefore 41/2 = 2. Option B misapplies Kepler's second law: it states a constant area rate for one planet, not equal area rates for different planets. Option C is the ratio of orbital speeds, since v ∝ r−1/2. Option D is the ratio of periods.
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Answer A
Answer: A
Common error: Kepler's second law compares different parts of one planet's orbit. It says nothing about two different planets having equal area rates.
58IBP2H-029-01-08Launch and escape from orbitPaper 1AHard1 mark
A satellite of mass m is in a circular orbit of radius r around a planet of mass M. What is the minimum energy needed to transfer it to a circular orbit of radius 2r?
Diagram not to scale
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The total orbital energy is E = −GMm/2r. The change is −GMm/4r − (−GMm/2r) = GMm/4r. Option A is the gain in potential energy alone and ignores the fall in kinetic energy. Option B adds the kinetic energy decrease instead of subtracting it. Option C assumes the potential energy gain equals the kinetic energy loss, but the kinetic energy loss is only half the potential energy gain.
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Answer D
Answer: D
Exam tip: Work with total energy, −GMm/2r, when changing orbits. The satellite slows down, but the total energy still rises.
A rocky 'super-Earth' exoplanet has the same mean density as the Earth but twice the Earth's radius. What is the ratio (escape speed from the exoplanet's surface)/(escape speed from the Earth's surface)?
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vesc = √(2GM/R). For the same density, M ∝ R3, so M/R ∝ R2 and vesc ∝ R. Doubling R doubles the escape speed, giving √(8/2) = 2. Option A keeps the mass constant. Option B treats M as proportional to R2. Option D forgets the square root.
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Answer C
Answer: C
Exam tip: When the density is fixed, replace M with (4/3)πR³ρ before deciding how a quantity scales with R.
60IBP2H-029-02-05Geostationary orbitsPaper 1AHard1 mark
An exoplanet has twice the mass of the Earth and takes two Earth days to rotate once on its axis. A satellite is placed in a synchronous (stationary) orbit around the exoplanet. What is the ratio radius of synchronous orbit around the exoplanet ÷ radius of geostationary orbit around the Earth?
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Equating the gravitational force to the centripetal force gives r3 = GMT2/4π2, so r ∝ (MT2)1/3. The ratio is (2 × 22)1/3 = 81/3 = 2.0. Option D omits the cube root. Option C takes a square root instead of a cube root. Option A ignores the change in mass, giving (22)1/3 ≈ 1.6.
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Answer B
Answer: B
Exam tip: Kepler's third law with the mass included, T² = 4π²r³/GM, handles changes of both planet mass and period in one step.
61IBP2H-029-02-07Mapping a field using potentialPaper 1AHard1 mark
Point X lies on the line joining the centres of the Earth and the Moon. At X the resultant gravitational field strength is zero. Which statement about the gravitational potential at X is correct?
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Both bodies contribute negative potentials, so the total is negative everywhere. This rules out options A and D. The field strength is g = −dV/dr, so g = 0 means the potential gradient is zero: V is at a turning point. Moving from X towards either body, V becomes more negative. X is therefore a maximum of V along the line. Option B mistakes this maximum for a minimum. Option A confuses zero field with zero potential.
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Answer C
Answer: C
Exam tip: Field strength is related to the gradient of the potential, not its value. Zero field means a flat potential, not zero potential.
62IBP3H-D1-20r32-05Launch and escape from orbitPaper 1AHard1 mark
A satellite of mass m moves in a circular orbit of radius 2R around a planet of radius R. The gravitational field strength at the surface of the planet is g. The satellite is moved to a circular orbit of radius 4R. The minimum energy that must be supplied to the satellite is
Diagram not to scale
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Total orbital energy is E = −GMm/2r and GM = gR². E at 2R is −mgR/4 and at 4R is −mgR/8, so the energy needed is mgR/8. mgR/4 is the increase in potential energy only, ignoring the fall in kinetic energy; 3mgR/8 uses a starting orbit of radius R; mgR/16 uses 1/2 twice.
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Answer B
Answer: B
Common error: Raising a satellite to a higher orbit changes both its potential and kinetic energy; use the total energy −GMm/2r.
A robotic probe is launched vertically from the surface of a planet of radius R with an initial speed equal to half of the escape speed. Ignore air resistance and the rotation of the planet. The maximum height of the probe above the surface of the planet is
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Initial kinetic energy per unit mass = ½(vesc/2)² = GM/4R. Energy conservation: GM/4R − GM/R = −GM/r, so r = 4R/3 from the centre and the height is 4R/3 − R = R/3. R/4 comes from assuming a uniform field (h = v²/2g); 4R/3 is the distance from the centre, not the height; 3R/4 results from subtracting R/4 from R, confusing the energy fraction with a distance.
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Answer C
Answer: C
Exam tip: For motion over distances comparable with the planet radius, use energy conservation with −GM/r, not constant g.
64IBP3H-D1-20r47-07Launch and escape from orbitPaper 1AHard1 mark
A satellite of mass m is in a circular orbit of radius r about a planet of mass M. Its engines are used to move it into a higher circular orbit of radius 2r. What is the minimum increase in its total energy that must be supplied?
Diagram not to scale
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The total energy in a circular orbit is −GMm/(2r). In the larger orbit it is −GMm/(4r), so the increase is GMm/(4r). The value GMm/(2r) is the increase in gravitational potential energy alone, ignoring that the kinetic energy decreases. The value 3GMm/(4r) wrongly adds the decrease in kinetic energy to the increase in potential energy. The value GMm/r is the magnitude of the potential energy in the original orbit.
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Answer C
Answer: C
Exam tip: Use total energy = −GMm/(2r) for circular orbits so that you do not have to track kinetic and potential energy separately.
A student uses published data for the four largest moons of Jupiter to test Kepler's third law. The mean orbital radius r of each moon was obtained from telescope measurements of its greatest angular distance from the planet, and the orbital period T from the times between successive eclipses of the moon by Jupiter. Each radius is known to within ±1 %; the uncertainties in the periods are negligible. The orbits may be treated as circular.
The student's hypothesis is that T²/r³ has the same value for every moon. 1 day = 8.64 × 104 s.
Moon
r / 108 m
T / days
Io
4.22
1.77
Europa
6.71
3.55
Ganymede
10.7
7.15
Callisto
18.8
16.7
(a)
The hypothesis.
(i)
Test the hypothesis, using the data for at least three of the moons.
(2)
(b)
The mass of Jupiter.
(i)
Determine the mass of Jupiter.
(2)
(ii)
Determine the absolute uncertainty in your answer to (b)(i).
(2)
(c)
A prediction.
(i)
Amalthea, a small inner moon of Jupiter, has an orbital radius of 1.81 × 108 m. Predict its orbital period, in days.
(1)
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Part (a)(i)
T²/r³ calculated with T in seconds for at least three moons: Io 3.11, Europa 3.11, Ganymede 3.12, Callisto 3.13 (× 10−16 s² m−3)
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Award only if at least three moons are used. Accept 2 s.f. Consistent other units (e.g. days² m⁻³) are acceptable.
The values agree to within about 1 % (less than the 3 % uncertainty in r³), so the hypothesis is supported
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The conclusion must be supported by the candidate's own values. Do not accept "they are roughly the same" without a comparison.
Part (b)(i)
Gravitational force provides the centripetal force: GMm/r² = m(2π/T)²r, so T²/r³ = 4π²/(GM) and M = 4π²/(G × T²/r³)
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M = 4π²/(6.67 × 10−11 × 3.12 × 10−16) = 1.90 × 1027 kg
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Allow ECF from (a). Accept 1.88–1.91 × 10²⁷ kg (any single moon's value of T²/r³ may be used).
Part (b)(ii)
M ∝ r³/T², so the percentage uncertainty in M is 3 × 1 % = 3 %
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Using 1 % (not multiplying by the power 3) scores 0 for this mark.
ΔM = 0.03 × 1.90 × 1027 ≈ 0.06 × 1027 kg, so M = (1.90 ± 0.06) × 1027 kg
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Allow ECF from (b)(i). The absolute uncertainty must be given to 1 s.f., with the value to the same decimal place.
Part (c)(i)
T = √(3.12 × 10−16 × (1.81 × 108)³) = 4.30 × 104 s = 0.498 days
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Allow ECF from (a) or (b)(i). Accept 0.49–0.51 days.
Answers: (a)(i) T²/r³ ≈ 3.12 × 10−16 s² m−3 for all four moons · (b)(i) 1.90 × 1027 kg · (b)(ii) ± 0.06 × 1027 kg · (c)(i) 0.50 days (the remaining parts are explanations — see the table above)
66D-1B-05Gravitational field strengthPaper 1BMedium7 marks
An astronaut measures the gravitational field strength g at the surface of the Moon. A steel ball is held by an electromagnet at a height s above a trapdoor switch. A timer starts when the current in the electromagnet is switched off and stops when the ball opens the trapdoor. Because the magnetic field takes a short time to decay, the ball is released slightly after the timer starts, by the same delay each time.
The fall time t is measured for six heights. The graph shows t against √s. The radius of the Moon is 1.74 × 106 m.
s / m
t / s
√s / m½
0.20
0.528
0.447
0.40
0.735
0.632
0.60
0.893
0.775
0.80
1.027
0.894
1.00
1.143
1.000
1.20
1.251
1.095
Graph drawn to scale.
(a)
The field strength.
(i)
Draw the line of best fit and determine its gradient, including its unit.
(2)
(ii)
Hence determine g at the surface of the Moon.
(1)
(b)
The delay and the Moon.
(i)
State the intercept of your line on the t axis and explain why the delay does not affect the value of g found in (a)(ii).
(2)
(ii)
Determine the mass of the Moon.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Straight line of best fit and gradient from a large triangle, e.g. (0.40, 0.476) to (1.10, 1.256)
✓ 1
Gradient = 1.11 s m−½
✓ 1
Accept 1.09–1.14. Unit required (accept s m⁻⁰·⁵).
Part (a)(ii)
s = ½gt² gives t = √(2/g) √s + delay, so gradient = √(2/g) and g = 2/1.11² = 1.61 m s−2
✓ 1
Allow ECF from (a)(i). Accept 1.54–1.68 m s⁻².
Part (b)(i)
Intercept ≈ 0.03 s (the release delay)
✓ 1
Accept 0.02–0.04 s. Allow ECF from the candidate's line.
The same delay is added to every time (a systematic error), so the line is shifted up without changing its gradient, and g depends only on the gradient
✓ 1
Part (b)(ii)
g = GM/R², so M = gR²/G = 1.61 × (1.74 × 106)²/(6.67 × 10−11)
✓ 1
Allow ECF from (a)(ii).
M = 7.3 × 1022 kg
✓ 1
Accept 7.0–7.6 × 10²² kg.
Answers: (a)(i) 1.11 s m−½ · (a)(ii) 1.61 m s−2 · (b)(i) ≈ 0.03 s · (b)(ii) 7.3 × 1022 kg (the remaining parts are explanations — see the table above)
67D-1B-11Mapping a field using potentialPaper 1BMedium7 marks
An orbiter surveying a dwarf planet uses radio tracking to determine the gravitational potential Vg at several distances r from the centre of the dwarf planet, which may be treated as a point mass. The results are shown in the table and in the graph of Vg against 1/r.
r / 105 m
6.0
7.5
9.0
11.0
14.0
18.0
Vg / 104 J kg−1
−10.00
−7.92
−6.52
−5.27
−4.06
−3.06
(1/r) / 10−6 m−1
1.667
1.333
1.111
0.909
0.714
0.556
Graph drawn to scale. The dashed line marks Vg = 0.
(a)
The mass.
(i)
Draw the line of best fit and determine its gradient, including its unit.
(2)
(ii)
Hence determine the mass of the dwarf planet.
(1)
(b)
Checking the data.
(i)
Extrapolate your line to find its intercept on the Vg axis. Explain what this intercept shows about the values of Vg.
(2)
(ii)
Use the values of Vg at r = 9.0 × 105 m and r = 11.0 × 105 m to estimate the gravitational field strength at r = 10.0 × 105 m. Compare your estimate with the value given by your answer to (a)(ii).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Straight line of best fit; gradient from a large triangle, e.g. (−10.0 − (−3.06)) × 104/((1.67 − 0.556) × 10−6)
✓ 1
Gradient = −6.2 × 1010 J m kg−1
✓ 1
Accept −6.0 to −6.4 × 10¹⁰. Unit required (accept N m² kg⁻¹ or m³ s⁻²).
Part (a)(ii)
Vg = −GM/r, so M = −gradient/G = 6.24 × 1010/6.67 × 10−11 = 9.4 × 1020 kg
✓ 1
Allow ECF from (a)(i). Accept 9.0–9.6 × 10²⁰ kg.
Part (b)(i)
Intercept ≈ +0.4 × 104 J kg−1
✓ 1
Accept +0.3 to +0.5 × 10⁴. Allow ECF from the candidate's line.
Vg should be zero at infinity (1/r = 0), so every value has been shifted by the same constant amount: a systematic (offset) error; it changes the intercept but not the gradient, so the mass found in (a)(ii) is unaffected
✓ 1
Both the systematic nature and the unaffected mass are needed.
Part (b)(ii)
g = ΔVg/Δr = (−5.27 − (−6.52)) × 104/(2.0 × 105) = 0.063 N kg−1 (towards the centre); the offset cancels in the difference
✓ 1
Accept 0.062–0.063.
GM/r² = 6.24 × 1010/(1.00 × 106)² = 0.062 N kg−1, which agrees with the estimate to within about 1 %
✓ 1
Allow ECF from (a)(ii). A comparison must be stated.
Answers: (a)(i) −6.2 × 1010 J m kg−1 · (a)(ii) 9.4 × 1020 kg · (b)(i) ≈ +0.4 × 104 J kg−1 · (b)(ii) 0.063 N kg−1; 0.062 N kg−1(the remaining parts are explanations — see the table above)
68D-1B-14Uniform field near the surfacePaper 1BMedium7 marks
A gravity model of Mars, built from the tracking of orbiting spacecraft, gives the gravitational potential difference ΔVg between the surface of Mars and a point at height h above it. Some values are shown in the table, together with values of h/ΔVg.
It can be shown that h/ΔVg = R²/(GM) + (R/(GM))h, where M and R are the mass and radius of Mars. The graph shows five of the six points.
h / km
100
250
500
750
1000
1500
ΔVg / MJ kg−1
0.362
0.868
1.62
2.29
2.88
3.87
(h/ΔVg) / 10−2 s² m−1
27.62
28.80
30.86
32.75
34.72
Graph drawn to scale (the vertical axis starts at 26).
(a)
The graph.
(i)
Complete the table.
(1)
(ii)
Plot the missing point, draw the line of best fit and determine its gradient and its intercept on the vertical axis, with units.
(2)
(b)
Mars and the uniform-field model.
(i)
Hence determine the gravitational field strength g0 at the surface of Mars and the radius R of Mars.
(2)
(ii)
A payload of mass 1200 kg is lifted from the surface to a height of 500 km. Determine the percentage by which the uniform-field estimate mg0h exceeds the actual increase in its gravitational potential energy.
69D-1B-06Escape speed and orbital speedPaper 1BHard7 marks
After its engine is shut down, a spacecraft moves directly away from the Earth. Radar tracking gives its speed v at several distances r from the centre of the Earth. The only force acting on the spacecraft is the gravitational force of the Earth.
The graph shows v² against 1/r.
r / 107 m
v / km s−1
(1/r) / 10−8 m−1
v² / 106 m² s−2
1.00
8.48
10.00
71.9
1.50
6.73
6.67
45.3
2.00
5.64
5.00
31.8
3.00
4.31
3.33
18.6
4.00
3.45
2.50
11.9
6.00
2.31
1.67
5.3
Graph drawn to scale. The dashed line marks v² = 0.
(a)
The gradient.
(i)
Draw the line of best fit and determine its gradient, including its unit.
(2)
(ii)
Use the conservation of energy to show that the gradient is equal to 2GM, where M is the mass of the Earth. Hence determine M.
(2)
(b)
Where the spacecraft goes.
(i)
Extrapolate your line to determine the value of v² when 1/r = 0. Deduce whether the spacecraft escapes from the Earth.
(2)
(ii)
Determine the greatest distance from the centre of the Earth that the spacecraft reaches.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Straight line of best fit and gradient from a large triangle, e.g. (71.9 − 5.3) × 106/((10.0 − 1.67) × 10−8)
✓ 1
Gradient = 8.0 × 1014 m³ s−2
✓ 1
Accept 7.8–8.2 × 10¹⁴. Unit required (accept N m² kg⁻¹ or J m kg⁻¹).
Part (a)(ii)
½mv² − GMm/r = constant, so v² = 2GM × (1/r) + constant: the gradient is 2GM
✓ 1
M = 8.0 × 1014/(2 × 6.67 × 10−11) = 6.0 × 1024 kg
✓ 1
Allow ECF from (a)(i). Accept 5.8–6.2 × 10²⁴ kg.
Part (b)(i)
Intercept ≈ −8 × 106 m² s−2
✓ 1
Accept −6 × 10⁶ to −10 × 10⁶. Allow ECF from the candidate's line.
v² cannot be negative, so the spacecraft never reaches 1/r = 0 (infinite distance): its total energy ½mv² − GMm/r is negative and it does not escape
✓ 1
The deduction must be consistent with the sign of the candidate's intercept.
Part (b)(ii)
v = 0 where the line cuts the 1/r axis: 1/r = 8.1 × 106/8.0 × 1014 = 1.0 × 10−8 m−1, so r = 1.0 × 108 m
✓ 1
Allow ECF from (a)(i) and (b)(i). Accept 0.8–1.3 × 10⁸ m.
Answers: (a)(i) 8.0 × 1014 m³ s−2 · (a)(ii) 6.0 × 1024 kg · (b)(i) ≈ −8 × 106 m² s−2; it does not escape · (b)(ii) 1.0 × 108 m (the remaining parts are explanations — see the table above)
70D-1B-08Determining the density of an asteroidPaper 1BHard7 marks
Surveys measure the rotation periods of thousands of asteroids from the regular variation in their brightness. The table shows, for asteroids in five ranges of diameter, the shortest rotation period Tmin observed in that range. Each value has an uncertainty of ±0.02 h.
Many asteroids are thought to be loose piles of rubble held together only by gravity. Such an asteroid is modelled as a uniform sphere of density ρ. A loose rock on its equator stays on the surface only if the gravitational field strength there is at least equal to the centripetal acceleration of the rock.
Diameter / km
0.3–1
1–3
3–10
10–30
30–100
Tmin / h
2.22
2.20
2.26
2.31
2.45
(a)
The spin limit.
(i)
Show that a rubble-pile asteroid of density ρ cannot rotate with a period shorter than √(3π/(Gρ)).
(2)
(ii)
Use the data to deduce the minimum density of rubble-pile asteroids.
(2)
(b)
Uncertainty and a deduction.
(i)
Determine the absolute uncertainty in your answer to (a)(ii).
(2)
(ii)
Some asteroids smaller than 0.2 km rotate with periods of about 0.50 h. Deduce what this suggests about these asteroids.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
At the limit GM/R² = ω²R = (4π²/T²)R
✓ 1
With M = (4/3)πR³ρ: (4/3)πGρ = 4π²/T², so T = √(3π/(Gρ)); a shorter period would need a larger centripetal acceleration than g can provide
✓ 1
Part (a)(ii)
The fastest rotator sets the limit, so use the shortest period, 2.20 h = 7.92 × 103 s
✓ 1
Using any other period or a mean scores [1 max] for the part if the rest is correct.
Allow ECF from (a)(i). Accept 2.2–2.3 × 10³ kg m⁻³.
Part (b)(i)
Percentage uncertainty in T = 0.02/2.20 = 0.9 %; ρ ∝ T−2, so the percentage uncertainty in ρ is 2 × 0.9 % = 1.8 %
✓ 1
Using 0.9 % for ρ scores 0 for this mark.
Δρ = 0.018 × 2.25 × 103 = 40 kg m−3, so ρ = (2.25 ± 0.04) × 103 kg m−3
✓ 1
Allow ECF from (a)(ii). Uncertainty to 1 s.f., value to the same decimal place.
Part (b)(ii)
As rubble piles they would need ρ ≥ 2.25 × 103 × (2.20/0.50)² ≈ 4 × 104 kg m−3, far more than any rock, so they cannot be held together by gravity alone: they must be single solid bodies held together by the forces between their particles
✓ 1
Allow ECF from (a)(ii). A numerical comparison (density or period limit) is required for the mark.
Answers: (a)(ii) 2.25 × 103 kg m−3 · (b)(i) ± 0.04 × 103 kg m−3(the remaining parts are explanations — see the table above)
71D-1B-13Newton's law of gravitationPaper 1BHard7 marks
In a torsion-balance experiment the gravitational force F between a large lead sphere of mass 1.50 kg and a small lead sphere of mass 20.0 g is determined for several centre-to-centre separations r. The absolute uncertainty in each value of F is ±0.10 × 10−10 N; the uncertainty in r is negligible.
At every separation the small sphere is also attracted in the opposite direction by a second large sphere on the other side of the balance. This opposing force is 4.0 % of the force due to the near sphere, and the tabulated values of F are the measured (resultant) forces. The graph shows five of the six points.
r / cm
4.5
5.0
5.5
6.0
7.0
8.5
F / 10−10 N
9.47
7.70
6.37
5.36
3.90
2.64
(1/r²) / m−2
494
400
331
278
204
Graph drawn to scale. Error bars show ±0.10 × 10⁻¹⁰ N.
(a)
The value of G.
(i)
Complete the table.
(1)
(ii)
Plot the missing point, draw the line of best fit and determine a value for G.
(2)
(b)
Uncertainty and a systematic error.
(i)
Draw the lines of maximum and minimum gradient consistent with the error bars, and hence state G with its absolute uncertainty.
(2)
(ii)
Compare your result with the accepted value of G, and explain the difference with reference to the second large sphere.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
1/0.085² = 138 m−2
✓ 1
Accept 138.4.
Part (a)(ii)
Point plotted with its error bar; gradient of the line of best fit = 1.93 × 10−12 N m²
✓ 1
Accept 1.88–1.98 × 10⁻¹² N m². Allow ECF from (a)(i).
G = gradient/(Mm) = 1.93 × 10−12/(1.50 × 0.0200) = 6.4 × 10−11 N m² kg−2
✓ 1
Allow ECF. Accept 6.3–6.6 × 10⁻¹¹.
Part (b)(i)
Maximum gradient ≈ 1.98 × 10−12 and minimum gradient ≈ 1.87 × 10−12 N m²
✓ 1
Accept each within ±0.03 × 10⁻¹². Both lines must pass through every error bar.
Gmax ≈ 6.6 × 10−11 and Gmin ≈ 6.2 × 10−11, so G = (6.4 ± 0.2) × 10−11 N m² kg−2
✓ 1
Uncertainty = half the range, to 1 s.f. Allow ECF from (a)(ii).
Part (b)(ii)
The accepted value, 6.67 × 10−11 N m² kg−2, lies outside the range found in (b)(i) (about 6.2–6.6 × 10−11), so the difference is not explained by the random uncertainties
✓ 1
The comparison must use the candidate's own range (ECF).
Every measured F is 4 % smaller than the force due to the near sphere, so the gradient and G are 4 % too small (a systematic error); corrected G = 6.42/0.96 = 6.7 × 10−11 N m² kg−2, in agreement with the accepted value
✓ 1
Accept the argument without the corrected value if the 4 % reduction of the gradient is clearly stated.
Answers: (a)(i) 138 m−2 · (a)(ii) 6.4 × 10−11 N m² kg−2 · (b)(i) (6.4 ± 0.2) × 10−11 N m² kg−2(the remaining parts are explanations — see the table above)
72D-2-17Field lines and Kepler's lawsPaper 2Easy13 marks
A planet of mass 4.8 × 1024 kg orbits a red dwarf star of mass 8.0 × 1029 kg in a circular orbit of radius 3.0 × 109 m. The radius of the star is 2.8 × 108 m and its luminosity is 1.0 × 1025 W.
(a)
The gravitational field of the star.
(i)
Outline why the star and the planet can be treated as point masses.
(1)
(ii)
Calculate the gravitational field strength due to the star at the position of the planet.
(1)
(iii)
Hence show that the orbital speed of the planet is about 1.3 × 105 m s−1.
(2)
(b)
Kepler's third law.
(i)
Determine the orbital period of the planet, in days.
(2)
(ii)
A second planet of the same star has an orbital period of 13 days. Determine its orbital radius.
(2)
(c)
(i)
Sketch the gravitational field lines around the star, treated as an isolated sphere.
(2)
(d)
Radiation from the star.
(i)
Calculate the intensity of the star's radiation at the orbit of the first planet.
(1)
(ii)
Estimate the mean surface temperature of this planet. Treat the planet as a black body with no atmosphere.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Both are (approximately) spherically symmetric, so their external fields are those of point masses at their centres; and/or their separation is much larger than the radius of either
✓ 1
Part (a)(ii)
g = GM/r² = 6.67 × 10−11 × 8.0 × 1029/(3.0 × 109)² = 5.9 N kg−1
✓ 1
Part (a)(iii)
g is the centripetal acceleration: g = v²/r
✓ 1
Allow ECF from (a)(ii).
v = √(gr) = √(5.93 × 3.0 × 109) = 1.33 × 105 m s−1
✓ 1
Full substitution or an answer to at least 3 s.f. is required.
Part (b)(i)
T = 2πr/v = 2π × 3.0 × 109/1.33 × 105 = 1.41 × 105 s
✓ 1
Allow ECF from (a)(iii).
= 1.64 days
✓ 1
Part (b)(ii)
T² ∝ r³, so r2 = r1(T2/T1)2/3
✓ 1
Allow ECF from (b)(i).
r2 = 3.0 × 109 × (13/1.64)2/3 = 1.2 × 1010 m
✓ 1
Accept 1.19–1.22 × 10¹⁰ m.
Part (c)(i)
Radial straight lines, evenly distributed around the star and meeting its surface at right angles
✓ 1
Arrows on the lines pointing towards the centre of the star
✓ 1
Part (d)(i)
b = L/(4πd²) = 1.0 × 1025/(4π × (3.0 × 109)²) = 8.8 × 104 W m−2
✓ 1
Part (d)(ii)
Power absorbed bπRp² = power emitted 4πRp²σT⁴, so T⁴ = b/(4σ)
✓ 1
Allow ECF from (d)(i).
T = (8.8 × 104/(4 × 5.67 × 10−8))1/4 ≈ 790 K
✓ 1
Accept 780–800 K.
Answers: (a)(ii) 5.9 N kg−1 · (a)(iii) 1.33 × 105 m s−1 · (b)(i) 1.64 days · (b)(ii) 1.2 × 1010 m · (d)(i) 8.8 × 104 W m−2 · (d)(ii) ≈ 790 K (the remaining parts are explanations — see the table above)
Model sketch
Model sketch: radial field lines, directed towards the centre, closer together near the star
A robotic lander is to land on Europa, a moon of Jupiter that has no atmosphere. Europa may be treated as a uniform sphere of mass 4.80 × 1022 kg and radius 1.56 × 106 m.
(a)
Europa's gravitational field.
(i)
Show that the gravitational field strength at the surface of Europa is about 1.3 N kg−1.
(1)
(b)
The uniform-field model.
(i)
The descent begins at a height of 15 km above the surface. Determine the percentage by which the gravitational field strength at this height is less than the value at the surface.
(3)
(ii)
State and explain whether the gravitational field may be treated as uniform during the descent.
(1)
(c)
Landing.
(i)
The engine is switched off when the lander is momentarily at rest 2.0 m above the surface. Determine the speed with which the lander reaches the surface.
(2)
(ii)
The landing legs are tested on Earth by dropping the lander from rest onto the ground. Determine the height from which it must be dropped on Earth so that it reaches the ground with the speed found in (c)(i). Ignore air resistance.
(2)
(iii)
Europa moves around Jupiter in a circular orbit of radius 6.71 × 108 m with a period of 3.55 days. Determine the gravitational field strength due to Jupiter at the position of Europa.
(2)
(iv)
Your answer to (c)(iii) is about one-fifth of your answer to (a)(i). Suggest why Jupiter's field does not affect the landing speed found in (c)(i).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
g = GM/R² = 6.67 × 10−11 × 4.80 × 1022/(1.56 × 106)² = 1.316 N kg−1
✓ 1
Full substitution or an answer to at least 3 s.f. is required.
Part (b)(i)
r = 1.56 × 106 + 1.5 × 104 = 1.575 × 106 m
✓ 1
The radius of Europa must be added to the height.
g = 6.67 × 10−11 × 4.80 × 1022/(1.575 × 106)² = 1.291 N kg−1
Yes: the field strength changes by only about 2 % over the descent (the height is much less than the radius of Europa), so g may be taken as constant
✓ 1
Allow ECF from (b)(i). The decision must be consistent with the candidate's percentage.
Part (c)(i)
Uniform field: v² = 2gs (or mgh = ½mv²)
✓ 1
v = √(2 × 1.32 × 2.0) = 2.3 m s−1
✓ 1
Allow ECF from (a)(i). Award [1 max] for 6.3 m s⁻¹, which uses g = 9.81 N kg⁻¹.
Part (c)(ii)
v² = 2gh with g = 9.81 N kg−1 (equivalently gEarthhEarth = gEuropa × 2.0 m)
✓ 1
Allow ECF from (c)(i).
h = 2.3²/(2 × 9.81) = 0.27 m
✓ 1
Accept 0.26–0.28 m (1.316 × 2.0/9.81 = 0.268 m).
Part (c)(iii)
The field strength of Jupiter equals the centripetal acceleration of Europa: gJ = 4π²r/T², with T = 3.55 × 8.64 × 104 = 3.07 × 105 s
✓ 1
ALT: M_J = 4π²r³/(GT²) followed by g = GM_J/r².
gJ = 4π² × 6.71 × 108/(3.07 × 105)² = 0.28 N kg−1
✓ 1
Accept 0.27–0.29 N kg⁻¹.
Part (c)(iv)
Europa and the lander are both in free fall towards Jupiter with the same acceleration, so Jupiter's field produces no acceleration of the lander relative to Europa's surface
✓ 1
Accept: Jupiter's field (almost) equally accelerates the lander and Europa. Do not accept "Jupiter is too far away".
Answers: (a)(i) 1.316 N kg−1 · (b)(i) 1.9 % · (c)(i) 2.3 m s−1 · (c)(ii) 0.27 m · (c)(iii) 0.28 N kg−1(the remaining parts are explanations — see the table above)
74IBP2H-029-00-09A lander on EuropaPaper 2Easy5 marks
A sports physiologist studies climbers on the summit of a mountain 8.85 km above sea level. Take the radius of the Earth as 6.37×106 m and the gravitational field strength at sea level as 9.8 N kg−1.
(a)
Define gravitational field strength.
(1)
(b)
Calculate the gravitational field strength at the summit.
(2)
(c)
Calculate the percentage reduction in the field strength at the summit compared with sea level.
(1)
(d)
A ski jumper on the same mountain falls through about 100 m. Outline why the gravitational field may be treated as uniform for this jump.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Gravitational force per unit mass acting on a small (point) test mass placed at that point
✓ 1
Part (b)
g = g0(R/(R+h))2, using g ∝ 1/r2
✓ 1
g = 9.8 × (6.37×106/6.37885×106)2 = 9.773 ≈ 9.77 N kg−1
✓ 1
Part (c)
Using the unrounded value: (9.8 − 9.773)/9.8 × 100 ≈ 0.28 %, equivalently [1 − (R/(R+h))2] × 100 = 0.28 % (accept 0.27–0.31 %, including 0.31 % from the rounded 9.77; ECF from (b))
✓ 1
Part (d)
The height change of about 100 m is negligible compared with the Earth's radius of 6.37×106 m, so g changes by only about 0.003 % over the jump
✓ 1
Exam tip: Use the ratio method g = g₀(R/r)². It avoids needing the Earth's mass. Keep unrounded values when finding small percentage changes.
75IBP2H-029-02-09Newton's law of gravitationPaper 2Easy6 marks
In a laboratory experiment, a small lead sphere of mass 0.75 kg is placed so that its centre is 0.23 m from the centre of a large lead sphere of mass 150 kg. Both spheres are uniform. Take G = 6.67 × 10−11 N m2 kg−2 and g = 9.81 N kg−1 at the Earth's surface.
(a)
State Newton's law of gravitation.
(2)
(b)
Calculate the gravitational force between the spheres.
(2)
(c)
Calculate the ratio gravitational force between the spheres ÷ weight of the small sphere at the Earth's surface.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The force between two point masses is attractive and proportional to the product of their masses . The force is inversely proportional to the square of their separation .
✓ 2
Part (b)
F = Gm1m2/r2 = 6.67 × 10−11 × 150 × 0.75 / 0.232 . F = 1.4 × 10−7 N (1.42 × 10−7 N unrounded) .
✓ 2
Part (c)
Weight = 0.75 × 9.81 = 7.36 N . Ratio = 1.42 × 10−7 / 7.36 = 1.9 × 10−8 . Allow error carried forward from the force calculated in the previous part. Using the rounded value 1.4 × 10−7 N also gives 1.9 × 10−8.
✓ 2
Exam tip: Uniform spheres can be treated as point masses at their centres, so always use the centre-to-centre distance. Carry unrounded values into later parts and round only the final answer.
76IBP3H-D1-20r32-09Orbital energyPaper 2Easy6 marks
A satellite used to monitor sea-surface temperatures for a fisheries agency orbits Earth in a circular orbit at an altitude of 8.0 × 105 m above the surface. Use G = 6.67 × 10−11 N m2 kg−2, mass of Earth = 5.97 × 1024 kg and radius of Earth = 6.37 × 106 m.
(a)
Show that the orbital speed v of a satellite in a circular orbit of radius r around a planet of mass M is given by v = √(GM/r).
(2)
(b)
Calculate the orbital speed of the satellite.
(2)
(c)
Calculate the orbital period of the satellite, in minutes, to three significant figures. Either the rounded or the unrounded value of the speed from (b) may be used.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
gravitational force provides the centripetal force: GMm/r² = mv²/r
✓ 1
cancel m and r and take the square root to reach v = √(GM/r)
✓ 1
Part (b)
r = 6.37 × 106 + 8.0 × 105 = 7.17 × 106 m
✓ 1
v = √(6.67 × 10−11 × 5.97 × 1024 / 7.17 × 106) = 7.45×103 m s−1
✓ 1
Part (c)
T = 2πr/v with r = 7.17 × 106 m (ECF from (b))
✓ 1
T = 6050 s = 101 minutes
✓ 1
Exam tip: Remember to add the radius of the planet to the altitude before using the orbital radius in any formula.
77IBP3H-D1-20r32-10Escape speed and orbital speedPaper 2Easy5 marks
A robotic sample-return craft is to launch a 15 kg capsule from the surface of the Moon. Use G = 6.67 × 10−11 N m2 kg−2, mass of the Moon = 7.35 × 1022 kg and radius of the Moon = 1.74 × 106 m. Ignore the rotation of the Moon.
(a)
Calculate the escape speed from the surface of the Moon.
(2)
(b)
Explain why the escape speed does not depend on the mass of the capsule.
(1)
(c)
Calculate the minimum kinetic energy the capsule needs at launch to escape from the Moon. Use the unrounded value of the escape speed.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
v = √(2GM/R)
✓ 1
v = √(2 × 6.67 × 10−11 × 7.35 × 1022 / 1.74 × 106) = 2.37×103 m s−1
✓ 1
Part (b)
both the kinetic energy needed and the gravitational potential energy to be overcome are proportional to the mass, so the mass cancels
✓ 1
Part (c)
Ek = ½mv² or GMm/R
✓ 1
Ek = ½ × 15 × (2373.8)² = 4.23×107 J
✓ 1
Common error: Escape speed is the minimum speed at launch for escape; it is not the speed needed to stay in orbit, which is smaller by a factor of √2.
78IBP3H-D1-20r47-09Escape speed and orbital speedPaper 2Easy5 marks
The Moon orbits Earth. Take the following data: mass of Earth = 5.97 × 1024 kg; mass of the Moon = 7.35 × 1022 kg; distance between their centres = 3.84 × 108 m; G = 6.67 × 10−11 N m2 kg−2. Treat both bodies as point masses.
(a)
State what is meant by the gravitational field strength at a point.
(1)
(b)
Calculate the gravitational force that Earth exerts on the Moon.
(2)
(c)
Calculate the gravitational field strength of Earth at the distance of the Moon.
(2)
Show mark scheme
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Mark
Notes
Part (a)
force per unit mass exerted on a small test mass placed at that point
✓ 1
Part (b)
M1 use of F = GMm/r2 with correct substitution
✓ 1
A1 F = 1.98 × 1020 N
✓ 1
Part (c)
M1 use of g = GM/r2 (or F/m using the answer to (b), ECF from (b))
✓ 1
A1 g = 2.70 × 10−3 N kg−1
✓ 1
Exam tip: Check that the two methods agree: dividing the force on the Moon by its mass must give the field strength of Earth at the Moon.
79IBP3H-D1-20r47-11A lander on EuropaPaper 2Easy6 marks
A sports scientist compares conditions at sea level with those at a high-altitude training camp at a height of 3.60 km above sea level. Take Earth to be a uniform sphere of mass 5.97 × 1024 kg and radius 6.37 × 106 m. G = 6.67 × 10−11 N m2 kg−2. Use unrounded values from earlier parts in later parts.
(a)
Calculate the gravitational field strength at sea level.
(2)
(b)
Calculate the gravitational field strength at the training camp.
(2)
(c)
Determine the percentage decrease in the field strength between sea level and the camp.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 use of g = GM/R2
✓ 1
A1 g = 9.81 N kg−1
✓ 1
Part (b)
M1 distance from the centre of Earth is R + h = 6.3736 × 106 m used in g = GM/r2
✓ 1
A1 g = 9.802 N kg−1 (accept 9.8 N kg−1)
✓ 1
Part (c)
M1 percentage decrease = (g0 − g1)/g0 × 100 using unrounded values (ECF from (a) and (b))
✓ 1
A1 0.113 % (accept 0.11 %)
✓ 1
Common error: Using the height above the surface instead of the distance from the centre of Earth in the inverse-square law gives a wildly wrong answer.
A star S orbits the massive black hole at the centre of our galaxy. Model the orbit as a circle of radius 1.0 × 103 AU with a period of 16 years. The black hole is 8.0 kpc from Earth.
1 year = 3.16 × 107 s; mass of the Sun = 2.0 × 1030 kg.
(a)
The mass of the black hole.
(i)
Show that the radius of the orbit is about 1.5 × 1014 m.
(1)
(ii)
Show that, for a body in a circular orbit of radius r around a mass M, T² = (4π²/GM)r³.
(2)
(iii)
Determine the mass of the black hole, in solar masses.
(2)
(b)
(i)
Determine the angle, in arcseconds, subtended at Earth by the radius of the orbit.
(2)
(c)
Escape from the black hole.
(i)
Estimate the radius of the region around the black hole from which the escape speed would be greater than the speed of light.
(2)
(ii)
State one assumption made in your estimate.
(1)
(d)
(i)
Calculate the orbital speed of S and comment on whether Newtonian mechanics is adequate for describing its orbit.
(2)
Show mark scheme
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Mark
Notes
Part (a)(i)
1.0 × 103 × 1.50 × 1011 m = 1.50 × 1014 m
✓ 1
The AU conversion from the data booklet must be seen.
r = 2 × 6.67 × 10−11 × 7.8 × 1036/(3.00 × 108)² = 1.2 × 1010 m
✓ 1
Allow ECF from (a)(iii).
Part (c)(ii)
Newtonian gravitation (and the escape-speed equation) still applies so close to the black hole / the black hole behaves as a point mass
✓ 1
Part (d)(i)
v = 2πr/T = 2π × 1.50 × 1014/(16 × 3.16 × 107) = 1.9 × 106 m s−1
✓ 1
Allow ECF from (a)(i).
v/c ≈ 0.006, so γ ≈ 1.00002: relativistic effects such as time dilation are negligible and the Newtonian model is adequate
✓ 1
A reasoned comparison with c is needed.
Answers: (a)(i) 1.50 × 1014 m · (a)(iii) 3.9 × 106 solar masses · (b)(i) 0.13″ · (c)(i) 1.2 × 1010 m · (d)(i) 1.9 × 106 m s−1(the remaining parts are explanations — see the table above)
The graph shows how the gravitational field strength g due to Mars varies with distance r from its centre, from the surface (r = 3.39 × 106 m) outwards. Ignore the rotation and the thin atmosphere of Mars.
A sample-return vehicle of mass 300 kg is to be launched from rest on the surface into a circular orbit of radius 4.0 × 106 m.
Graph drawn to scale. The dashed line marks the surface of Mars.
(a)
Mars and the orbit.
(i)
Use the graph to determine the mass of Mars.
(2)
(ii)
Show, using the graph, that the orbital speed of the vehicle will be about 3.3 km s−1.
(2)
(b)
The launch.
(i)
Determine the minimum energy needed to place the vehicle in this orbit.
(3)
(ii)
The propellant releases 1.3 × 107 J of energy per kilogram, and 12 % of this energy becomes mechanical energy of the vehicle. Estimate the minimum mass of propellant needed.
(2)
(iii)
Suggest why the actual mass of propellant needed is greater than your answer to (b)(ii).
(1)
(c)
(i)
Show that the minimum energy per unit mass needed to reach a circular orbit just above the surface of any planet is ½g0R, where g0 is the field strength at the surface and R is the radius of the planet.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
g at the surface = 3.7 N kg−1
✓ 1
Accept 3.6–3.8 N kg⁻¹.
M = gR²/G = 3.7 × (3.39 × 106)²/6.67 × 10−11 = 6.4 × 1023 kg
✓ 1
Accept 6.2–6.5 × 10²³ kg.
Part (a)(ii)
g = 2.7 N kg−1 at r = 4.0 × 106 m, and mg = mv²/r
✓ 1
The value of g must be read from the graph (accept 2.6–2.75 N kg⁻¹).
v = √(gr) = √(2.68 × 4.0 × 106) = 3.27 × 103 m s−1
✓ 1
Full substitution or an answer to at least 3 s.f. is required.
Part (b)(i)
Total energy at rest on the surface = −GMm/R
✓ 1
Total energy in orbit = −GMm/(2r) (= −½mv²)
✓ 1
Allow ECF from (a)(i) and (a)(ii).
ΔE = GMm(1/R − 1/(2r)) = 2.2 × 109 J
✓ 1
Accept 2.1–2.3 × 10⁹ J.
Part (b)(ii)
Energy from the propellant = 2.2 × 109/0.12 = 1.8 × 1010 J
✓ 1
Allow ECF from (b)(i).
Mass = 1.8 × 1010/1.3 × 107 = 1.4 × 103 kg
✓ 1
Part (b)(iii)
The propellant itself must be lifted and accelerated (the vehicle is much heavier than 300 kg during the burn) / energy is carried away by the exhaust gases
✓ 1
Part (c)(i)
Energy per unit mass changes from −GM/R to −GM/(2R), an increase of GM/(2R)
✓ 1
g0 = GM/R², so GM = g0R² and the energy per unit mass is ½g0R
✓ 1
Answers: (a)(i) 6.4 × 1023 kg · (a)(ii) 3.27 × 103 m s−1 · (b)(i) 2.2 × 109 J · (b)(ii) 1.4 × 103 kg (the remaining parts are explanations — see the table above)
82D-2-18Resultant field of two bodiesPaper 2Medium12 marks
A binary asteroid consists of a primary of mass 5.2 × 1011 kg and radius 390 m, and a small moon (the secondary) of mass 4.9 × 109 kg and radius 80 m. The distance between their centres is 1.20 km. Treat both bodies as uniform spheres and ignore all other masses.
Diagram not to scale
(a)
Field strengths.
(i)
Calculate the gravitational field strength at the surface of the primary due to the primary alone.
(1)
(ii)
A spacecraft is held stationary at point P, on the line joining the centres and 600 m from the centre of each body. Determine the magnitude and direction of the resultant gravitational field strength at P.
(2)
(iii)
Point S is on the surface of the secondary, on the line joining the centres, on the side facing the primary. Determine the resultant gravitational field strength at S.
(3)
(b)
Energy.
(i)
Calculate the gravitational potential energy of the two-body system.
(1)
(ii)
The spacecraft, of mass 550 kg, moves from P to the point on the surface of the primary that is nearest to the secondary. Determine the change in its gravitational potential energy.
(3)
(c)
(i)
The spacecraft hovers at P using an ion thruster that ejects xenon at a speed of 3.0 × 104 m s−1 relative to the spacecraft. Determine the mass of xenon ejected per hour.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
g = GM/R² = 6.67 × 10−11 × 5.2 × 1011/390² = 2.3 × 10−4 N kg−1
✓ 1
Part (a)(ii)
Primary: 9.63 × 10−5 N kg−1 towards the primary; secondary: 9.1 × 10−7 N kg−1 towards the secondary
✓ 1
Resultant = 9.5 × 10−5 N kg−1, towards the primary
✓ 1
Direction required.
Part (a)(iii)
S is 1200 − 80 = 1120 m from the centre of the primary
✓ 1
This step is required.
Secondary: GM2/80² = 5.1 × 10−5 N kg−1 towards its centre; primary: GM1/1120² = 2.8 × 10−5 N kg−1 towards the primary
✓ 1
Resultant = 2.3 × 10−5 N kg−1 towards the centre of the secondary
Thrust needed = weight at P = 550 × 9.54 × 10−5 = 5.2 × 10−2 N
✓ 1
Allow ECF from (a)(ii).
F = (Δm/Δt)v gives Δm/Δt = 5.2 × 10−2/3.0 × 104 = 1.7 × 10−6 kg s−1 = 6.3 × 10−3 kg per hour
✓ 1
Answers: (a)(i) 2.3 × 10−4 N kg−1 · (a)(ii) 9.5 × 10−5 N kg−1 towards the primary · (a)(iii) 2.3 × 10−5 N kg−1 · (b)(i) −1.4 × 108 J · (b)(ii) −17 J · (c)(i) 6.3 × 10−3 kg (the remaining parts are explanations — see the table above)
A space tug lowers a defunct satellite of mass 1500 kg from a circular orbit at an altitude of 800 km to a circular orbit at an altitude of 300 km, from which atmospheric drag will bring it down. The graph shows how the total energy E of the satellite in a circular orbit varies with the radius r of the orbit.
ME = 5.97 × 1024 kg, RE = 6.37 × 106 m.
Graph drawn to scale.
(a)
Lowering the orbit.
(i)
Show that the orbital speed of the satellite at an altitude of 800 km is about 7.5 km s−1.
(1)
(ii)
Use the graph to determine the minimum energy that the tug must remove from the satellite.
(2)
(iii)
Confirm the value of E at 800 km by calculation, using your answer to (a)(i).
(1)
(b)
Kinetic and potential energy.
(i)
Show that, when the radius of a circular orbit changes, the change in kinetic energy of the satellite is equal and opposite to the change in its total energy.
(2)
(ii)
Hence deduce the change in kinetic energy and the change in gravitational potential energy of the satellite when it is lowered.
(2)
(c)
The tug.
(i)
After releasing the satellite, the tug, of mass 900 kg, is in the 300 km orbit. Calculate the minimum energy it needs to escape from the Earth from this orbit.
(1)
(ii)
Explain why this is less than half of the energy the tug would need to escape from rest on the surface of the Earth.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
v = √(GM/r) = √(6.67 × 10−11 × 5.97 × 1024/7.17 × 106) = 7.45 × 103 m s−1
✓ 1
r = 7.17 × 10⁶ m must be used; full substitution or 3 s.f. required.
Part (a)(ii)
E = −4.17 × 1010 J at r = 7.17 × 106 m and −4.48 × 1010 J at r = 6.67 × 106 m
✓ 1
Accept readings within ±0.02 × 10¹⁰ J.
Energy removed = 3.1 × 109 J
✓ 1
Accept 2.7–3.5 × 10⁹ J.
Part (a)(iii)
E = −½mv² = −½ × 1500 × (7.45 × 103)² = −4.17 × 1010 J, in agreement with the graph
✓ 1
Allow ECF from (a)(i). E = −Ek for a circular orbit.
Part (b)(i)
Ek = GMm/(2r) and E = Ek + Ep = −GMm/(2r)
✓ 1
So Ek = −E at every radius, and ΔEk = −ΔE
✓ 1
Part (b)(ii)
ΔEk = +3.1 × 109 J (the satellite speeds up)
✓ 1
Allow ECF from (a)(ii).
ΔEp = ΔE − ΔEk = −6.2 × 109 J (twice the energy removed)
84D-2-21Escape speed and orbital speedPaper 2Medium11 marks
A proposed lunar mass driver is an electromagnetic track, 20 km long, laid horizontally on the Moon's surface. It accelerates payloads uniformly from rest. Treat the Moon as a non-rotating uniform sphere of mass 7.35 × 1022 kg and radius 1.74 × 106 m, with no atmosphere. ME = 5.97 × 1024 kg, RE = 6.37 × 106 m.
(a)
A very low orbit.
(i)
Calculate the gravitational field strength at the surface of the Moon.
(1)
(ii)
Show that a payload leaving the track horizontally at about 1.7 km s−1 could orbit the Moon just above its surface.
(2)
(iii)
Determine the period of this orbit.
(2)
(b)
Escape.
(i)
Calculate the escape speed from the surface of the Moon.
(1)
(ii)
Determine the acceleration the track must provide for a payload to leave it at the escape speed.
(2)
(c)
Moon versus Earth.
(i)
Determine the ratio (minimum energy per kilogram needed to escape from the Earth's surface)/(minimum energy per kilogram needed to escape from the Moon's surface).
(2)
(ii)
Suggest one reason why the real energy advantage of launching deep-space payloads from the Moon differs from this ratio.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
g = GM/R² = 6.67 × 10−11 × 7.35 × 1022/(1.74 × 106)² = 1.62 N kg−1
✓ 1
Part (a)(ii)
For a circular orbit of radius R the gravitational force provides the centripetal force: GMm/R² = mv²/R (or mg = mv²/R)
✓ 1
v = √(GM/R) = √(1.62 × 1.74 × 106) = 1.68 × 103 m s−1
a = (2.37 × 103)²/(2 × 2.0 × 104) = 1.4 × 102 m s−2
✓ 1
About 14g; accept 140–141 m s−2. Allow ECF from (b)(i).
Part (c)(i)
Energy per kg to escape = GM/R (= ½vesc²): Earth 6.25 × 107 J kg−1, Moon 2.82 × 106 J kg−1
✓ 1
Ratio = 22
✓ 1
Accept 21–23.
Part (c)(ii)
The payload leaving the Moon is still in the Earth's gravitational field (the Moon orbits the Earth), so extra energy is needed to leave the Earth–Moon system; or: launching from the Earth also requires energy to overcome air resistance
✓ 1
Any one sensible reason.
Answers: (a)(i) 1.62 N kg−1 · (a)(ii) 1.68 km s−1 · (a)(iii) 6.5 × 103 s · (b)(i) 2.37 km s−1 · (b)(ii) 1.4 × 102 m s−2 · (c)(i) 22 (the remaining parts are explanations — see the table above)
85D-2-35Escape speed and atmospheresPaper 2Medium13 marks
Titan, the largest moon of Saturn, has a thick atmosphere composed mainly of nitrogen (N2, molar mass 0.028 kg mol−1). Titan may be treated as a uniform sphere of mass 1.35 × 1023 kg and radius 2.57 × 106 m. The temperature at its surface is 94 K.
(a)
Titan's gravitational field.
(i)
Show that the gravitational field strength at the surface of Titan is about 1.4 N kg−1.
(1)
(ii)
Show that the escape speed from the surface of a body of radius R and surface gravitational field strength g0 is √(2g0R), and hence calculate the escape speed from Titan.
(2)
(b)
A planetary scientist's rule states that a body keeps a gas in its atmosphere for billions of years if its escape speed is more than six times the typical speed of the gas molecules. The typical speed is the speed of a molecule that has the average translational kinetic energy.
(i)
Show that the typical speed of the nitrogen molecules at the surface of Titan is about 290 m s−1.
(2)
(ii)
Deduce whether, according to the rule, Titan keeps its nitrogen.
(2)
(iii)
Determine the maximum surface temperature at which Titan could keep its nitrogen according to the rule.
(3)
(c)
Comparison with the Moon.
(i)
The Moon has an escape speed of 2.4 km s−1 and a daytime surface temperature of about 390 K. Use the rule to explain why the Moon cannot keep an atmosphere of nitrogen.
(2)
(ii)
The gravitational field strength at the surface of the Moon, 1.6 N kg−1, is greater than at the surface of Titan. Suggest why Titan nevertheless keeps an atmosphere while the Moon does not.
m = 0.028/6.02 × 1023 = 4.65 × 10−26 kg and Ek = (3/2)kBT = 1.5 × 1.38 × 10−23 × 94 = 1.95 × 10−21 J
✓ 1
v = √(2Ek/m) = 289 m s−1
✓ 1
Full substitution or an answer to at least 3 s.f. is required.
Part (b)(ii)
6 × 289 = 1.7 × 103 m s−1
✓ 1
This is less than the escape speed of 2.6 × 103 m s−1, so Titan keeps its nitrogen
✓ 1
Allow ECF from (a)(ii) and (b)(i).
Part (b)(iii)
Largest typical speed = vesc/6 = 2647/6 = 441 m s−1
✓ 1
Allow ECF from (a)(ii).
(3/2)kBT = ½mv², so T = mv²/(3kB)
✓ 1
T = 4.65 × 10−26 × 441²/(3 × 1.38 × 10−23) = 2.2 × 102 K
✓ 1
ALT: T ∝ v², so T = 94 × (441/289)² = 220 K. Accept 210–230 K.
Part (c)(i)
Typical speed = √(3kBT/m) = 589 m s−1
✓ 1
Allow ECF from (b)(i), e.g. 290 × √(390/94) = 590 m s⁻¹.
6 × 589 = 3.5 × 103 m s−1, which is greater than 2.4 km s−1, so the nitrogen escapes
✓ 1
Part (c)(ii)
Titan is much colder, so its molecules move more slowly; also the escape speed depends on g0R, and Titan's larger radius gives it the larger escape speed
✓ 1
Either point scores the mark.
Answers: (a)(i) 1.363 N kg−1 · (a)(ii) 2.65 × 103 m s−1 · (b)(i) 289 m s−1 · (b)(iii) 2.2 × 102 K (the remaining parts are explanations — see the table above)
87IBP2H-029-00-12Escape speed and orbital speedPaper 2Medium6 marks
During a lunar mission an astronaut strikes a golf ball from level ground. The ball leaves the surface at 25 m s−1 at 45° to the horizontal. There is no atmosphere. Mass of Moon = 7.35×1022 kg; radius of Moon = 1.74×106 m; G = 6.67×10−11 N m2 kg−2.
Diagram not to scale
(a)
Calculate the gravitational field strength at the surface of the Moon.
(2)
(b)
Estimate the maximum height reached by the ball and hence explain why the gravitational field may be treated as uniform throughout the flight of the ball.
(2)
(c)
Calculate the horizontal distance travelled by the ball before it lands.
(2)
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
✓ 1
Part (b)
Maximum height = (25 sin 45°)2/(2 × 1.62) ≈ 96 m (ECF from (a))
✓ 1
This is very small compared with the Moon's radius of 1.74×106 m, so g changes by only about 0.01 % and its direction is effectively constant over the flight
✓ 1
Part (c)
Use of range = v2 sin 2θ/g, or time of flight 2v sin θ/g = 21.8 s multiplied by the horizontal velocity 25 cos 45° = 17.7 m s−1
✓ 1
Range = 625/1.62 ≈ 386 m (ECF from (a))
✓ 1
Exam tip: To justify a uniform-field model, compare a calculated height or distance with the planet's radius. A bare statement that the ball stays near the surface earns less credit.
88IBP2H-029-00-14Orbital speedPaper 2Medium6 marks
A robotic arm is designed to collect rock from a small spherical asteroid of radius 250 m and uniform density 2.0×103 kg m−3. G = 6.67×10−11 N m2 kg−2.
(a)
Calculate the mass of the asteroid.
(1)
(b)
Calculate the gravitational field strength at the surface of the asteroid.
(2)
(c)
Determine the escape speed from the surface of the asteroid.
(2)
(d)
A fragment is knocked off the surface at 0.5 m s−1. State, with a reason, whether it returns to the surface.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M = (4/3)π(250)3 × 2.0×103 = 1.3×1011 kg
✓ 1
Part (b)
g = GM/R2 = 6.67×10−11 × 1.31×1011/2502 (ECF from (a))
✓ 1
g = 1.4×10−4 N kg−1
✓ 1
Part (c)
v = √(2GM/R)
✓ 1
v = √(2 × 6.67×10−11 × 1.31×1011/250) = 0.26 m s−1 (ECF from (a))
✓ 1
Part (d)
No, it does not return, because 0.5 m s−1 exceeds the escape speed of about 0.26 m s−1 (ECF from (c))
✓ 1
Common error: Escape speed does not depend on the direction of launch (provided the path does not hit the surface). Do not resolve the 0.5 m s⁻¹ into components.
89IBP2H-029-00-16Field lines and Kepler's lawsPaper 2Medium8 marks
A probe is placed in a circular orbit around a dwarf planet of radius 4.70×105 m. The orbital radius is 1.85×106 m and the orbital period is 17.6 h. G = 6.67×10−11 N m2 kg−2.
(a)
Show that, for a circular orbit, T2 = 4π2r3/GM, where M is the mass of the central body.
(2)
(b)
Determine the mass of the dwarf planet.
(2)
(c)
Calculate the mean density of the dwarf planet.
(2)
(d)
The probe is moved to a lower circular orbit of radius 1.00×106 m. Calculate its new orbital period.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Gravitational force provides the centripetal force: GMm/r2 = mv2/r
✓ 1
Substitutes v = 2πr/T and rearranges to T2 = 4π2r3/GM
✓ 1
Part (b)
T = 17.6 × 3600 = 6.336×104 s
✓ 1
M = 4π2(1.85×106)3/(6.67×10−11 × (6.336×104)2) = 9.3×1020 kg
✓ 1
Part (c)
V = (4/3)π(4.70×105)3 = 4.35×1017 m3
✓ 1
ρ = 9.33×1020/4.35×1017 = 2.1×103 kg m−3 (ECF from (b))
✓ 1
Part (d)
Uses T ∝ r3/2: T2 = 17.6 × (1.00/1.85)3/2
✓ 1
T2 = 7.0 h
✓ 1
Common error: Use the orbital radius (1.85×10⁶ m), not the planet's radius, in Kepler's third law. The planet's radius is needed only for its volume.
90IBP2H-029-01-11Mapping a field using potentialPaper 2Medium9 marks
A lunar relay probe is at point P on the line joining the centres of the Earth and the Moon. P is 3.00×108 m from the centre of the Earth and 8.4×107 m from the centre of the Moon. Mass of Earth = 5.97×1024 kg; mass of Moon = 7.35×1022 kg; Earth–Moon separation = 3.84×108 m; G = 6.67×10−11 N m2 kg−2.
Diagram not to scale
(a)
Calculate the magnitude and direction of the resultant gravitational field strength at P.
(3)
(b)
Determine the distance from the centre of the Earth of the point between the two bodies where the resultant field is zero.
(2)
(c)
Calculate the gravitational potential at P.
(2)
(d)
Outline why the gravitational potential is not zero at the zero-field point between the Earth and the Moon.
(2)
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Notes
Part (a)
gE = 6.67×10−11 × 5.97×1024/(3.00×108)2 = 4.42×10−3 N kg−1
✓ 1
gM = 6.67×10−11 × 7.35×1022/(8.4×107)2 = 6.95×10−4 N kg−1
✓ 1
resultant = 3.7×10−3 N kg−1 towards the Earth
✓ 1
Part (b)
x/(3.84×108 − x) = √(5.97×1024/7.35×1022) = 9.01
✓ 1
x = 3.46×108 m
✓ 1
Part (c)
V = −GME/rE − GMM/rM = −1.33×106 − 5.84×104
✓ 1
V = −1.39×106 J kg−1
✓ 1
Part (d)
Potential is a scalar, and both bodies give negative contributions that add rather than cancel
✓ 1
Field strength is the negative potential gradient, so zero field means V is at a maximum along the line, not that V is zero
✓ 1
Exam tip: Check directions before combining fields. Between two bodies their fields oppose each other. Potentials have no direction, and each contribution is negative, so they always add.
91IBP2H-029-01-12Variation of g with heightPaper 2Medium8 marks
Instruments on a spacecraft map the gravitational potential V around a gas-giant exoplanet at distances r from its centre. All the points lie outside the planet.
r / 108 m: 1.0, 2.0, 4.0, 8.0 V / MJ kg−1: −320, −160, −80, −40
Data from the table
(a)
Show that the data are consistent with V ∝ 1/r, and hence determine the mass of the planet.
(2)
(b)
Estimate the gravitational field strength at r = 3.0×108 m, using the potentials at 2.0×108 m and 4.0×108 m.
(2)
(c)
The true field strength at 3.0×108 m is 0.36 N kg−1. Explain why the estimate in the previous part is larger.
(2)
(d)
Calculate the minimum work needed to move a 2000 kg probe from r = 4.0×108 m to r = 8.0×108 m, ignoring any change in its kinetic energy.
(2)
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Notes
Part (a)
Vr = −3.2×1016 J m kg−1 for every row (constant product), so V ∝ 1/r
✓ 1
V = −GM/r, so GM = 3.2×1016 and M = 3.2×1016/6.67×10−11 = 4.8×1026 kg
✓ 1
Part (b)
g ≈ |ΔV/Δr| = (160 − 80)×106/(2.0×108)
✓ 1
g ≈ 0.40 N kg−1
✓ 1
Part (c)
The potential gradient is not constant. The field strength g ∝ 1/r2, so the gradient is steeper closer to the planet
✓ 1
The estimate is the average gradient over a wide interval, from 2.0 to 4.0 ×108 m. Because g falls faster than linearly with r, the extra field near 2.0×108 m more than offsets the weaker field near 4.0×108 m. The average therefore exceeds the value at the midpoint 3.0×108 m. Equivalently, the chord gradient matches the local gradient at about 2.8×108 m
✓ 1
Part (d)
W = mΔV = 2000 × (−40 − (−80))×106
✓ 1
W = 8.0×1010 J
✓ 1
Common error: Convert MJ kg⁻¹ to J kg⁻¹ (×10⁶) before calculating field strengths or work.
92IBP2H-029-01-14Escape speed and orbital speedPaper 2Medium8 marks
The gravitational field strength at the Earth's surface is 9.8 N kg−1. The radius of the Earth is 6.37×106 m. A space station orbits at a height of 420 km. A long-haul airliner cruises at a height of 11.0 km.
(a)
Show that the gravitational field strength at the space station is about 8.6 N kg−1.
(2)
(b)
Astronauts on the station describe themselves as weightless. Explain this, given your answer to (a).
(2)
(c)
Show that, for a height h much smaller than R, the fractional change in field strength is approximately \(\frac{\Delta g}{g_0}\approx-\frac{2h}{R}\). You may use the approximation \((1+x)^n\approx 1+nx\) for \(x\ll 1\).
(2)
(d)
Calculate the reduction in the weight of a 72 kg pilot when the airliner is at cruising height.
(2)
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Part (a)
g = g0R2/(R + h)2, so g = 9.8 × (6.37/6.79)2. Using GM = g0R2 with g = GM/r2 is equally valid
✓ 1
g = 8.63 N kg−1 ≈ 8.6 N kg−1
✓ 1
Part (b)
The astronauts and the station are both in free fall towards the Earth with the same acceleration, 8.6 m s−2, which is the field strength of 8.6 N kg−1
✓ 1
So there is no contact (normal) force between an astronaut and the station. It is this contact force that is felt as weight, so gravity still acts but nothing is felt
✓ 1
Part (c)
g = g0R2/(R + h)2 = g0(1 + h/R)−2
✓ 1
For h ≪ R, (1 + h/R)−2 ≈ 1 − 2h/R, so Δg/g0 ≈ −2h/R. Also accept differentiation: dg/dr = −2GM/r3, giving Δg/g0 ≈ −2h/R
93IBP2H-029-01-16Escape speed and orbital speedPaper 2Medium8 marks
A geophysics team uses a sensitive gravimeter to search for a buried ore body. The ore body is modelled as a sphere of radius 50 m. Its centre is 150 m below a flat surface. Its density is 3000 kg m−3 greater than that of the surrounding rock.
Diagram not to scale
(a)
Show that the excess mass of the ore body is about 1.6×109 kg.
(2)
(b)
Calculate the additional gravitational field strength at the point on the surface directly above the centre of the ore body.
(2)
(c)
Point Q is on the surface, 200 m horizontally from the point directly above the ore body. The gravimeter measures only vertical field. Determine the vertical component of the additional field at Q.
(3)
(d)
State the direction of the additional field at Q.
(1)
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Notes
Part (a)
V = (4/3)π(50)3 = 5.24×105 m3
✓ 1
Δm = 3000 × 5.24×105 = 1.57×109 kg ≈ 1.6×109 kg
✓ 1
Part (b)
Δg = GΔm/r2 = 6.67×10−11 × 1.57×109/1502
✓ 1
Δg = 4.7×10−6 N kg−1
✓ 1
Part (c)
distance to the centre = √(1502 + 2002) = 250 m
✓ 1
additional field = GΔm/2502 = 1.68×10−6 N kg−1 [allow error carried forward from the excess mass]
✓ 1
vertical component = 1.68×10−6 × 150/250 = 1.0×10−6 N kg−1
✓ 1
Part (d)
Towards the centre of the ore body: downward and horizontally towards the point above the ore body, at about 37° below the horizontal
✓ 1
Common error: At Q the additional field points along the line to the centre of the sphere. Resolve it into its vertical component before comparing it with a vertical measurement.
94IBP2H-029-02-10Escape speed and orbital speedPaper 2Medium8 marks
A satellite telephone call between two people is relayed through a geostationary satellite. Data: mass of Earth = 5.97 × 1024 kg radius of Earth = 6.37 × 106 m period of Earth's rotation = 8.6164 × 104 s G = 6.67 × 10−11 N m2 kg−2 speed of radio waves = 3.00 × 108 m s−1
(a)
Show that the radius of a geostationary orbit is about 4.2 × 107 m.
(3)
(b)
Explain why a geostationary orbit must lie in the plane of the equator.
(2)
(c)
Both callers are on the equator, almost directly below the satellite. One caller asks a question, and the other replies at once. Determine the minimum time between the question being asked and the reply being heard.
(3)
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Notes
Part (a)
Gravitational force provides the centripetal force: GMm/r2 = m(4π2/T2)r . Rearranging gives r3 = GMT2/4π2 = 6.67 × 10−11 × 5.97 × 1024 × (8.6164 × 104)2 / 4π2 . r = 4.215 × 107 m ≈ 4.2 × 107 m .
✓ 3
Part (b)
The gravitational force is directed towards the Earth's centre, so the centre of any orbit is the Earth's centre . Only an orbit in the equatorial plane keeps the satellite above a fixed point; an inclined orbit carries it north and south of the equator .
✓ 2
Part (c)
Height of satellite h = 4.215 × 107 − 6.37 × 106 = 3.58 × 107 m . ECF from (a). The signals travel up and down twice, a total distance of 4h = 1.43 × 108 m . t = 1.43 × 108 / 3.00 × 108 = 0.48 s .
✓ 3
Common error: The orbital radius is measured from the Earth's centre. Subtract the Earth's radius to find the distance a signal travels from the ground.
95IBP2H-029-02-12Gravitational potentialPaper 2Medium8 marks
A survey probe measures the gravitational potential V at various distances r from the centre of a spherical planet. The planet's radius is less than 4.0 × 106 m.
r / 106 m
4.0
5.0
6.0
8.0
10.0
V / MJ kg−1
−12.0
−9.6
−8.0
−6.0
−4.8
G = 6.67 × 10−11 N m2 kg−2.Data from the table
(a)
Show that the data are consistent with V ∝ 1/r.
(2)
(b)
Determine the mass of the planet.
(2)
(c)
Calculate the minimum work needed to move a 350 kg instrument package from r = 4.0 × 106 m to r = 8.0 × 106 m.
(2)
(d)
Estimate, using the potentials at 4.0 × 106 m and 6.0 × 106 m, the gravitational field strength at r = 5.0 × 106 m.
(2)
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Part (a)
Calculates V × r for at least three data points, e.g. 4.0 × 106 × (−12.0 × 106) . The product is constant at −4.8 × 1013 J m kg−1 in each case, so V ∝ 1/r .
✓ 2
Part (b)
V = −GM/r, so GM = −Vr = 4.8 × 1013 m3 s−2 . M = 4.8 × 1013 / 6.67 × 10−11 = 7.2 × 1023 kg . Allow use of any single data point. Allow ECF from (a).
Magnitude of g = ΔV/Δr = (−8.0 − (−12.0)) × 106 / (2.0 × 106) . g ≈ 2.0 N kg−1, directed towards the centre of the planet . Award full marks for g = −ΔV/Δr = −2.0 N kg−1 if the negative sign is explained as meaning towards the planet.
✓ 2
Exam tip: Field strength is minus the potential gradient. The minus sign shows that the field points towards lower potential, which here means towards the planet. Using the potentials on either side of a point gives a good estimate of the gradient at that point.
96IBP2H-029-02-14Resultant field of two bodiesPaper 2Medium7 marks
A deep-space probe is to be parked at the point X on the line joining the centres of the Earth and the Moon where the resultant gravitational force on it is zero. Data: mass of Earth = 5.97 × 1024 kg mass of Moon = 7.35 × 1022 kg Earth–Moon centre separation = 3.84 × 108 m G = 6.67 × 10−11 N m2 kg−2
Diagram not to scale
(a)
Explain why X is much closer to the Moon than to the Earth.
(2)
(b)
Show that X is about 3.46 × 108 m from the centre of the Earth.
(3)
(c)
Calculate the gravitational potential at X.
(2)
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Part (a)
Field strength is proportional to M/r2, and the Moon's mass is about 1/81 of the Earth's . For the fields to be equal and opposite, the distance to the Moon must be smaller by a factor of √81 ≈ 9 .
V = −GME/x − GMM/(d − x), with d − x = 3.84 × 107 m . V = −(1.15 × 106 + 1.28 × 105) = −1.28 × 106 J kg−1 . Allow error carried forward from the previous part, and accept answers in the range −1.28 to −1.29 × 106 J kg−1 arising from rounding.
✓ 2
Exam tip: Field strengths are vectors and can cancel. Potentials are scalars and must be added: because both contributions are negative, they never cancel.
97IBP2H-029-02-15Orbital energyPaper 2Medium7 marks
A 4.0 kg CubeSat orbits the Earth in a low circular orbit. Over 30 days, atmospheric drag lowers its altitude from 420 km to 410 km. Assume the orbit stays circular at all times. 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 drag force acts opposite to the CubeSat's velocity, yet the CubeSat's speed increases. Explain this.
(3)
(b)
Calculate the energy dissipated by drag during the 30 days.
(2)
(c)
Estimate the mean rate at which drag dissipates energy.
(2)
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Part (a)
Drag removes energy, so the total energy −GMm/2r becomes more negative and r decreases . Ek = GMm/2r (or v = √(GM/r)), so the kinetic energy and speed increase as r decreases . The potential energy falls by twice the gain in kinetic energy: half of the loss becomes kinetic energy and half is dissipated by drag. Alternatively: the path spirals slightly inwards, so gravity has a component along the velocity, and it does more positive work than the negative work done by drag .
✓ 3
Part (b)
r1 = 6.79 × 106 m and r2 = 6.78 × 106 m . Energy dissipated = (GMm/2)(1/r2 − 1/r1) ≈ 1.7 × 105 J (1.73 × 105 J) .
✓ 2
Part (c)
t = 30 × 24 × 3600 = 2.59 × 106 s . P = 1.73 × 105 J / 2.59 × 106 s ≈ 0.067 W. Using the rounded 1.7 × 105 J gives 0.066 W, which is equally acceptable. ECF from part 2 .
✓ 2
Common error: A drag force does not necessarily slow an orbiting body down. Lowering the orbit raises the speed, because v = √(GM/r).
98IBP3H-D1-20r32-11Orbital energyPaper 2Medium6 marks
A fashion house launches a promotional satellite of mass 450 kg into a circular orbit of radius 7.0 × 106 m around Earth. Use G = 6.67 × 10−11 N m2 kg−2 and mass of Earth = 5.97 × 1024 kg.
(a)
Calculate the kinetic energy of the satellite in this orbit.
(2)
(b)
Determine the total energy of the satellite in this orbit, taking the gravitational potential energy to be zero at infinity.
(2)
(c)
The satellite is later moved to a circular orbit of radius 1.0 × 107 m. Calculate the minimum energy that must be supplied. Use the unrounded value of the total energy in the original orbit.
(2)
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Part (a)
GMm/r² = mv²/r, so Ek = ½mv² = GMm/2r
✓ 1
Ek = 1.28×1010 J
✓ 1
Part (b)
Ep = −GMm/r = -2.56×1010 J
✓ 1
total energy = Ek + Ep = -1.28×1010 J
✓ 1
Part (c)
new total energy = −GMm/2r = −(6.67 × 10−11 × 5.97 × 1024 × 450)/(2 × 1.0 × 107) = −8.96 × 109 J
✓ 1
energy supplied = 3.84×109 J
✓ 1
Common error: Energy needed to change orbit is the difference in total energy, not just the change in potential energy.
99IBP3H-D1-20r32-12Potential gradientPaper 2Medium6 marks
A research satellite for a fisheries project measures the gravitational potential near an isolated, spherical planet. The potential is −4.44 × 107 J kg−1 at a distance of 9.0 × 106 m from the centre of the planet and −3.64 × 107 J kg−1 at a distance of 11.0 × 106 m. Use G = 6.67 × 10−11 N m2 kg−2.
(a)
Determine the average potential gradient between these two points.
(2)
(b)
State the relationship between the potential gradient and the gravitational field strength, and give the field strength at a distance of 1.0 × 107 m from the centre.
(2)
(c)
Estimate the mass of the planet using the field strength from (b).
100IBP3H-D1-20r32-13Escape speed and orbital speedPaper 2Medium7 marks
A robotic probe orbits the Moon at an altitude of 3.0 × 105 m. Use G = 6.67 × 10−11 N m2 kg−2, mass of the Moon = 7.35 × 1022 kg and radius of the Moon = 1.74 × 106 m.
(a)
Show that the escape speed from a point at distance r from the centre of a planet is √2 times the orbital speed of a circular orbit at that radius.
(3)
(b)
Calculate the orbital speed of the probe.
(2)
(c)
Calculate the minimum increase in speed required for the probe to escape from the Moon. Use the unrounded value of the orbital speed.
(2)
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Notes
Part (a)
orbital speed: GMm/r² = mv²/r gives v² = GM/r
✓ 1
escape: ½mvesc² = GMm/r (so that total energy is zero), giving vesc² = 2GM/r
✓ 1
vesc² = 2v² so vesc = √2 v
✓ 1
Part (b)
r = 1.74 × 106 + 3.0 × 105 = 2.04 × 106 m
✓ 1
v = √(GM/r) = 1.55×103 m s−1
✓ 1
Part (c)
vesc − v = (√2 − 1)v (ECF from (b))
✓ 1
increase = 6.42×102 m s−1
✓ 1
Common error: The probe already has orbital speed; only the extra speed needed to reach escape speed has to be supplied.
101IBP3H-D1-20r32-14Orbital energyPaper 2Medium7 marks
A communications satellite of mass 1200 kg is in a circular orbit of radius 7.20 × 106 m around Earth. Thin atmospheric drag slowly reduces the orbital radius to 7.00 × 106 m, the orbit remaining circular. Use G = 6.67 × 10−11 N m2 kg−2 and mass of Earth = 5.97 × 1024 kg.
(a)
Explain why the speed of the satellite increases even though the drag force does negative work on it.
(2)
(b)
Calculate the energy dissipated by drag as the radius decreases from 7.20 × 106 m to 7.00 × 106 m.
(3)
(c)
Calculate the decrease in the gravitational potential energy of the satellite. Use the unrounded value of the energy dissipated.
(2)
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Part (a)
a smaller orbital radius needs a larger orbital speed (v² = GM/r)
✓ 1
potential energy decreases by twice as much as the total energy decreases, so the kinetic energy rises
102IBP3H-D1-20r47-10A lander on EuropaPaper 2Medium8 marks
A lander is to touch down on Europa, a moon of Jupiter. Take Europa to be a uniform sphere of mass 4.80 × 1022 kg and radius 1.56 × 106 m, with no atmosphere. The mass of the lander is 1.80 × 103 kg. G = 6.67 × 10−11 N m2 kg−2. Use unrounded values from earlier parts in later parts.
(a)
Calculate the gravitational field strength at the surface of Europa.
(2)
(b)
Determine the weight of the lander on the surface of Europa.
(1)
(c)
During a test, the lander is released from rest at a height of 12.0 m above the surface. Assuming the field strength is constant over this height, calculate the speed with which it reaches the surface.
(2)
(d)
Calculate the escape speed from the surface of Europa.
(3)
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Part (a)
M1 use of g = GM/R2
✓ 1
A1 g = 1.32 N kg−1
✓ 1
Part (b)
A1 W = 1.80 × 103 × 1.316 = 2.37 × 103 N (accept 2.37–2.38 × 103 N; ECF from (a))
✓ 1
Part (c)
M1 use of v2 = 2gh (or equating mgh to ½mv2)
✓ 1
A1 v = 5.62 m s−1 (ECF from (a))
✓ 1
Part (d)
M1 kinetic energy at launch equals the gain in potential energy needed to reach infinity, ½mv2 = GMm/R
✓ 1
M1 v = √(2GM/R) with correct substitution
✓ 1
A1 v = 2.03 × 103 m s−1
✓ 1
Common error: Do not use v² = 2gh for the escape speed: the field strength falls with distance, so the constant-g approximation only holds near the surface.
103IBP3H-D1-20r47-12Potential gradientPaper 2Medium8 marks
The gravitational potential V outside Earth at two distances r from its centre is given below. At r = 7.00 × 106 m, V = −56.9 MJ kg−1. At r = 7.50 × 106 m, V = −53.1 MJ kg−1. Take the mass of Earth to be 5.97 × 1024 kg and G = 6.67 × 10−11 N m2 kg−2.
(a)
Define gravitational potential at a point.
(2)
(b)
Use the data to estimate the gravitational field strength at r = 7.25 × 106 m.
(3)
(c)
Calculate the gravitational field strength at r = 7.25 × 106 m using g = GM/r2.
(2)
(d)
Suggest why the two values of the field strength are not identical.
(1)
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Part (a)
work done per unit mass
✓ 1
in bringing a small test mass from infinity to the point
✓ 1
Part (b)
M1 field strength equals the magnitude of the potential gradient, g = ΔV/Δr
the gradient method gives the average field over the interval, while the field strength varies non-linearly with r (accept: potential data rounded to 3 s.f.; the chord is not the tangent at the midpoint)
✓ 1
Exam tip: The field strength is the negative of the gradient of the V–r relationship; for a numerical estimate, the change in V divided by the change in r gives the average field over that interval.
104IBP3H-D1-20r47-14Orbital energyPaper 2Medium7 marks
A communications satellite is placed in a geostationary orbit about Earth. Take the mass of Earth to be 5.97 × 1024 kg and G = 6.67 × 10−11 N m2 kg−2. The orbital period of the satellite is 8.62 × 104 s. Use unrounded values from earlier parts in later parts.
(a)
State two features of a geostationary orbit, apart from the period.
(2)
(b)
Show that the orbital radius of the satellite is about 4.2 × 107 m.
(3)
(c)
Calculate the orbital speed of the satellite.
(2)
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Part (a)
the orbit is circular
✓ 1
the orbit lies in the equatorial plane (accept: the satellite moves in the same direction as Earth's rotation; the satellite stays above a fixed point of the surface)
✓ 1
Part (b)
M1 equating gravitational force with centripetal force to obtain r3 = GMT2/(4π2)
✓ 1
M1 correct substitution of the data
✓ 1
A1 r = 4.216 × 107 m, which rounds to 4.2 × 107 m (a value to at least 3 s.f. must be shown)
✓ 1
Part (c)
M1 use of v = 2πr/T (or v = √(GM/r))
✓ 1
A1 v = 3.07 × 103 m s−1
✓ 1
Exam tip: A 'show that' answer must give the calculated value to more significant figures than the value quoted in the question.
A communications satellite of mass 1200 kg is released into a circular orbit of radius 7.00 × 106 m around the Earth. An ion thruster then exerts a small constant force of 0.20 N on the satellite, always in the direction of its velocity, until the radius of the orbit has increased to 1.20 × 107 m. The satellite spirals outwards very slowly, so at every instant its orbit may be treated as circular. Ignore the change in the mass of the satellite.
The graph shows how the orbital speed v of the satellite varies with the radius r of its orbit. Mass of the Earth = 5.97 × 1024 kg.
Graph drawn to scale.
(a)
The orbit.
(i)
Show that the orbital speed of the satellite at the start of the transfer is about 7.5 km s−1.
(1)
(ii)
Use the graph to determine the orbital period of the satellite when r = 1.00 × 107 m.
(2)
(b)
Energy.
(i)
Determine the increase in the total energy of the satellite during the transfer.
(2)
(ii)
Explain why the speed of the satellite decreases during the transfer, although the force exerted by the thruster always acts in the direction of its velocity.
(3)
(c)
Duration of the transfer.
(i)
Determine the distance travelled by the satellite along its spiral path during the transfer.
(2)
(ii)
Hence, using the graph, estimate the time taken for the transfer.
(2)
(iii)
State one assumption made in your answer to (c)(i).
(1)
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Part (a)(i)
v = √(GM/r) = √(6.67 × 10−11 × 5.97 × 1024/7.00 × 106) = 7.54 × 103 m s−1
✓ 1
Full substitution or an answer to at least 3 s.f. is required.
Part (a)(ii)
v read from the graph at r = 1.00 × 107 m: 6.3 km s−1
✓ 1
Accept 6.25–6.35 km s−1. ALT: T = 2π√(r³/GM) without the graph [1 max].
T = 2πr/v = 2π × 1.00 × 107/6.31 × 103 = 9.96 × 103 s (about 2.8 h)
✓ 1
Accept 9.8–10.1 × 103 s, consistent with the reading.
Part (b)(i)
Total energy in a circular orbit E = −GMm/(2r), so ΔE = (GMm/2)(1/r1 − 1/r2)
✓ 1
ALT: ΔE = −ΔEk = ½m(v1² − v2²) with speeds from the graph.
Accept 1.4–1.5 × 1010 J. Award [1 max] for 2.8 × 1010 J (change in potential energy only).
Part (b)(ii)
The thruster does positive work, so the total energy −GMm/(2r) increases (becomes less negative) and the radius of the orbit increases
✓ 1
In a circular orbit Ek = GMm/(2r) = −E, so as the total energy increases the kinetic energy decreases: the speed falls
✓ 1
Allow ECF from (b)(i). Accept v = √(GM/r), which decreases as r increases, as part of this mark only if linked to energy.
On the outward spiral the velocity has a small component away from the Earth, so the gravitational force has a component opposite to the velocity. The work done against gravity (the increase in potential energy, twice the work done by the thruster) is larger than the work done by the thruster
✓ 1
Accept: the potential energy increases by 2 × 1.42 × 1010 J while the thruster supplies only 1.42 × 1010 J. Do not accept "gravity is weaker further out" on its own.
Part (c)(i)
The work done by the thruster equals the increase in total energy: Fs = ΔE
✓ 1
Allow ECF from (b)(i).
s = 1.42 × 1010/0.20 = 7.1 × 1010 m
✓ 1
Accept 7.0–7.5 × 1010 m.
Part (c)(ii)
Average speed from the graph ≈ ½(7.54 + 5.76) km s−1 = 6.65 km s−1
✓ 1
Accept any reasonable mean speed in the range 6.4–6.8 km s−1 read from the graph.
t = 7.1 × 1010/6.65 × 103 = 1.1 × 107 s (about 4 months)
✓ 1
Allow ECF from (c)(i). Accept 1.0–1.15 × 107 s.
Part (c)(iii)
All the work done by the thruster becomes orbital (kinetic + gravitational potential) energy of the satellite / the force acts exactly along the velocity / the orbit is circular at every instant / the mass of the satellite is constant
✓ 1
Any one.
Answers: (a)(i) 7.54 × 103 m s−1 · (a)(ii) 9.96 × 103 s · (b)(i) 1.42 × 1010 J · (c)(i) 7.1 × 1010 m · (c)(ii) 1.1 × 107 s (the remaining parts are explanations — see the table above)
106D-2-06Gravitational potentialPaper 2Hard12 marks
The graph shows how the gravitational potential Vg varies with distance r from the centre of a planet of radius 3.0 × 106 m, from its surface outwards. The planet has no atmosphere and its rotation can be ignored.
Graph drawn to scale.
(a)
The field of the planet.
(i)
Use the graph to determine the mass of the planet.
(2)
(ii)
By drawing a tangent to the graph, determine the gravitational field strength at r = 6.0 × 106 m.
(2)
(iii)
Hence deduce the gravitational field strength at the surface of the planet.
(1)
(b)
A probe is launched vertically upwards from the surface with a speed of 3.5 km s−1.
(i)
Determine the maximum distance from the centre of the planet reached by the probe.
(3)
(ii)
A student calculates the maximum height from mgΔh = ½mv² using the surface value of g. Explain why the student's answer is too small.
(2)
(c)
(i)
Use the graph to determine the minimum speed with which the probe would have to be launched from the surface to escape from the planet.
(2)
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Part (a)(i)
Vg at the surface = −10.7 MJ kg−1
✓ 1
Accept −10.5 to −10.9 MJ kg⁻¹.
M = −VgR/G = 10.7 × 106 × 3.0 × 106/6.67 × 10−11 = 4.8 × 1023 kg
✓ 1
Accept 4.7–4.9 × 10²³ kg.
Part (a)(ii)
Tangent drawn at r = 6.0 × 106 m and its gradient found from a large triangle
✓ 1
g = |ΔVg/Δr| ≈ 0.89 N kg−1
✓ 1
Accept 0.75–1.05 N kg⁻¹.
Part (a)(iii)
g ∝ 1/r², and the surface is at half this distance, so gs = 4 × 0.89 = 3.6 N kg−1
✓ 1
Allow ECF from (a)(ii).
Part (b)(i)
Kinetic energy per unit mass at launch = ½v² = ½ × 3500² = 6.1 MJ kg−1
✓ 1
Vg at the highest point = −10.7 + 6.1 = −4.6 MJ kg−1
✓ 1
Allow ECF from (a)(i).
From the graph (or r = GM/|Vg|), r ≈ 7.0 × 106 m
✓ 1
Accept 6.8–7.3 × 10⁶ m.
Part (b)(ii)
Student: Δh = v²/(2gs) = 3500²/(2 × 3.6) ≈ 1.7 × 106 m, compared with 4.0 × 106 m from (b)(i)
✓ 1
Allow ECF from (a)(iii) and (b)(i).
ΔEp = mgΔh assumes a uniform field, but g falls with height (to about a fifth of the surface value at the top), so less energy is needed for each metre gained higher up and the probe rises further
✓ 1
Part (c)(i)
To escape, the total energy must be at least zero (Vg = 0 at infinity): ½mv² ≥ m|Vg| at the surface
✓ 1
Accept v = √(2GM/R) with GM/R = |Vg| from the graph.
v = √(2 × 10.7 × 106) = 4.6 × 103 m s−1
✓ 1
Allow ECF from (a)(i). Accept 4.6–4.7 × 10³ m s⁻¹.
Answers: (a)(i) 4.8 × 1023 kg · (a)(ii) 0.89 N kg−1 · (a)(iii) 3.6 N kg−1 · (b)(i) 7.0 × 106 m · (c)(i) 4.6 × 103 m s−1(the remaining parts are explanations — see the table above)
107D-2-20Mapping a field using potentialPaper 2Hard13 marks
The diagram shows equipotential lines in the plane containing the centres of a planet and its moon, which are 4.0 × 108 m apart. The values of the gravitational potential are in MJ kg−1, and the scale gives the distance x from the centre of the planet along the line joining the centres. Point A lies off this line; point B lies on it.
Equipotential lines (values in MJ kg−1) in the plane containing the centres of a planet and its moon. The scale gives the distance x from the centre of the planet in units of 108 m; small ticks on the line joining the centres are 0.2 × 108 m apart. The planet and moon are not drawn to scale.
(a)
Field lines.
(i)
Sketch, on the diagram, the gravitational field line that passes through point A. Indicate its direction.
(2)
(b)
Field strength from potential.
(i)
Estimate the magnitude of the gravitational field strength at B.
(3)
(ii)
Explain why the method used in (b)(i) gives only an estimate.
(1)
(c)
The point of zero field.
(i)
Identify the point N on the line joining the centres at which the resultant gravitational field strength is zero, and explain how the equipotential pattern shows this.
(2)
(ii)
Determine the ratio (mass of planet)/(mass of moon).
(2)
(d)
Energy.
(i)
A probe of mass 600 kg is at rest on the −4.0 MJ kg−1 equipotential near the planet. Calculate the minimum energy needed to move it to N.
(1)
(ii)
Determine the minimum speed with which the probe must leave the −4.0 MJ kg−1 equipotential, with its engines off, in order to reach the moon. Ignore the orbital motion of the moon.
(2)
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Part (a)(i)
Line crosses every equipotential it meets at right angles, curving to end on the planet
✓ 1
A line ending on the moon scores 0: from A the potential decreases most steeply towards the planet.
Arrow pointing towards the planet (towards lower potential)
✓ 1
Part (b)(i)
Along the line joining the centres, the −4.0 line crosses at x ≈ 1.04 × 108 m and the −3.0 line at x ≈ 1.41 × 108 m
✓ 1
Accept readings ±0.05 × 108 m.
ΔVg = 1.0 × 106 J kg−1 over Δx ≈ 0.37 × 108 m
✓ 1
g ≈ ΔVg/Δx ≈ 2.7 × 10−2 N kg−1
✓ 1
Accept 2.2–3.2 × 10−2 N kg−1.
Part (b)(ii)
The field strength (potential gradient) changes across the interval, so ΔVg/Δx gives only the average over the interval, not the value at B (and reading the positions from the diagram is imprecise)
✓ 1
Part (c)(i)
N is at x ≈ 3.0 × 108 m, where the −1.78 MJ kg−1 equipotential crosses itself (the two loops meet)
✓ 1
Accept 2.9–3.1 × 108 m.
Along the line the potential rises to a maximum at N, so the potential gradient, and therefore g = −ΔVg/Δr, is zero there
✓ 1
Part (c)(ii)
GMP/xN² = GMM/(4.0 × 108 − xN)²
✓ 1
MP/MM = (3.0/1.0)² = 9
✓ 1
Accept 8–10 consistent with the reading in (c)(i); allow ECF.
The probe need only just reach N (beyond N the resultant field points towards the moon), so ½mv² = mΔVg with ΔVg = 2.22 × 106 J kg−1
✓ 1
Or ½mv² = 1.33 × 109 J from (d)(i).
v = √(2 × 2.22 × 106) = 2.1 × 103 m s−1
✓ 1
Allow ECF from (d)(i).
Answers: (b)(i) ≈ 2.7 × 10−2 N kg−1 · (c)(i) x ≈ 3.0 × 108 m · (c)(ii) 9 · (d)(i) 1.3 × 109 J · (d)(ii) 2.1 km s−1(the remaining parts are explanations — see the table above)
108IBP2S-028-03-16Gravitational potentialPaper 2Hard8 marks
An asteroid-mining company plans a base on a spherical asteroid of uniform density 2.0×103 kg m−3 and radius 5.0 km. An astronaut working there can push off from the surface with a vertical speed of 4.0 m s−1. G = 6.67×10−11 N m2 kg−2
(a)
Show that the gravitational field strength at the surface is g = 4⁄3πGρR.
(2)
(b)
Calculate g at the surface.
(1)
(c)
Determine whether the astronaut can escape from the asteroid by jumping.
(2)
(d)
Determine the maximum height above the surface that the astronaut reaches.
(3)
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Part (a)
M = ρ × 4⁄3πR3
✓ 1
g = GM/R2 = G ρ 4⁄3πR3/R2 = 4⁄3πGρR
✓ 1
Part (b)
g = 4⁄3π × 6.67×10−11 × 2.0×103 × 5.0×103 = 2.8×10−3 m s−2
✓ 1
Part (c)
vesc = √(2GM/R) = √(2gR) = √(2 × 2.79×10−3 × 5.0×103) = 5.3 m s−1 (ECF from (b))
✓ 1
4.0 m s−1 < 5.3 m s−1, so the astronaut does not escape and falls back
✓ 1
Part (d)
Energy conservation: ½v2 − GM/R = −GM/r, using GM = gR2 = 7.0×104 N m2 kg−1
✓ 1
1/r = 1/R − v2/(2GM), giving r = 1.17×104 m (ECF from (b))
✓ 1
h = r − R = 6.7×103 m
✓ 1
Common error: Using h = v²/2g gives only about 2.9 km. When the height is comparable to the radius, g is not constant, so use gravitational potential energy instead.
109IBP2H-029-00-11Kepler's second lawPaper 2Hard8 marks
A sample-return mission targets an asteroid in an elliptical orbit around the Sun. At perihelion the asteroid is 1.0×1011 m from the Sun and moves at 4.70×104 m s−1. At aphelion it is 5.0×1011 m from the Sun. Mass of Sun = 1.99×1030 kg; G = 6.67×10−11 N m2 kg−2.
Diagram not to scale
(a)
State Kepler's second law.
(1)
(b)
Explain, with reference to angular momentum, why the asteroid moves fastest at perihelion.
(2)
(c)
Calculate the speed of the asteroid at aphelion.
(2)
(d)
The spacecraft intercepts the asteroid when it is 3.0×1011 m from the Sun. Determine the speed of the asteroid at this point.
(3)
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Part (a)
The line joining a planet (orbiting body) to the Sun sweeps out equal areas in equal time intervals
✓ 1
Part (b)
The gravitational force is directed towards the Sun, so it exerts no torque about the Sun and the asteroid's angular momentum is conserved
✓ 1
At perihelion and aphelion mvr is constant, so the smallest r corresponds to the greatest v
✓ 1
Part (c)
rpvp = rava
✓ 1
va = 4.70×104 × 1.0/5.0 = 9.4×103 m s−1
✓ 1
Part (d)
Energy conservation per unit mass: ½v2 − GM/r = constant
Common error: rv = constant applies only where the velocity is perpendicular to the radius, at perihelion and aphelion. At intermediate points use energy conservation instead.
110IBP2H-029-00-13Field lines and Kepler's lawsPaper 2Hard8 marks
A (hypothetical) crewed probe orbits a neutron star of mass 2.8×1030 kg at an orbital radius of 1.0×107 m. An astronaut of height 1.8 m floats inside the probe with her body aligned radially, feet towards the star. G = 6.67×10−11 N m2 kg−2.
(a)
Calculate the gravitational field strength due to the neutron star at the orbital radius.
(2)
(b)
Show that the difference between the field strength at her feet and at her head is approximately Δg = 2GMh/r3, where h is her height and h ≪ r.
(2)
(c)
Calculate Δg for the astronaut.
(2)
(d)
Explain why the astronaut does not feel the field strength calculated in (a), but does experience a stretching effect.
For h ≪ r, (1 + h/r)−2 ≈ 1 − 2h/r, so Δg ≈ (GM/r2)(2h/r) = 2GMh/r3. Alternatively, |Δg| ≈ |dg/dr| × h = 2GMh/r3.
✓ 1
Part (c)
Δg = 2 × 6.67×10−11 × 2.8×1030 × 1.8/(1.0×107)3
✓ 1
Δg = 0.67 N kg−1
✓ 1
Part (d)
The astronaut and the probe are in free fall together with the same acceleration (equal to the local g), so there is no contact force between them and she feels weightless
✓ 1
Her feet are pulled with a slightly larger g than her head; relative to her centre of mass, this difference (a tidal effect) stretches her body radially
✓ 1
Exam tip: Computing g at two radii that differ by only 1.8 m needs many significant figures. Use the binomial approximation (or dg/dr) to avoid subtracting two almost-equal numbers.
111IBP2H-029-00-15Orbital energyPaper 2Hard9 marks
A communications satellite is in a circular orbit in the Earth's equatorial plane. It moves in the same direction as the Earth rotates. An engineer at a ground station on the equator sees it pass directly overhead once every 6.00 hours. Mass of Earth = 5.97×1024 kg; G = 6.67×10−11 N m2 kg−2; rotation period of Earth = 23.93 h.
(a)
Explain why the time between overhead passes is longer than the orbital period of the satellite.
(2)
(b)
Show that the orbital period of the satellite is about 4.80 h.
(2)
(c)
Determine the orbital radius of the satellite.
(3)
(d)
Calculate the gravitational field strength due to the Earth at the position of the satellite.
(2)
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Part (a)
The satellite's angular speed is greater than the Earth's, and during one orbit the ground station is carried round in the same direction by the Earth's rotation
✓ 1
The satellite must therefore turn through more than 360° to be overhead again, which takes longer than one orbital period
✓ 1
Part (b)
The relative angular speed is ωsat − ωE, so 1/6.00 = 1/T − 1/23.93
✓ 1
1/T = 0.1667 + 0.0418 = 0.2085 h−1, giving T = 4.797 h ≈ 4.80 h
✓ 1
Part (c)
r3 = GMT2/4π2
✓ 1
T = 4.80 × 3600 = 1.728×104 s
✓ 1
r = (6.67×10−11 × 5.97×1024 × (1.728×104)2/4π2)1/3 = 1.44×107 m. Accept 1.44–1.45×107 m, which covers rounding of T
✓ 1
Part (d)
g = GM/r2, or g = 4π2r/T2
✓ 1
g = 3.98×1014/(1.44×107)2 = 1.9 N kg−1. Allow ECF from (c)
✓ 1
Exam tip: For a prograde orbit, subtract the angular speeds: 1/Tpass = 1/T − 1/TEarth. For a retrograde orbit you would add them. Keep extra significant figures in intermediate values so that later answers do not drift.
112IBP2H-029-01-09Gravitational field strengthPaper 2Hard7 marks
A probe descends vertically towards a planet of radius R and mass M, both unknown. An on-board accelerometer records the gravitational field strength g at different heights h above the surface:
h / 106 m: 0, 1.0, 2.0, 3.0, 4.0 g / N kg−1: 3.69, 2.21, 1.46, 1.04, 0.780
Take G = 6.67×10−11 N m2 kg−2.
(a)
Show that a graph of \(\frac{1}{\sqrt{g}}\) against h should be a straight line, and identify its gradient and intercept.
(2)
(b)
Using the data at h = 0 and h = 4.0×106 m, determine M.
(3)
(c)
Determine R.
(2)
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Part (a)
g = GM/(R + h)2
✓ 1
1/√g = h/√(GM) + R/√(GM), which is linear in h, with gradient 1/√(GM) and intercept R/√(GM)
GM = 4.28×1013, so M = 4.28×1013/6.67×10−11 ≈ 6.4×1023 kg
✓ 1
Part (c)
R = intercept × √(GM) = 0.521 × 6.54×106, or R = √(GM/g0) = √(4.28×1013/3.69) [allow error carried forward from the value of √(GM) found in the previous part]
✓ 1
R ≈ 3.4×106 m
✓ 1
Exam tip: When a relationship is an inverse square, rearrange it into straight-line form before using the data. The gradient and intercept then give the unknowns directly.
113IBP2H-029-01-10Determining the density of an asteroidPaper 2Hard8 marks
A small rubble-pile asteroid is modelled as a uniform sphere of radius 250 m and mean density 1200 kg m−3. A mission plans to land a rover on its surface. Ignore the asteroid's rotation.
(a)
Show that the mass of the asteroid is about 8×1010 kg.
(2)
(b)
Calculate the escape speed from the surface of the asteroid.
(2)
(c)
Outline why the escape speed does not depend on the mass of the object being launched.
(1)
(d)
A rover wheel kicks a pebble upward from the surface at 0.50 m s−1. Determine the speed of the pebble when it is very far from the asteroid.
(3)
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Part (a)
V = (4/3)π(250)3 = 6.54×107 m3
✓ 1
M = ρV = 1200 × 6.54×107 = 7.85×1010 kg ≈ 8×1010 kg
114IBP2H-029-01-13Orbital energyPaper 2Hard9 marks
A space tug transfers a 1500 kg communications satellite from a low circular orbit of radius 6.77×106 m to a circular orbit of radius 4.22×107 m around the Earth. The final orbit lies in the Earth's equatorial plane, and the satellite moves in the same sense as the Earth's rotation. Mass of Earth = 5.97×1024 kg; G = 6.67×10−11 N m2 kg−2.
(a)
Show that the orbital speed of a satellite in a circular orbit of radius r is \(v=\sqrt{\frac{GM}{r}}\).
(2)
(b)
Calculate the orbital speed in the final orbit. Use it to show that the satellite is geostationary.
(2)
(c)
The satellite moves more slowly in the final orbit. Explain why energy must nevertheless be supplied to transfer it.
(2)
(d)
Determine the minimum energy needed for the transfer.
(3)
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Part (a)
The gravitational force provides the centripetal force: GMm/r2 = mv2/r
✓ 1
Cancelling m and one factor of r gives v2 = GM/r, so v = √(GM/r)
✓ 1
Part (b)
v = √(3.98×1014 / 4.22×107) = 3.07×103 m s−1
✓ 1
T = 2πr/v = 8.6×104 s ≈ 24 h, which equals the Earth's rotation period; with the equatorial, same-sense orbit, the satellite stays above one point on the surface
✓ 1
Part (c)
The gravitational potential energy increases (becomes less negative) by GMm(1/r1 − 1/r2)
✓ 1
The kinetic energy (GMm/2r) decreases by only half of this amount, so the total energy increases
115IBP2H-029-01-15Escape speed and orbital speedPaper 2Hard7 marks
A proposed lunar mining base would use an electromagnetic mass driver to fire packages of ore vertically off the Moon. Mass of Moon = 7.35×1022 kg; radius of Moon = 1.74×106 m; surface field strength = 1.62 N kg−1. The Moon has no atmosphere, and its rotation can be ignored.
(a)
Calculate the escape speed from the surface of the Moon.
(2)
(b)
A package leaves the surface vertically at 2.00 km s−1. Determine the maximum height it reaches above the surface.
(3)
(c)
A student uses h = v2/2g with g = 1.62 N kg−1 and obtains 1.2×106 m. Explain why this underestimates the height.
116IBP2S-028-04-15Orbital energyPaper 2Hard9 marks
At the end of its working life, a 2500 kg geostationary communications satellite is boosted into a slightly higher circular 'graveyard' orbit. This keeps the geostationary orbit clear for new satellites. Initial orbital radius = 4.22×107 m (period 86 164 s) Graveyard orbital radius = 4.25×107 m Mass of Earth = 5.97×1024 kg G = 6.67×10−11 N m2 kg−2
(a)
Show that the total energy of a satellite of mass m in a circular orbit of radius r around the Earth (mass M) is −GMm/(2r).
(2)
(b)
Calculate the minimum energy needed to move the satellite into the graveyard orbit.
(2)
(c)
Explain why the satellite in the graveyard orbit drifts slowly westward as seen from the ground.
(2)
(d)
Determine the time, in days, for the satellite to drift once completely around the Earth relative to the ground.
Relative angular speed: 1/t = 1/T1 − 1/T2, so t = T1T2/(T2 − T1)
✓ 1
t ≈ 8.15×106 s ≈ 94 days; accept 94–95 days
✓ 1
Exam tip: When two orbital radii are very close, keep at least five significant figures in 1/r or in the period ratio. Otherwise rounding errors swamp the small difference.
117IBP2S-028-04-16Field lines and Kepler's lawsPaper 2Hard8 marks
A pair of white dwarf stars, each of mass M = 1.2×1030 kg, orbit their common centre of mass in a circle. Their centres are a distance d apart, and the orbital period is 600 s. Such systems are target sources for space-based gravitational-wave detectors. G = 6.67×10−11 N m2 kg−2
Diagram not to scale
(a)
Outline why the gravitational field strength at the midpoint between the two stars is zero.
(1)
(b)
Show that the orbital period T is given by T = π√(2d3/(GM)).
(3)
(c)
Calculate the separation d.
(2)
(d)
Determine the gravitational potential at the midpoint between the stars.
(2)
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Part (a)
The midpoint is equidistant from two equal masses, so the two fields there are equal in magnitude and opposite in direction, and they cancel
✓ 1
Part (b)
Each star moves in a circle of radius d/2 about the midpoint
118IBP2H-029-02-11Orbital energyPaper 2Hard8 marks
A 2400 kg weather satellite is placed in a circular parking orbit 3.0 × 105 m above the Earth's surface. It is later transferred to a geostationary orbit of radius 4.22 × 107 m. 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 total energy of a satellite of mass m in a circular orbit of radius r around a planet of mass M is −GMm/2r.
(3)
(b)
Calculate the minimum energy that must be supplied to transfer the satellite from the parking orbit to the geostationary orbit.
(3)
(c)
Energy is supplied to the satellite, yet it moves more slowly in the geostationary orbit. Explain this.
(2)
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Part (a)
Gravitational force provides the centripetal force: GMm/r2 = mv2/r, so Ek = ½mv2 = GMm/2r . Ep = −GMm/r . E = Ek + Ep = GMm/2r − GMm/r = −GMm/2r .
Orbital speed v = √(GM/r), so the speed and the kinetic energy decrease as r increases; the kinetic energy falls by an amount equal to the energy supplied, 6.0 × 1010 J . The gravitational potential energy rises by twice the energy supplied, 1.2 × 1011 J, so the energy supplied plus the kinetic energy lost together provide this gain in potential energy .
✓ 2
Exam tip: Measure the radius from the planet's centre: add the altitude to the planet's radius before using r.
119IBP2H-029-02-13A mission to MarsPaper 2Hard9 marks
Mission planners want a relay satellite that remains above one point on Mars's equator (an areostationary orbit). The moon Phobos moves in a circular orbit close to Mars's equatorial plane. It orbits in the same direction as Mars rotates. Data: mass of Mars = 6.42 × 1023 kg rotation period of Mars = 8.856 × 104 s orbital radius of Phobos = 9.38 × 106 m G = 6.67 × 10−11 N m2 kg−2
(a)
Determine the radius of an areostationary orbit.
(3)
(b)
Show that the orbital period of Phobos is about 8 hours. Hence explain how Phobos appears to move across the sky of an observer on the surface of Mars, compared with the daily motion of the stars.
(3)
(c)
A 1500 kg relay satellite starts in a circular orbit of the same radius as Phobos and is moved to the areostationary orbit. Calculate the minimum energy required. Ignore energy losses and any change in the satellite's mass.
(3)
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Part (a)
Equates gravitational force to the centripetal force to give r3 = GMT2/4π2 . r3 = 6.67 × 10−11 × 6.42 × 1023 × (8.856 × 104)2 / 4π2 = 8.51 × 1021 m3 . r = 2.04 × 107 m .
✓ 3
Part (b)
T = 2π√(r3/GM) = 2π√((9.38 × 106)3 / (6.67 × 10−11 × 6.42 × 1023)) . T = 2.76 × 104 s ≈ 7.7 h, which is about 8 h . This period is shorter than Mars's rotation period of about 24.6 h, and Phobos orbits in the same direction as the rotation. Phobos therefore overtakes the rotating surface, so it appears to rise in the west and set in the east, opposite to the motion of the stars .
✓ 3
Part (c)
The minimum energy is the change in total orbital energy: ΔE = (GMm/2)(1/r1 − 1/r2) . ΔE = (6.67 × 10−11 × 6.42 × 1023 × 1500 / 2) × (1/(9.38 × 106) − 1/(2.04 × 107)) . ΔE = 1.85 × 109 J . Allow error carried forward from the radius found in part (a).
✓ 3
Common error: The energy needed to change orbit is the change in total energy, −GMm/2r. Using only the change in potential energy, −GMm/r, gives twice the correct value.
120IBP2H-029-02-16Escape speed and orbital speedPaper 2Hard9 marks
A proposed space elevator is a cable anchored at a point on the equator and extending radially beyond geostationary orbit. The cable rotates with the Earth. A climber carries a 1000 kg payload up the cable from the surface to the geostationary radius of 4.22 × 107 m. Data: mass of Earth = 5.97 × 1024 kg, radius of Earth = 6.37 × 106 m, rotation period of Earth = 8.6164 × 104 s, G = 6.67 × 10−11 N m2 kg−2.
Diagram not to scale
(a)
Outline why a payload released from the cable at the geostationary radius stays above the same point on the equator.
(2)
(b)
Calculate the gain in gravitational potential energy of the payload.
(2)
(c)
The payload starts at the surface, moving with the rotating Earth. Determine the total increase in the mechanical energy (kinetic plus gravitational potential) of the payload between the surface and the geostationary radius. Measure the speeds relative to the Earth's centre, in a non-rotating frame.
(3)
(d)
A payload released from the cable well below the geostationary radius does not stay in a circular orbit at that radius. Explain this.
(2)
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Notes
Part (a)
At this radius, gravity alone provides exactly the centripetal force needed for an angular velocity equal to the Earth's, so no force from the cable is needed . Once released, the payload continues in a circular equatorial orbit with a period equal to the Earth's rotation period .
ω = 2π/(8.6164 × 104) = 7.29 × 10−5 rad s−1 . The speed rises from ωR to ωr, so ΔEk = ½mω2(r2 − R2) = 4.73 × 109 − 1.08 × 108 = 4.63 × 109 J . Total = 5.31 × 1010 + 4.63 × 109 = 5.77 × 1010 J . Allow error carried forward from the ΔEp value calculated earlier.
✓ 3
Part (d)
The payload moves at speed ωr, which is less than the circular orbital speed √(GM/r) at that radius . The gravitational force is greater than the centripetal force needed for a circle at that radius, so the payload moves inwards on an elliptical path, or falls to the Earth .
✓ 2
Exam tip: An object rotating with the Earth already has kinetic energy. Include the change in kinetic energy as well as the change in potential energy when finding the total energy gained.
121IBP3H-D1-20r32-16A mission to MarsPaper 2Hard11 marks
A robotic freight vehicle is to be sent from Earth towards the Moon. The distance between the centres of Earth and the Moon is 3.84 × 108 m. Use G = 6.67 × 10−11 N m2 kg−2, mass of Earth = 5.97 × 1024 kg, mass of the Moon = 7.35 × 1022 kg and radius of Earth = 6.37 × 106 m. Ignore the rotation and orbital motion of both bodies, and take the distance from the surface of Earth to the centre of the Moon as 3.84 × 108 − 6.37 × 106 m.
Diagram not to scale
(a)
Show that the point on the line between the centres at which the resultant gravitational field strength is zero lies at a distance x from the centre of Earth given by x/(d − x) = √(ME/MM), where d is the separation of the centres.
(3)
(b)
Calculate x, to three significant figures.
(2)
(c)
Calculate the gravitational potential at this point due to Earth and the Moon together. Use the unrounded value of x.
(2)
(d)
The vehicle is launched from the surface of Earth and is to arrive at this point with negligible speed. Calculate the minimum launch speed.
(4)
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Marking point
Mark
Notes
Part (a)
fields equal and opposite: GME/x² = GMM/(d − x)²
✓ 1
rearrange to (d − x)²/x² = MM/ME
✓ 1
square root: x/(d − x) = √(ME/MM)
✓ 1
Part (b)
√(ME/MM) = 9.012 and x = d × 9.012/(1 + 9.012)
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x = 3.46×108 m
✓ 1
Part (c)
V = −G(ME/x + MM/(d − x))
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V = -1.28×106 J kg−1
✓ 1
Part (d)
potential at the launch point includes both bodies: V = −G(ME/RE + MM/(d − RE))
✓ 1
Vlaunch = -6.25×107 J kg−1
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½v² = Vneutral − Vlaunch (ECF from (c))
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v = 1.11×104 m s−1
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Common error: Near the Moon the resultant field is the vector sum of two fields, but potentials simply add as scalars.
122IBP3H-D1-20r47-15Mapping a field using potentialPaper 2Hard11 marks
A spacecraft travels along the straight line joining the centre of Earth to the centre of the Moon. Take the following data: mass of Earth = 5.97 × 1024 kg; mass of the Moon = 7.35 × 1022 kg; distance between the centres = 3.84 × 108 m; G = 6.67 × 10−11 N m2 kg−2. Ignore the motion of both bodies and the radii of both bodies. Use unrounded values from earlier parts in later parts.
Diagram not to scale
(a)
Explain why there is a point on the line between Earth and the Moon at which the resultant gravitational field strength is zero.
(2)
(b)
Determine the distance of this point from the centre of Earth.
(4)
(c)
Calculate the gravitational potential at this point.
(3)
(d)
A small probe is launched from this point. Calculate the minimum speed it needs in order to escape from the Earth–Moon system.
(2)
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Marking point
Mark
Notes
Part (a)
the fields of Earth and the Moon point in opposite directions between the two bodies
✓ 1
the Earth's field decreases and the Moon's field increases as the point moves towards the Moon, so the magnitudes are equal at one point
✓ 1
Part (b)
M1 equating the field magnitudes, GME/x2 = GMM/(d − x)2
M1 potential is the sum of the potentials of Earth and the Moon, V = −G(ME/x + MM/(d − x))
✓ 1
M1 correct substitution of x = 3.456 × 108 m and d − x
✓ 1
A1 V = −1.28 × 106 J kg−1 (ECF from (b))
✓ 1
Part (d)
M1 ½v2 = −V (kinetic energy per unit mass equals the potential per unit mass change to reach infinity)
✓ 1
A1 v = 1.60 × 103 m s−1 (ECF from (c))
✓ 1
Common error: Potentials are scalars and add; do not subtract the Moon's potential from Earth's potential as if they were vectors pointing in opposite directions.
123IBP3H-D1-20r47-16Escape speed and orbital speedPaper 2Hard10 marks
A proposed electromagnetic launcher uses a vertical rail of length 2.0 km to accelerate a small capsule from rest to a speed at which it then coasts, without further thrust, to a maximum height of 1.50 × 106 m above the surface of Earth. Ignore air resistance and the rotation of Earth, and treat the acceleration along the rail as uniform. Earth has mass 5.97 × 1024 kg and radius 6.37 × 106 m; G = 6.67 × 10−11 N m2 kg−2. Use unrounded values from earlier parts in later parts and take the rail to end at the surface.
Diagram not to scale
(a)
Calculate the gravitational potential at the surface of Earth and at a height of 1.50 × 106 m above the surface.
(2)
(b)
Determine the minimum speed with which the capsule must leave the rail.
(3)
(c)
A student estimates the launch speed by assuming that the field strength stays constant at 9.8 N kg−1. Calculate the percentage by which this estimate exceeds your answer to (b).
(3)
(d)
Calculate the acceleration of the capsule along the rail.
(2)
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Marking point
Mark
Notes
Part (a)
A1 V at the surface = −GM/R = −6.25 × 107 J kg−1
✓ 1
A1 V at the maximum height = −GM/(R + h) = −5.06 × 107 J kg−1
✓ 1
Part (b)
M1 change in kinetic energy per unit mass equals the increase in potential, ½v2 = ΔV
✓ 1
A1 ΔV = 1.19 × 107 J kg−1 (ECF from (a))
✓ 1
A1 v = 4.88 × 103 m s−1
✓ 1
Part (c)
M1 use of v = √(2gh) with g = 9.8 N kg−1
✓ 1
A1 v = 5.42 × 103 m s−1
✓ 1
A1 percentage excess = 11.1 % (ECF from (b))
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Part (d)
M1 use of v2 = 2aL with L = 2.0 × 103 m
✓ 1
A1 a = 5.96 × 103 m s−2 (ECF from (b))
✓ 1
Exam tip: Compare a constant-g estimate with the full calculation by working out how much the field strength changes over the path; here it falls noticeably over 1500 km.
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