ExaminerPrep › Question bank › IAL Physics › 5.7 Gravitational fields and orbits
Edexcel International A Level Physics · Unit 5 · Thermodynamics, Radiation, Oscillations and Cosmology
40 free, high-quality past-paper-style questions on Gravitational fields and orbits for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.
Which statement correctly defines the gravitational field strength at a point?
| Answer | Mark | Notes |
|---|---|---|
| Gravitational field strength is defined as g = F/m: the force per unit mass on a small (test) mass at the point. It is a vector, measured in N kg−1. | ✓ 1 | Answer B |
Answer: B
Exam tip: Keep “per unit mass” in every definition: field strength is force per unit mass, potential is work done per unit mass.
Two small spheres attract each other with a gravitational force F. The distance between their centres is tripled.
What is the new gravitational force between them?
| Answer | Mark | Notes |
|---|---|---|
| F ∝ 1/r2. Tripling r multiplies r2 by 9, so the force becomes F/9. | ✓ 1 | Answer A |
Answer: A
Common error: Dividing by 3 instead of 9. The law of gravitation is an inverse square law.
Which statement describes a difference between the gravitational field of a point mass and the electric field of a point charge?
| Answer | Mark | Notes |
|---|---|---|
| Both fields follow inverse square laws, both are shown with field lines, and both potentials vary as 1/r. The difference is that mass is never negative, so gravity only attracts, while like charges repel and unlike charges attract. | ✓ 1 | Answer D |
Answer: D
Exam tip: Learn the comparison as pairs: same inverse square form for force and field, same 1/r form for potential; gravity attracts only.
An astronaut on Mars hangs a 2.50 kg mass from a calibrated newtonmeter. The newtonmeter reads 9.28 N.
State what is meant by a gravitational field.
Calculate the gravitational field strength at the surface of Mars.
The astronaut walks 200 m across the surface and repeats the measurement. Explain why the reading does not change measurably.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| a region (of space) in which a mass experiences a force | ✓ 1 | |
| Part (b) | ||
| g = F/m = 9.28 ÷ 2.50 | ✓ 1 | |
| = 3.71 N kg−1 | ✓ 1 | |
| Part (c) | ||
| 200 m is tiny compared with the radius of Mars, so the field over this region is (effectively) uniform / the distance from the centre is effectively unchanged | ✓ 1 | |
Exam tip: Give the unit of g as N kg−1 when it comes from F/m; m s−2 is equivalent but N kg−1 shows the definition.
The Earth has mass 5.97 × 1024 kg and the Moon has mass 7.35 × 1022 kg. The distance between their centres is 3.84 × 108 m. G = 6.67 × 10−11 N m2 kg−2.
Calculate the gravitational force between the Earth and the Moon.
Calculate the acceleration of the Moon towards the Earth.
State the size and direction of the gravitational force that the Moon exerts on the Earth, and explain why the Earth's acceleration is much smaller than the Moon's.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| F = Gm1m2/r2 = 6.67 × 10−11 × 5.97 × 1024 × 7.35 × 1022 ÷ (3.84 × 108)2 | ✓ 1 | |
| = 1.98 × 1020 N | ✓ 1 | |
| Part (b) | ||
| a = F/m = 1.98 × 1020 ÷ 7.35 × 1022 = 2.70 × 10−3 m s−2 (2.69 × 10−3 m s−2 from the rounded force; allow error carried forward) | ✓ 1 | |
| Part (c) | ||
| 1.98 × 1020 N, directed towards the Moon (Newton's third law pair) | ✓ 1 | |
| the Earth's mass is much larger (about 81 times), so for the same force its acceleration F/m is much smaller | ✓ 1 | |
Exam tip: The two bodies feel equal and opposite forces whatever their masses; the different accelerations come from a = F/m.
A spacecraft mapping the lunar surface travels in a circular path 100 km above the Moon. The Moon has mass 7.35 × 1022 kg and radius 1.74 × 106 m. G = 6.67 × 10−11 N m2 kg−2.
By equating two expressions for the force on the spacecraft, show that its speed is v = √(GM/r), where M is the mass of the Moon and r is the radius of the path.
Calculate the speed of the spacecraft.
Calculate the time for one complete circuit of the Moon, in minutes.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the gravitational force provides the centripetal force: GMm/r2 = mv2/r | ✓ 1 | |
| m and one r cancel: v2 = GM/r, so v = √(GM/r) | ✓ 1 | |
| Part (b) | ||
| r = 1.74 × 106 + 1.0 × 105 = 1.84 × 106 m | ✓ 1 | |
| v = √(6.67 × 10−11 × 7.35 × 1022 ÷ 1.84 × 106) = 1.63 × 103 m s−1 | ✓ 1 | |
| Part (c) | ||
| T = 2πr/v = 2π × 1.84 × 106 ÷ 1.632 × 103 = 7.08 × 103 s | ✓ 1 | |
| = 118 min | ✓ 1 | |
Common error: Forgetting to add the Moon's radius to the 100 km height. The radius of the path is measured from the centre of the Moon.
Figure 1 shows the gravitational field lines around an isolated planet. Points P and Q lie on the same field line.
Which statement is correct?
| Answer | Mark | Notes |
|---|---|---|
| Field lines show two things. The arrows give the direction of the force on a small mass: towards the planet. The spacing shows the field strength: the lines are closer together at P, so the field is stronger there. This agrees with g = GM/r2, which falls as r increases. | ✓ 1 | Answer A |
Answer: A
Exam tip: Read a field-line diagram twice: the arrows give the direction of the force on a small mass, and the spacing of the lines gives the field strength.
Figure 1 shows four patterns, A, B, C and D. In A, B and D the dot at the centre is an isolated point object.
Which pattern could represent both the gravitational field of an isolated point mass and the electric field of an isolated negative point charge?
| Answer | Mark | Notes |
|---|---|---|
| Both fields are radial. Gravity always attracts, so the field lines of a point mass point inwards. A negative charge attracts a positive test charge, so its field lines also point inwards: pattern B. | ✓ 1 | Answer B |
Answer: B
Exam tip: Field lines show the direction of the force on a small positive test charge or a small test mass. Both are pulled towards a negative charge and towards a mass.
Figure 1 shows four satellite orbits, W, X, Y and Z, around the Earth, seen from the side. R is the radius of the Earth. The arrows show the direction of motion of each satellite and of the Earth's surface.
Which orbit could be a geostationary orbit?
| Answer | Mark | Notes |
|---|---|---|
| A geostationary satellite must orbit in the plane of the equator, in the same direction as the Earth rotates, with a period of one day. A period of one day fixes the radius at about 4.2 × 107 m, about 6.6R. Only orbit Z meets all three conditions. | ✓ 1 | Answer D |
Answer: D
Exam tip: A geostationary orbit has three conditions: the equatorial plane, the same direction as the Earth's rotation, and a period of one day (radius about 6.6 Earth radii). Check each orbit against all three.
A lander on a small moon uses a force sensor to measure the weight W of several sample containers of known mass m. Figure 1 shows the results with a line of best fit. The moon has no atmosphere.
Explain how Figure 1 shows that the weight of an object at the surface is proportional to its mass.
Use Figure 1 to determine the gravitational field strength at the surface of the moon.
A tool is dropped from rest 1.5 m above the surface. Calculate the time it takes to reach the ground.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the line of best fit is a straight line through the origin | ✓ 1 | |
| Part (b) | ||
| g = W/m = gradient of the line, e.g. 5.0 ÷ 4.0 | ✓ 1 | |
| = 1.25 N kg−1 (allow 1.22–1.28) | ✓ 1 | |
| Part (c) | ||
| s = ½gt2, so t = √(2 × 1.5 ÷ 1.25) | ✓ 1 | |
| = 1.5 s (1.55 s; allow error carried forward from (b)) | ✓ 1 | |
Exam tip: Field strength in N kg−1 is the same as free-fall acceleration in m s−2: with no air resistance, weight is the only force and a = F/m = g.
Figure 1 shows how the gravitational field strength g varies with distance r from the centre of a planet, from the planet's surface outwards. G = 6.67 × 10−11 N m2 kg−2.
Use the labelled points to show that g is inversely proportional to r2.
Calculate the mass of the planet.
Explain why the curve starts at r = 5.0 × 106 m.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| each time r doubles (5.0 → 10.0 → 20.0 × 106 m), g falls to one quarter (12.0 → 3.0 → 0.75 N kg−1) | ✓ 1 | |
| so g × r2 is constant (3.0 × 1014 N m2 kg−1 for each point) | ✓ 1 | |
| Part (b) | ||
| g = GM/r2, so M = gr2/G = 12.0 × (5.0 × 106)2 ÷ 6.67 × 10−11 | ✓ 1 | |
| = 4.5 × 1024 kg (4.50 × 1024 kg) | ✓ 1 | |
| Part (c) | ||
| this is the radius of the planet (its surface); g = GM/r2 only applies outside the planet | ✓ 1 | |
Exam tip: To test an inverse square law from a graph, find pairs of points where r doubles: g should fall to one quarter. Checking that gr2 is constant is equally good.
Figure 1 shows the gravitational field strength g at the surface of four moons, P, Q, R and S. The radius of each moon is given below its bar. G = 6.67 × 10−11 N m2 kg−2.
Calculate the weight of a 70.0 kg astronaut standing on moon S.
Calculate the mass of moon Q.
A student says that a moon with a larger radius always has a larger surface field strength. Use Figure 1 to show that the student is wrong.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| W = mg = 70.0 × 1.8 = 126 N | ✓ 1 | |
| Part (b) | ||
| g = GM/R2, so M = gR2/G = 1.4 × (2.6 × 106)2 ÷ 6.67 × 10−11 | ✓ 1 | |
| = 1.42 × 1023 kg | ✓ 1 | |
| Part (c) | ||
| Q has the largest radius but S, with a smaller radius, has the greater field strength (1.8 N kg−1 compared with 1.4 N kg−1); g depends on M/R2, not on the radius alone | ✓ 1 | |
Exam tip: g = GM/R2 links three quantities. Rearranged as M = gR2/G, it gives the mass of a moon from its surface field strength and radius.
A planet has 3 times the mass of the Earth and twice the Earth's radius. The gravitational field strength at the Earth's surface is 9.81 N kg−1.
What is the gravitational field strength at the surface of the planet?
| Answer | Mark | Notes |
|---|---|---|
| g = GM/r2, so g ∝ M/r2. Factor = 3 ÷ 22 = 0.75, so g = 0.75 × 9.81 = 7.36 N kg−1. | ✓ 1 | Answer B |
Answer: B
Exam tip: For “a planet with x times the mass and y times the radius”, the field strength scales by x/y2. No need for G.
The Moon has mass 7.35 × 1022 kg and radius 1.74 × 106 m. G = 6.67 × 10−11 N m2 kg−2.
What is the gravitational potential at the Moon's surface?
| Answer | Mark | Notes |
|---|---|---|
| V = −GM/r = −6.67 × 10−11 × 7.35 × 1022 ÷ 1.74 × 106 = −2.82 × 106 J kg−1. It is negative because the potential is zero at infinity and gravity attracts. | ✓ 1 | Answer B |
Answer: B
Exam tip: Gravitational potential is never positive. Potential uses 1/r; field strength uses 1/r2.
Equipotential surfaces are drawn around an isolated planet at equal intervals of gravitational potential.
Which statement is correct?
| Answer | Mark | Notes |
|---|---|---|
| The field strength equals the potential gradient. Further from the planet the field is weaker, so a larger distance is needed for the same change in potential: the equipotentials get further apart. No work is done moving along an equipotential, because ΔV = 0 (the force is perpendicular to the motion). | ✓ 1 | Answer D |
Answer: D
Exam tip: Closely spaced equipotentials mean a strong field; widely spaced ones mean a weak field. Only a uniform field has equally spaced equipotentials.
A satellite moves in a circular orbit 630 km above the Earth's surface. The Earth has mass 5.97 × 1024 kg and radius 6.37 × 106 m. G = 6.67 × 10−11 N m2 kg−2.
What is the orbital speed of the satellite?
| Answer | Mark | Notes |
|---|---|---|
| r = 6.37 × 106 + 0.63 × 106 = 7.00 × 106 m. GMm/r2 = mv2/r gives v = √(GM/r) = √(6.67 × 10−11 × 5.97 × 1024 ÷ 7.00 × 106) = 7.54 × 103 m s−1. | ✓ 1 | Answer A |
Answer: A
Common error: Using the height above the surface as the orbital radius. r is measured from the centre of the Earth.
Two satellites, P and Q, move in circular orbits around the same planet. The orbital period of P is 8.0 times the orbital period of Q.
What is the ratio (orbital radius of P) ÷ (orbital radius of Q)?
| Answer | Mark | Notes |
|---|---|---|
| T2 ∝ r3, so r ∝ T2/3. Ratio = 8.02/3 = 4.0. | ✓ 1 | Answer C |
Answer: C
Exam tip: From T2 ∝ r3: square the period ratio, then take the cube root (82 = 64, ∛64 = 4).
Venus has radius 6.05 × 106 m. The gravitational field strength at its surface is 8.87 N kg−1. G = 6.67 × 10−11 N m2 kg−2.
Starting from Newton's law of gravitation, show that the gravitational field strength at a distance r from the centre of a planet of mass M is g = GM/r2.
Calculate the mass of Venus.
A spacecraft approaching Venus is 1.00 × 107 m from its centre. Calculate the gravitational field strength due to Venus at this point.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| force on a test mass m: F = GMm/r2 | ✓ 1 | |
| g = F/m, so m cancels to give g = GM/r2 | ✓ 1 | |
| Part (b) | ||
| M = gr2/G = 8.87 × (6.05 × 106)2 ÷ 6.67 × 10−11 | ✓ 1 | |
| = 4.87 × 1024 kg | ✓ 1 | |
| Part (c) | ||
| g ∝ 1/r2: 8.87 × (6.05 × 106 ÷ 1.00 × 107)2, or GM/r2 with the answer to (b) | ✓ 1 | |
| = 3.25 N kg−1 | ✓ 1 | |
Exam tip: A ratio is the quickest route in part (c): field strength scales with (old distance ÷ new distance)2.
A space probe measures the gravitational potential V at different distances r from the centre of a planet. The results are:
r / 107 m: 1.0, 2.0, 4.0, 8.0 — V / MJ kg−1: −38.0, −19.0, −9.5, −4.75.
G = 6.67 × 10−11 N m2 kg−2.
Use the data to show that V is inversely proportional to r.
Calculate the mass of the planet.
Calculate the minimum work needed to move a 1500 kg satellite from r = 1.0 × 107 m to r = 4.0 × 107 m.
Deduce the gravitational field strength at r = 2.0 × 107 m.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| calculates Vr for at least three readings | ✓ 1 | |
| all equal −3.8 × 1014 J m kg−1, so Vr is constant / each doubling of r halves V | ✓ 1 | |
| Part (b) | ||
| V = −GM/r, so M = |V|r/G = 3.8 × 1014 ÷ 6.67 × 10−11 | ✓ 1 | |
| = 5.70 × 1024 kg | ✓ 1 | |
| Part (c) | ||
| ΔV = −9.5 − (−38.0) = 28.5 MJ kg−1 | ✓ 1 | |
| W = mΔV = 1500 × 2.85 × 107 = 4.28 × 1010 J | ✓ 1 | |
| Part (d) | ||
| g = GM/r2 = |V|/r = 1.9 × 107 ÷ 2.0 × 107 = 0.95 N kg−1 | ✓ 1 | |
Exam tip: To test for inverse proportion, multiply each pair of values: if the product is constant, V ∝ 1/r.
A student says: “The gravitational field around a planet and the electric field around a charged sphere are really the same kind of field.”
Compare the gravitational field of a point mass with the electric field of a point charge. Your answer should include both similarities and differences.
| Answer | Mark | Notes |
|---|---|---|
| Indicative content, one mark for each point: | — | |
| both field strengths are defined as force per unit property: g = F/m and E = F/Q | ✓ 1 | |
| both forces obey an inverse square law with distance between centres | ✓ 1 | |
| both radial field strengths are ∝ 1/r2 and both potentials are ∝ 1/r | ✓ 1 | |
| gravitational forces are always attractive; electric forces can be attractive or repulsive | ✓ 1 | |
| gravitational potential is always negative; electric potential can be positive or negative | ✓ 1 | |
| for particles, the gravitational force is far weaker than the electric force / gravity cannot be shielded or cancelled | ✓ 1 |
Exam tip: In a compare question, write each point as a pair (“both …” or “gravitational … whereas electric …”). A list about only one field earns little.
Two alpha particles are 4.0 × 10−14 m apart. Each alpha particle has mass 6.64 × 10−27 kg and charge 3.20 × 10−19 C. G = 6.67 × 10−11 N m2 kg−2; k = 8.99 × 109 N m2 C−2.
Calculate the electric force between the alpha particles.
Calculate the gravitational force between them.
The electric force is about 1035 times larger. Explain why it is nevertheless gravity that keeps the Moon in orbit around the Earth.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| F = kQ1Q2/r2 = 8.99 × 109 × (3.20 × 10−19)2 ÷ (4.0 × 10−14)2 | ✓ 1 | |
| = 0.58 N (0.575 N), repulsive | ✓ 1 | |
| Part (b) | ||
| F = Gm2/r2 = 6.67 × 10−11 × (6.64 × 10−27)2 ÷ (4.0 × 10−14)2 | ✓ 1 | |
| = 1.8 × 10−36 N (1.84 × 10−36 N), attractive | ✓ 1 | |
| Part (c) | ||
| the Earth and the Moon are (almost exactly) electrically neutral, so the electric forces between them cancel / are negligible | ✓ 1 | |
| gravity only attracts, so the contributions of all their mass add up to a large force | ✓ 1 | |
Exam tip: Both forces vary as 1/r2, so their ratio does not depend on the separation. A ratio question needs no distance at all.
Two lead spheres are placed with their centres a distance r apart. The larger sphere has mass 12.0 kg. Figure 1 shows how the gravitational force F between the spheres varies with 1/r2. G = 6.67 × 10−11 N m2 kg−2.
What is the mass of the smaller sphere?
| Answer | Mark | Notes |
|---|---|---|
| F = Gm1m2 × (1/r2), so the gradient is Gm1m2. From the circled point, gradient = 30.0 × 10−9 ÷ 25.0 = 1.20 × 10−9 N m2. m2 = 1.20 × 10−9 ÷ (6.67 × 10−11 × 12.0) = 1.5 kg. | ✓ 1 | Answer B |
Answer: B
Exam tip: Match the law to y = mx before using a gradient: with F on the y-axis and 1/r2 on the x-axis, the gradient is Gm1m2.
Figure 1 shows equipotentials around a planet, labelled in MJ kg−1. A 1.0 kg mass is moved slowly along each of the paths A, B, C and D.
Along which path is the work done on the mass by the external force greatest?
| Answer | Mark | Notes |
|---|---|---|
| Work done = mΔV. A: −30 → −20, +10 MJ. B: along an equipotential, 0. C: −30 → −40, −10 MJ (the field does the work). D: −50 → −30, +20 MJ. Path D needs the most work even though it is shorter than path A, because closer to the planet the equipotentials are closer together. | ✓ 1 | Answer D |
Answer: D
Exam tip: Work depends only on the start and end potentials, never on the length of the path. Read the two labels and subtract.
Figure 1 shows how the gravitational potential V varies with distance r from the centre of a planet of radius 5.0 × 106 m. A space probe of mass 250 kg is momentarily at rest at point P. Ignore the effect of other bodies.
What is the minimum energy that must be given to the probe so that it can escape completely from the planet's gravitational field?
| Answer | Mark | Notes |
|---|---|---|
| To escape, the probe must reach infinity, where V = 0. At P, V = −12 MJ kg−1. Energy = m(0 − V) = 250 × 12 × 106 = 3.0 × 109 J. | ✓ 1 | Answer C |
Answer: C
Exam tip: Escaping means reaching the place where V = 0. The minimum energy is mass × (0 − starting potential), with the potential converted from MJ kg−1 to J kg−1.
Figure 1 shows two satellites, P and Q, in circular orbits around the same planet.
What is the ratio (orbital speed of Q) ÷ (orbital speed of P)?
| Answer | Mark | Notes |
|---|---|---|
| The gravitational force provides the centripetal force: GMm/r2 = mv2/r, so v = √(GM/r) and v ∝ 1/√r. Ratio = √(2.0 ÷ 3.2) = 0.79. The outer satellite moves more slowly. | ✓ 1 | Answer A |
Answer: A
Exam tip: Check the direction of your ratio: in a larger orbit the speed is smaller, so (speed of outer) ÷ (speed of inner) must be less than 1.
Figure 1 shows three small spheres, S, A and B, on a horizontal surface, seen from above. The lines joining S to A and S to B are at right angles. G = 6.67 × 10−11 N m2 kg−2.
What is the magnitude of the resultant gravitational force on S due to A and B?
| Answer | Mark | Notes |
|---|---|---|
| FA = 6.67 × 10−11 × 8.0 × 2.50 ÷ 0.202 = 3.335 × 10−8 N. FB = 6.67 × 10−11 × 6.0 × 2.50 ÷ 0.302 = 1.11 × 10−8 N. The forces are at right angles: resultant = √(FA2 + FB2) = 3.52 × 10−8 N. | ✓ 1 | Answer C |
Answer: C
Exam tip: Gravitational forces are vectors. Find each force separately, then combine them; for forces at right angles use Pythagoras.
Figure 1 shows a torsion balance, seen from above. A light rod hangs from a thin fibre at its centre and carries a small lead sphere at each end. A large lead sphere is placed beside each small sphere, on opposite sides of the rod, so that the gravitational attractions make the rod turn slightly. G = 6.67 × 10−11 N m2 kg−2.
Calculate the gravitational force between one small sphere and the large sphere beside it.
The two forces in (a) act at right angles to the rod, in opposite directions, one at each end. Calculate the moment of the couple they produce.
Each small sphere is also attracted by the large sphere at the other end of the rod. Show that this force is less than 1% of the force in (a).
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| F = Gm1m2/r2 = 6.67 × 10−11 × 160 × 0.750 ÷ 0.1002 | ✓ 1 | |
| = 8.00 × 10−7 N | ✓ 1 | |
| Part (b) | ||
| moment of a couple = one force × perpendicular distance between the forces | ✓ 1 | |
| = 8.00 × 10−7 × 1.80 = 1.44 × 10−6 N m | ✓ 1 | |
| Part (c) | ||
| distance = √(1.802 + 0.1002) = 1.80 m (1.803 m) | ✓ 1 | |
| force ratio = (0.100 ÷ 1.803)2 = 3.1 × 10−3, which is about 0.3%, less than 1% | ✓ 1 | |
Exam tip: In a ratio of two gravitational forces between the same masses, everything cancels except the distances: the ratio is (r1/r2)2.
In Figure 1, Graph 1 shows how the gravitational potential varies with distance r from the centre of a planet. Graph 2 shows how the electric potential varies with distance r from the centre of a positively charged metal sphere of radius 0.10 m. k = 8.99 × 109 N m2 C−2; e = 1.60 × 10−19 C.
Using the labelled points, state one similarity and one difference between the two graphs.
Calculate the charge on the sphere.
Explain why a gravitational potential can never be positive, although an electric potential can be.
A proton moves from r = 0.30 m to r = 0.60 m. Calculate the work done on the proton by the electric field.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| similarity: in both graphs, doubling r halves the potential (−40 → −20 and 6.0 → 3.0), so both potentials are ∝ 1/r | ✓ 1 | |
| difference: the gravitational potential is negative and increases towards zero, while this electric potential is positive and decreases towards zero | ✓ 1 | |
| Part (b) | ||
| V = kQ/r, so Q = Vr/k = 6.0 × 103 × 0.30 ÷ 8.99 × 109 | ✓ 1 | |
| = 2.0 × 10−7 C | ✓ 1 | |
| Part (c) | ||
| gravitational forces only attract, so work must always be done to move a mass out to infinity (where V = 0); electric forces can repel, so the field itself can do work moving a charge out to infinity | ✓ 1 | |
| Part (d) | ||
| W = QΔV = 1.60 × 10−19 × (6.0 − 3.0) × 103 = 4.8 × 10−16 J | ✓ 1 | |
Exam tip: Gravitational and electric potentials around a point source have the same 1/r form. The sign is the difference: gravity only attracts, so gravitational potential is always negative.
A space station moves anticlockwise in a circular orbit around the Earth, as seen in Figure 1. The figure shows its position at t = 0 and 15.0 minutes later. G = 6.67 × 10−11 N m2 kg−2.
Calculate the angular speed of the space station.
Calculate the orbital speed of the space station.
Calculate the mass of the Earth.
There is a resultant force on the space station, but its speed does not change. Explain why.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 58.0° = 58.0 × π/180 = 1.012 rad | ✓ 1 | |
| ω = 1.012 ÷ (15.0 × 60) = 1.12 × 10−3 rad s−1 | ✓ 1 | |
| Part (b) | ||
| v = ωr = 1.125 × 10−3 × 6.80 × 106 = 7.65 × 103 m s−1 | ✓ 1 | |
| Part (c) | ||
| GMm/r2 = mω2r, so M = ω2r3/G | ✓ 1 | |
| = (1.125 × 10−3)2 × (6.80 × 106)3 ÷ 6.67 × 10−11 = 5.96 × 1024 kg (allow 5.9–6.0 × 1024 kg) | ✓ 1 | |
| Part (d) | ||
| the gravitational force is always perpendicular to the velocity, so it does no work on the station: it changes only the direction of motion | ✓ 1 | |
Common error: Using the angle in degrees in ω = θ/t. Convert to radians, and the time to seconds, before dividing.
Figure 1 shows a geostationary satellite S, directly above ground station E on the equator, and a second ground station K on the equator. The Earth has mass 5.97 × 1024 kg and radius 6.37 × 106 m. G = 6.67 × 10−11 N m2 kg−2. Radio waves travel at 3.00 × 108 m s−1.
Calculate the gravitational field strength due to the Earth at the position of S.
A ground station on the equator can exchange signals with S only if the straight line from the station to S does not pass through the Earth. Calculate the greatest possible value of the angle KOE.
Calculate the distance d from K to S.
Calculate the time for a radio signal to travel from K to S and back to K.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| g = GM/r2 = 6.67 × 10−11 × 5.97 × 1024 ÷ (4.22 × 107)2 | ✓ 1 | |
| = 0.224 N kg−1 | ✓ 1 | |
| Part (b) | ||
| at the limit the line from the station to S just touches the Earth, so it is a tangent and the angle between it and the radius at the station is 90°: cos θ = R/r = 6.37 × 106 ÷ 4.22 × 107 | ✓ 1 | |
| θ = 81.3° | ✓ 1 | |
| Part (c) | ||
| cosine rule: d2 = r2 + R2 − 2rR cos 60.0° = (4.22 × 107)2 + (6.37 × 106)2 − 2 × 4.22 × 107 × 6.37 × 106 × 0.500 | ✓ 1 | |
| d = 3.94 × 107 m | ✓ 1 | |
| Part (d) | ||
| t = 2d/c = 2 × 3.94 × 107 ÷ 3.00 × 108 = 0.263 s | ✓ 1 | |
Exam tip: When a line from a point outside a circle just grazes the circle, it is a tangent, and a tangent meets the radius at 90°. Find that right angle before choosing between cos, sin and tan.
A moon moves in a circular orbit of radius 5.0 × 108 m around a planet. Its orbital period is 3.0 days. G = 6.67 × 10−11 N m2 kg−2.
What is the mass of the planet?
| Answer | Mark | Notes |
|---|---|---|
| T = 3.0 × 86 400 = 2.59 × 105 s. From GMm/r2 = m(2πr/T)2/r, M = 4π2r3/GT2 = 4π2 × (5.0 × 108)3 ÷ (6.67 × 10−11 × (2.59 × 105)2) = 1.1 × 1027 kg. | ✓ 1 | Answer C |
Answer: C
Common error: Leaving the period in days. With G in SI units, T must be in seconds.
A space probe of mass 500 kg is moved from a distance 2R from the centre of the Earth to a distance 3R, where R is the Earth's radius. The Earth has mass 5.97 × 1024 kg and radius 6.37 × 106 m. G = 6.67 × 10−11 N m2 kg−2.
What is the minimum work done on the probe?
| Answer | Mark | Notes |
|---|---|---|
| W = mΔV = m(−GM/3R − (−GM/2R)) = GMm/6R = 6.67 × 10−11 × 5.97 × 1024 × 500 ÷ (6 × 6.37 × 106) = 5.21 × 109 J. | ✓ 1 | Answer A |
Answer: A
Exam tip: Work done = mass × (final potential − initial potential). Write both potentials with their minus signs before subtracting.
Equipotential surfaces around the Earth are drawn at −25.0 MJ kg−1 and at −20.0 MJ kg−1. The Earth has mass 5.97 × 1024 kg and radius 6.37 × 106 m. G = 6.67 × 10−11 N m2 kg−2.
Calculate the distance from the centre of the Earth of the −25.0 MJ kg−1 equipotential.
Calculate the height of this equipotential above the Earth's surface.
The −20.0 MJ kg−1 equipotential is 1.99 × 107 m from the centre. Explain why the gap between the −20.0 and −25.0 MJ kg−1 equipotentials is larger than the gap between the −25.0 and −30.0 MJ kg−1 equipotentials.
A satellite of mass 800 kg moves from the −25.0 MJ kg−1 equipotential to the −20.0 MJ kg−1 equipotential. Calculate the increase in its gravitational potential energy and state the work done by gravity while it then moves along the −20.0 MJ kg−1 equipotential.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| r = GM/|V| = 6.67 × 10−11 × 5.97 × 1024 ÷ 2.50 × 107 | ✓ 1 | |
| = 1.59 × 107 m | ✓ 1 | |
| Part (b) | ||
| 1.59 × 107 − 6.37 × 106 = 9.56 × 106 m | ✓ 1 | |
| Part (c) | ||
| the field strength is the potential gradient / g = −ΔV/Δr | ✓ 1 | |
| further out the field is weaker, so a greater distance is needed for the same 5.0 MJ kg−1 change in potential | ✓ 1 | |
| Part (d) | ||
| mΔV = 800 × 5.0 × 106 = 4.0 × 109 J | ✓ 1 | |
| zero, because ΔV = 0 along an equipotential | ✓ 1 | |
Exam tip: Rearrange V = −GM/r as r = GM/|V| and convert MJ kg−1 to J kg−1 first.
A communications satellite is to be placed in an orbit around Mars so that it stays above the same point on the Martian surface. Mars has mass 6.42 × 1023 kg and radius 3.39 × 106 m, and rotates once every 8.86 × 104 s. G = 6.67 × 10−11 N m2 kg−2.
State two conditions, other than its radius, that the orbit must satisfy.
Show that the orbital radius must be about 2.0 × 107 m.
Calculate the height of the orbit above the surface of Mars.
A second satellite of twice the mass is to be placed in the same type of orbit. Explain whether it needs a different orbital radius.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| period equal to the rotation period of Mars, 8.86 × 104 s | ✓ 1 | |
| orbit in the plane of the equator of Mars, moving in the same direction as Mars rotates | ✓ 1 | |
| Part (b) | ||
| GMm/r2 = m(2π/T)2r, so r3 = GMT2/4π2 | ✓ 1 | |
| r3 = 6.67 × 10−11 × 6.42 × 1023 × (8.86 × 104)2 ÷ 4π2 = 8.51 × 1021 m3 | ✓ 1 | |
| r = 2.04 × 107 m | ✓ 1 | |
| Part (c) | ||
| 2.04 × 107 − 3.39 × 106 = 1.70 × 107 m | ✓ 1 | |
| Part (d) | ||
| no: the satellite's mass cancels in GMm/r2 = mω2r | ✓ 1 | |
| so r depends only on the mass of Mars and the period, which are the same | ✓ 1 | |
Exam tip: In a “show that”, give the answer to one more significant figure than the value in the question (2.04 × 107 m, not 2.0 × 107 m).
Astronomers observe four moons in circular orbits around a distant planet. Their results are:
orbital radius r / 108 m: 2.00, 3.50, 5.00, 8.00 — orbital period T / days: 0.514, 1.19, 2.03, 4.11.
G = 6.67 × 10−11 N m2 kg−2; 1 day = 8.64 × 104 s.
Use the data for the innermost and outermost moons to show that the results are consistent with T2 ∝ r3.
Use the data for the moon at 5.00 × 108 m to calculate the mass of the planet.
A fifth moon is discovered with an orbital radius of 1.20 × 109 m. Predict its orbital period in days.
The uncertainty in each orbital radius is ±1.0% and in each period is ±0.5%. Calculate the percentage uncertainty in the mass of the planet.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| T2/r3 = 0.5142 ÷ (2.00 × 108)3 and 4.112 ÷ (8.00 × 108)3 | ✓ 1 | |
| both = 3.30 × 10−26 days2 m−3, so the ratio is constant | ✓ 1 | |
| Part (b) | ||
| T = 2.03 × 8.64 × 104 = 1.754 × 105 s | ✓ 1 | |
| M = 4π2r3/GT2 = 4π2 × (5.00 × 108)3 ÷ (6.67 × 10−11 × (1.754 × 105)2) | ✓ 1 | |
| = 2.41 × 1027 kg (2.40 × 1027 kg if T is rounded to 1.754 × 105 s; accept 2.40–2.42 × 1027 kg) | ✓ 1 | |
| Part (c) | ||
| T ∝ r3/2: T = 2.03 × (1.20 × 109 ÷ 5.00 × 108)3/2 | ✓ 1 | |
| = 7.55 days | ✓ 1 | |
| Part (d) | ||
| M ∝ r3/T2: 3 × 1.0% + 2 × 0.5% = 4.0% | ✓ 1 | |
Exam tip: When a quantity is raised to a power, multiply its percentage uncertainty by that power, then add the contributions.
The centres of a planet X and its moon Y are 4.0 × 108 m apart. Figure 1 shows how the resultant gravitational field strength due to X and Y varies with distance from the centre of X, along the line joining their centres.
What is the ratio (mass of X) ÷ (mass of Y)?
| Answer | Mark | Notes |
|---|---|---|
| At N the resultant field is zero, so the two field strengths are equal: GMX/x2 = GMY/(d − x)2. N is 3.2 × 108 m from X and 0.8 × 108 m from Y, so MX/MY = (3.2/0.8)2 = 16. | ✓ 1 | Answer C |
Answer: C
Exam tip: The point of zero resultant field is always closer to the smaller mass. The mass ratio is the square of the ratio of the two distances.
Figure 1 shows how the square of the orbital period, T2, varies with the cube of the orbital radius, r3, for several moons of a planet. The moons move in circular orbits. G = 6.67 × 10−11 N m2 kg−2.
What is the mass of the planet?
| Answer | Mark | Notes |
|---|---|---|
| T2 = (4π2/GM)r3, so the gradient is 4π2/GM. Gradient = 6.0 × 1010 ÷ 5.0 × 1025 = 1.2 × 10−15 s2 m−3. M = 4π2 ÷ (6.67 × 10−11 × 1.2 × 10−15) = 4.9 × 1026 kg. | ✓ 1 | Answer D |
Answer: D
Common error: Dropping a power of ten from an axis label. Write each coordinate in full, here 5.0 × 1025 m3 and 6.0 × 1010 s2, before finding the gradient.
Figure 1 shows how the gravitational potential V varies with distance r from the centre of a planet. Point S is on the planet's surface. A tangent to the curve has been drawn at point T. G = 6.67 × 10−11 N m2 kg−2.
Explain why the gravitational potential is negative at every point.
Use the tangent at T to determine the gravitational field strength at r = 2.0 × 107 m.
Use point T to calculate the mass of the planet.
A probe of mass 1200 kg rests on the surface at S. Calculate the minimum energy needed for the probe to escape from the planet's gravitational field. Ignore the planet's rotation and any atmosphere.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the potential is zero at infinity and gravity attracts, so work must be done on a mass to move it from any point out to infinity | ✓ 1 | |
| Part (b) | ||
| g = magnitude of the potential gradient = (0 − (−40)) × 106 ÷ (4.0 × 107 − 0) | ✓ 1 | |
| g = 1.0 N kg−1 (allow 0.95–1.05) | ✓ 1 | |
| Part (c) | ||
| V = −GM/r, so M = |V|r/G = 20 × 106 × 2.0 × 107 ÷ 6.67 × 10−11 | ✓ 1 | |
| = 6.0 × 1024 kg | ✓ 1 | |
| Part (d) | ||
| energy = m(0 − VS) = 1200 × 80 × 106 | ✓ 1 | |
| = 9.6 × 1010 J | ✓ 1 | |
Exam tip: The field strength is the (negative) gradient of the V–r graph. Take the gradient from two well-separated points on the tangent, with MJ kg−1 and 107 m written in full.
A satellite of mass 800 kg is moved from one circular orbit around a planet to a higher circular orbit. Figure 1 shows how its gravitational potential energy Ep varies with its distance r from the centre of the planet. Points X and Y mark orbit 1 and orbit 2.
State the increase in the gravitational potential energy of the satellite.
Use point X and Ep = −GMm/r to show that GM for the planet is 3.0 × 1014 N m2 kg−1, where M is the mass of the planet.
Calculate the orbital speed of the satellite in orbit 1 and in orbit 2.
Calculate the change in the kinetic energy of the satellite.
Hence calculate the minimum energy that the satellite's engines must supply.
Calculate the orbital period of the satellite in orbit 2.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| −0.80 × 1010 − (−2.0 × 1010) = 1.2 × 1010 J | ✓ 1 | |
| Part (b) | ||
| GM = |Ep|r/m = 2.0 × 1010 × 1.2 × 107 ÷ 800 = 3.0 × 1014 N m2 kg−1 | ✓ 1 | |
| Part (c) | ||
| GMm/r2 = mv2/r, so v = √(GM/r); orbit 1: v = √(3.0 × 1014 ÷ 1.2 × 107) = 5.0 × 103 m s−1 | ✓ 1 | |
| orbit 2: v = √(3.0 × 1014 ÷ 3.0 × 107) = 3.16 × 103 m s−1 | ✓ 1 | |
| Part (d) | ||
| kinetic energy = ½ × 800 × (5.0 × 103)2 = 1.0 × 1010 J in orbit 1 and ½ × 800 × (3.16 × 103)2 = 4.0 × 109 J in orbit 2 | ✓ 1 | |
| a decrease of 6.0 × 109 J (allow error carried forward from (c)) | ✓ 1 | |
| Part (e) | ||
| increase in Ep − decrease in kinetic energy = 1.2 × 1010 − 6.0 × 109 = 6.0 × 109 J | ✓ 1 | |
| Part (f) | ||
| T = 2πr/v = 2π × 3.0 × 107 ÷ 3.16 × 103 = 5.96 × 104 s (allow 5.96–5.97 × 104 s) | ✓ 1 | |
Exam tip: Raising an orbit increases the potential energy but decreases the kinetic energy, because the orbital speed √(GM/r) is smaller in a larger orbit. The energy supplied is the change in the total energy, not the change in potential energy alone.
Five moons move in circular orbits around a planet of mass M. Figure 1 is a graph of lg (T/s) against lg (r/m), where T is the orbital period and r is the orbital radius of each moon. G = 6.67 × 10−11 N m2 kg−2.
Show that T2 = (4π2/GM)r3.
Explain why the graph is a straight line, and use the labelled points to show that its gradient is 1.5.
Use the point (9.00, 5.80) to calculate the mass of the planet.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the gravitational force provides the centripetal force: GMm/r2 = mv2/r | ✓ 1 | |
| substitutes v = 2πr/T: GM/r = 4π2r2/T2 | ✓ 1 | |
| rearranges to T2 = (4π2/GM)r3 | ✓ 1 | |
| Part (b) | ||
| taking logs: lg T = 1.5 lg r + lg (2π/√(GM)), which has the form y = mx + c with a constant intercept | ✓ 1 | |
| gradient = (5.80 − 4.30) ÷ (9.00 − 8.00) = 1.50 | ✓ 1 | |
| Part (c) | ||
| r = 109.00 = 1.0 × 109 m and T = 105.80 = 6.31 × 105 s | ✓ 1 | |
| M = 4π2r3/GT2 = 4π2 × (1.0 × 109)3 ÷ (6.67 × 10−11 × (6.31 × 105)2) = 1.49 × 1027 kg (allow 1.4–1.6 × 1027 kg) | ✓ 1 | |
Exam tip: For a power law y = kxn, a graph of lg y against lg x is a straight line of gradient n. Here T ∝ r3/2, so the gradient is 1.5.
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