Gravitational field of a point mass — Paper 4 · A Level Physics 9702

1 The definitions of electric potential and of gravitational potential at a point have some similarity. (a) State one similarity between these two definitions. … … [1] (b) Explain why values of gravitational potential are always negative whereas values of electric potential may be positive or negative. … … … … … [4]

5 marks

Mark scheme: 1 (a) either ratio of work done to mass/charge or work done moving unit mass/charge from infinity or both have zero potential at infinity B1 [1] (b) gravitational forces are (always attractive) B1 electric forces can be attractive or repulsive B1 for gravitational, work got out as masses come together /mass moves from infinity B1 for electric, work done on charges if same sign, work got out if opposite sign as charges come together B1 [4]

This question in 9702/41 Oct/Nov 2006

1 (a) Define gravitational field strength. … … [1] (b) A spherical planet has diameter 1.2 × 104 km. The gravitational field strength at the surface of the planet is 8.6 N kg–1. The planet may be assumed to be isolated in space and to have its mass concentrated at its centre. Calculate the mass of the planet. mass = … kg [3] (c) The gravitational potential at a point X above the surface of the planet in (b) is – 5.3 × 107 J kg–1. For point Y above the surface of the planet, the gravitational potential is – 6.8 × 107 J kg–1. (i) State, with a reason, whether point X or point Y is nearer to the planet. … … … [2] (ii) A rock falls radially from rest towards the planet from one point to the other. Calculate the final speed of the rock. speed = … m s–1 [2]

8 marks

Mark scheme: 1 (a) force per unit mass (ratio idea essential) B1 [1] (b) g = GM / R2 C1 8.6 × (0.6 × 107)2 = M × 6.67 × 10–11 C1 M = 4.6 × 1024 kg A1 [3] (c) (i) either potential decreases as distance from planet decreases or potential zero at infinity and X is closer to zero or potential α –1/r and Y more negative M1 so point Y is closer to planet. A1 [2] (ii) idea of ∆φ = ½v2 C1 (6.8 – 5.3) × 107 = ½v2 v = 5.5 × 103 ms–1 A1 [2]

This question in 9702/41 May/June 2009

1 (a) The Earth may be considered to be a uniform sphere of radius 6.38 × 103 km, with its mass concentrated at its centre. (i) Define gravitational field strength. … … [1] (ii) By considering the gravitational field strength at the surface of the Earth, show that the mass of the Earth is 5.99 × 1024 kg. [2] (b) The Global Positioning System (GPS) is a navigation system that can be used anywhere on Earth. It uses a number of satellites that orbit the Earth in circular orbits at a distance of 2.22 × 104 km above its surface. (i) Use data from (a) to calculate the angular speed of a GPS satellite in its orbit. angular speed = … rad s–1 [3] (ii) Use your answer in (i) to show that the satellites are not in geostationary orbits. For Examiner’s Use [3] (c) The planes of the orbits of the GPS satellites in (b) are inclined at an angle of 55° to the Equator. Suggest why the satellites are not in equatorial orbits. … … [1]

10 marks

Mark scheme: 1 (a) (i) force per (unit) mass ……(ratio idea essential) … B1 [1] (ii) g = GM / R2 … C1 9.81 = (6.67 × 10-11 × M) / (6.38 × 106)2 ……(all 3 s.f) … M1 M = 5.99 × 1024 kg … A0 [2] (b) (i) either GM = ω2r3 or gR2 = ω2r3 … C1 either 6.67 × 10-11 x 5.99 × 1024 = ω2 × (2.86 × 107)3 or 9.81 × (6.38 × 106)2 = ω2 × (2.86 × 107)3 … C1 ω = 1.3 × 10-4 rad s-1 … A1 [3] (use of r = 2.22 × 107m scores max 2 marks) (ii) period of orbit = 2π / ω … C1 = 4.8 × 104 s (= 13.4 hours) … A1 period for geostationary satellite is 24 hours (= 8.6 × 104 s) … A1 so no … A0 [3] (c) satellite can then provide cover at Poles … B1 [1] [Total: 10]

This question in 9702/42 Oct/Nov 2009

1 (a) Define gravitational potential at a point. … … … [2] (b) The Earth may be considered to be an isolated sphere of radius R with its mass concentrated at its centre. The variation of the gravitational potential φ with distance x from the centre of the Earth is shown in Fig. 1.1. distance x 0 R 2R 3R 4R 5R 0 –2.0 / 107 J kg–1 –4.0 –6.0 –8.0 Fig. 1.1 The radius R of the Earth is 6.4 × 106 m. (i) By considering the gravitational potential at the Earth’s surface, determine a value for the mass of the Earth. mass = … kg [3] (ii) A meteorite is at rest at infinity. The meteorite travels from infinity towards the For Earth. Examiner’s Use Calculate the speed of the meteorite when it is at a distance of 2R above the Earth’s surface. Explain your working. speed = … m s–1 [4] (iii) In practice, the Earth is not an isolated sphere because it is orbited by the Moon, as illustrated in Fig. 1.2. initial path of meteorite Moon Earth Fig. 1.2 (not to scale) The initial path of the meteorite is also shown. Suggest two changes to the motion of the meteorite caused by the Moon. 1. … … 2. … … [2]

11 marks

Mark scheme: 1 (a) work done moving unit mass M1 from infinity to the point A1 [2] (b) (i) at R, φ = 6.3 × 107 J kg–1 (allow ± 0.1 × 107) B1 φ = GM / R 6.3 × 107 = (6.67 × 10–11 × M) / (6.4 × 106) C1 M = 6.0 × 1024 kg (allow 5.95 → 6.14) A1 [3] Maximum of 2/3 for any value chosen for φ not at R (ii) change in potential = 2.1 × 107 J kg–1 (allow ± 0.1 × 107) C1 loss in potential energy = gain in kinetic energy B1 ½ mv 2 = φ m or ½ mv 2 = GM / 3R C1 ½ v 2 = 2.1 × 107 v = 6.5 × 103 m s–1 ………..……(allow 6.3 → 6.6) A1 [4] (answer 7.9 × 103 m s–1, based on x = 2R, allow max 3 marks) (iii) e.g. speed / velocity / acceleration would be greater B1 deviates / bends from straight path B1 [2] (any sensible ideas, 1 each, max 2)

This question in 9702/42 May/June 2010

1 (a) Define gravitational field strength. … … [1] (b) An isolated star has radius R. The mass of the star may be considered to be a point mass at the centre of the star. The gravitational field strength at the surface of the star is gs. On Fig. 1.1, sketch a graph to show the variation of the gravitational field strength of the star with distance from its centre. You should consider distances in the range R to 4R. 1.0gs 0.8gs gravitational field strength 0.6gs 0.4gs 0.2gs 0R 2R 3R 4R surface distance of star Fig. 1.1 [2] (c) The Earth and the Moon may be considered to be spheres that are isolated in space with their masses concentrated at their centres. The masses of the Earth and the Moon are 6.00 × 1024 kg and 7.40 × 1022 kg respectively. The radius of the Earth is RE and the separation of the centres of the Earth and the Moon is 60 RE, as illustrated in Fig. 1.2. RE Moon mass Earth 7.40 x 1022 kg mass 6.00 x 1024 kg 60 RE Fig. 1.2 (not to scale) (i) Explain why there is a point between the Earth and the Moon at which the For gravitational field strength is zero. Examiner’s Use … … … [2] (ii) Determine the distance, in terms of RE, from the centre of the Earth at which the gravitational field strength is zero. distance = … RE [3] (iii) On the axes of Fig. 1.3, sketch a graph to show the variation of the gravitational field strength with position between the surface of the Earth and the surface of the Moon. gravitational field strength 0 distance surface surface of Earth of Moon Fig. 1.3 [3]

11 marks

Mark scheme: 1 (a) force per unit mass (ratio idea essential) B1 [1] (b) graph: correct curvature M1 from (R,1.0 gS) & at least one other correct point A1 [2] (c) (i) fields of Earth and Moon are in opposite directions M1 either resultant field found by subtraction of the field strength or any other sensible comment A1 so there is a point where it is zero A0 [2] (allow FE = –FM for 2 marks) (ii) GME / x2 = GMM / (D – x)2 C1 (6.0 × 1024) / (7.4 × 1022) = x2 / (60RE – x)2 C1 x = 54 RE A1 [3] (iii) graph: g = 0 at least ⅔ distance to Moon B1 gE and gM in opposite directions M1 correct curvature (by eye) and gE > gM at surface A1 [3]

This question in 9702/41 Oct/Nov 2010

1 (a) Define gravitational field strength. … … [1] (b) An isolated star has radius R. The mass of the star may be considered to be a point mass at the centre of the star. The gravitational field strength at the surface of the star is gs. On Fig. 1.1, sketch a graph to show the variation of the gravitational field strength of the star with distance from its centre. You should consider distances in the range R to 4R. 1.0gs 0.8gs gravitational field strength 0.6gs 0.4gs 0.2gs 0R 2R 3R 4R surface distance of star Fig. 1.1 [2] (c) The Earth and the Moon may be considered to be spheres that are isolated in space with their masses concentrated at their centres. The masses of the Earth and the Moon are 6.00 × 1024 kg and 7.40 × 1022 kg respectively. The radius of the Earth is RE and the separation of the centres of the Earth and the Moon is 60 RE, as illustrated in Fig. 1.2. RE Moon mass Earth 7.40 x 1022 kg mass 6.00 x 1024 kg 60 RE Fig. 1.2 (not to scale) (i) Explain why there is a point between the Earth and the Moon at which the For gravitational field strength is zero. Examiner’s Use … … … [2] (ii) Determine the distance, in terms of RE, from the centre of the Earth at which the gravitational field strength is zero. distance = … RE [3] (iii) On the axes of Fig. 1.3, sketch a graph to show the variation of the gravitational field strength with position between the surface of the Earth and the surface of the Moon. gravitational field strength 0 distance surface surface of Earth of Moon Fig. 1.3 [3]

11 marks

Mark scheme: 1 (a) force per unit mass (ratio idea essential) B1 [1] (b) graph: correct curvature M1 from (R,1.0 gS) & at least one other correct point A1 [2] (c) (i) fields of Earth and Moon are in opposite directions M1 either resultant field found by subtraction of the field strength or any other sensible comment A1 so there is a point where it is zero A0 [2] (allow FE = –FM for 2 marks) (ii) GME / x2 = GMM / (D – x)2 C1 (6.0 × 1024) / (7.4 × 1022) = x2 / (60RE – x)2 C1 x = 54 RE A1 [3] (iii) graph: g = 0 at least ⅔ distance to Moon B1 gE and gM in opposite directions M1 correct curvature (by eye) and gE > gM at surface A1 [3]

This question in 9702/42 Oct/Nov 2010

1 (a) Define gravitational potential at a point. … … [1] (b) The gravitational potential φ at distance r from point mass M is given by the expression GM φ = – r where G is the gravitational constant. Explain the significance of the negative sign in this expression. … … … [2] (c) A spherical planet may be assumed to be an isolated point mass with its mass concentrated at its centre. A small mass m is moving near to, and normal to, the surface of the planet. The mass moves away from the planet through a short distance h. State and explain why the change in gravitational potential energy ΔEP of the mass is given by the expression ΔEP = mgh where g is the acceleration of free fall. … … … … … … [4] (d) The planet in (c) has mass M and diameter 6.8 × 103 km. The product GM for this planet For is 4.3 × 1013 N m2 kg–1. Examiner’s Use A rock, initially at rest a long distance from the planet, accelerates towards the planet. Assuming that the planet has negligible atmosphere, calculate the speed of the rock as it hits the surface of the planet. speed = … m s–1 [3]

10 marks

Mark scheme: 1 (a) work done in bringing unit mass from infinity (to the point) B1 [1] (b) gravitational force is (always) attractive B1 either as r decreases, object/mass/body does work or work is done by masses as they come together B1 [2] (c) either force on mass = mg (where g is the acceleration of free fall /gravitational field strength) B1 g = GM/r2 B1 if r @ h, g is constant B1 ∆EP = force × distance moved M1 = mgh A0 or ∆EP = m∆φ (C1) = GMm(1/r1 – 1/r2) = GMm(r2 – r1)/r1r2 (B1) if r2 ≈ r1, then (r2 – r1) = h and r1r2 = r2 (B1) g = GM/r2 (B1) ∆EP = mgh (A0) [4] (d) ½mv2 = m∆φ v2 = 2 × GM/r C1 = (2 × 4.3 × 1013) / (3.4 × 106) C1 v = 5.0 × 103 m s–1 A1 [3] (Use of diameter instead of radius to give v = 3.6 × 103 m s–1 scores 2 marks)

This question in 9702/41 May/June 2012

1 (a) Define gravitational potential at a point. … … [1] (b) The gravitational potential φ at distance r from point mass M is given by the expression GM φ = – r where G is the gravitational constant. Explain the significance of the negative sign in this expression. … … … [2] (c) A spherical planet may be assumed to be an isolated point mass with its mass concentrated at its centre. A small mass m is moving near to, and normal to, the surface of the planet. The mass moves away from the planet through a short distance h. State and explain why the change in gravitational potential energy ΔEP of the mass is given by the expression ΔEP = mgh where g is the acceleration of free fall. … … … … … … [4] (d) The planet in (c) has mass M and diameter 6.8 × 103 km. The product GM for this planet For is 4.3 × 1013 N m2 kg–1. Examiner’s Use A rock, initially at rest a long distance from the planet, accelerates towards the planet. Assuming that the planet has negligible atmosphere, calculate the speed of the rock as it hits the surface of the planet. speed = … m s–1 [3]

10 marks

Mark scheme: 1 (a) work done in bringing unit mass from infinity (to the point) B1 [1] (b) gravitational force is (always) attractive B1 either as r decreases, object/mass/body does work or work is done by masses as they come together B1 [2] (c) either force on mass = mg (where g is the acceleration of free fall /gravitational field strength) B1 g = GM/r2 B1 if r @ h, g is constant B1 ∆EP = force × distance moved M1 = mgh A0 or ∆EP = m∆φ (C1) = GMm(1/r1 – 1/r2) = GMm(r2 – r1)/r1r2 (B1) if r2 ≈ r1, then (r2 – r1) = h and r1r2 = r2 (B1) g = GM/r2 (B1) ∆EP = mgh (A0) [4] (d) ½mv2 = m∆φ v2 = 2 × GM/r C1 = (2 × 4.3 × 1013) / (3.4 × 106) C1 v = 5.0 × 103 m s–1 A1 [3] (Use of diameter instead of radius to give v = 3.6 × 103 m s–1 scores 2 marks)

This question in 9702/43 May/June 2012

1 (a) State what is meant by a gravitational field. … … … [2] (b) In the Solar System, the planets may be assumed to be in circular orbits about the Sun. Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in Fig. 1.1. radius of orbit / km Earth 1.50 × 108 Jupiter 7.78 × 108 Fig. 1.1 (i) State Newton’s law of gravitation. … … … … [3] (ii) Use Newton’s law to determine the ratio gravitational field strength due to the Sun at orbit of Earth . gravitational field strength due to the Sun at orbit of Jupiter ratio = … [3] (c) The orbital period of the Earth about the Sun is T. For Examiner’s (i) Use ideas about circular motion to show that the mass M of the Sun is given by Use 4π2R 3 M = GT 2 where R is the radius of the Earth’s orbit about the Sun and G is the gravitational constant. Explain your working. [3] (ii) The orbital period T of the Earth about the Sun is 3.16 × 107 s. The radius of the Earth’s orbit is given in Fig. 1.1. Use the expression in (i) to determine the mass of the Sun. mass = … kg [2]

13 marks

Mark scheme: 1 (a) region of space area / volume B1 where a mass experiences a force B1 [2] (b) (i) force proportional to product of two masses M1 force inversely proportional to the square of their separation M1 either reference to point masses or separation >> ‘size’ of masses A1 [3] (ii) field strength = GM / x2 or field strength ∝ 1 / x2 C1 ratio = (7.78 × 108)2 / (1.5 × 108)2 C1 = 27 A1 [3] (c) (i) either centripetal force = mRω2 and ω = 2π / T or centripetal force = mv2 / R and v = 2πR /T B1 gravitational force provides the centripetal force B1 either GMm / R2 = mRω2 or GMm / R2 = mv2 / R M1 M = 4π2R3 / GT2 A0 [3] (allow working to be given in terms of acceleration) (ii) M = {4π2 × (1.5 × 1011)3} / {6.67 × 10–11 × (3.16 × 107)2} C1 = 2.0 × 1030 kg A1 [2]

This question in 9702/41 May/June 2013

1 (a) Explain what is meant by a geostationary orbit. … … … … [3] (b) A satellite of mass m is in a circular orbit about a planet. The mass M of the planet may be considered to be concentrated at its centre. Show that the radius R of the orbit of the satellite is given by the expression GMT 2 R3 = 4π2 where T is the period of the orbit of the satellite and G is the gravitational constant. Explain your working. [4] (c) The Earth has mass 6.0 × 1024 kg. Use the expression given in (b) to determine the radius of the geostationary orbit about the Earth. radius = … m [3]

10 marks

Mark scheme: 1 (a) equatorial orbit / above equator B1 satellite moves from west to east / same direction as Earth spins B1 period is 24 hours / same period as spinning of Earth B1 [3] (allow 1 mark for ‘appears to be stationary/overhead’ if none of above marks scored) (b) gravitational force provides/is the centripetal force B1 GMm/R2 = mRω2 or GMm/R2 = mv2/R M1 ω = 2π /T or v = 2πR / T or clear substitution M1 clear working to give R3 = (GMT2 / 4π2) A1 [4] (c) R3 = 6.67 × 10–11 × 6.0 × 1024 × (24 × 3600)2 / 4π2 C1 = 7.57 × 1022 C1 R = 4.2 × 107 m A1 [3] (missing out 3600 gives 1.8 × 105 m and scores 2/3 marks)

This question in 9702/42 May/June 2013

1 (a) State what is meant by a gravitational field. … … … [2] (b) In the Solar System, the planets may be assumed to be in circular orbits about the Sun. Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in Fig. 1.1. radius of orbit / km Earth 1.50 × 108 Jupiter 7.78 × 108 Fig. 1.1 (i) State Newton’s law of gravitation. … … … … [3] (ii) Use Newton’s law to determine the ratio gravitational field strength due to the Sun at orbit of Earth . gravitational field strength due to the Sun at orbit of Jupiter ratio = … [3] (c) The orbital period of the Earth about the Sun is T. For Examiner’s (i) Use ideas about circular motion to show that the mass M of the Sun is given by Use 4π2R 3 M = GT 2 where R is the radius of the Earth’s orbit about the Sun and G is the gravitational constant. Explain your working. [3] (ii) The orbital period T of the Earth about the Sun is 3.16 × 107 s. The radius of the Earth’s orbit is given in Fig. 1.1. Use the expression in (i) to determine the mass of the Sun. mass = … kg [2]

13 marks

Mark scheme: 1 (a) region of space area / volume B1 where a mass experiences a force B1 [2] (b) (i) force proportional to product of two masses M1 force inversely proportional to the square of their separation M1 either reference to point masses or separation >> ‘size’ of masses A1 [3] (ii) field strength = GM / x2 or field strength ∝ 1 / x2 C1 ratio = (7.78 × 108)2 / (1.5 × 108)2 C1 = 27 A1 [3] (c) (i) either centripetal force = mRω2 and ω = 2π / T or centripetal force = mv2 / R and v = 2πR /T B1 gravitational force provides the centripetal force B1 either GMm / R2 = mRω2 or GMm / R2 = mv2 / R M1 M = 4π2R3 / GT2 A0 [3] (allow working to be given in terms of acceleration) (ii) M = {4π2 × (1.5 × 1011)3} / {6.67 × 10–11 × (3.16 × 107)2} C1 = 2.0 × 1030 kg A1 [2]

This question in 9702/43 May/June 2013

1 (a) Define gravitational potential at a point. … … … [2] (b) The Moon may be considered to be an isolated sphere of radius 1.74 × 103 km with its mass of 7.35 × 1022 kg concentrated at its centre. (i) A rock of mass 4.50 kg is situated on the surface of the Moon. Show that the change in gravitational potential energy of the rock in moving it from the Moon’s surface to infinity is 1.27 × 107 J. [1] (ii) The escape speed of the rock is the minimum speed that the rock must be given when it is on the Moon’s surface so that it can escape to infinity. Use the answer in (i) to determine the escape speed. Explain your working. speed = … m s–1 [2] (c) The Moon in (b) is assumed to be isolated in space. The Moon does, in fact, orbit the Earth. State and explain whether the minimum speed for the rock to reach the Earth from the surface of the Moon is different from the escape speed calculated in (b). … … … [2]

7 marks

Mark scheme: 1 (a) work done in moving unit mass M1 from infinity (to the point) A1 [2] (b) (i) gravitational potential energy = GMm / x energy = (6.67 × 10–11 × 7.35 × 1022 × 4.5) / (1.74 × 106) M1 energy = 1.27 × 107 J A0 [1] (ii) change in grav. potential energy = change in kinetic energy B1 ½ × 4.5 × v2 = 1.27 × 107 v = 2.4 × 103 m s–1 A1 [2] (c) Earth would attract the rock / potential at Earth(’s surface) not zero / <0 / at Earth, potential due to Moon not zero M1 escape speed would be lower A1 [2]

This question in 9702/41 Oct/Nov 2013

1 (a) Define gravitational potential at a point. … … … [2] (b) The Moon may be considered to be an isolated sphere of radius 1.74 × 103 km with its mass of 7.35 × 1022 kg concentrated at its centre. (i) A rock of mass 4.50 kg is situated on the surface of the Moon. Show that the change in gravitational potential energy of the rock in moving it from the Moon’s surface to infinity is 1.27 × 107 J. [1] (ii) The escape speed of the rock is the minimum speed that the rock must be given when it is on the Moon’s surface so that it can escape to infinity. Use the answer in (i) to determine the escape speed. Explain your working. speed = … m s–1 [2] (c) The Moon in (b) is assumed to be isolated in space. The Moon does, in fact, orbit the Earth. State and explain whether the minimum speed for the rock to reach the Earth from the surface of the Moon is different from the escape speed calculated in (b). … … … [2]

7 marks

Mark scheme: 1 (a) work done in moving unit mass M1 from infinity (to the point) A1 [2] (b) (i) gravitational potential energy = GMm / x energy = (6.67 × 10–11 × 7.35 × 1022 × 4.5) / (1.74 × 106) M1 energy = 1.27 × 107 J A0 [1] (ii) change in grav. potential energy = change in kinetic energy B1 ½ × 4.5 × v2 = 1.27 × 107 v = 2.4 × 103 m s–1 A1 [2] (c) Earth would attract the rock / potential at Earth(’s surface) not zero / <0 / at Earth, potential due to Moon not zero M1 escape speed would be lower A1 [2]

This question in 9702/42 Oct/Nov 2013

1 An isolated spherical planet has a diameter of 6.8 × 106 m. Its mass of 6.4 × 1023 kg may be assumed to be a point mass at the centre of the planet. (a) Show that the gravitational field strength at the surface of the planet is 3.7 N kg−1. [2] (b) A stone of mass 2.4 kg is raised from the surface of the planet through a vertical height of 1800 m. Use the value of field strength given in (a) to determine the change in gravitational potential energy of the stone. Explain your working. change in energy = … J [3] (c) A rock, initially at rest at infinity, moves towards the planet. At point P, its height above the surface of the planet is 3.5 D, where D is the diameter of the planet, as shown in Fig. 1.1. D 3.5 D path of rock P planet Fig. 1.1 Calculate the speed of the rock at point P, assuming that the change in gravitational potential energy is all transferred to kinetic energy. speed = … m s−1 [4]

9 marks

Mark scheme: 1 (a) g = GM / R2 C1 = (6.67 × 10–11 × 6.4 × 1023) / (3.4 × 106)2 = 3.7 N kg–1 A1 [2] (b) ∆EP = mg∆h because ∆h ≪ R (or 1800 m ≪ 3.4 × 106 m) g is constant B1 ∆EP = 2.4 × 3.7 × 1800 C1 = 1.6 × 104 J A1 [3] (use of g = 9.8 m s–2 max. 1 for explanation) (c) gravitational potential energy = (–)GMm / x C1 v2 = 2GM / x C1 x = 4D = 4 × 6.8 × 106 C1 v2 = (2 × 6.67 × 10–11 × 6.4 × 1023) / (4 × 6.8 × 106) = 3.14 × 106 v = 1.8 × 103 m s–1 A1 [4] (use of 3.5 D giving 1.9 × 103 m s–1, allow max. 3)

This question in 9702/41 Oct/Nov 2014

1 An isolated spherical planet has a diameter of 6.8 × 106 m. Its mass of 6.4 × 1023 kg may be assumed to be a point mass at the centre of the planet. (a) Show that the gravitational field strength at the surface of the planet is 3.7 N kg−1. [2] (b) A stone of mass 2.4 kg is raised from the surface of the planet through a vertical height of 1800 m. Use the value of field strength given in (a) to determine the change in gravitational potential energy of the stone. Explain your working. change in energy = … J [3] (c) A rock, initially at rest at infinity, moves towards the planet. At point P, its height above the surface of the planet is 3.5 D, where D is the diameter of the planet, as shown in Fig. 1.1. D 3.5 D path of rock P planet Fig. 1.1 Calculate the speed of the rock at point P, assuming that the change in gravitational potential energy is all transferred to kinetic energy. speed = … m s−1 [4]

9 marks

Mark scheme: 1 (a) g = GM / R2 C1 = (6.67 × 10–11 × 6.4 × 1023) / (3.4 × 106)2 = 3.7 N kg–1 A1 [2] (b) ∆EP = mg∆h because ∆h ≪ R (or 1800 m ≪ 3.4 × 106 m) g is constant B1 ∆EP = 2.4 × 3.7 × 1800 C1 = 1.6 × 104 J A1 [3] (use of g = 9.8 m s–2 max. 1 for explanation) (c) gravitational potential energy = (–)GMm / x C1 v2 = 2GM / x C1 x = 4D = 4 × 6.8 × 106 C1 v2 = (2 × 6.67 × 10–11 × 6.4 × 1023) / (4 × 6.8 × 106) = 3.14 × 106 v = 1.8 × 103 m s–1 A1 [4] (use of 3.5 D giving 1.9 × 103 m s–1, allow max. 3)

This question in 9702/42 Oct/Nov 2014

2 (a) On the axes of Fig. 2.1, sketch the variation with distance from a point mass of the gravitational field strength due to the mass. gravitational field strength 0 0 distance Fig. 2.1 [2] (b) On the axes of Fig. 2.2, sketch the variation with speed of the magnitude of the force on a charged particle moving at right-angles to a uniform magnetic field. force 0 0 speed Fig. 2.2 [2] (c) On the axes of Fig. 2.3, sketch the variation with time of the power dissipated in a resistor by a sinusoidal alternating current during two cycles of the current. power 0 0 time Fig. 2.3 [3]

7 marks

Mark scheme: 2 (a) smooth curve with decreasing gradient, not starting at x = 0 M1 end of line not at g = 0 or horizontal A1 [2] (b) straight line with positive gradient M1 line starts at origin A1 [2] (c) sinusoidal shape B1 only positive values and peak / trough height constant B1 4 ‘loops’ B1 [3] 5 4

This question in 9702/43 Oct/Nov 2014

1 (a) State Newton’s law of gravitation. … … … [2] (b) Some of the planets in the Solar System have several moons (satellites) that have circular orbits about the planet. The planet and each of its moons may be considered to be point masses. Show that the radius x of a moon’s orbit is related to the period T of the orbit by the expression 4π2x3 GM = 2 T where G is the gravitational constant and M is the mass of the planet. Explain your working. [3] (c) The planet Neptune has eight moons, each in a circular orbit of radius x and period T. The variation with T 2 of x3 for some of the moons is shown in Fig. 1.1. 5.0 4.0 x 3 / 1014 km3 3.0 2.0 1.0 0 0 0.1 0.2 0.3 0.4 T 2 / day2 Fig. 1.1 Use Fig. 1.1 and the expression in (b) to determine the mass of Neptune. mass = … kg [4]

9 marks

Mark scheme: 1 (a) (gravitational) force proportional to product of masses and inversely proportional to square of separation M1 either point masses or particles or ‘size’ ≪ separation A1 [2] (b) gravitational force provides the centripetal force B1 either GMm / x2 = mxω2 or mv2 / x M1 either ω = 2π / T or v = 2πx / T and working to GM = 4π2x 3 / T 2 A1 [3] (c) either use of gradient of graph or line through origin so can use single point or line shown extrapolated to origin B1 gradient = (4.5 × 1014) / 0.35 6.67 × 10–11 × M = 4π2 × (4.5 × 1014 × 109) / (0.35 × {24 × 3600}2) correct conversion for km3 and power of 10 C1 correct conversion for day2 C1 M = 1.02 × 1026 kg A1 [4]

This question in 9702/43 Oct/Nov 2015

1 (a) Define gravitational field strength. … … [1] (b) The nearest star to the Sun is Proxima Centauri. This star has a mass of 2.5 × 1029 kg and is a distance of 4.0 × 1013 km from the Sun. The Sun has a mass of 2.0 × 1030 kg. (i) State why Proxima Centauri may be assumed to be a point mass when viewed from the Sun. … … [1] (ii) Calculate 1. the gravitational field strength due to Proxima Centauri at a distance of 4.0 × 1013 km, field strength = … N kg–1 [2] 2. the gravitational force of attraction between the Sun and Proxima Centauri. force = … N [2] (c) Suggest quantitatively why it may be assumed that the Sun is isolated in space from other stars. … … … [2] [Total: 8]

8 marks

Mark scheme: 1 (a) force per unit mass B1 [1] (b) (i) radius/diameter/size (of Proxima Centauri) ≪ /is much less than 4.0 × 1013 km/separation (of Sun and star) or (because) it is a uniform sphere B1 [1] (ii) 1. field strength = GM / x2 = (6.67 × 10–11 × 2.5 × 1029) / (4.0 × 1013 × 103)2 C1 = 1.0 × 10–14 N kg–1 A1 [2] 2. force = field strength × mass = 1.0 × 10–14 × 2.0 × 1030 C1 or force = GMm / x2 = (6.67 × 10–11 × 2.5 × 1029 × 2.0 × 1030) / (4.0 × 1013 × 103)2 (C1) = 2.0 × 1016 N A1 [2] (c) force (of 2 × 1016 N) would have little effect on (large) mass of Sun B1 would cause an acceleration of Sun of 1.0 × 10–14 m s–2/very small/negligible acceleration B1 [2] or many stars all around the Sun (B1) net effect of forces/fields is zero (B1)

This question in 9702/42 Oct/Nov 2016

1 (a) Define gravitational field strength. … … [1] (b) The mass of a spherical comet of radius 3.6 km is approximately 1.0 × 1013 kg. (i) Assuming that the comet has constant density, calculate the gravitational field strength on the surface of the comet. field strength = … N kg–1 [2] (ii) A probe having a weight of 960 N on Earth lands on the comet. Using your answer in (i), determine the weight of the probe on the surface of the comet. weight = … N [2] (c) A second comet has a length of approximately 4.5 km and a width of approximately 2.6 km. Its outline is illustrated in Fig. 1.1. Fig. 1.1 Suggest one similarity and one difference between the gravitational fields at the surface of this comet and at the surface of the comet in (b). similarity: … … difference: … … [2] [Total: 7]

7 marks

Mark scheme: 1(a) force per unit mass B1 1(b)(i) g = GM / r 2 = (6.67 ×10–11 × 1.0 × 1013) / (3.6 × 103)2 C1 = 5.1 × 10–5 N kg–1 A1 1(b)(ii) mass = (960 / 9.81) kg weight on comet = (960 / 9.81) × 5.1 ×10–5 C1 = 5.0 × 10–3 N A1 1(c) similarity: e.g. both attractive/pointed towards the comet e.g. same order of magnitude B1 difference: e.g. radial/non-radial e.g. same (over surface)/varies (over surface) B1

This question in 9702/42 May/June 2017

3 (a) Define gravitational field strength. … … [1] (b) Explain why, for changes in vertical position of a point mass near the Earth’s surface, the gravitational field strength may be considered to be constant. … … … … [2] (c) The orbit of the Earth about the Sun is approximately circular with a radius of 1.5 × 108 km. The time period of the orbit is 365 days. Determine a value for the mass M of the Sun. Explain your working. M = … kg [5] [Total: 8]

8 marks

Mark scheme: 3(a) force per unit mass B1 3(b) changes in height much less than radius of Earth M1 so (radial) field lines are almost parallel or g = GM / R2 ≈ GM / (R + h)2 A1 Question Answer Marks 3(c) gravitational force provides/is centripetal force B1 GMm / r2 = mv2 / r C1 v = (2π × 1.5 × 1011) / (3600 × 24 × 365) = 2.99 × 104 (m s–1) C1 6.67 × 10–11M = 1.5 × 1011 × (2.99 × 104)2 C1 M = 2.0 × 1030 kg A1 or GMm / r2 = mrω2 (C1) ω = 2π / (3600 × 24 × 365) = 1.99 × 10–7 (rad s–1) (C1) 6.67 × 10–11M = (1.5 × 1011)3 × (1.99 × 10–7)2 (C1) M = 2.0 × 1030 kg (A1) or T2 = 4π2r3 / GM (C2) M = 4π2 × (1.5 × 1011)3 / ({3600 × 24 × 365}2 × 6.67 × 10–11) (C1) = 2.0 × 1030 kg (A1)

This question in 9702/41 Oct/Nov 2017

3 (a) Define gravitational field strength. … … [1] (b) Explain why, for changes in vertical position of a point mass near the Earth’s surface, the gravitational field strength may be considered to be constant. … … … … [2] (c) The orbit of the Earth about the Sun is approximately circular with a radius of 1.5 × 108 km. The time period of the orbit is 365 days. Determine a value for the mass M of the Sun. Explain your working. M = … kg [5] [Total: 8]

8 marks

Mark scheme: 3(a) force per unit mass B1 3(b) changes in height much less than radius of Earth M1 so (radial) field lines are almost parallel or g = GM / R2 ≈ GM / (R + h)2 A1 Question Answer Marks 3(c) gravitational force provides/is centripetal force B1 GMm / r2 = mv2 / r C1 v = (2π × 1.5 × 1011) / (3600 × 24 × 365) = 2.99 × 104 (m s–1) C1 6.67 × 10–11M = 1.5 × 1011 × (2.99 × 104)2 C1 M = 2.0 × 1030 kg A1 or GMm / r2 = mrω2 (C1) ω = 2π / (3600 × 24 × 365) = 1.99 × 10–7 (rad s–1) (C1) 6.67 × 10–11M = (1.5 × 1011)3 × (1.99 × 10–7)2 (C1) M = 2.0 × 1030 kg (A1) or T2 = 4π2r3 / GM (C2) M = 4π2 × (1.5 × 1011)3 / ({3600 × 24 × 365}2 × 6.67 × 10–11) (C1) = 2.0 × 1030 kg (A1)

This question in 9702/43 Oct/Nov 2017

1 (a) (i) State what is meant by a line of force in a gravitational field. … … … [1] (ii) By reference to the pattern of the lines of gravitational force near to the surface of the Earth, explain why the acceleration of free fall near to the Earth’s surface is approximately constant. … … … … … [3] (b) The Moon may be considered to be a uniform sphere that is isolated in space. It has radius 1.74 × 103 km and mass 7.35 × 1022 kg. (i) Calculate the gravitational field strength at the Moon’s surface. gravitational field strength = … N kg–1 [2] (ii) A satellite is in a circular orbit about the Moon at a height of 320 km above its surface. Calculate the time for the satellite to complete one orbit of the Moon. time = … s [3] [Total: 9]

9 marks

Mark scheme: 1(a)(i) either direction of force on a (small test) mass or direction of acceleration of a (small test) mass B1 1(a)(ii) Any three from: • the lines are radial • near the surface the lines are (approximately) parallel • parallel lines so constant field strength • constant field strength hence constant acceleration of free fall B3 1(b)(i) g = GM / R2 g = (6.67 × 10–11 × 7.35 × 1022) / (1.74 × 103 × 103)2 C1 g = 1.62 N kg–1 A1 1(b)(ii) either xω2 = GM / x2 and ω = 2π / T or v2 / x = GM / x2 and v = 2πr / T C1 (1.74 × 106 + 320 × 103)3 × 4π2 / T2 = (6.67 × 10–11 × 7.35 × 1022) C1 T2 = 7.04 × 107 T = 8400 s (8390) A1

This question in 9702/42 Feb/March 2018

1 (a) (i) A gravitational field may be represented by lines of gravitational force. State what is meant by a line of gravitational force. … … … [1] (ii) By reference to lines of gravitational force near to the surface of the Earth, explain why the gravitational field strength g close to the Earth’s surface is approximately constant. … … … … … … [3] (b) The Moon may be considered to be a uniform sphere of diameter 3.4 × 103 km and mass 7.4 × 1022 kg. The Moon has no atmosphere. During a collision of the Moon with a meteorite, a rock is thrown vertically up from the surface of the Moon with a speed of 2.8 km s–1. Assuming that the Moon is isolated in space, determine whether the rock will travel out into distant space or return to the Moon’s surface. [4] [Total: 8]

8 marks

Mark scheme: 1(a)(i) direction of force on a (small test) mass or path in which a (small test) mass will move B1 1(a)(ii) (at surface,) lines (of force) are radial B1 Earth has large radius/height above surface is small so lines are (approximately) parallel B1 parallel lines → constant field strength B1 1(b) (change in) KE of rock = (change in) PE or ½mv2 = GMm / R C1 (m)v2 = (m)(2 × 6.67 × 10–11 × 7.4 × 1022) / (1.7 × 103 × 103) C1 v = 2.4 × 103 m s–1 A1 correct conclusion based on comparison of v with 2.8 km s–1 B1 or (change in) KE of rock = (change in) PE (C1) (at infinity) EP = (6.67 × 10–11 × 7.4 × 1022 × m) / (1.7 × 103 × 103) = 2.9 × 106 m (C1) EK of rock = ½ × m × (2.8 × 103)2 = 3.9 × 106 m (A1) correct conclusion based on comparison of EK and EP values (B1) or Question Answer Marks (change in) KE of rock = (change in) PE or ½mv2 = GMm / R (C1) (m) (2800)2 = (m) (2 × 6.67 × 10–11 × 7.4 × 1022) / R (C1) R = 1.3 × 103 km (A1) correct conclusion based on comparison of R with 1.7 × 103 km (B1) or (change in) KE of rock = (change in) PE or ½mv 2 = GMm / R (C1) (m) (2800)2 = (m) (2 × 6.67 × 10–11 × M) / (1.7 × 106) (C1) M = 1.0 × 1023 kg (A1) correct conclusion based on comparison of M with 7.4 × 1022 kg (B1)

This question in 9702/42 May/June 2018

1 (a) (i) State what is meant by gravitational field strength. … … … [1] (ii) Explain why, at the surface of a planet, gravitational field strength is numerically equal to the acceleration of free fall. … … … [1] (b) An isolated uniform spherical planet has radius R. The acceleration of free fall at the surface of the planet is g. On Fig. 1.1, sketch a graph to show the variation of the acceleration of free fall with distance x from the centre of the planet for values of x in the range x = R to x = 4R. 1.00 g acceleration of free fall 0.75 g 0.50 g 0.25 g 0 0 R 2R 3R 4R x Fig. 1.1 [3] (c) The planet in (b) has radius R equal to 3.4 × 103 km and mean density 4.0 × 103 kg m–3. Calculate the acceleration of free fall at a height R above its surface. acceleration of free fall = … m s–2 [3] [Total: 8]

8 marks

Mark scheme: 1(a)(i) force per unit mass B1 1(a)(ii) acceleration = F / m, field strength = F / m, so equal B1 1(b) smooth curve between R and 4R with negative gradient of decreasing magnitude B1 line passing through (R, 1.00g) and (2R, 0.25g) B1 line ending at (4R, 0.0625g) B1 1(c) M = (4 / 3 × πR3)ρ C1 g = GM / (2R)2 C1 g = ⅓ × 6.67 × 10–11 × π × 3.4 × 106 × 4.0 × 103 = 0.95 m s–2 A1

This question in 9702/42 Oct/Nov 2018

1 (a) Two point masses are isolated in space and are separated by a distance x. State an expression relating the gravitational force F between the two masses to the magnitudes M and m of the masses. State the name of any other symbol used. … … … [1] (b) A spacecraft is to be put into a circular orbit about a spherical planet. The planet may be considered to be isolated in space. The mass of the planet, assumed to be concentrated at its centre, is 7.5 × 1023 kg. The radius of the planet is 3.4 × 106 m. (i) The spacecraft is to orbit the planet at a height of 2.4 × 105 m above the surface of the planet. At this altitude, there is no atmosphere. Show that the speed of the spacecraft in its orbit is 3.7 × 103 m s –1. [2] (ii) One possible path of the spacecraft as it approaches the planet is shown in Fig. 1.1. A 3.64 × 106 m B 5.00 × 107 m planet mass 7.5 × 1023 kg Fig. 1.1 (not to scale) The spacecraft enters the orbit at point A with speed 3.7 × 103 m s–1. At point B, a distance of 5.00 × 107 m from the centre of the planet, the spacecraft has a speed of 4.1 × 103 m s–1. The mass of the spacecraft is 650 kg. For the spacecraft moving from point B to point A, show that the change in gravitational potential energy of the spacecraft is 8.3 × 109 J. [3] (c) By considering changes in gravitational potential energy and in kinetic energy of the spacecraft, determine whether the total energy of the spacecraft increases or decreases in moving from point B to point A. A numerical answer is not required. … … … … [2] [Total: 8]

8 marks

Mark scheme: 1(a) (F =) GMm / x2, where G is the (universal) gravitational constant B1 1(b)(i) GMm / x2 = mv2 / x or v2 = GM / x C1 v 2 = (6.67 × 10–11 × 7.5 × 1023) / (3.4 × 106 + 240 × 103) so v = 3.7 × 103 m s–1 A1 1(b)(ii) potential energy = (–)GMm / x C1 EA = (–)(6.67 × 10–11 × 7.5 × 1023 × 650) / (3.64 × 106) or EB = (–)(6.67 × 10–11 × 7.5 × 1023 × 650) / (5.00 × 107) M1 correct substitution and subtraction EB – EA shown, leading to ∆Ep = 8.3 × 109 J A1 or φ = (–)GM / x and potential energy = mφ (C1) ∆φ = (6.67 × 10–11 × 7.5 × 1023) × [(1 / (3.64 × 106)) – (1 / (5.00 × 107))] ( = 1.27 × 107 J kg–1) (M1) ∆Ep = 1.27 × 107 × 650 = 8.3 × 109 J (A1) 1(c) kinetic energy or potential energy decreases B1 kinetic energy and potential energy decrease so total energy decreases B1

This question in 9702/41 May/June 2019

1 (a) Two point masses are isolated in space and are separated by a distance x. State an expression relating the gravitational force F between the two masses to the magnitudes M and m of the masses. State the name of any other symbol used. … … … [1] (b) A spacecraft is to be put into a circular orbit about a spherical planet. The planet may be considered to be isolated in space. The mass of the planet, assumed to be concentrated at its centre, is 7.5 × 1023 kg. The radius of the planet is 3.4 × 106 m. (i) The spacecraft is to orbit the planet at a height of 2.4 × 105 m above the surface of the planet. At this altitude, there is no atmosphere. Show that the speed of the spacecraft in its orbit is 3.7 × 103 m s –1. [2] (ii) One possible path of the spacecraft as it approaches the planet is shown in Fig. 1.1. A 3.64 × 106 m B 5.00 × 107 m planet mass 7.5 × 1023 kg Fig. 1.1 (not to scale) The spacecraft enters the orbit at point A with speed 3.7 × 103 m s–1. At point B, a distance of 5.00 × 107 m from the centre of the planet, the spacecraft has a speed of 4.1 × 103 m s–1. The mass of the spacecraft is 650 kg. For the spacecraft moving from point B to point A, show that the change in gravitational potential energy of the spacecraft is 8.3 × 109 J. [3] (c) By considering changes in gravitational potential energy and in kinetic energy of the spacecraft, determine whether the total energy of the spacecraft increases or decreases in moving from point B to point A. A numerical answer is not required. … … … … [2] [Total: 8]

8 marks

Mark scheme: 1(a) (F =) GMm / x2, where G is the (universal) gravitational constant B1 1(b)(i) GMm / x2 = mv2 / x or v2 = GM / x C1 v 2 = (6.67 × 10–11 × 7.5 × 1023) / (3.4 × 106 + 240 × 103) so v = 3.7 × 103 m s–1 A1 1(b)(ii) potential energy = (–)GMm / x C1 EA = (–)(6.67 × 10–11 × 7.5 × 1023 × 650) / (3.64 × 106) or EB = (–)(6.67 × 10–11 × 7.5 × 1023 × 650) / (5.00 × 107) M1 correct substitution and subtraction EB – EA shown, leading to ∆Ep = 8.3 × 109 J A1 or φ = (–)GM / x and potential energy = mφ (C1) ∆φ = (6.67 × 10–11 × 7.5 × 1023) × [(1 / (3.64 × 106)) – (1 / (5.00 × 107))] ( = 1.27 × 107 J kg–1) (M1) ∆Ep = 1.27 × 107 × 650 = 8.3 × 109 J (A1) 1(c) kinetic energy or potential energy decreases B1 kinetic energy and potential energy decrease so total energy decreases B1

This question in 9702/43 May/June 2019

1 (a) (i) State what is meant by a field of force. … … … [2] (ii) Define gravitational field strength. … … [1] (b) An isolated planet may be assumed to be a uniform sphere of radius 3.39 × 106 m with its mass of 6.42 × 1023 kg concentrated at its centre. Calculate the gravitational field strength at the surface of the planet. field strength = … N kg–1 [3] (c) Calculate the height above the surface of the planet in (b) at which the gravitational field strength is 1.0% less than its value at the surface of the planet. height = … m [3] [Total: 9]

9 marks

Mark scheme: 1(a)(i) region (of space) B1 where a particle experiences a force B1 1(a)(ii) force per unit mass B1 1(b) g = GM / R2 C1 = (6.67 × 10–11 × 6.42 × 1023) / (3.39 × 106)2 C1 = 3.73 N kg–1 A1 Question Answer Marks 1(c) 0.99 × 3.73 = (6.67 × 10–11 × 6.42 × 1023) / r2 C1 r = 3.41 × 106 (m) C1 height = (r – R) = 2 × 104 m A1 or 0.99 × 3.73 = (6.67 × 10–11 × 6.42 × 1023) / (R + h)2 (R + h)2 = 1.1596 × 1013 (C1) R + h = 3.41 × 106 (m) (C1) h = 2 × 104 m (A1) or 0.99 = (3.39 × 106)2 / r2 (C1) r = 3.41 × 106 (m) (C1) height = 2 × 104 m (A1)

This question in 9702/41 Oct/Nov 2020

1 (a) (i) State what is meant by a field of force. … … … [2] (ii) Define gravitational field strength. … … [1] (b) An isolated planet may be assumed to be a uniform sphere of radius 3.39 × 106 m with its mass of 6.42 × 1023 kg concentrated at its centre. Calculate the gravitational field strength at the surface of the planet. field strength = … N kg–1 [3] (c) Calculate the height above the surface of the planet in (b) at which the gravitational field strength is 1.0% less than its value at the surface of the planet. height = … m [3] [Total: 9]

9 marks

Mark scheme: 1(a)(i) region (of space) B1 where a particle experiences a force B1 1(a)(ii) force per unit mass B1 1(b) g = GM / R2 C1 = (6.67 × 10–11 × 6.42 × 1023) / (3.39 × 106)2 C1 = 3.73 N kg–1 A1 Question Answer Marks 1(c) 0.99 × 3.73 = (6.67 × 10–11 × 6.42 × 1023) / r2 C1 r = 3.41 × 106 (m) C1 height = (r – R) = 2 × 104 m A1 or 0.99 × 3.73 = (6.67 × 10–11 × 6.42 × 1023) / (R + h)2 (R + h)2 = 1.1596 × 1013 (C1) R + h = 3.41 × 106 (m) (C1) h = 2 × 104 m (A1) or 0.99 = (3.39 × 106)2 / r2 (C1) r = 3.41 × 106 (m) (C1) height = 2 × 104 m (A1)

This question in 9702/43 Oct/Nov 2020

1 (a) Define gravitational field strength. … … [1] (b) An isolated planet is a uniform sphere of radius 3.39 × 106 m. Its mass of 6.42 × 1023 kg may be considered to be a point mass concentrated at its centre. The planet rotates about its axis with a period of 24.6 hours. For an object resting on the surface of the planet at the equator, calculate, to three significant figures: (i) the gravitational field strength field strength = … N kg−1 [2] (ii) the centripetal acceleration acceleration = … m s−2 [2] (iii) the force per unit mass exerted on the object by the surface of the planet. force per unit mass = … N kg−1 [1] [Total: 6]

6 marks

Mark scheme: 1(a) (gravitational) force per unit mass B1 1(b)(i) g = GM / r2 C1 = (6.67 × 10–11 × 6.42 × 1023) / (3.39 × 106)2 = 3.73 N kg–1 A1 1(b)(ii) a = rω2 and ω = 2π / T or a = v2 / r and v = 2πr / T C1 a = 3.39 × 106 × (2π / (24.6 × 3600))2 = 0.0171 m s–2 A1 1(b)(iii) force per unit mass = 3.73 – 0.0171 = 3.71 N kg–1 A1

This question in 9702/42 May/June 2021

2 (a) Define gravitational potential. … … … [2] (b) The Earth E and the Moon M can both be considered as isolated point masses at their centres. The mass of the Earth is 5.98 × 1024 kg and the mass of the Moon is 7.35 × 1022 kg. The Earth and the Moon are separated by a distance of 3.84 × 108 m, as shown in Fig. 2.1. 3.84 × 108 m x P Earth E Moon M mass 5.98 × 1024 kg mass 7.35 × 1022 kg Fig. 2.1 (not to scale) P is a point, on the line joining the centres of E and M, where the resultant gravitational field strength is zero. Point P is at a distance x from the centre of the Earth. (i) Explain how it is possible for the gravitational field strength to be zero despite the presence of two large masses nearby. … … … [2] (ii) Show that x is approximately 3.5 × 108 m. [2] (iii) Calculate the gravitational potential φ at point P. φ = … J kg–1 [3] [Total: 9]

9 marks

Mark scheme: 2(a) work done per unit mass B1 (work done in) moving mass from infinity B1 2(b)(i) (gravitational) fields from the Earth and Moon are in opposite directions B1 (resultant is zero where gravitational) fields are equal (in magnitude) B1 2(b)(ii) g ∝ M / r2 C1 5.98 × 1024 / x2 = 7.35 × 1022 / (3.84 × 108 – x)2 leading to x = 3.5 × 108 (m) A1 2(b)(iii) φ (Earth) = (–)6.67 × 10–11 × (5.98 × 1024 / 3.5 × 108) and φ (Moon) = (–)6.67 × 10–11 × (7.35 × 1022 / 0.38 × 108) C1 φ = (–)6.67 × 10–11 × [(5.98 × 1024 / 3.5 × 108) + (7.35 × 1022 / 0.38 × 108)] C1 = – 1.3 × 106 J kg–1 A1

This question in 9702/41 Oct/Nov 2021

2 (a) State the relationship between gravitational potential and gravitational field strength. … … … [2] (b) A moon of mass M and radius R orbits a planet of mass 3M and radius 2R. At a particular time, the distance between their centres is D, as shown in Fig. 2.1. D x P planet moon mass 3M mass M radius 2R radius R Fig. 2.1 Point P is a point along the line between the centres of the planet and the moon, at a variable distance x from the centre of the planet. The variation with x of the gravitational potential φ at point P, for points between the planet and the moon, is shown in Fig. 2.2. φ 0 x 0 2R D – R Fig. 2.2 (i) Explain why φ is negative throughout the entire range x = 2R to x = D – R. … … … … [3] (ii) One of the features of Fig. 2.2 is that φ is negative throughout. Describe two other features of Fig. 2.2. 1. … … 2. … … [2] (iii) On Fig. 2.3, sketch the variation with x of the gravitational field strength g at point P between x = 2R and x = D – R. g 0 x 0 2R D – R Fig. 2.3 [3] [Total: 10]

10 marks

Mark scheme: 2(a) (gravitational) field strength equals (gravitational) potential gradient M1 reference to minus sign A1 2(b)(i) potential is zero at infinity B1 (gravitational) force is attractive B1 (test) mass getting closer (from infinity) loses potential energy B1 2(b)(ii) • potential at (surface of) planet is smaller than at (surface of) moon • potential gradient at (surface of) planet is smaller than at (surface of) moon • magnitude of potential varies inversely with distance from centre near the spheres • (point of) maximum potential is nearer to moon than planet Any two points, 1 mark each B2 2(b)(iii) sketch: one curve, starting with gradient of decreasing magnitude at 2R and finishing with gradient of increasing magnitude at D – R B1 field strength shown as zero (only) near the point of maximum potential B1 negative field strength near one sphere and positive field strength near the other B1

This question in 9702/42 Oct/Nov 2021

2 (a) Define gravitational potential. … … … [2] (b) The Earth E and the Moon M can both be considered as isolated point masses at their centres. The mass of the Earth is 5.98 × 1024 kg and the mass of the Moon is 7.35 × 1022 kg. The Earth and the Moon are separated by a distance of 3.84 × 108 m, as shown in Fig. 2.1. 3.84 × 108 m x P Earth E Moon M mass 5.98 × 1024 kg mass 7.35 × 1022 kg Fig. 2.1 (not to scale) P is a point, on the line joining the centres of E and M, where the resultant gravitational field strength is zero. Point P is at a distance x from the centre of the Earth. (i) Explain how it is possible for the gravitational field strength to be zero despite the presence of two large masses nearby. … … … [2] (ii) Show that x is approximately 3.5 × 108 m. [2] (iii) Calculate the gravitational potential φ at point P. φ = … J kg–1 [3] [Total: 9]

9 marks

Mark scheme: 2(a) work done per unit mass B1 (work done in) moving mass from infinity B1 2(b)(i) (gravitational) fields from the Earth and Moon are in opposite directions B1 (resultant is zero where gravitational) fields are equal (in magnitude) B1 2(b)(ii) g ∝ M / r2 C1 5.98 × 1024 / x2 = 7.35 × 1022 / (3.84 × 108 – x)2 leading to x = 3.5 × 108 (m) A1 2(b)(iii) φ (Earth) = (–)6.67 × 10–11 × (5.98 × 1024 / 3.5 × 108) and φ (Moon) = (–)6.67 × 10–11 × (7.35 × 1022 / 0.38 × 108) C1 φ = (–)6.67 × 10–11 × [(5.98 × 1024 / 3.5 × 108) + (7.35 × 1022 / 0.38 × 108)] C1 = – 1.3 × 106 J kg–1 A1

This question in 9702/43 Oct/Nov 2021

1 (a) The point P in Fig. 1.1 represents a point mass. On Fig. 1.1, draw lines to represent the gravitational field around P. P Fig. 1.1 [2] (b) A moon is in circular orbit around a planet. Explain why the path of the moon is circular. … … … … [2] (c) Many moons are in circular orbit about a planet. The angular velocity of a moon is ω when the orbit of the moon has a radius r about the planet. Fig. 1.2 shows the variation of r 3 with 1 / ω2 for these moons. 4 r3 / 1023 m3 3 2 1 0 0 1 2 3 4 5 6 1 2 / 107 rad–2 s2 ω Fig. 1.2 (i) Show that the mass M of the planet is given by the expression gradient M = G where G is the gravitational constant. [2] (ii) Use Fig. 1.2 and the expression in (c)(i) to show that the mass M of the planet is 1.0 × 1026 kg. [1] (iii) Determine the speed of a moon in orbit around the planet with an orbital radius of 1.2 × 108 m. speed = … m s–1 [3] [Total: 10]

10 marks

Mark scheme: 1(a) at least 4 straight radial lines to P B1 all arrows pointing along the lines towards P B1 1(b) Any 2 from: gravitational force provides the centripetal force (centripetal or gravitational) force has constant magnitude (centripetal or gravitational) force is perpendicular to velocity (of moon) / direction of motion (of moon) B2 1(c)(i) 2 2 GMm = mr r ω M1 3 2 r M= G ω and gradient = 3 2 r ω hence gradient M G = or r3 = GM × 1/ω2 so gradient = GM hence gradient M G = A1 1(c)(ii) M = 4.1 × 1023 / (6.0 × 107 × 6.67 × 10–11) = 1.0 × 1026 kg B1 Question Answer Marks 1(c)(iii) 2 2 GMm mv = r r 2 GM= v r C1 11 26 2 8 6.67 10 1.0 10 v = 1.2 10 − × × × × 2 7 1 v 5.6 10 m s− = × C1 1 v =7500 m s− A1

This question in 9702/42 Feb/March 2022

1 (a) (i) State Newton’s law of gravitation. … … … [2] (ii) Use Newton’s law of gravitation to show that the gravitational field strength g at a distance r away from a point mass M is given by GM g = . r 2 [2] (b) The Earth has a mass of 5.98 × 1024 kg and a radius of 6.37 × 106 m. The Moon has a mass of 7.35 × 1022 kg and a radius of 1.74 × 106 m. The Earth and the Moon can both be considered as point masses at their centres. Their centres are a distance of 3.84 × 108 m apart. (i) Show that the gravitational field strength at the surface of the Moon due to the mass of the Moon is 1.62 N kg–1. [1] (ii) Explain why there is a point X on the line between the centres of the Earth and the Moon where the resultant gravitational field strength due to the Earth and the Moon is zero. … … … [2] (iii) Calculate the distance x of point X from the centre of the Moon. x = … m [3] [Total: 10]

10 marks

Mark scheme: 1(a)(i) (gravitational) force is (directly) proportional to product of masses B1 force (between point masses) is inversely proportional to the square of their separation B1 1(a)(ii) g = F / m C1 F = GMm / r2 and so g = [GMm / r2] / m = GM / r2 A1 1(b)(i) g = (6.67  10–11  7.35  1022) / (1.74  106)2 = 1.62 N kg–1 A1 1(b)(ii) fields (due to Earth and the Moon) have equal magnitudes B1 fields (due to Earth and the Moon) are in opposite directions B1 1(b)(iii) distance of X from Earth = (3.84  108 – x) C1 (G ) 7.35  1022 / x2 = (G ) 5.98  1024 / (3.84  108 – x)2 C1 x = 3.8  107 m A1

This question in 9702/41 May/June 2022

1 (a) (i) State Newton’s law of gravitation. … … … [2] (ii) Use Newton’s law of gravitation to show that the gravitational field strength g at a distance r away from a point mass M is given by GM g = . r 2 [2] (b) The Earth has a mass of 5.98 × 1024 kg and a radius of 6.37 × 106 m. The Moon has a mass of 7.35 × 1022 kg and a radius of 1.74 × 106 m. The Earth and the Moon can both be considered as point masses at their centres. Their centres are a distance of 3.84 × 108 m apart. (i) Show that the gravitational field strength at the surface of the Moon due to the mass of the Moon is 1.62 N kg–1. [1] (ii) Explain why there is a point X on the line between the centres of the Earth and the Moon where the resultant gravitational field strength due to the Earth and the Moon is zero. … … … [2] (iii) Calculate the distance x of point X from the centre of the Moon. x = … m [3] [Total: 10]

10 marks

Mark scheme: 1(a)(i) (gravitational) force is (directly) proportional to product of masses B1 force (between point masses) is inversely proportional to the square of their separation B1 1(a)(ii) g = F / m C1 F = GMm / r2 and so g = [GMm / r2] / m = GM / r2 A1 1(b)(i) g = (6.67  10–11  7.35  1022) / (1.74  106)2 = 1.62 N kg–1 A1 1(b)(ii) fields (due to Earth and the Moon) have equal magnitudes B1 fields (due to Earth and the Moon) are in opposite directions B1 1(b)(iii) distance of X from Earth = (3.84  108 – x) C1 (G ) 7.35  1022 / x2 = (G ) 5.98  1024 / (3.84  108 – x)2 C1 x = 3.8  107 m A1

This question in 9702/43 May/June 2022

1 (a) (i) State what is indicated by the direction of the gravitational field line at a point in a gravitational field. … … [1] (ii) Explain, with reference to gravitational field lines, why the gravitational field near the surface of the Earth is approximately constant for small changes in height. … … … [2] (b) A large isolated uniform sphere has mass M and radius R. Point P lies on a straight line passing through the centre of the sphere, at a variable displacement x from the centre, as shown in Fig. 1.1. x P R uniform sphere, mass M Fig. 1.1 Fig. 1.2 shows the variation with x of the gravitational field g at point P due to the sphere for the values of x for which P is inside the sphere. 1.0Y g 0.5Y 0 – 3R – 2R – R 0 R 2R 3R x – 0.5Y – 1.0Y Fig. 1.2 The magnitude of the gravitational field at the surface of the sphere is Y. (i) Determine an expression for Y in terms of M and R. Identify any other symbols that you use. [2] (ii) Explain why, at the surface of the sphere, g always has the opposite sign to x. … … … [2] (iii) Complete Fig. 1.2 to show the variation of g with x for values of x, up to ±3R, for which point P is outside the sphere. [3] [Total: 10]

10 marks

Mark scheme: Question Answer Marks 1(a)(i) direction of the force acting on a (test) mass placed at the point B1 1(a)(ii) change in height negligible compared with radius (of Earth) B1 (so) field lines are (effectively) parallel B1 1(b)(i) Y = GM / R2 M1 G is the gravitational constant A1 1(b)(ii) gravitational force is (always) attractive B1 or gravitational force (always) acts towards the centre of the sphere force is in opposite direction to displacement B1 or at a point to the right of the centre, force acts to the left or at a point to the left of the centre, force acts to the right 1(b)(iii) sketch: smooth curve with decreasing positive gradient, starting at (R, –Y) and reaching 3R with g still negative B1 or smooth curve with increasing positive gradient, ending at (–R, Y) and reaching –3R with g still positive both of the above curves, in correct quadrants B1 curve passing through (2R, 0.25Y) and (3R, 0.11Y) B1

This question in 9702/41 Oct/Nov 2023

1 (a) (i) State what is indicated by the direction of the gravitational field line at a point in a gravitational field. … … [1] (ii) Explain, with reference to gravitational field lines, why the gravitational field near the surface of the Earth is approximately constant for small changes in height. … … … [2] (b) A large isolated uniform sphere has mass M and radius R. Point P lies on a straight line passing through the centre of the sphere, at a variable displacement x from the centre, as shown in Fig. 1.1. x P R uniform sphere, mass M Fig. 1.1 Fig. 1.2 shows the variation with x of the gravitational field g at point P due to the sphere for the values of x for which P is inside the sphere. 1.0Y g 0.5Y 0 – 3R – 2R – R 0 R 2R 3R x – 0.5Y – 1.0Y Fig. 1.2 The magnitude of the gravitational field at the surface of the sphere is Y. (i) Determine an expression for Y in terms of M and R. Identify any other symbols that you use. [2] (ii) Explain why, at the surface of the sphere, g always has the opposite sign to x. … … … [2] (iii) Complete Fig. 1.2 to show the variation of g with x for values of x, up to ±3R, for which point P is outside the sphere. [3] [Total: 10]

10 marks

Mark scheme: Question Answer Marks 1(a)(i) direction of the force acting on a (test) mass placed at the point B1 1(a)(ii) change in height negligible compared with radius (of Earth) B1 (so) field lines are (effectively) parallel B1 1(b)(i) Y = GM / R2 M1 G is the gravitational constant A1 1(b)(ii) gravitational force is (always) attractive B1 or gravitational force (always) acts towards the centre of the sphere force is in opposite direction to displacement B1 or at a point to the right of the centre, force acts to the left or at a point to the left of the centre, force acts to the right 1(b)(iii) sketch: smooth curve with decreasing positive gradient, starting at (R, –Y) and reaching 3R with g still negative B1 or smooth curve with increasing positive gradient, ending at (–R, Y) and reaching –3R with g still positive both of the above curves, in correct quadrants B1 curve passing through (2R, 0.25Y) and (3R, 0.11Y) B1

This question in 9702/43 Oct/Nov 2023

2 The Sun may be considered as a uniform sphere with a mass of 1.99 × 1030 kg and a surface temperature of 5780 K. A probe with a mass of 2.63 kg moves in a straight line towards the Sun. When it is at a distance x from the centre of the Sun, the probe measures the gravitational field strength g due to the Sun and the radiant flux intensity F of radiation from the Sun. (a) Define gravitational field. … … [1] (b) For the position of the probe where x = 1.47 × 1011 m: (i) calculate g g = … N kg–1 [2] (ii) determine the gravitational potential energy EP of the probe. EP = … J [2] (c) (i) Show that, for any particular value of x, the numerical values of g and F are related by 4πGM g = F L where M is the mass of the Sun, L is the luminosity of the Sun and G is the gravitational constant. [3] (ii) Fig. 2.1 shows the variation of g with F. 8 g / 10–3 N kg–1 4 0 0 0.5 1.0 1.5 2.0 F / 103 W m–2 Fig. 2.1 Determine a value for the luminosity L of the Sun. Give a unit with your answer. L = … unit … [2] (iii) Use your answer in (c)(ii) to determine the radius r of the Sun. r = … m [2] [Total: 12]

12 marks

Mark scheme: 2(a) force per unit mass B1 2(b)(i) g = GM / x2 C1 = (6.67  10–11  1.99  1030) / (1.47  1011)2 A1 = 6.14  10–3 N kg– 1 2(b)(ii) EP = – GMm / x C1 = – (6.67  10–11  1.99  1030  2.63) / (1.47  1011) = – 2.37  109 J A1 2(c)(i) F = L / 4x2 C1 (g = GM / x2 and so) x2 = GM / g M1 and x2 = L / 4F elimination of x and subsequent algebra shown leading to g = 4GMF / L A1 2(c)(ii) correct read-off of pair of values of g and F and full substitution of values of g, G, M and F into equation C1 e.g. L = (4  6.67  10–11  1.99  1030  1.83  103) / (8.0  10–3) L = 3.8  1026 W A1 2(c)(iii) L = 4r2T4 C1 3.8  1026 = (4  5.67  10–8  57804)  r2 r = 6.9  108 m A1

This question in 9702/42 Oct/Nov 2024

1 (a) Define gravitational field. … … [1] (b) The gravitational field strength g at a distance x from the centre of a uniform spherical planet of mass M is given by the expression GM g = x2 where G is the gravitational constant and distance x is greater than the radius of the planet. (i) Describe the pattern of the field lines outside the planet that represent the gravitational field due to the planet. … … … [2] (ii) Explain why, for small changes in vertical height near the surface of the planet, g may be assumed to be constant. … … … [2] (c) Assume that the Earth is a uniform sphere. For the Earth, the product GM is equal to 3.99 × 1014 m3 s–2. (i) Determine a value, to three significant figures, for the radius R of the Earth. R = … m [2] (ii) Calculate the gravitational potential at the Earth’s surface. Give a unit with your answer. gravitational potential = … unit … [2] (d) Explain why the gravitational potential energy of two point masses is always negative. … … … [2] [Total: 11]

11 marks

Mark scheme: Question Answer Marks 1(a) force per unit mass B1 1(b)(i) radial B1 towards (centre of) planet B1 1(b)(ii) (changes in) height (very) much smaller than radius B1 (radius + height)2  radius2 B1 or field lines are approximately parallel 1(c)(i) 9.81  R2 = 3.99  1014 C1 R = 6.38  106 m A1 1(c)(ii) gravitational potential = – (GM / R) C1 = – (3.99  1014) / (6.38  106) = – 6.25  107 J kg–1 A1 1(d) potential (energy) zero at infinite separation B1 (gravitational) force is attractive B1

This question in 9702/44 May/June 2025

3 (a) Define gravitational field at a point. … … [1] (b) Fig. 3.1 shows an isolated point mass of mass M. mass M P x Fig. 3.1 Point P is at distance x from the point mass. (i) By considering the force exerted by the point mass on a test mass of mass m placed at P, derive an equation for the gravitational field strength g at P, in terms of M and x. Identify any other symbols you use. [2] (ii) On Fig. 3.1, draw an arrow to indicate the direction of the gravitational field at P. [1] x (iii) Point Q is at distance from the point mass, on the opposite side of the mass from P, as 2 shown in Fig. 3.2. Q mass M P x 2 x Fig. 3.2 Compare the gravitational field at Q with that at P. … … … [2] (c) Two identical isolated uniform spheres X and Y each have radius R. The centres of the spheres are separated by distance L, as shown in Fig. 3.3. X Y P x L Fig. 3.3 Point P lies on the line joining the centres of X and Y, and is at a variable displacement x from the centre of sphere X. The gravitational field strength at the surface of each sphere is g0. On Fig. 3.4, sketch the variation with x of the gravitational field g at point P between x = R and x = L – R. g0 g 1 2 g0 0 R L / 2 L – R x – 1 2 g0 –g0 Fig. 3.4 [3] [Total: 9]

9 marks

Mark scheme: 3(a) force per unit mass B1 3(b)(i) F = GMm / x2 C1 g = F / m A1 g = [GMm / x2] / m = GM / x2 and G = gravitational constant 3(b)(ii) arrow drawn at P pointing directly towards the point mass B1 3(b)(iii) fields are in opposite directions B1 field strength at Q is four times the field strength at P B1 3(c) line starting at (R, –g0) and ending at (L – R, +g0) B1 line passing through (L / 2, 0) B1 curve becoming shallower from R to (L / 2) and then steeper from (L / 2) to (L – R) B1

This question in 9702/41 Oct/Nov 2025

3 (a) Define gravitational field at a point. … … [1] (b) Fig. 3.1 shows an isolated point mass of mass M. mass M P x Fig. 3.1 Point P is at distance x from the point mass. (i) By considering the force exerted by the point mass on a test mass of mass m placed at P, derive an equation for the gravitational field strength g at P, in terms of M and x. Identify any other symbols you use. [2] (ii) On Fig. 3.1, draw an arrow to indicate the direction of the gravitational field at P. [1] x (iii) Point Q is at distance from the point mass, on the opposite side of the mass from P, as 2 shown in Fig. 3.2. Q mass M P x 2 x Fig. 3.2 Compare the gravitational field at Q with that at P. … … … [2] (c) Two identical isolated uniform spheres X and Y each have radius R. The centres of the spheres are separated by distance L, as shown in Fig. 3.3. X Y P x L Fig. 3.3 Point P lies on the line joining the centres of X and Y, and is at a variable displacement x from the centre of sphere X. The gravitational field strength at the surface of each sphere is g0. On Fig. 3.4, sketch the variation with x of the gravitational field g at point P between x = R and x = L – R. g0 g 1 2 g0 0 R L / 2 L – R x – 1 2 g0 –g0 Fig. 3.4 [3] [Total: 9]

9 marks

Mark scheme: 3(a) force per unit mass B1 3(b)(i) F = GMm / x2 C1 g = F / m A1 g = [GMm / x2] / m = GM / x2 and G = gravitational constant 3(b)(ii) arrow drawn at P pointing directly towards the point mass B1 3(b)(iii) fields are in opposite directions B1 field strength at Q is four times the field strength at P B1 3(c) line starting at (R, –g0) and ending at (L – R, +g0) B1 line passing through (L / 2, 0) B1 curve becoming shallower from R to (L / 2) and then steeper from (L / 2) to (L – R) B1

This question in 9702/43 Oct/Nov 2025