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

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