Gravitational force between point masses — Paper 4 · A Level Physics 9702

1 The Earth may be considered to be a uniform sphere with its mass M concentrated at its centre. A satellite of mass m orbits the Earth such that the radius of the circular orbit is r. (a) Show that the linear speed v of the satellite is given by the expression ⎛GM⎞ v = . √ ⎝ r ⎠ [2] (b) For this satellite, write down expressions, in terms of G, M, m and r, for (i) its kinetic energy, kinetic energy = …………………………. [1] (ii) its gravitational potential energy, potential energy = …………………………. [1] (iii) its total energy. total energy = …………………………. [2] Use (c) The total energy of the satellite gradually decreases. State and explain the effect of this decrease on (i) the radius r of the orbit, … … … [2] (ii) the linear speed v of the satellite. … … … [2]

10 marks

Mark scheme: 1 (a) centripetal force is provided by gravitational force B1 mv2 / r = GMm / r2 B1 hence v = √(GM / r) A0 [2] (b) (i) EK (= ½mv2) = GMm / 2r B1 [1] (ii) EP = - GMm / r B1 [1] (iii) ET = - GMm / r + GMm / 2r C1 = - GMm / 2r. A1 [2] (c) (i) if ET decreases then - GMm / 2r becomes more negative or GMm / 2r becomes larger M1 so r decreases A1 [2] (ii) EK = GMm / 2r and r decreases M1 so (EK and) v increases A1 [2]

This question in 9702/41 May/June 2006

2 (a) An amount of 1.00 mol of Helium-4 gas is contained in a cylinder at a pressure of 1.02 × 105 Pa and a temperature of 27 °C. (i) Calculate the volume of gas in the cylinder. volume = … m3 [2] (ii) Hence show that the average separation of gas atoms in the cylinder is approximately 3.4 × 10–9 m. [2] (b) Calculate (i) the gravitational force between two Helium-4 atoms that are separated by a distance of 3.4 × 10–9 m, force = … N [3] Examiner’s Use (ii) the ratio weight of a Helium-4 atom . gravitational force between two Helium-4 atoms with separation 3.4 × 10–9 m ratio = … [2] (c) Comment on your answer to (b)(ii) with reference to one of the assumptions of the kinetic theory of gases. … … … [2]

11 marks

Mark scheme: 2 (a) (i) pV = nRT V = (8.31 × 300)/(1.02 × 105) … C1 = 0.0244 m3 (if uses Celsius, then 0/2) … A1 [2] (ii) volume occupied by one atom = 0.0244 / (6.02 × 1023) = 4.06 × 10-26 m3 … M1 separation ≈ 3√(4.06 × 10-26) … A1 = 3.44 × 10-9 m … A0 [2] (b) (i) F = GMm / r2 … C1 = (6.67 × 10-11 × {4 × 1.66 × 10-27}2) / (3.44 × 10-9)2 … C1 = 2.49 × 10-46 N … A1 [3] (ii) ratio = (4 × 1.66 × 10-27 × 9.8) / 2.49 × 10-46 … C1 = 2.6 × 1020 … A1 [2] (c) assumption that forces between atoms are negligible … B1 comment e.g. ratio shows gravitational force to be very small e.g. force is very much less than weight e.g. if there are forces, they are not gravitational … B1 [2] GCE A/AS LEVEL – October/November 2007 9702 04

This question in 9702/41 Oct/Nov 2007

1 A spherical planet has mass M and radius R. The planet may be assumed to be isolated in space and to have its mass concentrated at its centre. The planet spins on its axis with angular speed ω, as illustrated in Fig. 1.1. mass m R equator of planet pole of planet Fig. 1.1 A small object of mass m rests on the equator of the planet. The surface of the planet exerts a normal reaction force on the mass. (a) State formulae, in terms of M, m, R and ω, for (i) the gravitational force between the planet and the object, … [1] (ii) the centripetal force required for circular motion of the small mass, … [1] (iii) the normal reaction exerted by the planet on the mass. … [1] (b) (i) Explain why the normal reaction on the mass will have different values at the equator and at the poles. … … … [2] (ii) The radius of the planet is 6.4 × 106 m. It completes one revolution in 8.6 × 104 s. For Calculate the magnitude of the centripetal acceleration at Examiner’s Use 1. the equator, acceleration = … m s–2 [2] 2. one of the poles. acceleration = … m s–2 [1] (c) Suggest two factors that could, in the case of a real planet, cause variations in the acceleration of free fall at its surface. 1. … … 2. … … [2]

10 marks

Mark scheme: 1 (a) (i) F = GMm / R2 B1 [1] (ii) F = mRω2 B1 [1] (iii) reaction force = GMm / R2 – mRω2 (allow e.c.f.) B1 [1] (b) (i) either value of R in expression Rω2 varies or mRω2 no longer parallel to GMm / R2 / normal to surface B1 becomes smaller as object approaches a pole / is zero at pole B1 [2] (ii) 1. acceleration = 6.4 × 106 × (2π / {8.6 × 104})2 C1 = 0.034 m s–2 A1 [2] 2. acceleration = 0 A1 [1] (c) e.g. ‘radius’ of planet varies density of planet not constant planet spinning nearby planets / stars (any sensible comments, 1 mark each, maximum 2) B2 [2]

This question in 9702/41 Oct/Nov 2008

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) State Newton’s law of gravitation. … … … [2] (b) The Earth may be considered to be a uniform sphere of radius R equal to 6.4 × 106 m. A satellite is in a geostationary orbit. (i) Describe what is meant by a geostationary orbit. … … … … [3] (ii) Show that the radius x of the geostationary orbit is given by the expression gR2 = x3ω2 where g is the acceleration of free fall at the Earth’s surface and ω is the angular speed of the satellite about the centre of the Earth. [3] (iii) Determine the radius x of the geostationary orbit. radius = … m [3]

11 marks

Mark scheme: 1 (a) F ∝ Mm / R2 …..…(words or explained symbols) … M1 either M and m are point masses or R >> diameter of masses …(do not allow ‘size’) … A1 [2] (b) (i) equatorial orbit … B1 period 24 hours / same angular speed … B1 from west to east / same direction of rotation … B1 [3] (allow one of the last two marks for ‘always overhead’ if 2nd or 3rd marks not scored) (ii) gravitational force provides centripetal force / gives rise to centripetal acceleration ….(in ‘words’) … B1 GM / x2 = xω2 … M1 g = GM / R2 … M1 to give gR2 = x3ω2 … A0 [3] (iii) ω = 2π / (24 × 3600) = 7.27 × 10-5 rad s-1 … C1 9.81 × (6.4 × 106)2 = x3 × (7.27 × 10-5)2 … C1 x3 = 7.6 × 1022 x = 4.2 × 107 m … A1 [3] (use of g = 10 m s-2, loses 1 mark but once only in the Paper) [Total: 11]

This question in 9702/41 Oct/Nov 2009

4 (a) Explain what is meant by the potential energy of a body. For Examiner’s … Use … … [2] 2 (b) Two deuterium ( 1 H) nuclei each have initial kinetic energy EK and are initially separated by a large distance. The nuclei may be considered to be spheres of diameter 3.8 × 10–15 m with their masses and charges concentrated at their centres. The nuclei move from their initial positions to their final position of just touching, as illustrated in Fig. 4.1. 2 2 initially 1 H 1 H kinetic energy EK kinetic energy EK 3.8 × 10–15 m 2 2 finally 1 H 1 H at rest Fig. 4.1 (i) For the two nuclei approaching each other, calculate the total change in 1. gravitational potential energy, energy = … J [3] 2. electric potential energy. energy = … J [3] (ii) Use your answers in (i) to show that the initial kinetic energy EK of each nucleus For is 0.19 MeV. Examiner’s Use [2] (iii) The two nuclei may rebound from each other. Suggest one other effect that could happen to the two nuclei if the initial kinetic energy of each nucleus is greater than that calculated in (ii). … … [1]

11 marks

Mark scheme: 4 (a) ability to do work B1 as a result of the position/shape, etc. of an object B1 [2] (b) (i) 1 ∆Egpe = GMm / r C1 = (6.67 × 10–11 × {2 × 1.66 × 10–27}2) / (3.8 × 10–15) C1 = 1.93 × 10–49 J A1 [3] 2 ∆Eepe = Qq / 4πε0r C1 = (1.6 × 10–19)2 / (4π × 8.85 × 10–12 × 3.8 × 10–15) C1 = 6.06 × 10–14 J A1 [3] (ii) idea that 2EK = ∆Eepe – ∆Egpe B1 EK = 3.03 × 10–14 J = (3.03 × 10–14) / 1.6 × 10–13 M1 = 0.19 MeV A0 [2] (iii) fusion may occur / may break into sub-nuclear particles B1 [1]

This question in 9702/41 May/June 2010

1 A planet of mass m is in a circular orbit of radius r about the Sun of mass M, as illustrated in Fig. 1.1. planet mass m Sun mass M r Fig. 1.1 The magnitude of the angular velocity and the period of revolution of the planet about the Sun are x and T respectively. (a) State (i) what is meant by angular velocity, … … … [2] (ii) the relation between x and T. … [1] (b) Show that, for a planet in a circular orbit of radius r, the period T of the orbit is given by the expression T 2 = cr 3 where c is a constant. Explain your working. [4] (c) Data for the planets Venus and Neptune are given in Fig. 1.2. For Examiner’s Use planet r / 108 km T / years Venus 1.08 0.615 Neptune 45.0 Fig. 1.2 Assume that the orbits of both planets are circular. (i) Use the expression in (b) to calculate the value of T for Neptune. T = … years [2] (ii) Determine the linear speed of Venus in its orbit. speed = … km s–1 [2]

11 marks

Mark scheme: 1 (a) (i) rate of change of angle / angular displacement M1 swept out by radius A1 [2] (ii) ω × T = 2π B1 [1] (b) centripetal force is provided by the gravitational force B1 either mr(2π/T)2 = GMm/r 2 or mrω 2 = GMm/r 2 M1 r 3 × 4π2 = GM × T 2 A1 GM/4π2 is a constant (c) A1 T 2 = cr 3 A0 [4] (c) (i) either T 2 = (45/1.08)3 × 0.6152 or T 2 = 0.30 × 453 C1 T = 165 years A1 [2] (ii) speed = (2π × 1.08 × 108) / (0.615 × 365 × 24 × 3600) C1 = 35 km s–1 A1 [2]

This question in 9702/43 Oct/Nov 2010

1 (a) Newton’s law of gravitation applies to point masses. (i) State Newton’s law of gravitation. … … … [2] (ii) Explain why, although the planets and the Sun are not point masses, the law also applies to planets orbiting the Sun. … … [1] (b) Gravitational fields and electric fields show certain similarities and certain differences. State one aspect of gravitational and electric fields where there is (i) a similarity, … … [1] (ii) a difference. … … … [2]

6 marks

Mark scheme: 1 (a) (i) force proportional to product of masses B1 force inversely proportional to square of separation B1 [2] (ii) separation much greater than radius / diameter of Sun / planet B1 [1] (b) (i) e.g. force or field strength ∝ 1 / r 2 potential ∝ 1 / r B1 [1] (ii) e.g. gravitational force (always) attractive B1 electric force attractive or repulsive B1 [2]

This question in 9702/41 May/June 2011

1 (a) State what is meant by a field of force. … … [1] (b) Gravitational fields and electric fields are two examples of fields of force. State one similarity and one difference between these two fields of force. similarity: … … difference: … … … [3] (c) Two protons are isolated in space. Their centres are separated by a distance R. Each proton may be considered to be a point mass with point charge. Determine the magnitude of the ratio force between protons due to electric field . force between protons due to gravitational field ratio = … [3]

7 marks

Mark scheme: 1 (a) region (of space) where a particle / body experiences a force B1 [1] (b) similarity: e.g. force ∝ 1 / r 2 potential ∝ 1 / r B1 [1] difference: e.g. gravitation force (always) attractive B1 electric force attractive or repulsive B1 [2] (c) either ratio is Q1Q2 / 4πε0m1m2G C1 = (1.6 × 10–19)2 / 4π × 8.85 × 10–12 × (1.67 × 10–27)2 × 6.67 × 10–11 C1 = 1.2 × 1036 A1 [3] or FE = 2.30 × 10–28 × R –2 (C1) FG = 1.86 × 10–64 × R –2 (C1) FE / FG = 1.2 × 1036 (A1)

This question in 9702/42 May/June 2011

1 (a) State what is meant by a field of force. … … [1] (b) Gravitational fields and electric fields are two examples of fields of force. State one similarity and one difference between these two fields of force. similarity: … … difference: … … … [3] (c) Two protons are isolated in space. Their centres are separated by a distance R. Each proton may be considered to be a point mass with point charge. Determine the magnitude of the ratio force between protons due to electric field . force between protons due to gravitational field ratio = … [3]

7 marks

Mark scheme: 1 (a) region (of space) where a particle / body experiences a force B1 [1] (b) similarity: e.g. force ∝ 1 / r 2 potential ∝ 1 / r B1 [1] difference: e.g. gravitation force (always) attractive B1 electric force attractive or repulsive B1 [2] (c) either ratio is Q1Q2 / 4πε0m1m2G C1 = (1.6 × 10–19)2 / 4π × 8.85 × 10–12 × (1.67 × 10–27)2 × 6.67 × 10–11 C1 = 1.2 × 1036 A1 [3] or FE = 2.30 × 10–28 × R –2 (C1) FG = 1.86 × 10–64 × R –2 (C1) FE / FG = 1.2 × 1036 (A1)

This question in 9702/43 May/June 2011

1 (a) A moon is in a circular orbit of radius r about a planet. The angular speed of the moon in its orbit is ω. The planet and its moon may be considered to be point masses that are isolated in space. Show that r and ω are related by the expression r 3ω 2 = constant. Explain your working. [3] (b) Phobos and Deimos are moons that are in circular orbits about the planet Mars. Data for Phobos and Deimos are shown in Fig. 1.1. period of rotation radius of orbit moon about Mars / m / hours Phobos 9.39 × 106 7.65 Deimos 1.99 × 107 Fig. 1.1 (i) Use data from Fig. 1.1 to determine For Examiner’s 1. the mass of Mars, Use mass = … kg [3] 2. the period of Deimos in its orbit about Mars. period = … hours [3] (ii) The period of rotation of Mars about its axis is 24.6 hours. Deimos is in an equatorial orbit, orbiting in the same direction as the spin of Mars about its axis. Use your answer in (i) to comment on the orbit of Deimos. … … [1]

10 marks

Mark scheme: 1 (a) gravitational force provides the centripetal force B1 GMm/r 2 = mrω2 (must be in terms of ω) B1 r 3ω2 = GM and GM is a constant B1 [3] (b) (i) 1. for Phobos, ω = 2π/(7.65 × 3600) C1 = 2.28 × 10–4 rad s–1 (9.39 × 106)3 × (2.28 × 10–4)2 = 6.67 × 10–11 × M C1 M = 6.46 × 1023 kg A1 [3] 2. (9.39 × 106)3 × (2.28 × 10–4)2 = (1.99 × 107)3 × ω2 C1 ω = 7.30 × 10–5 rad s–1 C1 T = 2π/ω = 2π/(7.30 × 10–5) = 8.6 × 104 s = 23.6 hours A1 [3] (ii) either almost ‘geostationary’ or satellite would take a long time to cross the sky B1 [1]

This question in 9702/41 Oct/Nov 2011

1 (a) State Newton’s law of gravitation. … … … [2] (b) The Earth and the Moon may be considered to be isolated in space with their masses concentrated at their centres. The orbit of the Moon around the Earth is circular with a radius of 3.84 × 105 km. The period of the orbit is 27.3 days. Show that (i) the angular speed of the Moon in its orbit around the Earth is 2.66 × 10–6 rad s–1, [1] (ii) the mass of the Earth is 6.0 × 1024 kg. [2] (c) The mass of the Moon is 7.4 × 1022 kg. For Examiner’s (i) Using data from (b), determine the gravitational force between the Earth and the Use Moon. force = … N [2] (ii) Tidal action on the Earth’s surface causes the radius of the orbit of the Moon to increase by 4.0 cm each year. Use your answer in (i) to determine the change, in one year, of the gravitational potential energy of the Moon. Explain your working. energy change = … J [3]

10 marks

Mark scheme: 1 (a) force proportional to product of masses and inversely proportional to square of separation (do not allow square of distance/radius) M1 either point masses or separation @ size of masses A1 [2] (b) (i) ω = 2π / (27.3 × 24 × 3600) or 2π / (2.36 x 106) M1 = 2.66 × 10–6 rad s–1 A0 [1] (ii) GM = r3ω2 or GM = v2r C1 M = (3.84 × 105 × 103)3 × (2.66 × 10–6)2 / (6.67 × 10–11) M1 = 6.0 × 1024 kg A0 [2] (special case: uses g = GM/r2 with g = 9.81, r = 6.4 × 106 scores max 1 mark) (c) (i) grav. force = (6.0 × 1024) × (7.4 × 1022) × (6.67 × 10–11)/(3.84 × 108)2 C1 = 2.0 × 1020 N (allow 1 SF) A1 [2] (ii) either ∆EP = Fx because F constant as x ! radius of orbit B1 ∆EP = 2.0 × 1020 × 4.0 × 10–2 C1 = 8.0 × 1018 J (allow 1 SF) A1 [3] or ∆EP = GMm/r1 – GMm/r2 C1 Correct substitution B1 8.0 × 1018 J A1 (∆EP = GMm/r1 + GMm/r2 is incorrect physics so 0/3) 2 2

This question in 9702/42 May/June 2012

1 (a) State Newton’s law of gravitation. … … … [2] (b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M. For this planet, the product GM is 4.00 × 1014 N m2 kg–1, where G is the gravitational constant. The planet may be assumed to be isolated in space. (i) By considering the gravitational force on the satellite and the centripetal force, show that the kinetic energy EK of the satellite is given by the expression GMm EK = . 2r [2] (ii) The satellite has mass 620 kg and is initially in a circular orbit of radius 7.34 × 106 m, as illustrated in Fig. 1.1. initial orbit 7.34 × 106 m 7.30 × 106 m new orbit Fig. 1.1 (not to scale) Resistive forces cause the satellite to move into a new orbit of radius 7.30 × 106 m. For Examiner’s Determine, for the satellite, the change in Use 1. kinetic energy, change in kinetic energy = … J [2] 2. gravitational potential energy. change in potential energy = … J [2] (iii) Use your answers in (ii) to explain whether the linear speed of the satellite increases, decreases or remains unchanged when the radius of the orbit decreases. … … … [2]

10 marks

Mark scheme: 1 (a) force is proportional to the product of the masses and inversely proportional to the square of the separation M1 either point masses or separation >> size of masses A1 [2] (b) (i) gravitational force provides the centripetal force B1 mv2/r = GMm/r2 and EK = ½mv2 M1 hence EK = GMm/2r A0 [2] (ii) 1. ∆EK = ½ × 4.00 × 1014 × 620 × ({7.30 × 106}–1 – {7.34 × 106}–1) C1 = 9.26 × 107 J (ignore any sign in answer) A1 [2] (allow 1.0 × 108 J if evidence that EK evaluated separately for each r) 2. ∆EP = 4.00 × 1014 × 620 × ({7.30 × 106}–1 – {7.34 × 106}–1) C1 = 1.85 × 108 J (ignore any sign in answer) A1 [2] (allow 1.8 or 1.9 × 108 J) (iii) either (7.30 × 106)–1 – (7.34 × 106)–1 or ∆EK is positive / EK increased M1 speed has increased A1 [2]

This question in 9702/41 Oct/Nov 2012

1 (a) State Newton’s law of gravitation. … … … [2] (b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M. For this planet, the product GM is 4.00 × 1014 N m2 kg–1, where G is the gravitational constant. The planet may be assumed to be isolated in space. (i) By considering the gravitational force on the satellite and the centripetal force, show that the kinetic energy EK of the satellite is given by the expression GMm EK = . 2r [2] (ii) The satellite has mass 620 kg and is initially in a circular orbit of radius 7.34 × 106 m, as illustrated in Fig. 1.1. initial orbit 7.34 × 106 m 7.30 × 106 m new orbit Fig. 1.1 (not to scale) Resistive forces cause the satellite to move into a new orbit of radius 7.30 × 106 m. For Examiner’s Determine, for the satellite, the change in Use 1. kinetic energy, change in kinetic energy = … J [2] 2. gravitational potential energy. change in potential energy = … J [2] (iii) Use your answers in (ii) to explain whether the linear speed of the satellite increases, decreases or remains unchanged when the radius of the orbit decreases. … … … [2]

10 marks

Mark scheme: 1 (a) force is proportional to the product of the masses and inversely proportional to the square of the separation M1 either point masses or separation >> size of masses A1 [2] (b) (i) gravitational force provides the centripetal force B1 mv2/r = GMm/r2 and EK = ½mv2 M1 hence EK = GMm/2r A0 [2] (ii) 1. ∆EK = ½ × 4.00 × 1014 × 620 × ({7.30 × 106}–1 – {7.34 × 106}–1) C1 = 9.26 × 107 J (ignore any sign in answer) A1 [2] (allow 1.0 × 108 J if evidence that EK evaluated separately for each r) 2. ∆EP = 4.00 × 1014 × 620 × ({7.30 × 106}–1 – {7.34 × 106}–1) C1 = 1.85 × 108 J (ignore any sign in answer) A1 [2] (allow 1.8 or 1.9 × 108 J) (iii) either (7.30 × 106)–1 – (7.34 × 106)–1 or ∆EK is positive / EK increased M1 speed has increased A1 [2]

This question in 9702/42 Oct/Nov 2012

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) State Newton’s law of gravitation. … … … [2] (b) A star and a planet are isolated in space. The planet orbits the star in a circular orbit of radius R, as illustrated in Fig. 1.1. t planet star mass M R Fig. 1.1 The angular speed of the planet about the star is ω. By considering the circular motion of the planet about the star of mass M, show that ω and R are related by the expression R 3ω2 = GM where G is the gravitational constant. Explain your working. [3] (c) The Earth orbits the Sun in a circular orbit of radius 1.5 × 108 km. The mass of the Sun For is 2.0 × 1030 kg. Examiner’s A distant star is found to have a planet that has a circular orbit about the star. The radius Use of the orbit is 6.0 × 108 km and the period of the orbit is 2.0 years. Use the expression in (b) to calculate the mass of the star. mass = … kg [3]

8 marks

Mark scheme: 1 (a) force proportional to product of the two masses and inversely proportional to the square of their separation M1 either reference to point masses or separation >> ‘size’ of masses A1 [2] (b) gravitational force provides the centripetal force B1 GMm / R2 = mRω2 M1 where m is the mass of the planet A1 GM = R3ω2 A0 [3] (c) ω = 2π / T C1 either Mstar / MSun = (Rstar / RSun)3 × (TSun / Tstar)2 Mstar = 43 × (½)2 × 2.0 × 1030 C1 = 3.2 × 1031 kg A1 [3] or Mstar = (2π)2 Rstar3 / GT2 (C1) = {(2π)2 × (6.0 × 1011)3} / {6.67 × 10–11 × (2 × 365 × 24 × 3600)2} (C1) = 3.2 × 1031 kg (A1)

This question in 9702/43 Oct/Nov 2013

1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet. (a) A stone, travelling at speed v, is in a circular orbit of radius r about the planet, as illustrated in Fig. 1.1. stone v planet r Fig. 1.1 Show that the speed v is given by the expression GM v = r where G is the gravitational constant. Explain your working. [2] (b) A second stone, initially at rest at infinity, travels towards the planet, as illustrated in Fig. 1.2. stone V0 planet x Fig. 1.2 (not to scale) The stone does not hit the surface of the planet. (i) Determine, in terms of the gravitational constant G and the mass M of the planet, the speed V0 of the stone at a distance x from the centre of the planet. Explain your working. You may assume that the gravitational attraction on the stone is due only to the planet. [3] (ii) Use your answer in (i) and the expression in (a) to explain whether this stone could enter a circular orbit about the planet. … … … … [2]

7 marks

Mark scheme: 1 (a) gravitational force provides/is the centripetal force B1 GMm / r2 R mv2 / r M1 v R √(GM / r) A0 [2] allow gravitational field strength provides/is the centripetal acceleration (B1) GM / r2 R v2 / r (M1) (b) (i) kinetic energy increase/change R loss / change in (gravitational) potential energy B1 ½mV02 R GMm / x C1 V02 R 2GM / x V0 R √(2GM / x) A1 [3] (max. 2 for use of r not x) (ii) V0 is (always) greater than v (for x = r) M1 so stone could not enter into orbit A1 [2] (expressions in (a) and (b)(i) must be dimensionally correct)

This question in 9702/42 May/June 2014

4 A helium nucleus contains two protons. In a model of the helium nucleus, each proton is considered to be a charged point mass. The separation of these point masses is assumed to be 2.0 × 10−15 m. (a) For the two protons in this model, calculate (i) the electrostatic force, electrostatic force = … N [2] (ii) the gravitational force. gravitational force = … N [2] (b) Using your answers in (a), suggest why (i) there must be some other force between the protons in the nucleus, … … … … [3] (ii) this additional force must have a short range. … … … [2]

9 marks

Mark scheme: 4 (a) (i) FE R Q1Q2 / 4πε0r2 C1 = = = = R 8.99 × 109 × (1.6 × 10–19)2 / (2.0 × 10–15)2 R 58 N A1 [2] (ii) FG R Gm1m2 / r2 C1 = = = R 6.67 × 10–11 × (1.67 × 10–27)2 / (2.0 × 10–15)2 = = = R 4.7 × 10–35 N A1 [2] (b) (i) force of repulsion (much) greater than force of attraction B1 must be some other force of attraction M1 to hold nucleus together A1 [3] (Do not allow if FG > FE in (a) or one of the forces not calculated in (a)) (ii) outside nucleus there is repulsion between protons B1 either attractive force must act only in nucleus or if not short range, all nuclei would stick together B1 [2]

This question in 9702/42 May/June 2014

1 (a) State Newton’s law of gravitation. … … … [2] (b) The planet Neptune has eight moons (satellites). Each moon orbits Neptune in a circular path of radius r with a period T. Assuming that Neptune and each moon behave as point masses, show that r and T are related by the expression 4π2r3 GMN = 2 T where G is the gravitational constant and MN is the mass of Neptune. [3] (c) Data for the moon Triton that orbits Neptune and for the moon Oberon that orbits the planet Uranus are given in Fig. 1.1. planet moon radius of orbit period of orbit r /105 km T / days Neptune Triton 3.55 5.9 Uranus Oberon 5.83 13.5 Fig. 1.1 Use the expression in (b) to determine the ratio mass of Neptune . mass of Uranus ratio = … [3]

8 marks

Mark scheme: 1 (a) (gravitational) force proportional to product of masses and inversely proportional to square of separation M1 reference to either point masses or particles or ‘size’ much less than separation A1 [2] (b) gravitational force provides/is the centripetal force B1 GMNm / r 2 = mrω2 (or mv 2 / r) M1 2π / T (or v = 2πr / T) leading to GMN = 4π2r 3 / T 2 A1 [3] (c) MN / MU = (3.55 / 5.83)3 × (13.5 / 5.9)2 x3 factor correct C1 T2 factor correct C1 ratio = 1.18 (allow 1.2) A1 alternative method: mass of Neptune = 1.019 × 1026 kg (C1) mass of Uranus = 8.621 × 1025 kg (C1) ratio = 1.18 (A1) [3]

This question in 9702/41 May/June 2015

1 (a) The Earth may be considered to be a uniform sphere of radius 6.37 × 103 km with its mass of 5.98 × 1024 kg concentrated at its centre. The Earth spins on its axis with a period of 24.0 hours. (i) A stone of mass 2.50 kg rests on the Earth’s surface at the Equator. 1. Calculate, using Newton’s law of gravitation, the gravitational force on the stone. gravitational force = … N [2] 2. Determine the force required to maintain the stone in its circular path. force = … N [2] (ii) The stone is now hung from a newton-meter. Use your answers in (i) to determine the reading on the meter. Give your answer to three significant figures. reading = … N [2] (b) A satellite is orbiting the Earth. For an astronaut in the satellite, his sensation of weight is caused by the contact force from his surroundings. The astronaut reports that he is ‘weightless’, despite being in the Earth’s gravitational field. Suggest what is meant by the astronaut reporting that he is ‘weightless’. … … … … [3]

9 marks

Mark scheme: 1 (a) (i) 1. F = Gm1m2 / x2 = (6.67 × 10–11 × 2.50 × 5.98 × 1024) / (6.37 × 106)2 M1 = 24.6 N (accept 2 s.f. or more) A1 [2] 2. F = mxω2 or F = mv 2 / x and v = ωx (accept x or r for distance) C1 = 2.50 × 6.37 × 106 × (2π / 24 × 3600)2 = 0.0842 N (accept 2 s.f. or more) A1 [2] (ii) reading = 24.575 – 0.0842 B1 = 24.5 N (accept only 3 s.f.) A1 [2] (b) gravitational force provides the centripetal force M1 gravitational force is ‘equal’ to the centripetal force (accept Gm1m2 / x2 = mxω2 or FC = FG) M1 ‘weight’/sensation of weight/contact force/reaction force is difference between FG and FC which is zero A1 [3] 3 1

This question in 9702/42 May/June 2015

1 (a) State Newton’s law of gravitation. … … … [2] (b) The planet Neptune has eight moons (satellites). Each moon orbits Neptune in a circular path of radius r with a period T. Assuming that Neptune and each moon behave as point masses, show that r and T are related by the expression 4π2r3 GMN = 2 T where G is the gravitational constant and MN is the mass of Neptune. [3] (c) Data for the moon Triton that orbits Neptune and for the moon Oberon that orbits the planet Uranus are given in Fig. 1.1. planet moon radius of orbit period of orbit r /105 km T / days Neptune Triton 3.55 5.9 Uranus Oberon 5.83 13.5 Fig. 1.1 Use the expression in (b) to determine the ratio mass of Neptune . mass of Uranus ratio = … [3]

8 marks

Mark scheme: 1 (a) (gravitational) force proportional to product of masses and inversely proportional to square of separation M1 reference to either point masses or particles or ‘size’ much less than separation A1 [2] (b) gravitational force provides/is the centripetal force B1 GMNm / r 2 = mrω2 (or mv 2 / r) M1 2π / T (or v = 2πr / T) leading to GMN = 4π2r 3 / T 2 A1 [3] (c) MN / MU = (3.55 / 5.83)3 × (13.5 / 5.9)2 x3 factor correct C1 T2 factor correct C1 ratio = 1.18 (allow 1.2) A1 alternative method: mass of Neptune = 1.019 × 1026 kg (C1) mass of Uranus = 8.621 × 1025 kg (C1) ratio = 1.18 (A1) [3]

This question in 9702/43 May/June 2015

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) State Newton’s law of gravitation. … … … … [2] (b) A satellite of mass m has a circular orbit of radius r about a planet of mass M. It may be assumed that the planet and the satellite are uniform spheres that are isolated in space. Show that the linear speed v of the satellite is given by the expression GM v = r where G is the gravitational constant. Explain your working. [2] (c) Two moons A and B have circular orbits about a planet, as illustrated in Fig. 1.1. vA B A vB rA rB planet Fig. 1.1 (not to scale) Moon A has an orbital radius rA of 1.3 × 108 m, linear speed vA and orbital period TA. Moon B has an orbital radius rB of 2.2 × 1010 m, linear speed vB and orbital period TB. (i) Determine the ratio vA 1. , vB ratio = … [2] TA 2. . TB ratio = … [3] (ii) The planet spins about its own axis with angular speed 1.7 × 10–4 rad s–1. Moon A is always above the same point on the planet’s surface. Determine the orbital period TB of moon B. TB = … s [2] [Total: 11]

11 marks

Mark scheme: 1 (a) force proportional to product of the (two) masses and inversely proportional to the square of their separation M1 either reference to point masses or separation << ‘size’ of masses A1 [2] (b) gravitational force provides / is the centripetal force B1 GMm / r 2 = mv 2 / r or GMm / r 2 = mrω 2 and v = rω and algebra leading to v = (GM / r )1 / 2 B1 [2] (c) (i) 1. vA / vB = (rB / rA)1 / 2 = (2.2 × 1010 / 1.3 × 108)1 / 2 C1 = 13 (13.0) A1 [2] 2. v = 2πr / T or v ∝ r / T or vT / r = constant C1 TA / TB = (rA / rB) × (vB / vA) = (1.3 × 108 / 2.2 × 1010) × (1 / 13) C1 = 4.5 (4.54) × 10–4 A1 or T 2 = 4π2r 3 / GM or T 2 ∝ r 3 or T 2 / r 3 = constant (C1) TA / TB = (rA3 / rB3)1 / 2 = [(1.3 × 108)3 / (2.2 × 1010)3]1 / 2 (C1) = 4.5 (4.54) × 10–4 (A1) [3] (ii) T = 2π / 1.7 ×10–4 = 3.70 × 104 s C1 TB = 3.70 × 104 / 4.54 × 10–4 = 8.1 × 107 s A1 [2] If identifies TA as TB then 0 / 2

This question in 9702/42 Feb/March 2016

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) Explain how a satellite may be in a circular orbit around a planet. … … … [2] (b) The Earth and the Moon may be considered to be uniform spheres that are isolated in space. The Earth has radius R and mean density ρ. The Moon, mass m, is in a circular orbit about the Earth with radius nR, as illustrated in Fig. 1.1. Earth radius R Moon nR Fig. 1.1 The Moon makes one complete orbit of the Earth in time T. Show that the mean density ρ of the Earth is given by the expression 3πn3 ρ = 2 . GT [4] (c) The radius R of the Earth is 6.38 × 103 km and the distance between the centre of the Earth and the centre of the Moon is 3.84 × 105 km. The period T of the orbit of the Moon about the Earth is 27.3 days. Use the expression in (b) to calculate ρ. ρ = … kg m–3 [3] [Total: 9]

9 marks

Mark scheme: 1(a) gravitational force (of attraction between satellite and planet) B1 provides / is centripetal force (on satellite about the planet) B1 1(b) M = (4/3) × πR3ρ B1 ω = 2π / T or v = 2πnR / T B1 GM / (nR)2 = nRω2 or v 2 / nR M1 substitution clear to give ρ = 3πn3 / GT 2 A1 1(c) n = (3.84 × 105) / (6.38 × 103) = 60.19 or 60.2 C1 ρ = 3π × 60.193 / [(6.67 × 10–11) × (27.3 × 24 × 3600)2] C1 ρ = 5.54 × 103 kg m–3 A1

This question in 9702/41 May/June 2017

1 (a) Explain how a satellite may be in a circular orbit around a planet. … … … [2] (b) The Earth and the Moon may be considered to be uniform spheres that are isolated in space. The Earth has radius R and mean density ρ. The Moon, mass m, is in a circular orbit about the Earth with radius nR, as illustrated in Fig. 1.1. Earth radius R Moon nR Fig. 1.1 The Moon makes one complete orbit of the Earth in time T. Show that the mean density ρ of the Earth is given by the expression 3πn3 ρ = 2 . GT [4] (c) The radius R of the Earth is 6.38 × 103 km and the distance between the centre of the Earth and the centre of the Moon is 3.84 × 105 km. The period T of the orbit of the Moon about the Earth is 27.3 days. Use the expression in (b) to calculate ρ. ρ = … kg m–3 [3] [Total: 9]

9 marks

Mark scheme: 1(a) gravitational force (of attraction between satellite and planet) B1 provides / is centripetal force (on satellite about the planet) B1 1(b) M = (4/3) × πR3ρ B1 ω = 2π / T or v = 2πnR / T B1 GM / (nR)2 = nRω2 or v 2 / nR M1 substitution clear to give ρ = 3πn3 / GT 2 A1 1(c) n = (3.84 × 105) / (6.38 × 103) = 60.19 or 60.2 C1 ρ = 3π × 60.193 / [(6.67 × 10–11) × (27.3 × 24 × 3600)2] C1 ρ = 5.54 × 103 kg m–3 A1

This question in 9702/43 May/June 2017

1 (a) State Newton’s law of gravitation. … … … [2] (b) The planet Jupiter and one of its moons, Io, may be considered to be uniform spheres that are isolated in space. Jupiter has radius R and mean density ρ. Io has mass m and is in a circular orbit about Jupiter with radius nR, as illustrated in Fig. 1.1. Jupiter radius R density ρ Io nR Fig. 1.1 The time for Io to complete one orbit of Jupiter is T. Show that the time T is related to the mean density ρ of Jupiter by the expression 2 3πn3 ρT = G where G is the gravitational constant. [4] (c) (i) The radius R of Jupiter is 7.15 × 104 km and the distance between the centres of Jupiter and Io is 4.32 × 105 km. The period T of the orbit of Io is 42.5 hours. Calculate the mean density ρ of Jupiter. ρ = … kg m–3 [3] (ii) The Earth has a mean density of 5.5 × 103 kg m–3. It is said to be a planet made of rock. By reference to your answer in (i), comment on the possible composition of Jupiter. … … [1] [Total: 10]

10 marks

Mark scheme: 1(a) force proportional to product of masses and inversely proportional to square of separation B1 idea of force between point masses B1 1(b) mass of Jupiter (M) = (4 / 3) πR3ρ B1 ω = 2π / T or v = 2πnR / T B1 (m)ω2x = GM(m) / x2 or (m)v2 / x = GM(m) / x2 M1 substitution and correct algebra leading to ρT2 = 3πn3 / G A1 1(c)(i) n = (4.32 × 105) / (7.15 × 104) or n = 6.04 C1 ρ × (42.5 × 3600)2 = (3π × 6.043) / (6.67 × 10–11) C1 ρ = 1.33 × 103 kg m–3 A1 1(c)(ii) Jupiter likely to be a gas/liquid (at high pressure) [allow other sensible suggestions] B1

This question in 9702/42 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) State Newton’s law of gravitation. … … … … [2] (b) A distant star is orbited by several planets. Each planet has a circular orbit with a different radius. (i) Each planet orbits at constant speed. Explain whether the planets are in equilibrium. … … … [1] (ii) The radius of the orbit of a planet is R and the orbital period is T. Data for some of the planets are given in Fig. 1.1. planet R / m T 2 / s2 c 9.6 × 1010 2.5 × 1011 e 4.0 × 1011 1.8 × 1013 g 2.1 × 1012 2.6 × 1015 Fig. 1.1 The relationship between R and T is given by the expression R 3 = kT 2. 1. Show that the constant k is given by the expression GM k = 4π2 where G is the gravitational constant and M is the mass of the star. [3] 2. Use data from Fig. 1.1 for the three planets and the expression for k to calculate the mass M of the star. M = … kg [3] [Total: 9]

9 marks

Mark scheme: 1(a) force proportional to product of masses and inversely proportional to square of separation B1 idea of force between point masses B1 1(b)(i) velocity changes/direction of motion changes/there is an acceleration/there is a resultant force so not in equilibrium B1 1(b)(ii)1. gravitational force equals/is centripetal force C1 GMm / R2 = mRω2 and ω = 2π / T or Gm / R2 = mv2 / R and v = 2πr / T or GMm / R2 = mR (2π / T)2 M1 convincing algebra leading to k = GM / 4π2 A1 1(b)(ii)2. correct use of R3 / T2 for one planet (c gives 3.54 × 1021; e and g both give 3.56 × 1021) C1 3.5(5) × 1021 = (6.67 × 10–11 × M) / 4π2 M = 2.1 × 1033 kg A1 two or three values of R3 / T2 correctly calculated and used in a valid way to find a value for M based on more than one k B1

This question in 9702/41 May/June 2018

1 (a) State Newton’s law of gravitation. … … … … [2] (b) A distant star is orbited by several planets. Each planet has a circular orbit with a different radius. (i) Each planet orbits at constant speed. Explain whether the planets are in equilibrium. … … … [1] (ii) The radius of the orbit of a planet is R and the orbital period is T. Data for some of the planets are given in Fig. 1.1. planet R / m T 2 / s2 c 9.6 × 1010 2.5 × 1011 e 4.0 × 1011 1.8 × 1013 g 2.1 × 1012 2.6 × 1015 Fig. 1.1 The relationship between R and T is given by the expression R 3 = kT 2. 1. Show that the constant k is given by the expression GM k = 4π2 where G is the gravitational constant and M is the mass of the star. [3] 2. Use data from Fig. 1.1 for the three planets and the expression for k to calculate the mass M of the star. M = … kg [3] [Total: 9]

9 marks

Mark scheme: 1(a) force proportional to product of masses and inversely proportional to square of separation B1 idea of force between point masses B1 1(b)(i) velocity changes/direction of motion changes/there is an acceleration/there is a resultant force so not in equilibrium B1 1(b)(ii)1. gravitational force equals/is centripetal force C1 GMm / R2 = mRω2 and ω = 2π / T or Gm / R2 = mv2 / R and v = 2πr / T or GMm / R2 = mR (2π / T)2 M1 convincing algebra leading to k = GM / 4π2 A1 1(b)(ii)2. correct use of R3 / T2 for one planet (c gives 3.54 × 1021; e and g both give 3.56 × 1021) C1 3.5(5) × 1021 = (6.67 × 10–11 × M) / 4π2 M = 2.1 × 1033 kg A1 two or three values of R3 / T2 correctly calculated and used in a valid way to find a value for M based on more than one k B1

This question in 9702/43 May/June 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 separated by a distance x in a vacuum. State an expression for the force F between the two masses M and m. State the name of any other symbol used. … … … [1] (b) A small sphere S is attached to one end of a rod, as shown in Fig. 1.1. thread rod small sphere S 8.0 cm view from side Fig. 1.1 (not to scale) The rod hangs from a vertical thread and is horizontal. The distance from the centre of sphere S to the thread is 8.0 cm. A large sphere L is placed near to sphere S, as shown in Fig. 1.2. large sphere L initial position of rod 6.0 cm final position of rod θ 1.2 mm small sphere S 8.0 cm thread view from above Fig. 1.2 (not to scale) There is a force of attraction between spheres S and L, causing sphere S to move through a distance of 1.2 mm. The line joining the centres of S and L is normal to the rod. (i) Show that the angle θ through which the rod rotates is 1.5 × 10–2 rad. [1] (ii) The rotation of the rod causes the thread to twist. The torque T (in N m) required to twist the thread through an angle β (in rad) is given by T = 9.3 × 10–10 × β. Calculate the torque in the thread when sphere L is positioned as shown in Fig. 1.2. torque = … N m [1] (c) The distance between the centres of spheres S and L is 6.0 cm. The mass of sphere S is 7.5 g and the mass of sphere L is 1.3 kg. (i) By equating the torque in (b)(ii) to the moment about the thread produced by gravitational attraction between the spheres, calculate a value for the gravitational constant. gravitational constant = … N m2 kg–2 [3] (ii) Suggest why the total force between the spheres may not be equal to the force calculated using Newton’s law of gravitation. … … [1] [Total: 7]

7 marks

Mark scheme: 1(a) (F =) GMm / x2, where G is the (universal) gravitational constant B1 1(b)(i) angle = (1.2 × 10–3) / (8.0 × 10–2) = 1.5 × 10–2 (rad) B1 1(b)(ii) torque = 1.5 × 10–2 × 9.3 × 10–10 = 1.4 × 10–11 N m A1 1(c)(i) force × 8.0 × 10–2 = 1.4 × 10–11 C1 (G × 1.3 × 7.5 × 10–3 × 8.0 × 10–2) / (6.0 × 10–2)2 = 1.4 × 10–11 C1 G = 6.4 × 10–11 N m2 kg–2 A1 1(c)(ii) Any one from: • law applies only to point masses/spheres are not point masses • radii of spheres not small compared with separation • spheres may not be uniform • the masses are not isolated • force between L and rod • spheres may be charged/may be electrostatic force (between spheres) B1

This question in 9702/42 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) State Newton’s law of gravitation. … … … [2] (b) A geostationary satellite orbits the Earth. The orbit of the satellite is circular and the period of the orbit is 24 hours. (i) State two other features of this orbit. 1. … … 2. … … [2] (ii) The radius of the orbit of the satellite is 4.23 × 104 km. Determine a value for the mass of the Earth. Explain your working. mass = … kg [4] [Total: 8]

8 marks

Mark scheme: 1(a) force proportional to product of masses and inversely proportional to square of separation B1 idea of (gravitational) force between point masses B1 1(b)(i) above the equator B1 from west to east B1 1(b)(ii) gravitational force provides/is the centripetal force B1 GM / r2 = r (2π / T)2 C1 (6.67 × 10–11 × M) = {(4.23 × 107)3 × 4π2} / (24 × 3600)2 C1 M = 6.0 × 1024 kg A1

This question in 9702/41 Oct/Nov 2019

1 (a) State Newton’s law of gravitation. … … … [2] (b) The astronomer Johannes Kepler showed that the period T of rotation of a planet about the Sun is related to its mean distance R from the centre of the Sun by the expression R 3 2 = k T where k is a constant. Use Newton’s law to show that, for planets in circular orbits about the Sun of mass M, the constant k is given by GM k = 2 4π where G is the gravitational constant. Explain your working. [4] (c) A satellite is in a circular orbit about Mars. The radius of the orbit of the satellite is 4.38 × 106 m. The orbital period is 2.44 hours. Use the expressions in (b) to calculate a value for the mass of Mars. mass = … kg [2] [Total: 8]

8 marks

Mark scheme: 1(a) force proportional to product of masses and inversely proportional to square of separation B1 idea of (gravitational) force between point masses B1 1(b) gravitational force provides/is the centripetal force B1 GM / R2 = Rω2 or GM / R2 = v2 / R M1 ω = 2π / T or v = 2πR / T M1 algebra leading to R3/T2 = GM / 4π2 A1 1(c) (6.67 × 10–11 × M) / 4π2 = (4.38 × 106)3 / (2.44 × 3600)2 C1 M = 6.45 × 1023 kg A1

This question in 9702/42 Oct/Nov 2019

1 (a) State Newton’s law of gravitation. … … … [2] (b) A geostationary satellite orbits the Earth. The orbit of the satellite is circular and the period of the orbit is 24 hours. (i) State two other features of this orbit. 1. … … 2. … … [2] (ii) The radius of the orbit of the satellite is 4.23 × 104 km. Determine a value for the mass of the Earth. Explain your working. mass = … kg [4] [Total: 8]

8 marks

Mark scheme: 1(a) force proportional to product of masses and inversely proportional to square of separation B1 idea of (gravitational) force between point masses B1 1(b)(i) above the equator B1 from west to east B1 1(b)(ii) gravitational force provides/is the centripetal force B1 GM / r2 = r (2π / T)2 C1 (6.67 × 10–11 × M) = {(4.23 × 107)3 × 4π2} / (24 × 3600)2 C1 M = 6.0 × 1024 kg A1

This question in 9702/43 Oct/Nov 2019

1 (a) State what is meant by a gravitational force. … … [1] (b) A binary star system consists of two stars S1 and S2, each in a circular orbit. The orbit of each star in the system has a period of rotation T. Observations of the binary star from Earth are represented in Fig. 1.1. S1 S1 S2 S2 T t = 0 t = — 4 S2 S1 S2 S1 T 3T t = — t = — 2 4 S1 S2 t = T Fig. 1.1 (not to scale) Observed from Earth, the angular separation of the centres of S1 and S2 is 1.2 × 10–5 rad. The distance of the binary star system from Earth is 1.5 × 1017 m. Show that the separation d of the centres of S1 and S2 is 1.8 × 1012 m. [1] (c) The stars S1 and S2 rotate with the same angular velocity ω about a point P, as illustrated in Fig. 1.2. d P S1 S2 x Fig. 1.2 (not to scale) Point P is at a distance x from the centre of star S1. The period of rotation of the stars is 44.2 years. (i) Calculate the angular velocity ω. ω = … rad s–1 [2] (ii) By considering the forces acting on the two stars, show that the ratio of the masses of the stars is given by mass of S1 d – x = . mass of S2 x [2] (iii) The mass M1 of star S1 is given by the expression GM1 = d 2 (d – x) ω 2 where G is the gravitational constant. The ratio in (ii) is found to be 1.5. Use data from (b) and your answer in (c)(i) to determine the mass M1. M1 = … kg [3] [Total: 9]

9 marks

Mark scheme: 1(a) force acting between two masses or force on mass due to another mass or force on mass in a gravitational field B1 1(b) arc length = rθ d = 1.5 × 1017 × 1.2 × 10–5 = 1.8 × 1012 m A1 1(c)(i) ω = 2π / T C1 = 2π / (44.2 × 365 × 24 × 3600) = 4.5 × 10–9 rad s–1 A1 1(c)(ii) gravitational forces are equal or centripetal force about P is the same C1 M1xω2 = M2(d – x)ω2 so M1 / M2 = (d – x) / x A1 1(c)(iii) x = 0.4d C1 6.67 × 10–11 × M1 = (1.0 – 0.4) × (1.8 × 1012)3 × (4.5 × 10–9)2 C1 M1 = 1.1 × 1030 kg A1

This question in 9702/41 May/June 2020

1 (a) State what is meant by a gravitational force. … … [1] (b) A binary star system consists of two stars S1 and S2, each in a circular orbit. The orbit of each star in the system has a period of rotation T. Observations of the binary star from Earth are represented in Fig. 1.1. S1 S1 S2 S2 T t = 0 t = — 4 S2 S1 S2 S1 T 3T t = — t = — 2 4 S1 S2 t = T Fig. 1.1 (not to scale) Observed from Earth, the angular separation of the centres of S1 and S2 is 1.2 × 10–5 rad. The distance of the binary star system from Earth is 1.5 × 1017 m. Show that the separation d of the centres of S1 and S2 is 1.8 × 1012 m. [1] (c) The stars S1 and S2 rotate with the same angular velocity ω about a point P, as illustrated in Fig. 1.2. d P S1 S2 x Fig. 1.2 (not to scale) Point P is at a distance x from the centre of star S1. The period of rotation of the stars is 44.2 years. (i) Calculate the angular velocity ω. ω = … rad s–1 [2] (ii) By considering the forces acting on the two stars, show that the ratio of the masses of the stars is given by mass of S1 d – x = . mass of S2 x [2] (iii) The mass M1 of star S1 is given by the expression GM1 = d 2 (d – x) ω 2 where G is the gravitational constant. The ratio in (ii) is found to be 1.5. Use data from (b) and your answer in (c)(i) to determine the mass M1. M1 = … kg [3] [Total: 9]

9 marks

Mark scheme: 1(a) force acting between two masses or force on mass due to another mass or force on mass in a gravitational field B1 1(b) arc length = rθ d = 1.5 × 1017 × 1.2 × 10–5 = 1.8 × 1012 m A1 1(c)(i) ω = 2π / T C1 = 2π / (44.2 × 365 × 24 × 3600) = 4.5 × 10–9 rad s–1 A1 1(c)(ii) gravitational forces are equal or centripetal force about P is the same C1 M1xω2 = M2(d – x)ω2 so M1 / M2 = (d – x) / x A1 1(c)(iii) x = 0.4d C1 6.67 × 10–11 × M1 = (1.0 – 0.4) × (1.8 × 1012)3 × (4.5 × 10–9)2 C1 M1 = 1.1 × 1030 kg A1

This question in 9702/43 May/June 2020

1 (a) State Newton’s law of gravitation. … … … [2] (b) Planets have been observed orbiting a star in another solar system. Measurements are made of the orbital radius r and the time period T of each of these planets. The variation with R3 of T2 is shown in Fig. 1.1. 2.6 2.4 2.2 T2 / year2 2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 R3 / 1034 m3 Fig. 1.1 The relationship between T and R is given by T2 = 4π2R3 GM where G is the gravitational constant and M is the mass of the star. Determine the mass M. M = … kg [3] (c) A rock of mass m is also in orbit around the star in (b). The radius of the orbit is r. (i) Explain why the gravitational potential energy of the rock is negative. … … … … [3] (ii) Show that the kinetic energy Ek of the rock is given by Ek = GMm . 2r [2] (iii) Use the expression in (c)(ii) to derive an expression for the total energy of the rock. [2] [Total: 12]

12 marks

Mark scheme: 1(a) (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(b) correct read offs from the graph with correct power of ten for R3 C1 ( ) 2 34 2 11 4 1.2 10 6.67 10 2.4 365 24 3600 M π − × × × = × × × × × C1 30 3.0 10 kg = × A1 1(c)(i) potential energy is zero at infinity B1 (gravitational) forces are attractive B1 work must be done on the rock to move it to infinity B1 1(c)(ii) 2 2 2 GMm mv GM GM OR v OR v r r r r = = = M1 use of ½ mv2 (e.g. multiplication by ½ m) leading to 2 GMm r A1 1(c)(iii) Ep = φ m and φ = GM r − or p GMm E r − = Total energy = Ek + Ep C1 Total energy 2 GMm GMm r r − = + 2 GMm r − = A1

This question in 9702/42 Feb/March 2021

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) State the equation for the gravitational force F between two point masses m1 and m2 that are separated by a distance r. State the meaning of any other symbols you use. [2] (b) A satellite is in a circular orbit of radius R around a planet of mass M. Show that the period T of the orbit is given by T 2 = kR3 where k is a constant that depends on the value of M. Explain your reasoning. [3] (c) A satellite is in a circular orbit around the Earth with a period of 24 hours. The mass of the Earth is 6.0 × 1024 kg. (i) Calculate the radius of the orbit. radius = … m [2] (ii) State the two other conditions that must be met for the orbit to be geostationary. 1 … … 2 … … [2] [Total: 9]

9 marks

Mark scheme: Question Answer Marks 1(a) F = (Gm1m2) / r 2 M1 where G is the gravitational constant A1 1(b) gravitational force provides the centripetal force B1 mR 2 = GMm / R 2 and = 2 / T M1 or mv 2 / R = GMm / R 2 and v = 2R / T or 42mR / T 2 = GMm / R 2 correct completion of algebra to get T2 = (42 / GM) R3, with identification of (42 / GM) as k A1 1(c)(i) (24  3600)2 = (42  R3) / (6.67  10–11  6.0  1024) C1 R = 4.2  107 m A1 1(c)(ii) (orbit) must be above the Equator B1 (direction) must be from west to east B1

This question in 9702/41 Oct/Nov 2022

1 (a) State the equation for the gravitational force F between two point masses m1 and m2 that are separated by a distance r. State the meaning of any other symbols you use. [2] (b) A satellite is in a circular orbit of radius R around a planet of mass M. Show that the period T of the orbit is given by T 2 = kR3 where k is a constant that depends on the value of M. Explain your reasoning. [3] (c) A satellite is in a circular orbit around the Earth with a period of 24 hours. The mass of the Earth is 6.0 × 1024 kg. (i) Calculate the radius of the orbit. radius = … m [2] (ii) State the two other conditions that must be met for the orbit to be geostationary. 1 … … 2 … … [2] [Total: 9]

9 marks

Mark scheme: Question Answer Marks 1(a) F = (Gm1m2) / r 2 M1 where G is the gravitational constant A1 1(b) gravitational force provides the centripetal force B1 mR 2 = GMm / R 2 and = 2 / T M1 or mv 2 / R = GMm / R 2 and v = 2R / T or 42mR / T 2 = GMm / R 2 correct completion of algebra to get T2 = (42 / GM) R3, with identification of (42 / GM) as k A1 1(c)(i) (24  3600)2 = (42  R3) / (6.67  10–11  6.0  1024) C1 R = 4.2  107 m A1 1(c)(ii) (orbit) must be above the Equator B1 (direction) must be from west to east B1

This question in 9702/43 Oct/Nov 2022

1 (a) State Newton’s law of gravitation. … … … [2] (b) A satellite is in a circular orbit around a planet. The radius of the orbit is R and the period of the orbit is T. The planet is a uniform sphere. Use Newton’s law of gravitation to show that R and T are related by 4π2R 3 = GMT 2 where M is the mass of the planet and G is the gravitational constant. [2] (c) The Earth may be considered to be a uniform sphere of mass 5.98 × 1024 kg and radius 6.37 × 106 m. A geostationary satellite is in orbit around the Earth. Use the expression in (b) to determine the height of the satellite above the Earth’s surface. height = … m [3] (d) Another satellite is in a circular orbit around the Earth with the same orbital radius and period as the satellite in (c). (i) Calculate the angular speed of the satellite in this orbit. Give a unit with your answer. angular speed = … unit … [2] (ii) Despite having the same orbital period, the orbit of this satellite is not geostationary. Suggest two ways in which the orbit of this satellite could be different from the orbit of the satellite in (c). 1 … … 2 … … [2] [Total: 11]

11 marks

Mark scheme: 1(a) (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(b) GMm / R2 = mR2 M1  = 2 / T and algebra leading to 42R3 = GMT2 A1 or GMm / R2 = mv2 / R (M1) v = 2R / T and algebra leading to 42R3 = GMT2 (A1) 1(c) 42  R3 = 6.67  10–11  5.98  1024  (24  60  60)2 (R = 4.22  107 m) C1 h = R – (6.37  106) C1 h = (4.22  107) – (6.37  106) = 3.6  107 m A1 1(d)(i)  = 2 / T C1 = 2 / (24  60  60) = 7.3  10–5 rad s–1 A1 1(d)(ii) orbit is from east to west B1 orbit is not equatorial / orbit is polar B1

This question in 9702/42 May/June 2023

1 (a) State Newton’s law of gravitation. … … … [2] (b) A planet may be considered as a uniform sphere. A satellite is in circular orbit of period T around the planet at a height h above the surface. The height of the orbit can be adjusted by use of the satellite’s rocket engines. 2 Fig. 1.1 shows the variation with h of T 3 . 1600 1200 2 2 T 3 /s 3 800 400 0 0 2 4 6 8 10 12 h / 106 m Fig. 1.1 (i) By reference to forces, explain why the orbit of the satellite is circular. … … … [2] (ii) Use Newton’s law of gravitation to show that h and T are related by GA 2 (h + B)3 = T 4π2 where G is the gravitational constant and A and B are constants that depend on the properties of the planet. [3] (iii) Use the gradient and intercept of the line in Fig. 1.1 to determine values for A and B. Give units with your answers. A = … unit … B = … unit … [5] [Total: 12]

12 marks

Mark scheme: Question Answer Marks 1(a) (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(b)(i) (gravitational) force acts perpendicular to direction of motion B1 gravitational force provides centripetal acceleration B1 1(b)(ii) (F =) GMm / x2 = mx2 and = 2 / T M1 or GMm / x2 = 42mx / T2 completion of algebra leading to x3 = GMT2 / 42 A1 clear indication that B = radius of planet and that A = mass (of planet) B1 1(b)(iii) gradient = 3√(42 / GA) C1 e.g. (1280 – 360) / (12  106) = 3√(42 / [6.67  10–11  A]) C1 A = 1.3  1024 kg A1 intercept = gradient  B C1 e.g. 360 = ((1280 – 360)  B) / (12  106) A1 B = 4.7  106 m

This question in 9702/41 Oct/Nov 2024

1 (a) State Newton’s law of gravitation. … … … [2] (b) A planet may be considered as a uniform sphere. A satellite is in circular orbit of period T around the planet at a height h above the surface. The height of the orbit can be adjusted by use of the satellite’s rocket engines. 2 Fig. 1.1 shows the variation with h of T 3 . 1600 1200 2 2 T 3 /s 3 800 400 0 0 2 4 6 8 10 12 h / 106 m Fig. 1.1 (i) By reference to forces, explain why the orbit of the satellite is circular. … … … [2] (ii) Use Newton’s law of gravitation to show that h and T are related by GA 2 (h + B)3 = T 4π2 where G is the gravitational constant and A and B are constants that depend on the properties of the planet. [3] (iii) Use the gradient and intercept of the line in Fig. 1.1 to determine values for A and B. Give units with your answers. A = … unit … B = … unit … [5] [Total: 12]

12 marks

Mark scheme: Question Answer Marks 1(a) (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(b)(i) (gravitational) force acts perpendicular to direction of motion B1 gravitational force provides centripetal acceleration B1 1(b)(ii) (F =) GMm / x2 = mx2 and = 2 / T M1 or GMm / x2 = 42mx / T2 completion of algebra leading to x3 = GMT2 / 42 A1 clear indication that B = radius of planet and that A = mass (of planet) B1 1(b)(iii) gradient = 3√(42 / GA) C1 e.g. (1280 – 360) / (12  106) = 3√(42 / [6.67  10–11  A]) C1 A = 1.3  1024 kg A1 intercept = gradient  B C1 e.g. 360 = ((1280 – 360)  B) / (12  106) A1 B = 4.7  106 m

This question in 9702/43 Oct/Nov 2024

2 (a) State Newton’s law of gravitation. … … … [2] (b) One of the basic assumptions of the kinetic theory of gases is that there are no forces exerted between the molecules of the gas except during collisions. State two other basic assumptions of the kinetic theory of gases. 1 … … 2 … … [2] (c) Hydrogen gas consists of molecules that each have a mass of 3.34 × 10–27 kg. Hydrogen may be considered to be an ideal gas. A spherical balloon contains 0.0160 mol of hydrogen gas at a temperature of 282 K. At this temperature, the volume of gas in the balloon is 1.87 × 10– 4 m3. (i) Determine the pressure of the gas. pressure = … Pa [2] (ii) Estimate the average separation of the hydrogen molecules in the gas. average separation = … m [2] (d) (i) Use your answer in (c)(ii) to calculate the average gravitational force between adjacent molecules in hydrogen gas. average force = … N [2] (ii) By considering the weight of a molecule, suggest with a reason whether your answer in (d)(i) is consistent with the assumption of the kinetic theory of gases that there are no forces exerted between molecules. … … … [1] [Total: 11]

11 marks

Mark scheme: 2(a) (gravitational) force is (directly) proportional to product of masses B1 force (between point masses) is inversely proportional to the square of their separation B1 2(b) Any two points from: B2 • molecules are in continuous random motion • molecules have negligible volume compared with volume of gas • collisions (involving molecules) are (perfectly) elastic • collisions (of molecules) are instantaneous 2(c)(i) pV = nRT C1 p = (0.0160  8.31  282) / (1.87  10–4) A1 = 2.01  105 Pa 2(c)(ii) number of molecules = 0.0160  6.02  1023 C1 separation = 3√[(1.87  10–4) / (0.0160  6.02  1023)] A1 = 2.7  10–9 m (allow any answer that is 3 10–9 m to one significant figure) 2(d)(i) F = 6.67  10–11  (3.34  10–27)2 / (2.7  10–9)2 C1 = 1.0  10–46 N A1 2(d)(ii) numerical comparison between 10–46 N (F) and 10–26 N (the weight of molecule) leading to a conclusion that the assumption B1 is supported

This question in 9702/42 Oct/Nov 2025

1 (a) State Newton’s law of gravitation. … … … [2] (b) A binary star consists of star A, of mass 4.0 × 1030 kg, and star B, of mass 2.0 × 1030 kg, separated by a distance of 3.3 × 1012 m. The stars are both in circular orbit around their common centre of gravity X, as shown in Fig. 1.1. orbit of A 3.3 × 1012 m star B star A X RA RB orbit of B Fig. 1.1 The radius RB of the orbit of star B is double the radius RA of the orbit of star A. (i) Use Newton’s law of gravitation to calculate the magnitude of the gravitational force exerted by each star on the other. force = … N [2] (ii) Calculate the centripetal acceleration of star A. acceleration = … m s–2 [1] (iii) Use your answer in (b)(ii) to determine the period of the orbit of star A. period = … s [3] (iv) By placing a tick (✓) in each row, complete Table 1.1 to show how the quantities indicated for star B compare with the same quantities for star A. Table 1.1 B less than A B equal to A B greater than A centripetal acceleration linear speed period [3] [Total: 11]

11 marks

Mark scheme: Question Answer Marks 1(a) (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(b)(i) force = (6.67  10–11  4.0  1030  2.0  1030) / (3.3  1012)2 C1 = 4.9  1025 N A1 1(b)(ii) a = F / m = (4.9  1025) / (4.0  1030) A1 = 1.2  10–5 m s–2 1(b)(iii) a = r2 and = 2 / T C1 or a = v2 / r and v = (2r / T) r = (3.3  1012) / 3 C1 T = 2 [(3.3  1012) / (3  1.2  10–5)]½ A1 = 1.9  109 s 1(b)(iv) acceleration: B greater than A B1 linear speed: B greater than A B1 period: B equal to A B1

This question in 9702/44 Oct/Nov 2025