Gravitational fields

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 4×π2×1.2×1034 C1 M = 6.67×10 −11×2.4×( 365×24×3600 )2 = 3.0×1030 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) GMm mv2 GM GM M1 = O R v2 = O R v = r2 r r r GMm A1 use of ½ mv2 (e.g. multiplication by ½ m) leading to 2r 1(c)(iii) −GM −GMm C1 Ep = φ m and φ = or E = r p r Total energy = E + E k p GMm −GMm −GMm A1 Total energy= + = 2r r 2r © UCLES 2021 Page 8 of 19

1(a) force per unit mass B1 1(b) GMm / r 2 = mrω 2 and ω = 2π/T C1 or GMm / r 2 = mv2 / r and v = 2πr / T 6.67 × 10–11 × 6.0 × 1024 = r3 × [2π / (94 × 60)]2 C1 r = 6.9 × 106 m A1 1(c)(i) r3ω2 = constant or r3 / T2 = constant C1 r3 / (6.9 × 106)3 = (150 / 94)2 so r = 9.4 × 106 m A1 or GMT2/4π2 = r3 and clear that M is 6.0 × 1024 (C1) 6.67 × 10–11 × 6.0 × 1024 = r3 × [2π / (150 × 60)]2 (A1) so r = 9.4 × 106 m 1(c)(ii) separation increases so (potential energy) increases B1 or movement is against gravitational force so (potential energy) increases 1(c)(iii) potential energy = (–)GMm / r C1 ΔE = 6.67 × 10–11 × 6.0 × 1024 × 1200 × [(6.9 × 106)–1 – (9.4 × 106)–1] C1 P = 1.9 × 1010 J A1 © UCLES 2021 Page 8 of 18

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 A1 = 3.73 N kg–1 1(b)(ii) a = rω2 and ω = 2π / T C1 or a = v2 / r and v = 2πr / T a = 3.39 × 106 × (2π / (24.6 × 3600))2 A1 = 0.0171 m s–2 1(b)(iii) force per unit mass = 3.73 – 0.0171 A1 = 3.71 N kg–1 © UCLES 2021 Page 8 of 19

1(a) force per unit mass B1 1(b) GMm / r 2 = mrω 2 and ω = 2π/T C1 or GMm / r 2 = mv2 / r and v = 2πr / T 6.67 × 10–11 × 6.0 × 1024 = r3 × [2π / (94 × 60)]2 C1 r = 6.9 × 106 m A1 1(c)(i) r3ω2 = constant or r3 / T2 = constant C1 r3 / (6.9 × 106)3 = (150 / 94)2 so r = 9.4 × 106 m A1 or GMT2/4π2 = r3 and clear that M is 6.0 × 1024 (C1) 6.67 × 10–11 × 6.0 × 1024 = r3 × [2π / (150 × 60)]2 (A1) so r = 9.4 × 106 m 1(c)(ii) separation increases so (potential energy) increases B1 or movement is against gravitational force so (potential energy) increases 1(c)(iii) potential energy = (–)GMm / r C1 ΔE = 6.67 × 10–11 × 6.0 × 1024 × 1200 × [(6.9 × 106)–1 – (9.4 × 106)–1] C1 P = 1.9 × 1010 J A1 © UCLES 2021 Page 8 of 18

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 A1 leading to x = 3.5 × 108 (m) 2(b)(iii) φ (Earth) = (–)6.67 × 10–11 × (5.98 × 1024 / 3.5 × 108) C1 and φ (Moon) = (–)6.67 × 10–11 × (7.35 × 1022 / 0.38 × 108) 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 © UCLES 2021 Page 8 of 15

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 B2 • 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 2(b)(iii) sketch: one curve, starting with gradient of decreasing magnitude at 2R and finishing with gradient of increasing magnitude B1 at D – R 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 © UCLES 2021 Page 8 of 19

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 A1 leading to x = 3.5 × 108 (m) 2(b)(iii) φ (Earth) = (–)6.67 × 10–11 × (5.98 × 1024 / 3.5 × 108) C1 and φ (Moon) = (–)6.67 × 10–11 × (7.35 × 1022 / 0.38 × 108) 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 © UCLES 2021 Page 8 of 15

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: B2 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) 1(c)(i) GMm M1 = mrω2 r2 r3ω2 gradient A1 M= and gradient = r3ω2 hence M= G G or gradient r3 = GM × 1/ω2 so gradient = GM hence M= G 1(c)(ii) M = 4.1 × 1023 / (6.0 × 107 × 6.67 × 10–11) = 1.0 × 1026 kg B1 © UCLES 2022 Page 6 of 17 1(c)(iii) GMm mv2 C1 = r2 r GM = v2 r 6.67×10 −11×1.0×1026 C1 v2= 1.2×108 v2 =5.6×107m s −1 v =7500 m s −1 A1 Question Answer Marks

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 A1 and so g = [GMm / r2] / m = GM / r2 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 © UCLES 2022 Page 7 of 16

1(a)(i) work (done) per unit mass B1 work (done on mass) in moving mass from infinity (to the point) B1 1(a)(ii) E = ϕm B1 P E = (– GM / r)  m = – GMm / r P or ϕ = – GM / r and E = ϕm = – GMm / r P 1(b)(i) E = 6.67  10–11  1.99  1030  2.20  1014  [1 / (6.38  1010) – 1 / (8.44  1011)] C1 P = 4.23  1023 J A1 1(b)(ii) (gravitational) force is attractive so decrease B1 or (gravitational) force does work so decrease 1(b)(iii) E = ½m(v 2 – v 2) C1 P 2 1 4.23  1023 = ½  2.20  1014  (v2 – 341002) C1 v (= 70800 m s–1) = 70.8 km s–1 A1 1(c) both PE and KE equations include m, so path is unchanged B1 © UCLES 2022 Page 7 of 16

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 A1 and so g = [GMm / r2] / m = GM / r2 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 © UCLES 2022 Page 7 of 16

1(a) F = (Gm m ) / r 2 M1 1 2 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 © UCLES 2022 Page 6 of 15

1(a) F = (Gm m ) / r 2 M1 1 2 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 © UCLES 2022 Page 6 of 15

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 C1 (R = 4.22  107 m) h = R – (6.37  106) C1 h = (4.22  107) – (6.37  106) A1 = 3.6  107 m 1(d)(i)  = 2 / T C1 = 2 / (24  60  60) A1 = 7.3  10–5 rad s–1 1(d)(ii) orbit is from east to west B1 orbit is not equatorial / orbit is polar B1 © UCLES 2023 Page 6 of 15

1(a) (gravitational) potential is zero at infinity B1 (gravitational force between two masses is attractive so) either work is done on a mass to move it away from another mass B1 or work is done on a mass to move it to infinity 1(b)(i) M = (–) gradient / G C1 e.g. M = (1.76  108) / (3.0  10–8  6.67  10–11) = 8.8  1025 kg A1 1(b)(ii) either GMm / r2 = mr2 and  = 2 / T C1 or GMm / r2 = mv2 / r and v = 2r / T or GMm / r2 = 42mr / T2 R3 = 6.67  10–11  8.8  1025  (0.72  24  60  60)2 / 42 C1 R = 8.3  107 m A1 1(b)(iii) E = (GMm / r) – ½mv2 C1 kinetic energy = (½  1200  84002) potential energy = (–)[(6.67  10–11  8.8  1025  1200) / (8.3  107)] C1 E = [(6.67  10–11  8.8  1025  1200) / (8.3  107)] A1 – (½  1200  84002) = 4.3  1010 J © Cambridge University Press & Assessment 2024 Page 6 of 14

1(a) work done per unit mass B1 work done moving mass from infinity (to the point) B1 1(b)(i) potential is zero at infinity B1 work is done by (two) masses in moving them closer together B1 or work is done on (two) masses in moving them apart 1(b)(ii) magnitude of potential shown as 4 B1 potential negative and shown as a multiple of – [potential = –4 if fully correct] B1 1(b)(iii) field strength at X:  / 4R A1 field strength at Y: 4 / R A1 potential energy at X: –M A1 potential energy at Y: –8M A1 © Cambridge University Press & Assessment 2024 Page 6 of 15

1(a) work done per unit mass B1 work done moving mass from infinity (to the point) B1 1(b)(i) potential is zero at infinity B1 work is done by (two) masses in moving them closer together B1 or work is done on (two) masses in moving them apart 1(b)(ii) magnitude of potential shown as 4 B1 potential negative and shown as a multiple of – [potential = –4 if fully correct] B1 1(b)(iii) field strength at X:  / 4R A1 field strength at Y: 4 / R A1 potential energy at X: –M A1 potential energy at Y: –8M A1 © Cambridge University Press & Assessment 2024 Page 6 of 15

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 © Cambridge University Press & Assessment 2024 Page 6 of 15

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 © Cambridge University Press & Assessment 2024 Page 6 of 15

1(a) work done per unit mass B1 work (done in) moving mass from infinity (to the point) B1 1(b)(i) evidence of addition of 3.4 × 106 to 1.7 × 106 or 6.8 × 106 C1 GM × 122 / (5.1 × 106) or GM × 122 / (10.2 × 106) C1 6.67 × 10–11 × M × 122 × [(5.1 × 106)–1 – (10.2 × 106)–1] = 5.1 × 108 A1 leading to M = 6.4 × 1023 kg 1(b)(ii)  = (–) (6.67 × 10–11 × 6.4 × 1023) / (3.4 × 106) C1 = –1.3 × 107 J kg–1 A1 1(c)(i) Mars takes (just under) 25 hours to rotate once on its axis B1 1(c)(ii) orbit is equatorial B1 or orbit is in same direction as direction of rotation of Mars © Cambridge University Press & Assessment 2025 Page 8 of 19

1(a) work done per unit mass B1 work (done in) moving mass from infinity (to the point) B1 1(b)(i) evidence of addition of 3.4 × 106 to 1.7 × 106 or 6.8 × 106 C1 GM × 122 / (5.1 × 106) or GM × 122 / (10.2 × 106) C1 6.67 × 10–11 × M × 122 × [(5.1 × 106)–1 – (10.2 × 106)–1] = 5.1 × 108 A1 leading to M = 6.4 × 1023 kg 1(b)(ii)  = (–) (6.67 × 10–11 × 6.4 × 1023) / (3.4 × 106) C1 = –1.3 × 107 J kg–1 A1 1(c)(i) Mars takes (just under) 25 hours to rotate once on its axis B1 1(c)(ii) orbit is equatorial B1 or orbit is in same direction as direction of rotation of Mars © Cambridge University Press & Assessment 2025 Page 8 of 19

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, –g ) and ending at (L – R, +g ) B1 0 0 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 © Cambridge University Press & Assessment 2025 Page 10 of 17

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, –g ) and ending at (L – R, +g ) B1 0 0 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 © Cambridge University Press & Assessment 2025 Page 10 of 17