Gravitational field worksheet answers for A Level physics (free PDF download)

Problem 1 Key relationships

1. F = G m₁ m₂ ÷ r².
2. g = G M ÷ r².
3. V = −G M ÷ r.
4. v = √(G M ÷ r) and T² = 4π² r³ ÷ (G M).
F = Gm₁m₂/r²; g = GM/r²; V = −GM/r; v = √(GM/r).

Problem 2 Two people close together

1. F = G m₁ m₂ ÷ r² = 6.67 × 10⁻¹¹ × 70 × 70 ÷ 1.0² = 3.27 × 10⁻⁷ N.
2. Weight = 70 × 9.81 = 687 N, so the force is about 5 × 10⁻¹⁰ of the weight, which is negligible.
F = 3.3 × 10⁻⁷ N.

Problem 3 Surface gravity of the Earth

1. g = G M ÷ R² = 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ ÷ (6.37 × 10⁶)² = 3.98 × 10¹⁴ ÷ 4.06 × 10¹³ = 9.81 N/kg.
2. g = F ÷ m and F = m a for a freely falling mass, so a = g. Hence N/kg and m/s² are equivalent.
g = 9.81 N/kg.

Problem 4 Weight on the Moon

1. g = 6.67 × 10⁻¹¹ × 7.35 × 10²² ÷ (1.74 × 10⁶)² = 4.90 × 10¹² ÷ 3.03 × 10¹² = 1.62 N/kg.
2. Moon: 60 × 1.62 = 97 N. Earth: 60 × 9.81 = 589 N.
3. The mass is unchanged at 60 kg.
g = 1.62 N/kg; 97 N; 589 N; 60 kg.

Problem 5 Inverse-square thinking

1. F ∝ 1/r², so the force is F ÷ 9.
2. F ∝ m₁, so the force is 2F.
3. 2 × 2 ÷ 2² = 1, so the force stays F.
4. 1 ÷ (½)² = 4, so the force is 4F.
F/9; 2F; F; 4F.

Problem 6 Potential at the surface

1. V = −G M ÷ R = −3.98 × 10¹⁴ ÷ 6.37 × 10⁶ = −6.25 × 10⁷ J/kg.
2. The potential is lower than at infinity, so work must be done on the mass to remove it.
3. ΔE = m × (0 − V) = 2.0 × 6.25 × 10⁷ = 1.25 × 10⁸ J.
V = −6.25 × 10⁷ J/kg; 1.25 × 10⁸ J.

Problem 7 Moons of Jupiter

1. T = 1.77 × 86 400 = 1.53 × 10⁵ s.
2. M = 4π² r³ ÷ (G T²) = 39.48 × (4.22 × 10⁸)³ ÷ (6.67 × 10⁻¹¹ × (1.529 × 10⁵)²) = 2.97 × 10²⁷ ÷ 1.56 = 1.90 × 10²⁷ kg.
M = 1.9 × 10²⁷ kg.

Problem 8 Half the surface gravity

1. g ∝ 1/r², so g ÷ g₀ = R² ÷ r² = ½ gives r² = 2R² and r = √2 R.
2. h = r − R = (√2 − 1) × 6.37 × 10⁶ = 0.414 × 6.37 × 10⁶ = 2.64 × 10⁶ m.
h = 2.64 × 10⁶ m.

Problem 9 Reading a g–r graph

1. At r = 2R, g = 2.45 N/kg.
2. In units of R², g r² = 9.81 × 1 = 9.81; 2.45 × 4 = 9.80; 1.09 × 9 = 9.81. The product is constant, so g ∝ 1/r².
3. ΔV = G M (1/R − 1/3R) = 6.25 × 10⁷ × (2/3) = 4.17 × 10⁷ J/kg, which equals the area under the graph between R and 3R.
ΔV = 4.17 × 10⁷ J/kg.

Problem 10 Orbit of the space station

1. r = 6.37 × 10⁶ + 4.1 × 10⁵ = 6.78 × 10⁶ m.
2. v = √(G M ÷ r) = √(3.98 × 10¹⁴ ÷ 6.78 × 10⁶) = √(5.87 × 10⁷) = 7.66 × 10³ m/s.
3. T = 2π r ÷ v = 2π × 6.78 × 10⁶ ÷ 7663 = 5560 s = 92.7 min.
v = 7.66 km/s; T = 92.7 min.

Problem 11 Kepler on the planets

1. T = r^1.5 = 1.52 × √1.52 = 1.52 × 1.233 = 1.87 years.
2. r³ = T² = 5776, so r = 5776^(1/3) = 17.9 AU.
T = 1.87 years; r = 17.9 AU.

Problem 12 Weighing the Sun

1. The Sun’s pull supplies the centripetal force: G M m ÷ r² = m (4π² r ÷ T²), so M = 4π² r³ ÷ (G T²).
2. M = 39.48 × (1.50 × 10¹¹)³ ÷ (6.67 × 10⁻¹¹ × (3.156 × 10⁷)²) = 1.332 × 10³⁵ ÷ 6.64 × 10⁴ = 2.01 × 10³⁰ kg.
3. 2.01 × 10³⁰ ÷ 5.97 × 10²⁴ = 3.4 × 10⁵, so the Sun is about 340 000 times as massive as the Earth.
M = 2.01 × 10³⁰ kg.

Problem 13 Leaving the Earth and Moon

1. v = √(2 G M ÷ R) = √(2 × 3.98 × 10¹⁴ ÷ 6.37 × 10⁶) = 1.12 × 10⁴ m/s.
2. v = √(2 × 4.90 × 10¹² ÷ 1.74 × 10⁶) = √(5.63 × 10⁶) = 2.37 × 10³ m/s.
3. KE = G M m ÷ R = 6.25 × 10⁷ × 500 = 3.13 × 10¹⁰ J.
11.2 km/s; 2.37 km/s; 3.13 × 10¹⁰ J.

Problem 14 Sun against Moon

1. Ratio = (M_S ÷ M_M) × (d_M ÷ d_S)² = 2.71 × 10⁷ × (2.56 × 10⁻³)² = 177.
2. Tides depend on the difference in the pull across the Earth, which falls as 1/d³. The Moon is much closer, so its difference is larger.
Sun’s pull is about 177 times larger.

Problem 15 No gravity in orbit?

1. g = 3.98 × 10¹⁴ ÷ (6.77 × 10⁶)² = 8.69 N/kg.
2. 8.69 ÷ 9.81 × 100 % = 88.6 %.
3. The station and astronauts fall together with the same acceleration, so there is no contact force between them and they feel weightless.
g = 8.69 N/kg (89 %).

Problem 16 The neutral point

1. G M_E ÷ x² = G M_M ÷ (d − x)².
2. Rearranging, (d − x)² ÷ x² = M_M ÷ M_E, so x ÷ (d − x) = √(M_E ÷ M_M) = √81.2 = 9.01.
3. x = 9.01 (d − x) gives x = 9.01 d ÷ 10.01 = 0.900 × 3.84 × 10⁸ = 3.46 × 10⁸ m from the Earth’s centre.
x = 3.46 × 10⁸ m.

Problem 17 Energy to reach orbit

1. KE = G M m ÷ 2r = 3.98 × 10¹⁴ × 1200 ÷ (1.6 × 10⁷) = 2.99 × 10¹⁰ J.
2. E_total = −G M m ÷ 2r = −2.99 × 10¹⁰ J.
3. At rest on the ground E = −G M m ÷ R = −3.98 × 10¹⁴ × 1200 ÷ 6.37 × 10⁶ = −7.50 × 10¹⁰ J.
4. Energy supplied = −2.99 × 10¹⁰ − (−7.50 × 10¹⁰) = 4.51 × 10¹⁰ J.
KE = 2.99 × 10¹⁰ J; energy supplied = 4.51 × 10¹⁰ J.

Problem 18 Binary stars

1. The stars have equal masses, so the centre of mass is at the midpoint and both circle it with the same ω.
2. G M² ÷ d² = M ω² (d ÷ 2), so ω² = 2 G M ÷ d³.
3. ω² = 2 × 6.67 × 10⁻¹¹ × 2.0 × 10³⁰ ÷ (3.0 × 10¹¹)³ = 9.88 × 10⁻¹⁵, so ω = 9.94 × 10⁻⁸ rad/s and T = 2π ÷ ω = 6.32 × 10⁷ s = 2.0 years.
4. v = ω × d ÷ 2 = 9.94 × 10⁻⁸ × 1.5 × 10¹¹ = 1.49 × 10⁴ m/s.
T = 2.0 years; v = 14.9 km/s.

Problem 19 Orbit decay paradox

1. v₁ = √(3.98 × 10¹⁴ ÷ 6.78 × 10⁶) = 7664 m/s and v₂ = √(3.98 × 10¹⁴ ÷ 6.68 × 10⁶) = 7720 m/s.
2. E = −G M m ÷ 2r: E₁ = −1.468 × 10¹⁰ J and E₂ = −1.490 × 10¹⁰ J. The total energy fell by 2.2 × 10⁸ J.
3. Gravitational potential energy falls by twice as much (4.4 × 10⁸ J), so half of it becomes extra kinetic energy (2.2 × 10⁸ J) and the other half is the energy dissipated by drag. The kinetic energy rises, so the speed rises.
v₁ = 7.66 km/s; v₂ = 7.72 km/s; total energy lost 2.2 × 10⁸ J.

Problem 20 Atmosphere of the Moon

1. N₂: v = √(3 × 1.38 × 10⁻²³ × 300 ÷ 4.65 × 10⁻²⁶) = √(2.67 × 10⁵) = 517 m/s. H₂: v = √(3 × 1.38 × 10⁻²³ × 300 ÷ 3.34 × 10⁻²⁷) = 1930 m/s.
2. Earth: 11 200 ÷ 6 = 1870 m/s. Moon: 2370 ÷ 6 = 395 m/s.
3. Earth: nitrogen (517 m/s) is below 1870 m/s so it stays; hydrogen (1930 m/s) is above it and escapes. Moon: both are above 395 m/s, so the Moon cannot keep an atmosphere.
517 m/s and 1930 m/s; Earth keeps nitrogen but loses hydrogen; the Moon loses both.