A2 · Paper 4 practice · Gravitational fields

The pull of a mass, from surface to orbit.

A full set of ten original structured questions in the style of Paper 4, covering the whole of Gravitational fields: field strength and Newton's law, orbits, and potential and energy. Each question is linked to the lesson it draws on; attempt them all, then reveal the worked solutions.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level Paper 4 style. They are not from past papers. They test the same concepts and skills the syllabus rewards. Take G = 6.67 × 10⁻¹¹ N m² kg⁻².

Keep these straight

One mass, one field.

Newton's law: F = Gm₁m₂ / r², with uniform spheres treated as point masses at their centres.
Field strength and potential: g = GM / r² and ϕ = −GM / r, so g is the gradient of ϕ.
Orbits: gravity is the centripetal force, GMm / r² = mv² / r, giving v = √(GM / r) and T = 2π√(r³ / GM).

Paper 4

[13 marks]

Field & Newton's law →

A planet P has mass M = 4.8 × 10²⁴ kg and radius R = 5.6 × 10⁶ m. A satellite of mass m = 320 kg is in a circular orbit at a height h = 2.0 × 10⁶ m above the surface of P, as shown in Fig. 1.1.

Fig. 1.1

(a) State Newton's law of gravitation. [2]

(b) Show that the gravitational field strength at the surface of P is about 10 N kg⁻¹. [2]

(c) For the orbiting satellite, calculate:

(i) the orbital radius r, [1]

(ii) the gravitational force acting on it, [2]

(iii) its orbital speed, [2]

(iv) its period of orbit. [2]

(d) State and explain whether the orbital speed would change if a satellite of larger mass were placed in the same orbit. [2]

(a)The gravitational force between two point masses is proportional to the product of the masses and inversely proportional to the square of their separation ✓, directed along the line joining them (attractive) ✓
(b)g = GM / R² = (6.67 × 10⁻¹¹ × 4.8 × 10²⁴) / (5.6 × 10⁶)² ✓ = 3.20 × 10¹⁴ / 3.14 × 10¹³ = 10.2 N kg⁻¹, about 10 N kg⁻¹ ✓
(c)(i)r = R + h = 5.6 × 10⁶ + 2.0 × 10⁶ = 7.6 × 10⁶ m ✓
(c)(ii)F = GMm / r² = (6.67 × 10⁻¹¹ × 4.8 × 10²⁴ × 320) / (7.6 × 10⁶)² ✓ = 1.8 × 10³ N ✓
(c)(iii)v = √(GM / r) = √(3.20 × 10¹⁴ / 7.6 × 10⁶) ✓ = √(4.21 × 10⁷) = 6.5 × 10³ m s⁻¹ ✓
(c)(iv)T = 2πr / v = (2π × 7.6 × 10⁶) / 6.5 × 10³ ✓ = 7.4 × 10³ s (about 2.0 h) ✓
(d)No change ✓. Gravity provides the centripetal force: GMm / r² = mv² / r, so v = √(GM / r); the satellite mass cancels and v depends only on M and r ✓

Paper 4

[12 marks]

Orbits & field strength →

A satellite is in a geostationary orbit around the Earth, which has mass M = 6.0 × 10²⁴ kg. The orbit is shown in Fig. 2.1.

Fig. 2.1

(a) State two features of a geostationary orbit. [2]

(b) Show that the radius of the orbit is about 4.2 × 10⁷ m. [3]

(c) Calculate the orbital speed of the satellite. [2]

(d) Calculate the gravitational field strength of the Earth at this orbital radius. [2]

(e) Explain why a satellite in a lower circular orbit cannot remain above the same point on the Earth's surface. [2]

(f) State one practical application that the geostationary orbit makes possible. [1]

(a)Any two: period of 24 h (equal to Earth's rotation period) ✓; orbit lies in the equatorial plane ✓; moves in the same direction (west to east) as the Earth's rotation, so it stays above one fixed point.
(b)Gravity is the centripetal force: GMm / r² = mr(2π/T)², so r³ = GMT² / 4π² ✓. With T = 24 h = 8.64 × 10⁴ s, r³ = (6.67 × 10⁻¹¹ × 6.0 × 10²⁴ × (8.64 × 10⁴)²) / 4π² ✓ = 7.57 × 10²², r = 4.2 × 10⁷ m ✓
(c)v = 2πr / T = (2π × 4.23 × 10⁷) / 8.64 × 10⁴ ✓ = 3.1 × 10³ m s⁻¹ ✓
(d)g = GM / r² = (6.67 × 10⁻¹¹ × 6.0 × 10²⁴) / (4.23 × 10⁷)² ✓ = 0.22 N kg⁻¹ ✓
(e)A smaller radius gives a shorter period (T² ∝ r³), so T < 24 h ✓; the satellite then completes an orbit faster than the Earth turns, so it moves relative to the ground and cannot stay above one point ✓
(f)Communications or television relay using a fixed (non-tracking) ground aerial ✓

Paper 4

[13 marks]

Gravitational potential →

A space probe of mass m = 1200 kg moves in the gravitational field of a planet of mass M = 3.0 × 10²⁵ kg. Fig. 3.1 shows how the gravitational potential φ varies with distance r from the centre of the planet.

Fig. 3.1

(a) Define gravitational potential at a point. [2]

(b) Calculate the gravitational potential at a point r₁ = 8.0 × 10⁶ m from the centre of the planet. [2]

(c) Calculate the gravitational potential energy of the probe at this point. [2]

(d) The probe moves outward from r₁ = 8.0 × 10⁶ m to r₂ = 2.0 × 10⁷ m. Calculate the change in its potential energy, and state whether work is done on or by the probe. [3]

(e) Explain why values of gravitational potential are always negative. [2]

(f) State how the potential φ changes as r increases without limit. [2]

(a)The gravitational potential at a point is the work done per unit mass ✓ in bringing a small test mass from infinity to that point ✓
(b)φ = −GM / r = −(6.67 × 10⁻¹¹ × 3.0 × 10²⁵) / 8.0 × 10⁶ ✓ = −2.5 × 10⁸ J kg⁻¹ ✓
(c)Eₕ = mφ = 1200 × (−2.5 × 10⁸) ✓ = −3.0 × 10¹¹ J ✓
(d)φ₂ = −GM / r₂ = −1.0 × 10⁸ J kg⁻¹, so Eₕ₂ = 1200 × (−1.0 × 10⁸) = −1.2 × 10¹¹ J ✓. ΔEₕ = −1.2 × 10¹¹ − (−3.0 × 10¹¹) = +1.8 × 10¹¹ J ✓. The potential energy increases, so work is done on the probe (against the gravitational field) ✓
(e)The zero of potential is defined at infinity ✓; gravity is attractive, so the field does positive work on a mass brought in from infinity, leaving the potential below zero everywhere, hence negative ✓
(f)As r increases, φ = −GM / r increases (becomes less negative) ✓ and tends toward zero as r tends to infinity ✓

Paper 4

[10 marks]

Field & Newton's law →

This question is about gravitational field strength and Newton’s law of gravitation. Take G = 6.67 × 10⁻¹¹ N m² kg⁻², and for the Earth M = 6.0 × 10²⁴ kg and R = 6.4 × 10⁶ m.

(a) Define gravitational field strength. [2]

(b) State Newton’s law of gravitation. [2]

(c) Show that the field strength at distance r from a point mass M is g = GM / r². [2]

(d) Calculate the gravitational field strength at the Earth’s surface. [2]

(e) Calculate the weight of a 70 kg astronaut at the Earth’s surface. [1]

(f) State and explain how g changes if the astronaut moves to twice the distance from the Earth’s centre. [1]

(a)The gravitational force per unit mass acting on a small test mass placed at the point ✓✓
(b)The gravitational force between two point masses is proportional to the product of the masses and inversely proportional to the square of their separation ✓✓
(c)g = F/m = (GMm/r²)/m = GM/r² ✓✓
(d)g = GM/R² = (6.67 × 10⁻¹¹ × 6.0 × 10²⁴)/(6.4 × 10⁶)² ✓ = 9.8 N kg⁻¹ ✓
(e)W = mg = 70 × 9.8 = 6.9 × 10² N ✓
(f)g ∝ 1/r², so at twice the distance g falls to one quarter (about 2.4 N kg⁻¹) ✓

Paper 4

[10 marks]

Field & Newton's law →

Planet P has twice the mass of the Earth and twice its radius, as shown in Fig. 5.1. The Earth has surface field strength 9.8 N kg⁻¹.

Fig. 5.1

(a) Write the expression for the gravitational field strength at the surface of a planet of mass M and radius R. [2]

(b) Show that the surface field strength of Planet P is half that of the Earth. [3]

(c) Calculate the surface field strength of Planet P. [1]

(d) A 5.0 kg mass is taken from the Earth’s surface to the surface of Planet P. Compare its weight in the two places. [2]

(e) Explain why surface field strength depends on both the mass and the radius of a planet. [2]

(a)g = GM/R² ✓✓
(b)gₚ = G(2M)/(2R)² = 2GM/4R² = ½(GM/R²) ✓✓ = ½ gₐ ✓
(c)gₚ = ½ × 9.8 = 4.9 N kg⁻¹ ✓
(d)On Earth W = 5.0 × 9.8 = 49 N ✓; on Planet P W = 5.0 × 4.9 = 25 N (half) ✓
(e)g = GM/R²: a larger mass increases the field, while a larger radius places the surface further from the centre and reduces it through the inverse-square dependence ✓✓

Paper 4

[10 marks]

Orbits & field strength →

A satellite orbits the Earth (M = 6.0 × 10²⁴ kg) in a circular orbit of radius r = 7.0 × 10⁶ m.

(a) State what provides the centripetal force on the satellite. [1]

(b) Show that the orbital speed is given by v = √(GM/r). [2]

(c) Calculate the orbital speed. [2]

(d) Calculate the orbital period. [2]

(e) State and explain how the orbital speed differs for a satellite in a higher orbit. [2]

(f) State the direction of the satellite’s acceleration. [1]

(a)The gravitational attraction of the Earth on the satellite ✓
(b)Gravity provides the centripetal force: GMm/r² = mv²/r ✓, so v² = GM/r and v = √(GM/r) ✓
(c)v = √(6.67 × 10⁻¹¹ × 6.0 × 10²⁴ / 7.0 × 10⁶) ✓ = 7.6 × 10³ m s⁻¹ ✓
(d)T = 2πr/v = (2π × 7.0 × 10⁶)/(7.56 × 10³) ✓ = 5.8 × 10³ s (about 97 min) ✓
(e)v ∝ 1/√r, so a satellite in a higher orbit moves more slowly ✓✓
(f)Directed toward the centre of the Earth (centripetal) ✓

Paper 4

[10 marks]

Orbits & field strength →

This question is about Kepler’s third law for circular orbits. The Moon orbits the Earth at radius r = 3.8 × 10⁸ m with a period of 27.3 days.

(a) State Kepler’s third law for circular orbits. [1]

(b) Show that for a circular orbit T² = 4π²r³ / GM. [3]

(c) Use the Moon’s orbital data to estimate the mass of the Earth. [3]

(d) State one assumption made in this calculation. [1]

(e) Two satellites orbit at radii r and 4r. State and explain the ratio of their periods. [2]

(a)The square of the orbital period is proportional to the cube of the orbital radius (T² ∝ r³) ✓
(b)GMm/r² = mrω² = mr(2π/T)² ✓; rearranging, T² = 4π²r³/GM ✓✓
(c)M = 4π²r³/(GT²); T = 27.3 × 86 400 = 2.36 × 10⁶ s ✓; M = 4π² × (3.8 × 10⁸)³ / (6.67 × 10⁻¹¹ × (2.36 × 10⁶)²) ✓ = 5.8 × 10²⁴ kg ✓
(d)The orbit is circular, or the Earth’s mass greatly exceeds the Moon’s ✓
(e)T ∝ r³⁄², so the ratio is 4³⁄² = 8 ✓: the outer satellite has a period eight times the inner one ✓

Paper 4

[11 marks]

Gravitational potential →

This question is about gravitational potential and potential energy near the Earth (M = 6.0 × 10²⁴ kg).

(a) Define gravitational potential at a point. [2]

(b) Explain why gravitational potential is always negative. [2]

(c) Write the expression for the gravitational potential energy of a mass m at distance r from a mass M. [1]

(d) Calculate the gravitational potential energy of a 500 kg satellite at r = 7.0 × 10⁶ m. [2]

(e) Calculate the work done to raise the satellite to r = 1.4 × 10⁷ m. [3]

(f) State how the gravitational potential varies with distance r. [1]

(a)The work done per unit mass in bringing a small test mass from infinity to the point ✓✓
(b)The zero of potential is taken at infinity, and gravity is attractive, so the field does positive work bringing a mass inward, leaving the potential below zero everywhere ✓✓
(c)Eₙ = −GMm/r ✓
(d)Eₙ = −(6.67 × 10⁻¹¹ × 6.0 × 10²⁴ × 500)/(7.0 × 10⁶) ✓ = −2.9 × 10¹⁰ J ✓
(e)At r = 1.4 × 10⁷ m (double), Eₙ = −1.43 × 10¹⁰ J ✓; work done = ΔEₙ = (−1.43 × 10¹⁰) − (−2.86 × 10¹⁰) ✓ = +1.4 × 10¹⁰ J ✓
(f)φ = −GM/r, so its magnitude varies as 1/r, rising toward zero as r increases ✓

Paper 4

[12 marks]

Gravitational potential →Orbits & field strength →

A satellite of mass 500 kg orbits the Earth (M = 6.0 × 10²⁴ kg) in a circular orbit of radius r = 7.0 × 10⁶ m. This question connects the orbital motion of Q6 with the energy ideas of Q8.

(a) Write expressions, in terms of G, M, m and r, for the kinetic energy and the gravitational potential energy of the satellite. [2]

(b) Show that the total energy of the satellite is E = −GMm / 2r. [2]

(c) Calculate the kinetic energy, the potential energy and the total energy of this satellite. [3]

(d) The satellite is moved to a higher orbit. State and explain what happens to its kinetic energy and to its total energy. [3]

(e) Explain how a satellite in a higher orbit can move more slowly yet have a greater total energy. [2]

(a)KE = ½mv² = GMm/2r (using v² = GM/r) ✓; Eₙ = −GMm/r ✓
(b)E = KE + Eₙ = GMm/2r − GMm/r = −GMm/2r ✓✓
(c)KE = +1.4 × 10¹⁰ J ✓; Eₙ = −2.9 × 10¹⁰ J ✓; total E = −1.4 × 10¹⁰ J ✓
(d)At larger r the speed is smaller, so the kinetic energy decreases ✓; the total energy E = −GMm/2r increases (becomes less negative) ✓✓
(e)The gain in potential energy is larger than the loss in kinetic energy, so the total energy rises toward zero even though the satellite is slower ✓✓

Paper 4

[12 marks]

Gravitational potential →Field & Newton's law →

Fig. 10.1 shows how the gravitational potential φ near a planet varies with distance r from its centre. This question connects potential (Q8) with field strength (Q4).

Fig. 10.1

(a) State the relationship between gravitational field strength and gravitational potential. [2]

(b) Explain how the field strength at a point can be obtained from the φ-r graph. [1]

(c) Compare how field strength and potential each vary with distance from a point mass, referring to their different powers of r. [3]

(d) State the relationship between gravitational field lines and equipotential surfaces. [2]

(e) The potential at the planet’s surface is −6.3 × 10⁷ J kg⁻¹. Calculate the energy needed to remove a 2.0 kg mass completely from the surface to infinity. [3]

(f) State the value of the gravitational potential at infinity. [1]

(a)The field strength is the negative gradient of the potential: g = −dφ/dr ✓✓
(b)It is the magnitude of the gradient (steepness) of the φ-r graph at that point ✓
(c)Field strength varies as 1/r² (an inverse-square fall-off, steeper) ✓; potential varies as 1/r (a gentler fall-off) ✓; so the potential reaches further before becoming negligible ✓
(d)Field lines are perpendicular to equipotential surfaces ✓ and point from high to low potential (toward the planet) ✓
(e)Energy = m(φ∞ − φₙ) = 2.0 × (0 − (−6.3 × 10⁷)) ✓✓ = 1.3 × 10⁸ J ✓
(f)Zero ✓

Mark this once you have attempted all ten questions and checked your working against the solutions. Revealing the solutions alone does not count.