A2 · Practice questions · Gravitational potential

Zero at infinity, so every value is negative.

Six original Cambridge-style questions on the definition of gravitational potential, φ = −GM / r, potential energy E = −GMm / r and the energy needed to move a mass in a field.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

Keep these straight

Potential is work per unit mass from infinity.

Definition: the gravitational potential at a point is the work done per unit mass in bringing a small test mass from infinity to that point; for a point mass φ = −GM / r, in J kg⁻¹.
Sign: the zero is at infinity and gravity is attractive, so the field does the work on the way in and φ is negative everywhere, rising toward zero as r increases.
Potential energy: E = mφ = −GMm / r; moving outward makes E less negative (work done on the mass), and E = mgh is only the near-surface approximation.

Analysis

[2 marks]

State the difference between gravitational potential and gravitational potential energy, and give the SI unit of each.

Gravitational potential is the work done per unit mass in bringing a small test mass from infinity to the point; it is a property of the point in the field, with unit J kg⁻¹. ✓
Gravitational potential energy is the work done in bringing a particular mass m from infinity to the point, E = mφ, so it depends on the mass as well as the point; its unit is the joule, J. ✓

Calculation

[3 marks]

The Earth has mass 6.0 × 10²⁴ kg and radius 6.4 × 10⁶ m. Calculate the gravitational potential at the Earth's surface, and hence the minimum energy needed to move a probe of mass 2400 kg from the surface to a point very far from the Earth.

φ = −GM / R = −(6.67 × 10⁻¹¹ × 6.0 × 10²⁴) / 6.4 × 10⁶ = −6.25 × 10⁷ J kg⁻¹. ✓
The potential at a very large distance is zero, so each kilogram must gain 6.25 × 10⁷ J of potential energy to get there. ✓
Minimum energy = mΔφ = 2400 × 6.25 × 10⁷ = 1.5 × 10¹¹ J. ✓

Calculation

[3 marks]

A meteoroid starts from rest at a very large distance from the Moon and falls freely onto the Moon's surface. The Moon has mass 7.35 × 10²² kg and radius 1.74 × 10⁶ m, and it has no atmosphere. Calculate the speed of the meteoroid as it reaches the surface.

Change in potential from infinity to the surface: Δφ = 0 − (−GM / R) = GM / R = (6.67 × 10⁻¹¹ × 7.35 × 10²²) / 1.74 × 10⁶ = 2.82 × 10⁶ J kg⁻¹. ✓
The potential energy lost per kilogram becomes kinetic energy per kilogram: ½v² = 2.82 × 10⁶. ✓
v = √(2 × 2.82 × 10⁶) = 2.4 × 10³ m s⁻¹. ✓

Analysis

[3 marks]

A student says that a space probe moving away from a planet "loses potential energy, because the potential gets smaller as the distance increases". Explain the error in this statement.

The potential φ = −GM / r is negative; as r increases its magnitude falls but its value rises toward zero, so the potential increases, not decreases. ✓
The potential energy E = mφ = −GMm / r therefore also becomes less negative, which is an increase in potential energy. ✓
This is consistent with work having to be done on the probe against the attractive field to move it outward; the probe gains, not loses, potential energy. ✓

Calculation

[2 marks]

Planet X has twice the mass and twice the radius of planet Y. Determine the ratio of the gravitational potential at the surface of X to that at the surface of Y, and state whether the surface field strengths are also in this ratio.

φ = −GM / R, so φ_X / φ_Y = (2M / 2R) / (M / R) = 1: the surface potentials are equal. ✓
No: g = GM / R² gives g_X / g_Y = 2 / 4 = ½, because potential depends on 1 / R but field strength on 1 / R². ✓

Calculation

[4 marks]

The Earth has mass 6.0 × 10²⁴ kg and radius 6.4 × 10⁶ m. A probe of mass 1500 kg is moved from the Earth's surface to a distance of 2.0 × 10⁷ m from the Earth's centre.

(a) Calculate the gravitational potential at the Earth's surface. (b) Calculate the gravitational potential at 2.0 × 10⁷ m from the centre. (c) Calculate the increase in gravitational potential energy of the probe. (d) Explain why using E = mgh with g = 9.81 m s⁻² would give a much larger, and incorrect, answer.

(a) φ₁ = −GM / R = −(6.67 × 10⁻¹¹ × 6.0 × 10²⁴) / 6.4 × 10⁶ = −6.25 × 10⁷ J kg⁻¹. ✓
(b) φ₂ = −GM / r = −4.00 × 10¹⁴ / 2.0 × 10⁷ = −2.0 × 10⁷ J kg⁻¹. ✓
(c) ΔE = m(φ₂ − φ₁) = 1500 × (−2.0 × 10⁷ + 6.25 × 10⁷) = 1500 × 4.25 × 10⁷ = 6.4 × 10¹⁰ J. ✓
(d) E = mgh assumes g is constant over the height h = 1.36 × 10⁷ m, giving 2.0 × 10¹¹ J, but g falls as 1 / r² and is only about 1 N kg⁻¹ at the far end, so the true work done is much less. ✓

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