OCR A Level Physics A: Gravitational potential, potential energy and escape velocity (H556)

This study guide covers section 5.4.4 (Gravitational potential and energy), statements (a) to (e), of the OCR Advanced GCE in Physics A specification (H556), Version 3.0 (March 2026), for first assessment in 2017. H556 is the full A level. Field strength and Newton’s law (5.4.1 and 5.4.2) and planetary motion (5.4.3) are taught elsewhere, so here you meet only potential, potential energy, the force-distance graph and escape velocity.

Related pages: the revision notes, the practice questions, the OCR A Level Physics hub, the printable checklist and the free diagnostics. For the general idea of work and energy transfer, see work, energy, power and efficiency.

What this unit covers

Statement	What you must be able to do
5.4.4(a)	Define gravitational potential as the work done in bringing unit mass from infinity to a point, and state that it is zero at infinity
5.4.4(b)	Use Vg = -GM/r for a point mass M and calculate changes in gravitational potential
5.4.4(c)	Use a force-distance graph for a point or spherical mass; work done is the area under the graph
5.4.4(d)	Use E = mVg = -GMm/r for gravitational potential energy at distance r from a point mass M
5.4.4(e)	Find and use escape velocity, including predicting whether atoms escape from a planet’s atmosphere

Data used in the worked examples (treat them as data given in a question, not things to memorise): G = 6.67 × 10⁻¹¹ N m² kg⁻², Earth mass 5.97 × 10²⁴ kg and radius 6.37 × 10⁶ m, Moon mass 7.35 × 10²² kg and radius 1.74 × 10⁶ m, Mars mass 6.42 × 10²³ kg and radius 3.39 × 10⁶ m, g = 9.81 N kg⁻¹ at the surface of the Earth.

5.4.4(a) Gravitational potential

The gravitational potential Vg at a point is the work done per unit mass in bringing a small test mass from infinity to that point. It is a scalar quantity and its unit is J kg⁻¹.

Gravitational potential is defined to be zero at infinity, where the pull of the mass is negligible. Gravity is always attractive, so you do not have to push a mass to bring it in from infinity: the field does positive work on it. The work done by you (the external agent) is therefore negative, and every point has a negative potential. The closer you get to the mass, the more negative the potential becomes.

Two points to keep clear:

• Potential is a property of the point in the field, not of the test mass, because the mass is divided out.
• Potential is not potential energy. Potential is per kilogram. Potential energy is for a particular mass.

5.4.4(b) Vg = -GM/r and changes in potential

For a point mass M, or a spherical mass whose density depends only on distance from the centre, the potential at distance r from the centre is

Vg = -GM/r

Because r is in the denominator, the potential rises towards zero as r grows. Doubling r halves the size of Vg. A sketch of Vg against r is a curve in the lower right-hand quadrant that starts steeply negative at small r and flattens towards zero without ever reaching it.

For a change in position, subtract the potentials: ΔVg = Vg(final) - Vg(initial). Moving away from the mass gives a positive ΔVg (you move to a less negative value). Moving towards the mass gives a negative ΔVg. The energy needed for a mass m to make the move is m × ΔVg.

For a planet, r is measured from its centre, so a point at height h above the surface has r = R + h.

Worked example 1: potential at a planet’s surface

Calculate the gravitational potential at the surface of Mars.

Vg = -GM/r = -(6.67 × 10⁻¹¹ × 6.42 × 10²³) / (3.39 × 10⁶) Vg = -4.282 × 10¹³ / 3.39 × 10⁶ = -1.26 × 10⁷ J kg⁻¹

Worked example 2: work to change orbit height

An 800 kg satellite is lifted from the surface of Mars to a point 500 km above the surface. Calculate the work that must be done on it.

r(final) = 3.39 × 10⁶ + 5.0 × 10⁵ = 3.89 × 10⁶ m Vg(final) = -4.282 × 10¹³ / 3.89 × 10⁶ = -1.101 × 10⁷ J kg⁻¹ ΔVg = -1.101 × 10⁷ - (-1.263 × 10⁷) = +1.624 × 10⁶ J kg⁻¹ Work done = m ΔVg = 800 × 1.624 × 10⁶ = 1.30 × 10⁹ J

Keep an extra digit in the potentials, because they are close and early rounding would damage the difference.

Link to mgh

Near a planet’s surface the field is almost uniform, so mgh is a good approximation to the exact change. For a 1000 kg object raised 10 km above the Earth, the exact method gives m × GMe × (1/Re - 1/(Re + 10⁴)) = 9.80 × 10⁷ J, while mgh = 1000 × 9.81 × 10⁴ = 9.81 × 10⁷ J. The two differ by about 0.1%. The approximation worsens as h becomes a larger fraction of R.

5.4.4(c) The force-distance graph

The magnitude of the gravitational force between a point mass M and a test mass m is F = GMm/r². Plotted against r it falls steeply at first and then flattens. The same graph shape applies to a spherical mass, with r measured from its centre and starting at the surface radius R if the test mass is outside the sphere.

The work done by a force that varies with distance is the area under the force-distance graph. Between r₁ and r₂ the area equals GMm(1/r₁ - 1/r₂), which is the same as m × ΔVg. This is the link between the graph and the potential: area under the graph = m × (change in potential).

For a curve you may be asked to estimate the area by counting squares or by trapezia. The gravitational curve is bowed in towards the axes, so straight lines joining points on it lie above the curve and trapezia give an overestimate.

Worked example 3: area under the graph

A 1000 kg probe is moved away from the centre of Mars. The force on it is calculated at five distances.

r / 10⁶ m	3.39	5.085	6.78	8.475	10.17
F / N	3726	1656	931.5	596.2	414.0

Estimate the work done in moving from r = 3.39 × 10⁶ m to 10.17 × 10⁶ m using trapezia, then compare with the exact value.

Each strip has width 1.695 × 10⁶ m. The sum of the trapezia is 1.695 × 10⁶ × [(3726 + 1656)/2 + (1656 + 931.5)/2 + (931.5 + 596.2)/2 + (596.2 + 414.0)/2] = 1.695 × 10⁶ × 5254 ≈ 8.9 × 10⁹ J.

Exact: W = GMm(1/r₁ - 1/r₂) = 4.282 × 10¹³ × 1000 × (1/3.39 × 10⁶ - 1/10.17 × 10⁶) = 8.42 × 10⁹ J.

The trapezium estimate is about 6% too high, as expected for a curve bowed towards the axes. Using narrower strips would reduce the error.

5.4.4(d) Gravitational potential energy

The gravitational potential energy of a mass m at distance r from a point mass M is

E = mVg = -GMm/r

This is the work done by an external agent bringing the mass from infinity to that point. Its value is zero at infinity and negative everywhere else, because the field does the work as the mass falls in. A mass at rest in the field has less energy than one at infinity, so it is bound to the larger mass.

Changes in potential energy are what matter in problems: ΔE = m ΔVg = GMm(1/r₁ - 1/r₂). Gain of potential energy means the mass moves outwards. Loss means it moves inwards, and the energy appears as kinetic energy if there are no other forces (conservation of energy).

Worked example 4: falling to the Moon

A small body is released from rest at a distance of 2 × 1.74 × 10⁶ m = 3.48 × 10⁶ m from the centre of the Moon. Calculate its speed on reaching the surface, ignoring any other forces.

Vg(surface) = -GM/R = -2.818 × 10⁶ J kg⁻¹ Vg(start) = -GM/2R = -1.409 × 10⁶ J kg⁻¹ Decrease in potential energy per kilogram = 1.409 × 10⁶ J kg⁻¹ ½v² = 1.409 × 10⁶, so v = √(2.818 × 10⁶) = 1.68 × 10³ m s⁻¹

The mass of the falling body cancels, which is why every body falls in the same way in a gravitational field.

5.4.4(e) Escape velocity

Escape velocity is the minimum speed at which an object must be launched from a point so that it can reach infinity (and arrive there with zero kinetic energy) with no further propulsion.

At infinity the total energy is zero. For the object to just reach infinity, its total energy at launch must also be zero:

½mv² - GMm/r = 0, so v = √(2GM/r)

Features of this result:

• The mass m of the object cancels, so the escape velocity is the same for a grain of dust and a spacecraft.
• It depends on the mass M and the distance r from the centre, not on the direction of launch (if air resistance and the rotation of the planet are ignored and the path does not hit anything).
• At the surface g = GM/R², so GM = gR² and v = √(2gR).
• It is a speed, since only its size matters.

Worked example 5: escape speeds

Mars: v = √(2 × 4.282 × 10¹³ / 3.39 × 10⁶) = √(2.526 × 10⁷) = 5.03 × 10³ m s⁻¹.

Earth, using g: v = √(2 × 9.81 × 6.37 × 10⁶) = 1.12 × 10⁴ m s⁻¹. Using GM gives the same value.

Earth, from a point 2000 km above the surface: r = 6.37 × 10⁶ + 2.0 × 10⁶ = 8.37 × 10⁶ m, so v = √(2 × 3.982 × 10¹⁴ / 8.37 × 10⁶) = 9.75 × 10³ m s⁻¹. Escape is easier from higher up.

Predicting whether atoms escape from an atmosphere

Escape velocity can be used to judge whether a planet keeps a gas. Compare the escape speed with the speeds of the gas atoms or molecules in the upper atmosphere. If typical speeds are well below the escape speed, the planet keeps the gas. If they are close to or above it, atoms escape over time.

Worked example 6: the Moon has an escape speed of v = √(2 × 4.902 × 10¹² / 1.74 × 10⁶) = 2.37 × 10³ m s⁻¹. A light gas whose atoms have a mean speed of 1.9 × 10³ m s⁻¹ at the surface has a mean speed of about 80% of the escape speed. Atoms move at a range of speeds, so a significant fraction exceed the mean and can escape. This is one reason the Moon has almost no atmosphere.

Common errors

• Writing a positive potential, or dropping the minus sign. Negative values are expected.
• Using the height above the surface as r. Use r = R + h from the centre.
• Taking ΔVg as initial minus final. Take final minus initial.
• Using mgh over a large height change, where the field is not uniform.
• Forgetting to square-root in v = √(2GM/r), or using the planet’s diameter instead of radius.

Try the practice questions next, and tick off each statement on the checklist.

Official syllabus

OCR Advanced GCE in Physics A (H556), Version 3.0 (March 2026), for first assessment in 2017, published by OCR. Section 5.4.4, Gravitational potential and energy.