OCR A Level Physics A: Gravitational fields, Newton's law and planetary motion (H556)

This guide covers sections 5.4.1 (Point and spherical masses), 5.4.2 (Newton’s law of gravitation) and 5.4.3 (Planetary motion) 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. Gravitational potential and energy (5.4.4) is a separate topic and is not taught here.

See the OCR A Level Physics hub and the printable checklist. Pair this page with the revision notes and the practice questions, and test yourself with the free diagnostics. Circular motion and centripetal force are needed for the orbit work.

Use G = 6.67 × 10⁻¹¹ N m² kg⁻² unless told otherwise.

What you must be able to do

Statement	Content
5.4.1(a)	Gravitational fields are due to objects having mass
5.4.1(b)	Model a spherical object as a point mass at its centre
5.4.1(c)	Use field lines to map gravitational fields
5.4.1(d)	Gravitational field strength, g = F/m
5.4.1(e)	Gravitational fields as one of several forms of field giving a force (link to section 6.2)
5.4.2(a)	Newton’s law, F = −GMm/r², for two point masses
5.4.2(b)	g = −GM/r² for a point mass
5.4.2(c)	g uniform near the Earth’s surface; equal to the acceleration of free fall
5.4.3(a)	Kepler’s three laws of planetary motion
5.4.3(b)	Gravity provides the centripetal force on a planet
5.4.3(c)	T² = (4π²/GM) r³, and you must be able to derive it
5.4.3(d)	T² ∝ r³ applied to systems other than the Solar System
5.4.3(e)	Geostationary orbit and the uses of geostationary satellites

5.4.1 Fields due to mass

Mass creates a gravitational field: any other mass in the region feels a force (5.4.1(a)). Every object with mass has a field, but only planets and stars have fields strong enough to notice.

The point-mass model (5.4.1(b)). Outside a spherical object, the field is the same as if all its mass were at its centre. Measure the distance r from the centre, not from the surface. A satellite 400 km above a planet of radius R is at r = R + 400 km.

Field lines (5.4.1(c)). A field line shows which way a small mass placed on it would be pulled, and the arrows point that way.

Situation	Pattern of lines
Single point or spherical mass	Straight lines pointing radially inwards. They are closer together nearer the mass, so the field is stronger there.
Small region near a planet’s surface	Parallel, equally spaced, vertical lines: a uniform field
Two nearby masses	Lines curve between them and never cross. Midway between equal masses there is a point where the fields cancel.

Lines never cross because the field at a point has one direction.

Gravitational field strength (5.4.1(d)). It is the force per unit mass felt by a small test mass:

g = F/m, in N kg⁻¹

It is a vector, directed towards the mass that creates the field. The test mass must be small so that it does not disturb the field.

Worked example 1. A 6.0 kg instrument on a planet’s surface experiences a gravitational force of 52.8 N. Find g and the free-fall acceleration.

g = F/m = 52.8 / 6.0 = 8.8 N kg⁻¹

The direction is towards the planet’s centre, and free fall is at 8.8 m s⁻².

Link to other fields (5.4.1(e)). Gravitational, electric and magnetic fields each give a force on an object with the right property. Gravity acts on mass. Electric fields act on charge, with field strength defined as force per unit charge in the same way: E = F/Q (section 6.2). Both point-source fields are radial and weaken with distance. They differ in one key way: gravity only attracts, because mass has one sign, while electric forces can attract or repel.

5.4.2 Newton’s law of gravitation

Newton’s law (5.4.2(a)). Two point masses attract each other with a force proportional to the product of the masses and inversely proportional to the square of their separation:

F = −GMm/r²

G is the gravitational constant. The minus sign is a sign convention. If r is measured outwards from M, the force on m points inwards, so it is negative. In a calculation, find the magnitude GMm/r² and state the direction in words (towards M) unless you are told to keep the sign.

The forces on the two masses are equal and opposite (Newton’s third law), whatever their sizes. Doubling r divides F by 4. Doubling either mass doubles F.

Worked example 2. The Sun has mass 1.99 × 10³⁰ kg and Jupiter 1.90 × 10²⁷ kg, and their centres are 7.78 × 10¹¹ m apart. Find the force each exerts on the other.

F = GMm/r² = 6.67e-11 × 1.99e30 × 1.90e27 / (7.78e11)²
  = 4.17 × 10²³ N

Each body pulls the other with this force.

Field strength of a point mass (5.4.2(b)). Put a test mass m at distance r from M. Then g = F/m, so:

g = −GM/r²

The test mass cancels, so g depends only on the source mass and the distance. The sign shows that g points inwards, and g at 2r is a quarter of g at r.

Worked example 3. Mercury has mass 3.30 × 10²³ kg and radius 2.44 × 10⁶ m. Find g at its surface.

g = GM/R² = 6.67e-11 × 3.30e23 / (2.44e6)² = 3.70 N kg⁻¹

Worked example 4. At what height above the Earth’s surface is g one quarter of its surface value? (Earth radius 6.37 × 10⁶ m, surface g 9.81 N kg⁻¹.)

g ∝ 1/r², so g is a quarter when r is doubled: r = 2R. The height is h = r − R = R = 6.37 × 10⁶ m, where g = 9.81 / 4 = 2.45 N kg⁻¹.

Uniform close to the surface (5.4.2(c)). Near the Earth’s surface g changes very little with height, so it is treated as uniform. The numerical value of g in N kg⁻¹ is the same as the acceleration of free fall in m s⁻², because a = F/m = g. The units N kg⁻¹ and m s⁻² are equivalent.

Worked example 5. Show that treating g as uniform is reasonable over 100 m. Use the Earth data above.

g at the surface:       GM/R²        = 9.8134 N kg⁻¹
g at 100 m up:          GM/(R+100)²  = 9.8131 N kg⁻¹
fractional decrease     = 0.0031 %

A 100 m change in height alters g by about three parts in a hundred thousand, far smaller than the uncertainty in most experiments. The field lines are effectively parallel and equally spaced.

5.4.3 Planetary motion

Kepler’s laws (5.4.3(a)).

1. Each planet moves in an elliptical orbit with the Sun at one focus.
2. The line joining a planet to the Sun sweeps out equal areas in equal times.
3. The square of the orbital period is directly proportional to the cube of the orbit’s radius: T² ∝ r³.

For a circular orbit, r is the orbital radius. For an ellipse it becomes the semi-major axis (half the longest width). The second law means a planet moves fastest when closest to the Sun and slowest when furthest away. In a short time dt the area swept is about ½ r v dt. At the two points on the orbit where velocity is perpendicular to the line to the Sun, equal areas per second gives:

v₁r₁ = v₂r₂

Worked example 6. A comet passes the Sun at 1.2 × 10¹¹ m with speed 5.4 × 10⁴ m s⁻¹. At its furthest point it is 6.0 × 10¹¹ m away. Find its speed there.

v₂ = v₁r₁ / r₂ = 5.4e4 × 1.2e11 / 6.0e11 = 1.08 × 10⁴ m s⁻¹

Gravity as the centripetal force (5.4.3(b)). For a planet of mass m in a circular orbit of radius r around a star of mass M, the gravitational force is the only force towards the centre, so it provides the centripetal force.

Derivation of T² = (4π²/GM) r³ (5.4.3(c)). This is required from first principles. Write each step.

Gravitational force = centripetal force
GMm/r² = mv²/r
GM/r  = v²
Orbital speed v = 2πr/T (circumference ÷ period)
GM/r  = 4π²r²/T²
T²    = (4π²/GM) r³

The orbiting mass m cancels, so the period does not depend on it. M is the mass of the body being orbited. Since 4π²/GM is a constant for one system, T² ∝ r³, which is Kepler’s third law. A graph of T² against r³ is a straight line through the origin with gradient 4π²/GM.

Using Kepler’s third law in other systems (5.4.3(d)). The law applies to any set of bodies orbiting one central mass: moons around a planet, planets around another star, a satellite around the Earth. Each system has its own constant. Within one system, T₂ = T₁ (r₂/r₁)^(3/2) compares two orbits without knowing M.

Worked example 7. Jupiter’s moon Io orbits at 4.22 × 10⁸ m with period 1.77 days. Find Jupiter’s mass.

T = 1.77 × 86 400 = 1.529 × 10⁵ s
M = 4π²r³ / (G T²)
  = 4π² × (4.22e8)³ / (6.67e-11 × (1.529e5)²)
  = 1.90 × 10²⁷ kg

Worked example 8. Another moon of the same planet has orbital radius 1.9 × 10⁸ m and period 2.8 days. Find the period of a moon at 4.8 × 10⁸ m.

T₂ = 2.8 × (4.8 / 1.9)^1.5 = 11 days  (11.2 days)

Geostationary orbit (5.4.3(e)). A geostationary satellite stays above the same point on the Earth’s surface. To do this it must:

• orbit directly above the equator
• move in the same direction as the Earth rotates (west to east)
• have a period of 24 hours, equal to the Earth’s rotation period
• be in a circular orbit, at one particular height.

Worked example 9. Predict the radius and height of a geostationary orbit. (Earth: M = 5.97 × 10²⁴ kg, equatorial radius 6.37 × 10⁶ m.)

T = 24 × 3600 = 86 400 s
r³ = GM T² / 4π²  = 6.67e-11 × 5.97e24 × (86 400)² / 4π²
r  = 4.22 × 10⁷ m
h  = r − R = 4.22e7 − 6.37e6 = 3.59 × 10⁷ m  (about 36 000 km)
v  = 2πr/T = 3.07 × 10³ m s⁻¹

Newtonian theory predicts where the orbit must be. Only one radius gives a 24-hour period, so all geostationary satellites share the same orbit height.

Uses and issues. Because the satellite stays fixed in the sky, a ground dish can point at it without tracking. Uses include television broadcasting, telephone links and observing weather over a fixed region. Drawbacks: a signal takes at least 2 × 3.59 × 10⁷ / 3.00 × 10⁸ ≈ 0.24 s to go up and back, and coverage near the poles is poor. Satellites are expensive to build and launch, and debris from collisions or failures endangers others in the same orbit.

Common errors

• Using the height above the surface instead of r, the distance from the centre, in GM/r².
• Forgetting to square r, or squaring only the number in standard form, such as (6.37 × 10⁶)² written as 6.37 × 10¹².
• Cancelling the wrong mass in the derivation. The orbiting mass cancels. The central mass M stays.
• Leaving the period in days or hours. Convert to seconds.
• Saying “gravity is zero in orbit”. The field is what holds the satellite in orbit.

Official syllabus

OCR Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), for first assessment in 2017, published by OCR. Sections 5.4.1 Point and spherical masses, 5.4.2 Newton’s law of gravitation and 5.4.3 Planetary motion.