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

Quick recall for sections 5.4.1 to 5.4.3 of the OCR Advanced GCE in Physics A specification (H556), Version 3.0 (March 2026), for first assessment in 2017: fields due to mass, Newton’s law of gravitation and planetary motion. H556 is the full A level. Full explanations and worked examples are in the study guide. Gravitational potential and energy (5.4.4) is a separate topic.

Use the OCR A Level Physics hub and the printable checklist to track progress. Check your recall with the practice questions and the free diagnostics.

Definitions and key facts

Term	Meaning (statement)
Gravitational field	Region where a mass experiences a force due to other mass (5.4.1(a))
Gravitational field strength, g	Force each kilogram of a small test mass feels, g = F/m, in N kg⁻¹; a vector pointing to the source (5.4.1(d))
Point-mass model	A sphere acts as if all its mass is at its centre (5.4.1(b))
Field line	Gives the direction a small mass is pulled; closer lines mean a stronger field (5.4.1(c))
Uniform field	Parallel, equally spaced lines; a good model close to the Earth’s surface (5.4.2(c))
Geostationary orbit	Stays above one point on the equator, period 24 h (5.4.3(e))

Formulae

Relationship	Notes
g = F/m	N kg⁻¹ is equivalent to m s⁻²
F = −GMm/r²	Negative sign: force on m is towards M. Use the magnitude and state the direction
g = −GM/r²	Depends on M and r only, not on the test mass
v = 2πr/T	Circular orbit speed from period
T² = (4π²/GM) r³	Gradient of T² against r³ is 4π²/GM
T₂ = T₁ (r₂/r₁)^(3/2)	Same central body only
v₁r₁ = v₂r₂	Only at the two points where velocity is perpendicular to the line to the centre

Field lines

• Single sphere: radial lines pointing inwards, closer together near the surface.
• Near the surface: parallel, equally spaced, vertical lines.
• Lines never cross. Arrows point towards the mass.
• Link to section 6.2: electric field strength is force per unit charge, E = F/Q. Both fields are radial for a point source and weaken with distance. Gravity only attracts. Electric forces can attract or repel.

Method boxes

Force or field strength from Newton’s law

1. Find r from the centre of the body, not the surface. Add the planet radius to any height.
2. Convert to metres and kilograms.
3. Substitute into GMm/r² (force) or GM/r² (field strength).
4. Give the answer with units and direction (towards the centre).

Deriving T² = (4π²/GM) r³

1. Gravitational force provides the centripetal force: GMm/r² = mv²/r.
2. Cancel m and one r: GM/r = v².
3. Substitute v = 2πr/T: GM/r = 4π²r²/T².
4. Rearrange: T² = (4π²/GM) r³.

Kepler’s three laws

1. Orbits are ellipses with the Sun at one focus.
2. The line from the Sun to a planet sweeps equal areas in equal times.
3. T² ∝ r³.

Geostationary orbit: conditions and radius

1. Above the equator, moving west to east, circular, period 24 h.
2. Set T = 86 400 s in r³ = GMT²/4π².
3. For the Earth, r ≈ 4.22 × 10⁷ m, which is about 3.59 × 10⁷ m above the surface.

Small worked reminders

Triton has mass 2.14 × 10²² kg and radius 1.35 × 10⁶ m. A probe is 1.5 × 10⁶ m above its surface.

r = 1.35e6 + 1.5e6 = 2.85 × 10⁶ m
g = GM/r² = 6.67e-11 × 2.14e22 / (2.85e6)² = 0.176 N kg⁻¹
Surface value: 6.67e-11 × 2.14e22 / (1.35e6)² = 0.783 N kg⁻¹

The distance from the centre is 2.11 times the surface radius, so g drops by 2.11² = 4.46 times: 0.783 / 4.46 = 0.176.

A moon orbits a planet at radius 1.8 × 10⁸ m with period 1.1 days. Find the planet’s mass.

T = 1.1 × 86 400 = 9.50e4 s
M = 4π²r³ / (G T²) = 4π² × (1.8e8)³ / (6.67e-11 × (9.50e4)²)
  = 3.8 × 10²⁶ kg

Check the steps in this order: convert the period to seconds, cube the radius, square the period, then divide. The moon’s own mass is not needed, and a graph of T² against r³ for several moons would give a straight line through the origin.

Must-know distinctions

• g and G. g is field strength at a place and varies with r. G is a universal constant.
• r and h. r is distance from the centre. h is height above the surface. For a satellite, r = R + h.
• Force and field strength. Force F depends on both masses and is measured in N. Field strength g depends only on the source mass and distance, and is measured in N kg⁻¹.
• Kepler’s second and third laws. The second compares speeds at different points on one orbit. The third compares whole orbits of different bodies around the same central mass.
• Mass of the orbiting body and mass of the central body. Only the central body appears in T² = (4π²/GM) r³.
• Geostationary and geosynchronous. Geostationary also needs an equatorial orbit so that the satellite stays above one point.
• Uses of geostationary satellites. Television and telephone communications and weather observation of a fixed region. The disadvantages are a signal delay of about 0.24 s and poor coverage near the poles.

Quick self-test

1. Define gravitational field strength.
2. Two masses of 1500 kg and 800 kg are 20 m apart. Calculate the force between them, using G = 6.67 × 10⁻¹¹ N m² kg⁻².
3. A moon has mass 1.35 × 10²³ kg and radius 2.57 × 10⁶ m. Calculate g at its surface.
4. The surface field strength of a planet is 12.0 N kg⁻¹. State g at a distance of three radii from its centre.
5. A satellite orbits the Earth at radius 1.0 × 10⁷ m. Calculate its period (Earth mass 5.97 × 10²⁴ kg).
6. A planet orbits at four times the radius of another planet around the same star. State the ratio of the periods.
7. State why the mass of a satellite does not appear in the formula for its orbital period.
8. A comet is 1.5 × 10¹¹ m from the Sun at speed 3.6 × 10⁴ m s⁻¹. Calculate its speed at 4.5 × 10¹¹ m, at the other end of its orbit.
9. A graph of T² against r³ for moons of a planet has gradient 1.0 × 10⁻¹⁴ s² m⁻³. Find the planet’s mass.
10. State three conditions a satellite must meet to be geostationary.

Answers

1. The force per unit mass experienced by a small test mass at that point; g = F/m.
2. F = GMm/r² = 6.67 × 10⁻¹¹ × 1500 × 800 / 20² = 2.0 × 10⁻⁷ N.
3. g = GM/R² = 6.67 × 10⁻¹¹ × 1.35 × 10²³ / (2.57 × 10⁶)² = 1.36 N kg⁻¹.
4. g ∝ 1/r², so g = 12.0 / 3² = 1.33 N kg⁻¹.
5. T = 2π √(r³/GM) = 2π √((1.0e7)³ / (6.67e-11 × 5.97e24)) = 9.96 × 10³ s (about 2.8 hours).
6. T ∝ r^(3/2), so the ratio is 4^1.5 = 8: the outer planet’s period is 8 times longer.
7. The satellite’s mass appears in both the gravitational force and the centripetal force expression, so it cancels.
8. v₂ = v₁r₁/r₂ = 3.6e4 × 1.5e11 / 4.5e11 = 1.2 × 10⁴ m s⁻¹.
9. Gradient = 4π²/GM, so M = 4π² / (G × gradient) = 39.5 / (6.67e-11 × 1.0e-14) = 5.9 × 10²⁵ kg.
10. Orbit above the equator, move in the same direction as the Earth rotates, and have a period of 24 hours. It must also orbit at the radius that gives this period.

Where marks are usually lost

• Putting the height above the planet into GM/r² where the distance from its centre belongs.
• Squaring the coefficient but not the power of ten in r².
• Writing the minus sign in F = −GMm/r² in an answer and then giving the magnitude as negative with no direction.
• Giving the period in days when the formula needs seconds.
• A derivation that does not state that gravity provides the centripetal force, or that does not show where v = 2πr/T comes from.
• Saying a geostationary satellite is “stationary”, or giving a polar orbit as geostationary.
• Cancelling the central mass instead of the satellite mass.
• Drawing field lines that cross, or without arrows, or equally spaced for a point mass.
• Unit slips: N kg⁻¹ for field strength, N for force.

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.