Gravitational Fields: Complete A Level H2 Physics Cheatsheet
A Level Physics (H2), Chapter 8 · 9478 Topic 8 · Read time: ~11 minutes
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Key Formulas
Newton's law of gravitation and field strength
Every point mass attracts every other with a force F = Gm₁m₂/r², along the line joining them. Spherical bodies act as point masses at their centres.
Gravitational field strength is the force per unit mass: g = F/m = GM/r². It follows an inverse square law.
Near the Earth's surface, g is roughly constant and equal to the acceleration of free fall, 9.81 m s⁻².
Trap: r is measured from the centre of the planet, not its surface. For a satellite at height h, r = R + h.
Gravitational potential and potential energy
Gravitational potential (φ) at a point is the work done per unit mass by an external force in bringing a small test mass from infinity to that point. φ = −GM/r.
It is negative because gravity attracts: the external force does negative work bringing a mass in. φ rises towards zero as r increases.
The potential energy of two point masses is U = −GMm/r. Work done moving a mass between two points = m × Δφ.
g = −dφ/dr: the field strength is the negative of the gradient of a φ-r graph.
Trick: For any large change in height, use the change in potential, not mgΔh. mgΔh only works where g is constant, close to the surface.
Orbits and escape velocity
For a circular orbit, gravity provides the centripetal force: GMm/r² = mv²/r, so v = √(GM/r). The satellite's own mass cancels.
With v = 2πr/T this gives T² ∝ r³: further orbits are slower and take longer.
Geostationary satellites orbit above the equator, in the same direction as Earth's rotation, with a 24-hour period, so they stay above one point. They are used for communications and weather monitoring.
Escape velocity: the minimum launch speed to reach infinity. Kinetic energy must supply the energy to go from −GMm/R to zero: ½mv² = GMm/R, so v = √(2GM/R), about 11 km s⁻¹ for Earth.
Trap: Escape velocity does not depend on the object's mass or launch direction. It is √2 times the circular orbital speed at the same radius.
Worked Example
A 650 kg satellite orbits Earth in a circular orbit of radius 7.20 × 10⁶ m. Mass of Earth = 5.97 × 10²⁴ kg, G = 6.67 × 10⁻¹¹ N m² kg⁻².
(a) Calculate the gravitational field strength at this radius. [2]
g = GM/r² = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴) ÷ (7.20 × 10⁶)²
= 7.68 N kg⁻¹
(b) Calculate the orbital speed and period. [3]
v = √(GM/r) = 7.44 × 10³ m s⁻¹
T = 2πr / v = (2π × 7.20 × 10⁶) ÷ (7.44 × 10³)
= 6.08 × 10³ s (about 1.69 hours)
(c) Calculate the gravitational potential energy of the satellite in this orbit. [2]
U = −GMm/r = −(6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × 650) ÷ 7.20 × 10⁶
= −3.59 × 10¹⁰ J
(d) Explain why this satellite cannot be geostationary. [1]
A geostationary orbit needs a 24-hour period. This one is 1.69 hours, so it must be much further out to be geostationary.
Why this question is a good test of the topic: it chains g, v, T and U from the same G M/r relationships, and the last part checks the meaning of geostationary rather than formula recall.
Frequently Asked Questions
Why is gravitational potential negative?
It is defined as zero at infinity. Gravity is attractive, so bringing a mass closer means the external force does negative work, giving negative potential.
What is the link between g and φ?
g = −dφ/dr. The field strength equals the negative of the potential gradient.
Does a satellite's orbital speed depend on its mass?
No. v = √(GM/r) depends only on the central mass and the orbital radius.
What makes an orbit geostationary?
A 24-hour period, orbiting above the equator in the same direction as Earth's rotation.
How is escape velocity derived?
By setting the kinetic energy at launch equal to the energy needed to reach infinity: ½mv² = GMm/R, so v = √(2GM/R).
Gravitational potential still confusing?
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