A Level Physics: Gravitational Fields — Revision Notes

Condensed for the final weeks. For the full explanation, use the Gravitational Fields study guide.

The core equations

Newton's law        F = G M m / r^2
field strength      g = F / m = G M / r^2        N/kg
potential           phi = -G M / r               J/kg
potential energy    Ep = -G M m / r              J

Everything is measured from the centre of the body, not the surface. For a satellite at height h above a planet of radius R, use r = R + h.

G = 6.67 × 10⁻¹¹ N m² kg⁻² — the gravitational constant, the same everywhere in the universe. It is printed on the data sheet (page 2 of the written papers), so copy it accurately from there, including its units, rather than relying on memory.

Field lines represent the field: they point in the direction of the force on a small test mass, and are drawn closer together where the field is stronger.

Worked example: Earth's mass = 5.97 x 10^24 kg, radius = 6.37 x 10^6 m.
g = GM/r^2 = (6.67 x 10^-11 x 5.97 x 10^24) / (6.37 x 10^6)^2 = 9.8 N/kg

This confirms g = 9.81 m/s² used throughout AS mechanics is not an arbitrary constant, but a direct consequence of Newton’s law of gravitation applied at the Earth’s surface.

Why potential is negative

Gravitational potential is defined as zero at infinity. Since gravity is always attractive, work must be done on a mass to move it to infinity — so at any finite distance the potential is negative.

Potential increases (becomes less negative) as r increases, approaching zero as r approaches infinity but never becoming positive for a real, finite separation.

Field strength vs potential

	Field strength g	Potential φ
Type	Vector	Scalar
Formula	GM/r²	−GM/r
Falls off as	1/r²	1/r
Units	N/kg	J/kg
At infinity	0	0

Relationship: g = −dφ/dr — field strength is the negative gradient of the potential. See the Gravitational Fields study guide for the full derivations and worked reasoning behind every equation above.

Orbits

For a circular orbit, gravity provides the entire centripetal force needed to keep the satellite on its curved path:

G M m / r^2 = m v^2 / r     ->    v = sqrt(G M / r)

and using v = 2 pi r / T:

T^2 = (4 pi^2 / G M) r^3      <- Kepler's third law

So T² ∝ r³. The orbiting mass cancels — orbital speed and period depend only on the central mass and the radius, which is exactly why every satellite at a given radius above a given planet, regardless of its own mass, keeps the same period.

Geostationary orbit: period exactly 24 hours, orbiting west to east, directly above the Equator, radius ≈ 42 000 km from Earth’s centre — all three conditions together are what let the satellite stay fixed above the same point on the ground as the Earth turns beneath it.

Escape velocity

Escape velocity is the minimum launch speed for which total energy (kinetic plus gravitational potential) just reaches zero, allowing the object to reach infinity with nothing left over:

1/2 m v^2 = G M m / r     ->    v_esc = sqrt(2 G M / r)

Independent of the escaping object’s mass, since m cancels from both sides of the energy equation, leaving only the central mass M and the starting distance r.

Exam traps

• Using height above the surface instead of distance from the centre.
• Dropping the minus sign on potential or potential energy.
• Treating potential as a vector — it is a scalar, so potentials simply add.
• Confusing g (1/r²) with φ (1/r).
• Forgetting that geostationary requires equatorial, west-to-east and 24 hours — all three.
• Assuming a heavier satellite orbits faster; mass cancels.
• Miscopying the value or units of G (6.67 × 10⁻¹¹ N m² kg⁻²) from the data sheet — it is always supplied, so take care to copy it, and its units, accurately.
• Drawing field lines further apart where the field is actually stronger — closer lines always mean a stronger field, never the reverse.

Self-test

1. Write Newton’s law of gravitation and state what r measures.
2. Why is gravitational potential always negative?
3. Derive T² ∝ r³ for a circular orbit.
4. State the three conditions for a geostationary orbit.
5. Does a more massive satellite need a greater orbital speed at the same radius?
6. Where do you find the value of the gravitational constant G, and why should you still write down its units carefully?
7. A planet has mass 6.4 × 10²³ kg and radius 3.4 × 10⁶ m. Calculate the gravitational field strength at its surface.

Answers: 1. F = GMm/r², where r is the distance between the centres of the two masses. 2. Potential is defined as zero at infinity and gravity is attractive, so work must be done on a mass to move it to infinity; at any finite separation the potential is therefore negative. 3. GMm/r² = mv²/r gives v² = GM/r; substituting v = 2πr/T gives 4π²r²/T² = GM/r, so T² = (4π²/GM)r³. 4. Period of 24 hours, orbiting west to east, directly above the Equator. 5. No — the satellite’s mass cancels, so orbital speed depends only on the central mass and the orbital radius. 6. On the data sheet (it is not required from memory); its units, N m² kg⁻², must still be copied correctly since they are needed to check the working. 7. g = GM/r² = (6.67 × 10⁻¹¹ × 6.4 × 10²³) ÷ (3.4 × 10⁶)² ≈ 3.7 N/kg.