Newton's Law of Gravitation — Quick Summary

Mr Toogood's Physics · Gravitational fields

AQA 3.7.2.1 / 3.7.2.2
F = Gm₁m₂/r²
Newton's law
G = 6.67×10⁻¹¹
N m² kg⁻²
g = F/m
Field strength
g = GM/r²
Radial field

Newton's law of gravitation

The equal and opposite gravitational force between two objects of different mass

Even with unequal masses, the force on each object is equal and opposite (Newton's 3rd law).

Gravity is universal: every particle attracts every other particle along the line joining them. F∝m (Newton's 2nd law in action); F∝1/r² — the same inverse-square spreading as light over an expanding sphere (A=4πr²).

F = Gm₁m₂/r²
  • F in N; G=6.67×10⁻¹¹ N m²kg⁻²; m₁,m₂ in kg; r = distance between centres, in m.
Common slip: forgetting r is squared gives a wildly oversized force — always sanity-check the size of your answer.

The gravitational constant, G

G is tiny, so the force between everyday objects is negligible — only significant when at least one mass is planet- or moon-scale. Cavendish (1798) measured G using a torsion balance: the tiny twist of a suspended rod, caused by attraction between small and large lead spheres, gave the force and hence G — nicknamed "weighing the Earth," since it let Earth's mass be calculated for the first time.

Gravitational field strength, g

g = F/m

g is force per unit mass — a property of the location, not the test mass placed there: a bigger test mass feels a bigger force, but also needs a bigger force for the same acceleration (F=ma) — these effects cancel exactly. At Earth's surface, g=9.81 N kg⁻¹.

Because g is mass-independent, every object at the same point accelerates identically — a feather and hammer dropped together on the Moon (no air resistance) hit the ground at the same time.

g in a radial field

Substituting F=mg into Newton's law and cancelling the test mass gives:

g = GM/r²
  • Half the mass, same radius → g halves.
  • Double the distance → g quarters (inverse square, not inverse).
Common slip: doubling distance doesn't halve g — it quarters it. Always square the scale factor.
A test mass at the point of zero net gravitational force between two source masses

Between two equal masses, net force is zero exactly at the midpoint.

For unequal masses (e.g. Earth–Moon), the zero-force point satisfies GM_E/x²=GM_M/y², giving x/y=√(M_E/M_M). For Earth–Moon this ratio ≈9.0, so the point sits 9/10 of the way from Earth to the Moon (~3.5×10⁸ m from Earth).

Radial vs. uniform fields

A radial field appearing uniform when viewed close to a planet's surface

Zoomed in near a surface, diverging field lines look almost parallel.

Close to a planet's surface, field lines diverge so little over typical heights that the field looks uniform — g barely changes. Even at ISS altitude (r=6770 km), g≈8.7 N kg⁻¹, only slightly below the surface value.

9.81 N kg⁻¹ is a surface-only approximation. For satellites, high altitude, or other planets, always use g=GM/r² with the correct r.

Worked example: exoplanet radius

Substituting M=ρV=(4/3)πR³ρ into g_s=GM/R² and cancelling R² gives:

g_s = (4/3)πGρR

Rearranging for R and substituting g_s=6.60 N kg⁻¹, ρ=4.00×10³ kg m⁻³:

R ≈ 5.91×10⁶ m  (3 s.f., matching the given data)

Exam essentials

Key equations

  • F=Gm₁m₂/r²
  • G=6.67×10⁻¹¹ N m²kg⁻²
  • g=F/m   g=GM/r²

Field line rules

  • Arrows point towards the source (always attractive).
  • Closer lines = stronger field; lines never cross.
  • Radiate symmetrically for a spherical mass; extend to infinity, never reaching zero.

Common slips

  • r is squared — sanity-check the magnitude of your answer.
  • Double the distance → g quarters, not halves.
  • 9.81 N kg⁻¹ only applies near Earth's surface — use g=GM/r² for altitude or other planets.