Physics I: Mechanics › Oscillations & gravitation › Newton's gravitation

F = G Mm/r²Say it: “the gravitational force equals G times big-M times little-m over r squared”

Newton's law of gravitation

One law for the falling apple and the orbiting Moon — every mass attracts every other mass, weakening with the square of the distance.

Notation on this page: G = 6.674×10−11 N·m²/kg² is the universal constant, M and m the two masses, r the centre-to-centre distance.

Where it comes from

Newton's leap (in the Principia, 1687) was unifying two motions: an apple falling 5 m and the Moon “falling” around the Earth. Both, he argued, are pulled by the same gravity — and comparing their accelerations reveals the distance law. The Moon is ~60 Earth radii away, and its centripetal acceleration is ~1/3600 of g:

aMoon / g ≈ 1/3600 = 1/60²60× farther → 60²× weaker — the inverse square, read off the sky
Before reading on: if gravity weakened as 1/r instead of 1/r², what fraction of g would the Moon feel at 60 Earth radii? Compare with the observed 1/3600 before reading the verdict.

1/r would give 1/60 of g — sixty times stronger than observed. The data demands 1/60² = 1/3600, so gravity falls as the inverse square. Newton then argued the force must involve both masses symmetrically (his third law: the Earth pulls the Moon exactly as hard as the Moon pulls the Earth), giving F ∝ Mm/r² — and G, measured by Cavendish in 1798, turns the proportion into an equation.

Derivation

Newton derived the inverse square from Kepler's third law plus circular motion. Follow the algebra: Kepler's T² ∝ r³ goes in, 1/r² comes out.

F
=
mv²/r,   v = 2πr/T
Step 1 — circular orbit. Gravity provides the centripetal force; orbital speed is circumference over period.
F
=
4π²mr/T²
Step 2 — substitute v. m(2πr/T)²/r = 4π²mr/T². The force needed to hold the orbit.
F
∝
mr/r³ = m/r²
Step 3 — Kepler's third law. T² ∝ r³, so 1/T² ∝ 1/r³. The r on top cancels one power: F ∝ m/r².
F
=
G Mm/r²
Step 4 — symmetrise. Newton's third law: the force must treat both masses alike, so M joins m. G (Cavendish, 1798) fixes the constant. ∎

Why must M appear? Step 3 gives F ∝ m/r² for the orbiter — but the force the planet feels is equal and opposite, so it must be proportional to M too. The only symmetric form is Mm.

How to use it

The procedure, every time:

  1. Find r centre-to-centre. For a person on Earth: r = Earth's radius (6.371×106 m), not zero, not your height.
  2. Use SI throughout. kg, metres, G = 6.674×10−11. Mixing km with m is the classic destroyer.
  3. Think in ratios. Double r → F/4. Triple r → F/9. Half r → 4F. The inverse square halves every scaling.
  4. Spheres act as points. Outside a spherical body, use its total mass at its centre (shell theorem).
  5. Forces add as vectors. Multiple masses? Compute each F = GMm/r² and add the vectors — not the magnitudes.

The scaling law

F′/F = (r/r′)²double the distance → quarter the force — no G neededSay it: “the force ratio equals the distance ratio, squared and flipped”
Common mistake: using the distance from the surface (e.g. satellite altitude 400 km as r). r is centre-to-centre: 400 km altitude means r = 6.371×106 + 4.0×105 = 6.771×106 m.

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — your own weight: 70 kg person on Earth

  1. Inventory. M = 5.972×1024 kg (Earth), m = 70 kg, r = 6.371×106 m (Earth's radius, centre-to-centre).
  2. Substitute. F = GMm/r² = 6.674×10−11 × 5.972×1024 × 70 / (6.371×106)².
  3. Evaluate. = 2.790×1016 / 4.059×1013 ≈ 687 N.
  4. Cross-check. 70 kg × 9.8 m/s² = 686 N ✓ — Newton's law reproduces your bathroom scale. The tiny gap is rounding.
Common mistake: r = 0 (“I'm on the Earth”) → division by zero, infinite force. You are one Earth radius from Earth's centre — r is never zero for extended bodies.
Your turn — a 60 kg person on Earth. F?

Answer: ≈ 589 N. Scale Example 1: F ∝ m, so 687 × 60/70 ≈ 589 N. Check: 60 × 9.8 = 588 N ✓.

Example 2 — the Moon's pull on you: 70 kg, Moon overhead

  1. Inventory. M = 7.348×1022 kg (Moon), m = 70 kg, r = 3.844×108 m (Earth–Moon distance).
  2. Substitute. F = 6.674×10−11 × 7.348×1022 × 70 / (3.844×108)².
  3. Evaluate. = 3.433×1014 / 1.478×1017 ≈ 2.32×10−3 N.
  4. Sanity check. Two millinewtons — about the weight of a mosquito. The Moon moves oceans only because there are oceans of water ✓
Common mistake: expecting a noticeable tug (“the Moon causes tides!”). Tides come from the difference in the Moon's pull across Earth's diameter, not the pull itself — and even that is ~10−7 g.
Your turn — the Earth–Moon gravitational force (both bodies)?

Answer: ≈ 1.98×1020 N. F = 6.674×10−11 × 5.972×1024 × 7.348×1022/(3.844×108)² ≈ 1.98×1020 N — the force holding the Moon in orbit, equal and opposite on both bodies.

Example 3 — scaling: your weight at 2 Earth radii from Earth's centre

  1. Use ratios. F ∝ 1/r²: r′ = 2r → F′ = F/2² = F/4.
  2. Evaluate. 687/4 ≈ 172 N — you'd weigh ~17.5 kg-equivalent out there.
  3. Check the meaning. At the ISS (~400 km up, r ≈ 1.06R) it's g′ ≈ 0.89g — astronauts float from freefall, not from zero gravity ✓
Common mistake: “twice as far, half the weight.” Inverse-square: twice the distance is a quarter the force. Say “squared” every time you scale.
Your turn — at 3 Earth radii from the centre, your 687 N weight becomes?

Answer: ≈ 76 N. F′ = 687/3² = 687/9 ≈ 76.3 N. Triple the distance, one-ninth the force.

Before reading on: two 1.0 kg masses sit 10 cm apart on a lab bench. Estimate their gravitational attraction — a newton? a millinewton? far less? — before computing.

Example 4 — judgment call: the Cavendish regime

  1. Substitute. F = 6.674×10−11 × 1.0 × 1.0 / (0.10)² = 6.674×10−11/0.01.
  2. Evaluate. F ≈ 6.67×10−9 N — seven nanonewtons, a billion times weaker than your weight.
  3. The lesson. This is why gravity feels absent between everyday objects and why Cavendish needed a torsion fibre to detect it in 1798. G's tininess is the phenomenon.
Common mistake: “1 kg masses, so the force must be ~10 N.” Mass alone means nothing — G = 10−11 crushes every lab-scale product of masses. Always let G set the scale.
Your turn — two 2.0 kg masses, 20 cm apart. F?

Answer: ≈ 6.67×10−9 N — the same! F = 6.674×10−11 × 4/0.04 = 6.67×10−9 N. Doubling both masses (×4) and doubling the distance (÷4) cancel exactly — the inverse square at work.

Memorization tips

  • Chant it: “G M m over r squared.” Both masses on top, distance squared below.
  • G ≈ 6.67×10−11: “two-thirds, ten-to-the-minus-eleven.” Tiny G, tiny everyday forces.
  • The scaling mantra: ×2 distance → ÷4 force; ×3 → ÷9. Never “half.”
  • r is centre-to-centre. For a person on Earth, r = 6.371×106 m — tattoo it on the problem before computing.
  • The weight audit: any Earth-surface gravity computation should land near mg. If F = GMm/r² gives 70 kg a force far from ~686 N, recheck r.
  • Symmetry: the force on m from M equals the force on M from m. Newton's third law is built into the Mm product.

Final challenge

Five mixed questions — computations, scalings, and the traps, all in one. Score 5/5 and Newton's gravitation is yours.

← Back to the Physics I formula sheet

How to learn a formula here

  1. Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
  2. Work the examples with the answers covered, then uncover one step at a time and compare.
  3. Finish with the final challenge — five mixed questions including the classic traps.
  4. Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

Why is G so incredibly small?

Nobody knows why — it's just how our universe is tuned. G = 6.674×10−11 N·m²/kg² means gravity is absurdly weak between everyday masses; only planet-sized accumulations of mass make it noticeable. Its smallness is why you need Cavendish's delicate torsion balance to measure it.

Is r measured from the surfaces or the centres?

From centre to centre. For spheres (planets, stars), the shell theorem says all the mass acts as if concentrated at the centre — as long as you're outside the body. Using surface-to-surface distance is a classic error.

Does the Earth pull on me as hard as I pull on the Earth?

Yes — Newton's third law. The forces are equal and opposite (about 687 N for a 70 kg person). You don't notice Earth's acceleration because a = F/m with Earth's enormous mass gives an imperceptible acceleration.

Does gravity work inside the Earth?

The simple F = GMm/r² with Earth's full mass only works outside. Inside, only the mass below you pulls (the shell above cancels), so gravity weakens linearly toward the centre for uniform density.

How did Cavendish measure G?

In 1798 he hung small lead balls from a torsion fibre and measured the tiny twist caused by nearby large lead balls. The twist gave the force (~10−7 N), and F = GMm/r² then gave G. It was called ‘weighing the Earth’ because G plus g yields Earth's mass.

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