Newton's Law of Gravitation

Treat two bodies as points m₁ and m₂ a distance r apart. Each pulls the other along the line joining them with F = G m₁ m₂/r². The constant G = 6.674 × 10⁻¹¹ N m² kg⁻², equivalently m³ kg⁻¹ s⁻², was measured in a laboratory a century after Newton wrote the law, and its smallness sets the scale of everything human-sized: two 70 kg students standing 1.0 m apart attract with 3.3 × 10⁻⁷ N, the weight of a 33 μg grain. Earth pulls one of them with 687 N, two billion times more, only because Earth's mass is 5.97 × 10²⁴ kg. Gravity dominates at large scales for a structural reason: there is no negative mass, so nothing screens it and every kilogram adds. In vector form, with r̂ the unit vector from m₁ to m₂, the force on m₂ is −(G m₁ m₂/r²) r̂; the minus sign is what makes the law attractive.

The law is symmetric in the two masses, so it never produces one force; it produces a pair. The force on m₂ due to m₁ and the force on m₁ due to m₂ share the magnitude G m₁ m₂/r², point in opposite directions and act on different bodies: Newton's third law is built into the formula. What differs is the response, a = F/m. Earth and the Moon pull each other with (6.674 × 10⁻¹¹)(5.97 × 10²⁴)(7.35 × 10²²)/(3.84 × 10⁸)² = 1.99 × 10²⁰ N. The Moon accelerates at 2.70 × 10⁻³ m s⁻² and Earth at 3.33 × 10⁻⁵ m s⁻², a ratio of 81 that is simply the mass ratio, and both move about their common centre of mass 4670 km from Earth's centre, inside Earth. Notice too that a body's acceleration toward another, G M/r², does not contain its own mass, which is why a hammer and a feather fall together on the airless Moon.

With more than two bodies, work out every pair as if the others were absent and add the forces as vectors. A third mass never changes the pull between the first two; nothing shields gravity. Components are the working language. Put equal masses M at (0, +a) and (0, −a) and a probe m on the x-axis at distance x. Each pull has magnitude G M m/(x² + a²) and slopes toward its source, so the y-components cancel and the x-components add: F(x) = 2 G M m x/(x² + a²)³⁄², directed toward the origin. At x = 0 the pulls cancel completely. Far away, x ≫ a, F tends to 2 G M m/x², the pull of one point of mass 2M; at x = 10a it is already within 1.5% of that. In between lies a maximum: dF/dx ∝ (a² − 2x²)/(x² + a²)⁵⁄² vanishes at x = a/√2, where F = 4 G M m/(3√3 a²) ≈ 0.770 G M m/a². Adding magnitudes would miss the cancellation entirely.

Newton needed one more result before r could mean the distance to Earth's centre. A uniform thin spherical shell attracts a mass outside it exactly as if the shell's mass sat at its centre, and exerts no net force on a mass anywhere inside it. The inside statement has a short argument: a narrow double cone from the mass cuts the shell in two patches whose areas grow as the square of their distances while their pulls fall as the inverse square, so opposite patches cancel, and only because the exponent is exactly 2. Build a planet from nested shells, each uniform though density may change with depth, and every spherically symmetric body acts outside as a point at its centre. So a 75 kg person weighs (6.674 × 10⁻¹¹)(5.97 × 10²⁴)(75)/(6.371 × 10⁶)² = 736 N, using 6371 km rather than the distance to the ground. Inside, only the shells beneath pull; for a uniform sphere that mass is M r³/R³, so F = G M m r/R³, which halves, rather than quadruples, halfway to the centre.

Because F ∝ 1/r², ratios need neither G nor the masses: F₂/F₁ = (r₁/r₂)². Doubling a separation quarters the force, and tripling it leaves a ninth. Measure r from the centre every time. A satellite 420 km up is at r = 6791 km, so its pull is (6371/6791)² = 0.88 of its launch-pad value; gravity has barely weakened, and why the crew float is a question about free fall, taken up with circular orbits. The exponent itself is an empirical claim, and Newton tested it. The Moon is 3.844 × 10⁸ m, or 60.3 Earth radii, away, so an inverse-square pull should give it an acceleration of 9.82/60.3² = 2.70 × 10⁻³ m s⁻². Its measured centripetal acceleration, 4π²r/T² with T = 27.32 days, is 2.72 × 10⁻³ m s⁻², agreement within 1%. An exponent of 2.1 would predict 1.79 × 10⁻³ m s⁻², a third too small, which shows how sharply a distance ratio of 60 tests the square.

Superposition is exact for any shape; the centre shortcut is exact only for spheres. For anything else, sum over the pieces and measure how far off the shortcut is. Let a uniform rod of mass M and length L lie along a line, with m on that line a distance d beyond one end. A slice ds at distance s holds mass (M/L) ds, so F = ∫ G m (M/L) ds/s² from s = d to d + L, which is (G M m/L)[1/d − 1/(d + L)] = G M m/[d(d + L)]. Putting all of M at the rod's centre gives G M m/(d + L/2)², which the exact result exceeds by the factor 1 + L²/[4d(d + L)]: by 12.5% at d = L, by only 0.23% at d = 10L. The error shrinks roughly as (size/distance)², and its sign depends on direction: side-on, a two-lobed body pulls less than a point would. The centre of mass, the right place to hang weight in uniform g, is not a centre of attraction for a rod, a dumbbell or a lumpy asteroid seen close up.