Universal gravitation

Newton's law of universal gravitation describes the attraction between masses with one rule that works for falling objects, planets and satellites. The force weakens with distance but never suddenly switches off.

The universal gravitational force

F = Gm₁m₂/r²

G = 6.67430 × 10⁻¹¹ N m² kg⁻². The distance r is measured between the centers of mass for point-like or spherically symmetric bodies.

Each mass pulls on the other with the same force magnitude in the opposite direction. Their accelerations can differ because a = F/m.

Gravity follows an inverse-square law. Doubling the separation between two point masses reduces the force to one quarter.

Gravitational field and weight

Instead of recalculating the force for every test mass, define the gravitational field g⃗ = F⃗/m. Outside a spherical mass M:

g = GM/r²

Near Earth's surface, this is approximately 9.81 N kg⁻¹, numerically the same as 9.81 m s⁻². Weight is then W = mg.

Orbiting astronauts are not beyond gravity. They feel weightless because the spacecraft and everything inside it are accelerating together in free fall. The local gravitational field can still be substantial even when a scale reading is near zero.

Spherical bodies and the role of radius

Outside a spherically symmetric body, its gravitational effect is the same as if all its mass were concentrated at its center. This is why altitude calculations must use distance from the planet's center, not simply height above the surface.

Inside a uniform spherical body, the simple GM/r² expression using the total mass no longer applies; only the mass enclosed within radius r contributes to the net inward field in the same way.

This result is a consequence of the shell theorem: a spherical shell attracts an external mass as if the shell's mass were concentrated at its center, while the net gravitational field everywhere inside an ideal uniform shell is zero.

Orbits are falling trajectories

An orbit is continuous free fall. Tangential motion carries the object forward while gravity continually bends its velocity toward the central body.

For a circular orbit, gravity supplies the centripetal acceleration. Equating GMm/r² with mv²/r gives v = √(GM/r). Higher circular orbits have lower orbital speed, even though their orbital period is longer.

Gravitational potential energy and escape

When large changes in distance matter, use the zero of gravitational potential energy at infinite separation:

U = −GMm/r

The negative sign means a bound pair has less energy than the separated masses at infinity. The minimum launch speed that reaches infinity with zero final speed, ignoring atmosphere and rotation, is v_esc = √(2GM/R).

Gravitational force and potential energy are two descriptions of the same interaction. Force tells how momentum changes locally; potential energy makes it easier to compare states. For a circular orbit, the total mechanical energy is negative, E = −GMm/(2r), showing that the satellite is gravitationally bound.

Worked examples

1. Effect of doubling distance

Two masses attract with force F at separation r. What is the force at 2r?

Solution

Because F ∝ 1/r², F' = F/(2²) = F/4.

2. Gravitational field at altitude

At a distance 2R from Earth's center, how does gravitational field strength compare with its value at the surface R?

Solution

g ∝ 1/r², so g(2R) = g(R)/4. This location is one Earth radius above the surface.

3. Circular-orbit speed

Derive the orbital speed of a small satellite in a circular orbit of radius r around mass M.

Solution

Gravity supplies the centripetal force: GMm/r² = mv²/r. Cancel m and one r:

v = √(GM/r). The satellite's own mass does not affect the ideal orbital speed.