Gravitational potential energy and escape speed

To bring a mass m from very far away (infinity) to a distance r from a mass M, gravity does the work. Taking the potential energy to be zero at infinity, the gravitational potential energy is V = –GMm/r.

Why negative? Gravity attracts, so the masses lose energy as they come together. V is largest (zero) at infinity and becomes more negative the closer they get. For three or more masses, add V over every pair.

The gravitational potential at a point is the potential energy per unit mass: –GM/r, in J/kg. It is a scalar, so the potentials of several masses simply add. (Potential belongs to one mass; potential energy needs two.)

For small heights the exact formula gives back the familiar one: ΔV = GMm[1/R – 1/(R + h)] ≈ GMmh/R² = mgh. For large heights you must use ½mv² – GMm/R = ½mv′² – GMm/r, never mgh.

How fast must you throw a body so that it never comes back? Its total energy must be at least zero: ½mv² – GMm/R = 0. This gives the escape speed v_e = √(2GM/R) = √(2gR).

• For the earth, v_e ≈ 11.2 km/s; for the moon, only about 2.3 km/s, which is why the moon has no air.
• It does not depend on the body's mass, nor on the direction it is thrown in (as long as it misses the planet).
• From a height h, use R + h: from 1600 km up, v_e = 11.2 × √(6400/8000) ≈ 10 km/s.

Since v_e = R√((8/3)πGρ), a planet with the same density but twice the radius has twice the escape speed.

Between two attracting bodies, such as the earth and the moon, a projectile need not climb all the way to infinity. It only has to reach the neutral point, where the two pulls balance; after that the other body pulls it on.