We know that the magnitude of the gravitational force is given by:

F = -GmM/r²

Use the connection between force and potential energy to determine the general form of gravitational potential energy. U = mgh applies only for a uniform field, so it does not apply here where the field goes as 1/r².

F = -dU/dr

ΔU = - ∫ F dr

This gives U = -GmM/r, if we define the potential energy to be zero at r = infinity. This is what we do - you are NOT free to define the zero anywhere you want - it is pre-defined to be zero at infinity.

Does it matter that the potential energy is negative everywhere? Not at all - all that matters is how the potential energy changes. If a mass moves from close to an object to further away, the potential energy changes from a larger negative number to a smaller one - this is an increase, as we expect.

Is this consistent with the mgh we used for potential energy near the surface of the Earth? Yes. If you move an object up a height h from ground level, the potential energy changes as follows:

ΔU = U_f - U_i = -GmM/(R+h) - -GmM/R.

1/(R+h) = 1/[R(1+h/R)] = (1/R)*(1 - h/R) when h is small compared to R.

Substituting this in gives:

ΔU = -GmM/R + GmMh/R² + GmM/R = GmMh/R²

We showed previously that g = GM/R², so:

ΔU = mgh.

If you throw an object up in the air, it generally comes back down. How fast would you have to throw it so it never came back down? Ignore air resistance. The minimum speed required to escape from a planet's gravitational pull is known as the escape speed.

There is a negative potential energy associated with an object of mass m being at a planet's surface. For a planet of mass M and radius R, that potential energy is:

U = -GmM/R.

How much kinetic energy is needed for the object to escape from the planet's gravitational pull?

We know the potential energy at infinity is zero, so the minimum kinetic energy must be:

K_escape = +GmM/R.

We're applying conservation of energy here:

U_i + K_i = U_f + K_f

The total initial energy must be at least zero for the object to escape. If it's negative, when the kinetic energy hits zero there will still be some negative potential energy, so the object will be a finite distance away and will be pulled back to the planet.

If K_escape = GmM/R then:

½ mv² = GmM/R

v_escape = (2GM/R)^½

For the Earth this works out to 11.2 km/s.