Question

Gravitational Potential Energy is given by the formula

GMm/ R+h

- GmM/ (R+h)²

- GmM(R+h)²

- GMm / √R+h​

Answer 1

Answer:

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 = Uf - Ui = -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.