If you have 2 stars of mass $M$ and seperated by a distance $2R$, and a planet of mass $m$ is equidistant between them, the potential energy of the planet is given by:
$$E_p = \frac{-2GMm}{R}$$
This is the sum of the potential energy the planet has due to each star. However, considering that gravitational potential energy is just a measure of how much kinetic energy would be gained were the planet to fall towards the source of the field, shouldn't the potential energy be 0?
The planet is effectively stuck between the 2 stars. It will be at rest on the line joining the centres of the two stars, unable to move because the net force on it would be 0. Therefore, I do not understand why the net potential energy is not also 0.
Best Answer
The potential diagram on the left also shows the gravitational field lines.
The point labelled $S$ is a saddle point (neutral point) as for as the potential diagram is concerned and as such the gradient of the potential and hence the force on a mass at that position is zero.
As soon as the mass moves away from position $S$ there is a force on it as there is a potential gradient.
The right hand diagram shows the variation of potential in "3D".
What you will notice is that around position $S$ the potential is varying.
There is no reason why you could not say that the potential energy of your mass when placed at position $S$ is zero although it is more usual to make infinity the zero of potential.
However you will need to do work, ie raise the potential energy of your mass, if you move your mass away from the two masses and work will be done for you or your mass will gain kinetic energy if it moves towards one of the two masses. The important thing to note is that even if the potential gradient (or force) is zero that does not mean that the potential is necessarily zero.
All it means is that at that position the potential is not varying with position.
The potential diagram also shows that placing a mass at position $S$ results in an unstable equilibrium in that a small displacement of the mass will result in the mass moving away from position $S$.