Why doesn't gravitational potential energy cancel out?

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.

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 $S$.

• After reading your answer I realised that my problem came from a misunderstanding of what the conventional way of writing potential energy (which I have used in my question) actually means. Namely, it is the energy needed to reach infinity from that point. Of course, work needs to be done to move the planet from it's location to an infinite distance away from all mass, so using this convention its potential energy cannot be 0. – Pancake_Senpai Nov 16 '17 at 18:55

An object is at rest if the net force acting on it is zero. It has noting to do with the specific value of the potential energy at the point. In fact, you have the freedom to shift the potential energy function by a constant without changing the physics. This is because the physics is related to forces, and force is the derivative of the potential. In your case, if the planet is at distance $r$ from one of the stars, the potential energy is given by

$$E_{\rm p}=-\frac{GMm}{r}-\frac{GMm}{2R-r}$$

The force that acts on the body at $r$ is the derivative of the above potential fuction

$$F=-\frac{\partial E_{\rm p}}{\partial r}=\frac{GMm}{r^2}-\frac{GMm}{(2R-r)^2}$$

and you can clearly see that $F(R)=0$ as you stated.

The potential energy is negative because, if you attached a string to the planet and pulled it very far from the two stars, then you would do work doing so, and in the limit the work you would do would be $2GMm/R$. And if we arbitrarily define the zero of gravitational potential energy as the potential energy infinitely far from other masses, which is the conventional way it's defined, then the potential energy of the planet in its initial position must be $-2GMm/R$.

(Note that you can define the zero wherever you like: it is only differences that matter. Defining the zero as being at infinite separation is usually done because it's a nice, universal, choice.)

Your question is unclear: energy is defined only to a constant. What matters is how the potential energy changes as the position of the planet changes. So your question might be: why is the derivative of the potential energy zero with respect to the position of the planet?

The planet is like a ball at the very top of a mountain, where the ball cannot decide whether to roll down one side or the other.

That's what the zero of the derivative of the potential energy means. The planet finds both directions (towards star A or B) equally appealing. Zero means it cannot decide between one or the other.

The weird thing is that the planet is at a maximum (on the line joining the two stars, see @Farcher's answer about saddle points). If something happens to nudge one side or the other, then it will suddenly find that it increasingly prefers one star over the other. Suddenly, the derivative of the potential energy is no longuer zero. So, even though the planet cannot decide on its own, where somebody to force the decision for it (say a solar flare from one star), then it's gonna roll with it.

To be accurate, the planet is at the L1 Lagrange point of the system.

protected by Qmechanic♦Nov 16 '17 at 9:59

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