# Gravitational potential energy questions

1. Jul 20, 2009

### andytrust

Hello,
I haven't learned a lot of physics and math, but I don't know how I misunderstand the following concepts:

Why is the gravitational potential energy of a system of two bodies described in one term instead of two? If each body in a two-body system effects a gravitational force on the other one, wouldn't both objects have their own gravitational potential energy (albeit they would be the same value), instead of sharing one energy?

My second question: The derivation for the escape velocity of an object at the Earth's surface (as found for instance on Wikipedia here: http://en.wikipedia.org/wiki/Escape_velocity#Derivation_using_G_and_M), seems to me to assume that the Earth remains static while the object is moving away from it. If the object were sufficiently large, wouldn't the Earth and its gravitational field move after it, making it harder for the object to escape the Earth's gravitational field, so that it would not reach infinity? Why doesn't this happen?

2. Jul 20, 2009

### diazona

I guess you could say that each object has its own gravitational potential energy that is half of the total. But that'd be more complicated than just having the one potential energy for the pair together.

Actually, this might be a better line of reasoning: in many cases, one of the two bodies will be stationary, or at least so massive (compared to the other) that it effectively doesn't move. Example: the Earth and a satellite. In those cases, it makes sense to say that one body (the smaller one) has its own individual potential energy. Essentially, you assign all the potential energy of the system to the smaller body. But in the general case, both bodies are free to move, and then it's impossible to define a potential energy for one body alone based only on that body's position. You have to incorporate the position of both bodies. So it just makes sense to assign the potential energy to the system of the two bodies, and not try to split it up between them.

Plus, the formula just works

You're right, the derivation does assume that the Earth remains fixed in place. Escape velocity is really only a meaningful concept in a static gravitational field - which in practice means, only for small things like rockets.

3. Jul 20, 2009

### rcgldr

The number of bodies doesn't matter. The gravitational potential energy difference between two points a and b, GPE(b) - GPE(a) is the negative of the work peformed by gravity when the object moves from a to b. For a finite number of finite sized bodies, GPE for one of those bodies at some point 'p', relative to the center of mass of the other bodies can be defined by defining GPE(∞) = 0, and then defining the potential energy at a point 'p' with respect to ∞, to be GPE(p) = GPE(∞) - GPE(p).

This derivation ignores the mass of that object. The true escape velocity of a two body system, would require that the initial kinetic energy at point p be greater than or equal to GPE(p).

Last edited: Jul 20, 2009
4. Jul 22, 2009

### andytrust

Thank you for clarifying this. I guess my question is: does it mean that when gravitational potential energy is calculated (I guess this also applies for electrical potential energy), you have to assume the object(s) doing the work remain stationary (in an inertial frame of reference)?
If that is so, does the idea of gravitational potential energy have any meaning in a system where all the bodies are moving?