# USAPhO F=ma 2010 exam # 17 (Gravitational potential energy)

1. Jan 24, 2014

### Agrasin

1. The problem statement, all variables and given/known data

Four masses m are arranged at the vertices of a tetrahedron of side length a. What is the gravitational potential energy of this arrangement?

2. Relevant equations

U = -Gm2 / r

3. The attempt at a solution

The potential energy is the negative of the work done by gravity in this case, I believe. Basically, how much work would gravity do in order to assemble this tetrahedron, pulling the masses from infinity? Then take the negative of that.

To put 1 mass in place, 0 * Gm2 / r

To add the 2nd mass, 1 * Gm2 / r

To add the 3rd mass, (2 * Gm2 / r) divided by 2 because the work is shared by two masses doing the pulling. So, 1 * Gm2 / r

To add the 4th mass, (3 * Gm2 / r) / 3 with similar reasoning. So, 1 * Gm2 / r

Add it all together because energy is a scalar. And then take the negative.
-3 * Gm2 / r

However, the answer is -6 * Gm2 / r

2. Jan 24, 2014

### tiny-tim

Welcome to PF!

Hi Agrasin! Welcome to PF!
What do you mean by "shared"?

Why does it make any difference?

Don't we want the total work done, no matter who does it?​

3. Jan 24, 2014

### Staff: Mentor

Hi Agrasin. Welcome to Physics Forums.

There's no "sharing" of the work done. Each new body is is being moved from infinity to a location in space where the existing field (caused by other bodies) is determining the potential. You only want to consider the work done by some theoretical agency in bringing the new body to that spot.

As a crude metaphor, imagine that some "hand" is moving the body very slowly from infinity to that spot. The work done by that "hand" is the work you're after. In the case of assembling masses which are mutually attractive, the "hand" actually has to restrain the body from accelerating all the way; the force it needs to apply is opposite to the direction of motion, hence the work done is negative.

[Ah! Beaten to the punch by tiny-tim.]