# Why does the Gravitational Potential energy formula indicate that heavier masses in the same point in the field have lower potentials?

$$U = -\frac{Gmm_E}{r}$$

Intuitively, it'd appear that the further away two objects are, the greater their gravitational potential, and this is confirmed by the gravitational potential formula. The greater r is, the closer the value would be to zero, where zero is the maximum point.

However, the same formula also seems to say that if two objects are the exact same distance from a body like earth, but have different masses, the one with the lower mass would have a greater gravitational potential. The heavier mass would have the higher absolute magnitude but would be the smaller number due to the negative sign. But since more energy is required to move that heavier object the same distance, you'd expect it to have a higher gravitational potential.

Why isn't that reflected by the formula, or have I just interpreted it wrong? From my understanding, the greater the mass and the greater the radius between the two objects, the greater their gravitational potential. But it seems that the formula only yields one of those as being true.

Thanks

• Heavier masses have a more negative potential. So you have to do more work to push them to infinity (zero). – BowlOfRed Oct 23 '15 at 6:23
• @BowlOfRed that should be an answer (preferably with a couple more sentences of explanation, but it is technically a valid answer itself) – David Z Oct 23 '15 at 6:28

You've already got the idea of having the zero potential energy point being at an "infinite" distance. From there we see that the potential energy decreases as the masses approach each other. If we let them fall to that point, the reduction in $GPU$ is exactly matched by an increase in $KE$.
If the $KE$ were dissipated, it would take work to move the objects back to an infinite distance apart. The amount of work necessary would greater if the objects were closer together to start, or if the objects were more massive.