Gravitational Potential of a Sphere vs Gravitational Binding Energy of a Sphere

My question is about two equations regarding uniform spheres that I've run into:

1. $\quad V=\frac{GM}{r},$

and

1. $\quad U = \frac{3}{5}\frac{GM^2}{r}.$

1) On one hand, $V$ is unknown to me, and is described (in Solved Problems in Geophysics) as "the gravitational potential of a sphere of mass M." I also found it online called "the potential due to a uniform sphere."

2) On the other hand, $U$ is what I've seen before and I know it by the descriptions "sphere gravitational potential energy" or "gravitational binding energy."

My understanding is that $U$ is the amount of energy required to build the sphere piece by piece from infinity. I also recognize $GMm/r$ as the gravitational potential between two masses.

Can someone explain the difference between these concepts? How can $GM/r$ be the "gravitational potential of a sphere"? Isn't that what $U$ is?

There is a mistake in one of your formulas, $U=\frac{3 G M^2}{5 R}$ with $R$ equal to the sphere radius is the energy required to blow every tiny shred of the sphere apart so that its pieces no longer interact gravitationally, as you said, while $V$ as given above with $r$ equal to distance from the sphere center describes how the sphere interacts with other (celestial) bodies, i.e test particles moving in the sphere's gravitational field feel $V$.

To elaborate: the gravitational field around a point mass and around an object that's spherically symmetric is the same outside of the object due to symmetry considerations, which is why $V$ agrees with the formula for the gravitational potential between 2 masses.

• Well then perhaps my confusion is this... What is the difference between $GM/r$ and $GMm/r$ ? The second equation is what I have always considered to be the equation for gravitational potential energy. – equant Mar 1 '13 at 1:21
• Well if you check the units, only the second one is actually an energy. However it's useful to consider this quantity per unit (of the test) mass since gravity affects all objects of the same mass equally, so we call both the gravitational potential depending on context. I think people usually call the one without the factor of $m$ "potential" and the other one "potential energy" to distinguish the two though. – alexarvanitakis Mar 1 '13 at 1:26

For a ball with uniform/constant mass-density $\rho$, one has

$$\tag{1} \frac{M}{R^3}~=~\frac{4\pi\rho}{3}~=~\frac{m}{r^3}.$$

The gravitational potential is

$$\tag{2} V(r)~=~-\frac{Gm}{r}.$$

The infinitesimal gravitational potential self-energy is

$$\tag{3} dU ~=~ V(r)dm~\stackrel{(2)}{=}~-\frac{Gm}{r} dm~\stackrel{(1)}{=}~-\frac{GM^{\frac{1}{3}}}{R} m^{\frac{2}{3}}~dm.$$

Hence the integrated gravitational potential self-energy is

$$\tag{4} U~=~\int \! dU ~\stackrel{(3)}{=}~ -\frac{GM^{\frac{1}{3}}}{R} \int_0^M m^{\frac{2}{3}}~dm~=~-\frac{3}{5}\frac{GM^2}{R}.$$

• Thank you very much for this quantitative answer, it was very helpful. – equant Mar 1 '13 at 7:21

I don't know what $U = \frac{3}{5}\frac{GM2}{r}$ in your question is, but V seems to be just gravitational potential energy, the usual formula is $U=-\frac{GMm}{r}$ which, in your case it seems that $M=m$, and U is simply replaced by V and the terms on the right positive instead of negative because the definition is flipped (instead of the energy required to bring something from the sphere to infinite distance away, reverse it to bringing something from infinite distance away to the sphere so you flip the sign from '-' to '+' on the right side of the equation $U=-\frac{GMm}{r}$)

It seems that your book just isn't using the convention of U=gravitational potential energy (it chose V instead).

• Regardless of conventions, gravitational (and electrostatic) potential is proportional to $\frac{1}{r}$. – alexarvanitakis Mar 1 '13 at 1:12