# Confusion over the gravitational potential energy inside a sphere

I'm confused over the concept of gravitational potential energy inside a sphere. I understand that the gravitational potential energy inside the sphere is supposed to be a constant $U = \frac{GMm}{R}$ where $R$ is the radius of the sphere.

What I am confused about is how this applies in conservation of energy equations. For instance, the classic "Drill a hole through the planet". In that instance, GPE at the surface is converted to KE at the core and then back to GPE as it continues through. But, if the object still has gravitational potential energy, this equation doesn't hold. What am I missing concept wise?

• 'r' in that equation is your distance from the center of the sphere, not the radius of the sphere itself. – Sparr Mar 3 '14 at 15:43
• A note on mathematical terminology: A sphere $S^2$ is a $2$-dimensional object, while the corresponding $3$-dimensional object is called a ball $B^3$. In other words: the boundary of a ball is a sphere. Gravity will depend on whether we consider a massive planet or just a shell. Related: physics.stackexchange.com/q/18446/2451 , physics.stackexchange.com/q/7346/2451 and links therein. – Qmechanic Mar 3 '14 at 15:47

Your equation is incorrect. The gravitational potential is $$\phi(r)=-GM\frac{3a^2-r^2}{2a^3}$$ when you're inside a uniform sphere of radius $a$ with total mass $M$. This is a quadratic potential in $r$, which is why it gives rise to harmonic exchange of energy when you oscillate between the planet surface and the core.
• Can you explain what is $a$ and $r$ just to be clear. I think you are talking about a hollow shell, and the poster is talking about a solid sphere. – ja72 Mar 3 '14 at 17:23
• Good, concise answer. You can even check consistency with the expression for external points by plugging in $r=a$, and you find -GM/a. Which is correct. – Alan Rominger Mar 3 '14 at 17:55