# What's wrong with my calculation of gravitational potential for a uniform sphere?

This is really embarrassing, but I'm not quite sure where I'm going wrong here... Why is this calculation of the gravitational potential inside a sphere with uniform mass distribution incorrect?

Set-Up

Let's say the sphere has mass $M$ and radius $R$ (and uniform mass density $\mu$), and what we want to find is the potential at any distance $r$ from the center of the sphere, where $r<R$. We normalize the potential to zero at infinity.

Calculation

The potential $\phi(r)$ is equal to the potential right outside of the sphere, plus the potential difference between some point inside the sphere and a point right outside.

$$\phi(r)=\phi_0-\int_R^r \frac{\mu G}{r}dV$$

(Sorry for using $r$ for the upper limit of the integral as well as for the variable in the integrand. Hopefully this doesn't cause confusion.)

Now to figure out the different aspects of the above equation equation. The potential right outside of the sphere is:

$$\phi_0=-\frac{MG}{R}$$

The differential volume element can be expressed as the constant-potential spherical shell's surface area times the shell's differential width: $$dV=4\pi r^2 dr$$ And one final detail, the mass density of the sphere: $$\mu=\frac{3M}{4\pi R^3}$$

Using this information,

$$\phi(r)=-\frac{MG}{R}-\frac{3MG}{R^3}\int_R^r r dr$$

$$\phi(r)=-\frac{MG}{R^3}\left[R^2+\frac{3r^2}{2}-\frac{3R^2}{2}\right]$$

$$\phi(r)=-\frac{MG}{2R^3}(3r^2-R^2)$$

Conclusion

This result disagrees with a few places I've visited, like this one, which states that the correct result (in terms of the variables I've used) is

$$\phi(r)=-\frac{MG}{2R^3}(3R^2-r^2)$$

Both results give the same potential at $r=R$, obviously, but my result starts to look ridiculous for values like $r=R/2$.

The only part in my calculation that seems sketchy to me is that first equation, where I talk about the potential difference at points inside and outside of the sphere; I don't know if it's correct to be dividing by $r$ in the integrand... Or maybe I just made a stupid algebra mistake somewhere in there.

Where did I go wrong?

• Just a little issue with language, setting the potential at infinity as zero is not normalizing but just setting the boundary conditions. It reflects a belief. – Ignacio Vergara Kausel Jul 11 '14 at 9:20
• My Guess would be that you got the boundary conditions wrong. – krismath Jul 11 '14 at 10:00
• Your $r$ isn't what you think it is. To calculate the potential using the integral method you are using, $r$ would have to be the distance between the source point and the observation point. Your $r$ is the distance between the shells that contain $r$ and $R$. @AlphaCentauri has the best advice. – garyp Jul 11 '14 at 13:37
• You can look at the correct description and solution of this problem in the document of this link (pages 34-35): http://astrowww.phys.uvic.ca/~tatum/celmechs/celm5.pdf – user67930 Dec 18 '14 at 5:18
• @Oscar Iglesias Clotas : What is name of a book having this chapter ? – Murtuza Vadharia Dec 18 '14 at 5:33

The problem in your calculation lies in your first equation which is simply wrong. What you are doing, according to your equation, is to somehow calculate and subtract the potential of the shell outside of $r$. However, this is not how you obtain the potential at the point $r$.
What you should do instead is integrate the force acting on a test mass $m$ from $R$ to $r$: $$\Delta\phi=\phi(r)-\phi(R)=-\int_R^r \frac{F_{\rm G}(r')}{m}\,\mathrm{d}r'\,.$$ Here, you have to use the fact mentioned before that only the mass inside of $r'$ contributes to $F(r')$. If you do this, you should obtain the correct result already stated in your question.
Hint: The issue in a nutshell is Newton's shell theorem: For a given radial position $r$ only mass parts further in contribute to the gravitational binding, while the effects from mass parts further out cancel because of spherical symmetry.
The safest is therefore to integrate the potential energy from the center $r=0$ and outwards. If one tries to integrate from the outside and inwards, it is easy to not properly remove effects from mass parts further out.