Gravitational potential inside a long but finite cylinder Suppose a cylinder of length $\ell$, radius $R$ with a bore hole through its long axis of radius $r$. How can the gravitational potential (not the gravitational field) inside the cylinder be derived.
I thought of using Gauss's law of gravity, but it only talks about the gravitational field and if $\ell$ is large enough, the field goes to zero. But this does not mean the potential is zero, its just constant. Is there a simple formula for it?
 A: First I was afraid that I would not find solutions to the integrals but sagemath.org came to the rescue, so here is my humble attempt. (Sorry for the ugly artwork.)

Let the cylinder have length $L$, an outer radius of $R$ and a bore hole radius $r$. For a start lets concentrate on the gravitational potential for a point $\vec{x}$ in the middle of the bore hole. The volume integral to be solved goes over the while cylinder:
$$
V(\vec{x}) = \left(\int_0^{l_1}+\int_0^{L-l_1}\right) \int_r^R \int_0^{2\pi} 
\frac{G\cdot\rho(\vec{r})}{|\vec{x}-\vec{q}|} d\alpha\, da\,dr
$$
where $\vec{q}=(l,a,\alpha)$ is a point with a long axis coordinate of $l$, a cylinder radius coordinate of $a$ at rotation angle $\alpha$. We have to integrals involving the long axis to account for the left and the right part of the cyclinder when $\vec{x}$ is somewhere inside the cylinder.
The distance $|\vec{x}-\vec{q}|$ is then just $\sqrt{l^2+a^2}$ independent of $\alpha$, which is why the inner integral from $0$ to $2\pi$ simplifies to factor of $2\pi$. Assuming a fixed mass density $\rho$, the integral is then:
$$
V(\vec{x}) = 2\pi G\rho \cdot
\int_r^R 
\left(\int_0^{l_1}+\int_0^{L-l_1}\right) 
\frac{1}{\sqrt{l^2+a^2}} da\,dr
$$
Luckily I came across sagemath.org which quickly solves the inner integral for the left side $0\dots l_1$:
innerint(a, l1)=definite_integral(1/sqrt(l^2+a^2), l, 0, l1).subs({sqrt(a^2): a});

(I am leaving out a few variable definitions here).
The result is
$$I_1(a,l_1) = \operatorname{asinh}(l_1/a)$$
Now we still need
$$V(\vec{x}) = 2\pi G\rho \cdot
\int_r^R I(a,l_1) + I(a,L-l_1) da
$$
Again sagemath helped me with the integration. The result for the first summand only look like this:
\begin{align}
I_2(l_1, r, R) &=
R\cdot {\rm arcsinh}\left(\frac{l_{1}}{R}\right) - r {\rm arcsinh}\left(\frac{l_{1}}{r}\right)\\
 &+ \frac{1}{2} \, l_{1} \log\left(R + \sqrt{R^{2} + l_{1}^{2}}\right) - \frac{1}{2} \, l_{1} \log\left(-R + \sqrt{R^{2} + l_{1}^{2}}\right) \\
&- \frac{1}{2} \, l_{1} \log\left(r + \sqrt{l_{1}^{2} + r^{2}}\right) + \frac{1}{2} \, l_{1} \log\left(-r + \sqrt{l_{1}^{2} + r^{2}}\right)
\end{align}
The complete result for a point $\vec{x}$ located on the cylinder axis a distance $l_1$ into the cylinder would then be
$$V(\vec{x}) =  2\pi G\rho \cdot (\,I_2(l_1, r, R)+I_2(L-l_1, r, R)\,) .
$$
If I did get my constants all right, $R=4m$, $r=0.02m$ and $L=1000m$ should have generate a potential along the cylinder axis like this:

One can imagine, that the longer the cylinder gets, the closer comes the potential to a constant along the axis of the cylinder.
A: I found a paper by Y Chen and A Cook : The Gravitational Field Inside a Long Hollow Cylinder of Finite Length which seems to be free access and may be useful.  No simple formula and not zero.  You only need look at section 2, as the rest refers to a more involved calculation to aid in determining limits for non-Newtonian effects experimentally.
To see simply why it won't be zero consider just the axis of the cylinder.
Pick a spot in the middle of the cylinder's length along the axis.  Assuming equal density throughout it is clear that the force from the section on the left will be balanced the section on the right.
But now move away from the middle of the length and stay on the axis.
Because the cylinder is of finite length there is clearly less mass on one side of our point than the other.  The section that's smaller has it's force balanced by just part of the section on the other side of the point.  The point away from the exact center, even along the axis, must feel a net force.
And of course at the end of the cylinder a point will feel force acting in only direction with nothing to balance it because there is nothing on one side of that point.
