How to calculate the bulge in a swimming pool surface caused by a local point mass? Imagine a swimming pool of an arbitrary width, small on a planetary scale. At a height above its usual surface, a superdense mass is suspended. How do I calculate the distortion in the surface of the small body of water caused by a given mass (or vice versa) in close proximity to the water? Approximations are acceptable.
 A: In general, the surface of a fluid at rest will follow the contours of equipotential surfaces.  For swimming pools, where we can use the approximation $U=mgh$, those are horizontal planes; the $1/r$ potential from the sphere just adds to it.
If you choose a coordinate system where the water's unperturbed surface is in the $z=0$ plane and the suspended mass $M$ is at position $(x,y,z)=(0,0,h)$, then the potential energy of a little blob of water with mass $m$ is
$$
U = mgz - \frac{GMm}{\sqrt{(h-z)^2 + x^2 + y^2}}
$$
where $g=GM_\text{earth}/R_\text{earth}^2$ is the gravitational acceleration near Earth's surface.  If you choose $U$ equal to any constant and $(h-z)\approx h$ (that is: the "bulge" is small), then it's trivial to solve for the equipotential surfaces $z(x,y)$.  In this example the $U=0$ equipotential --- that is, the water's surface --- is shifted upwards by a $\Delta z$ obeying
$$
\frac{\Delta z}{h} = \frac{M/h^2}{M_\text{earth}/R_\text{earth}^2}
\underset{\text{sphere}}{\approx} \frac{\rho h}{\rho_\text{Earth} R_\text{Earth}}
$$
where the final $\approx$ assumes that the attracting mass is a sphere that comes pretty close to the liquid surface.  Your intuition that the effect is small was good: for a sphere of any ordinary density $\rho$, the bulge can be no taller than $\Delta z/h \approx h/R_\text{earth}$.
The half-maximum of the bulge is a circle with radius $h\sqrt 3$: the bulge gets lower and broader as a point mass moves away.
A: Assume that the surface of the water is an equipotential surface; i.e., that the gravitational potential is the same at each point on the surface. If it were not, the water would move.
The total gravitational potential is the sum of the potential due to the Earth and the potential due to the superdense point mass.
Doing the actual calculation is inappropriate since we don’t provide complete solutions to homework-like problems.
