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I found this interesting problem in Introduction to Classical Mechanics with Problems and Solutions by David Morin:

Given a point $P$ in space, and given a piece of malleable material of constant density, how should you shape and place the material in order to create the largest possible gravitational field at $P$?

Any ideas?

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This website seems to provide the solution star.tau.ac.il/QUIZ/02/A02.02.html –  Qmechanic Jul 24 '11 at 17:56
@Qmechanic: Thanks! Could you make your comment an answer so that I could accept it? –  Bernhard Heijstek Jul 24 '11 at 18:02
What is a "large" field in this context? BTW, has this curve (or the solid) a special name? –  Georg Jul 24 '11 at 19:11

4 Answers 4

up vote 4 down vote accepted

This website seems to provide the solution http://star.tau.ac.il/QUIZ/02/A02.02.html

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The most surprising thing to me is how small the improvement is over a simple sphere. –  Ted Bunn Jul 24 '11 at 21:11

My first guess is to shape the material into a unit-radius sphere and place it so that P is on its surface- this way, you don't 'lose' any gravitational field due either to the 1/r^2 term, or to any asymmetric considerations in the direction of the vector pointing from the source.

In other words- here we have a maximally symmetric source, and we are sitting right on top of it. I'd be interested to see if anything more exotic comes up, though!

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Hmmm...that's fun.

It is important to note that we seek the largest possible field, and not the lowest potential versus infinity which might yield a lower result.

So let's consider some obvious cases (all of which I will reduce to a dependence only on $\rho$ and $M$.).

  • A nearby sphere

    We form the material into a sphere of volume $V=M/\rho$ and radius $R = 3/4 V^{1/3}/\pi$, place it next to $p$ and compute the field as $F_G = G M/R^2 = (3/4) G M V^{-2/3} = (3/4) G M^{1/3} \rho^{-2/3}$

  • A nearby plane segment

    We form the material into a disk of thickness $T$ radius $R = 10 t$ (claiming that this is sufficient for a "infinite place" approximation near the center), so that $V = 100 t^3$so that $t = (V/100)^{1/3} = [M/(100 \rho)]^{1/3}$. We place the it such that $p$ borders center on one side. The surface density is $\sigma = \rho * t = 100^{-2/3} \rho^{2/3} M^{1/3}$ and the field is $F_G = 2 \pi G \sigma = 2 \pi 100^{-2/3} G M^{1/3} \rho^{-2/3}$.

Note that so far, both distributions result in a form $F_G = C \times G M^{1/3} \rho^{-2/3}$, so we need only compare the constants. Well, $2 \pi 100^{-2/3} = 2 \pi (0.046) = 0.29$, so the spherical distribution is better than the flat one.

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If you want to minimize the potential, rather than maximizing the field strength, the best you can do is to pack the material into a sphere, with the point $P$ at the center. After all, for any other shape, it'd be possible to move a mass element from larger to smaller distance, reducing the potential. –  Ted Bunn Jul 24 '11 at 21:10

The trivial solution is compact the mass to a point object, I guess. In that case the gravity at point P is infinite, because the distance is zero. I'm not sure if this violates the 'constant density' principle though.

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