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Let us consider the following problem:  it is necessary to find the shape of the body with fixed mass and density which at large distances compared with its characteristic dimensions would give the greatest deviations from the strength of the gravitational field of the material point with the same mass placed in the center of the mass of the object. I understand that in this case the dipole moment is zero and it is necessary to maximize all the components of the quadruple moment tensor of the object. It is naturally that it is necessary to use a variation calculus. But I can’t make specific calculations. Please, help me with a solution to this problem.

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The biggest part of your problem is finding out the right formulation. I formulate it like this: let there be a body of mass $M$, volume $V$, and a constant density $\rho = M/V$ that is within a ball $B_R$ of radius $R$ around its center of mass. What shape, holding the parameters above fixed, maximizes the mass quadrupole $Q$ of the body?

Let us pick spherical coordinates $r,\vartheta,\varphi$. Let me assume axisymmetry for now, and you can verify later that it cannot extremize the problem further. We then use the formula for the mass quadrupole in terms of spherical harmonics, divide it by the total mass, and find out that we are then maximizing the following expression with respect to $V$

$$\frac{Q}{M} = \frac{\int_V r^2\frac{1}{2}(3 \cos^2 \vartheta-1) 2 \pi r \sin \vartheta dr d\vartheta}{\int_V 2 \pi r \sin \vartheta dr d\vartheta}$$

where I have cancelled the factor $\rho$ in both the expressions but let the geometric factors uncancelled for better transparency.

Technically, I do not immediately see how to make a tidy expression $\delta Q/ \delta V$ while keeping the body within $B_R$, but I know the solution to the problem because it is obvious from the structure of the extrema of the harmonic $r^2(3 \cos^2 \vartheta-1)/2$. The extrema do not exist within the domain of finite volumes $V$. However, there are two asymptotic extrema corresponding to an extremely prolate ($Q>0$) and extremely oblate ($Q<0$) body. They are respectively the infinitesimal volumes around two points at the poles $\vartheta = 0,\pi$ at a distance $R$ from the center, and an infinitely thin ring at $\vartheta=\pi/2$ of radius $R$.

answered Nov 28, 2018 by (1,645 points)

Dear Dr. Void! Thank you so much for your very elegant solution to this problem. I guessed about these asymptotic cases, since I know that the field strength at the center of the ring is zero and that binary stars have a large quadrupole moment. But what can you say about an infinitely thin rod with a length equal to twice the radius of the sphere?

@reterty No need for the titles, we are not at a university here :) I believe the rod of the same mass would have a three times smaller quadrupole then the two point masses. This is because of the $r^2$ factor in the harmonic and it integrating out to $R^3/3$ along the rod.

Dear Void! Your solution is valid for any symmetrical system with the polar axis of symmetry with order greater than 2 (3, 4, 5...).  So, the curve for $\theta=\pi/2$ can be more complex then circle. But I have a problem with its construction again:). Another words we can try to find the closed curve  with the $Q<Q_{\rm ring}$$\theta$

The formula above (Q/M) is just a number. In order to figure our the body shape, one has to admit a variable density (from zero to infinity) in the integrand: $\rho=\rho(r,\theta,\varphi)$.

Not really infinity since a dot shape is excluded and density is always positive. I agree with Vladimir that the spherical density has a meaning here. I upvoted the above answer elegance. However, it is not very physical in the strict conditions of the question.

The alternative reterty answer with many bodies doesn't correspond to the question: 1 body, 1 density without pressure consideration and explicitely, since it is related to a cosmic observation, stability. The interesting question is, given a matter resistance and a density, what is the biggest body able to be stable without sphericity. Don't forget the 3D-pressure.

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