Answer to the Question 02/02

MAXIMAL GRAVITY

The question was:

Consider a celestial body (planet) of mass M and volume V consisting of material of uniform density M/V. If the body is spherical, then its radius is R=(3V/4{pi})1/3, and the free-fall acceleration on its surface is g=GM/R2=GM(4{pi}/3V)2/3, where G is the gravitational constant. Consider all possible shapes of the body. What is the largest g, that can be achieved at one point on a surface, and what is the shape of the body for which such acceleration is achieved?




(5/2002) The problem has been solved (9/2/2002) by Eva Sandlokk (e-mail eva.sandlokk@prodigy.net), (18/2/2002) by Kirk T McDonald from Henry Laboratories, Princeton, NJ (e-mail mcdonald@puphed.princeton.edu) - see his solution in PDF format, as well as by Mattew Horton (1/4/2002) (e-mail matthew.horton@ssmb.com), and (2/4/2002) Daniel Gulotta from Illinois Mathematics and Science Academy (e-mail dgulotta@imsa.edu). While direct analytical solution of the problem is possible, a very interesting approach was suggested (2/4/2002) by Horst Schneider from the Physics Department in Martin-Luther-Universitat Halle-Wittenberg in Germany (e-mail schneider@physik.uni-halle.de); the approach begins with numerical optimization, and eventually arrives to an analytical solution - see files 1 and 2. It was also partially solved (26/3/2002) by Jared Daniel Kaplan from Stanford (e-mail jaredk@Stanford.edu).

The answer: The optimal body has axial symmetry and has the shape depicted at the picture below. The green dot denotes the point of maximal free-fall acceleration. The body can be described by relating distance r from the point at which the maximal acceleration is found as a function of the angle {theta} from the axis of symmetry (vertical black line).

r2=R2cos({theta}).

The acceleration at the green point will be g=(4/5)(15/4)1/3{pi}2/3M/V2/3. This is only 2.6% larger than the gravity on the surface of a spherical planet.



The solution: Assume for a moment that we already know that the body should have axial symmetry. Consider a narrow ring of matter lying on the surface of the body. (Center of the ring coincides with the axis of symmetry and its plane is perpendicular to that axis. It is depicted on the figure below.) If the mass of the ring is dM and the distance between the points of the ring and the "green point" is r, then the contribution of the ring to free-fall acceleration will be G*{dM}cos({theta})/r2.



If we remove this ring of material from the surface and place it somewhere else on the surface, it will have different r and different angle {theta}. Since the shape is optimal, it is "stationary", i.e. infinitesimal changes should not change the acceleration. This will happen only if cos({theta})/r2 is constant. Consequently, r2=R2cos({theta}). The constant R denotes the maximal diameter of the body. Once the shape of the body is given, we can easily find (by an elementary integration*) that its volume V=(4/15){pi}R3, while the acceleration at point g is given** by the expression which has been presented above.

Using similar argument you can now show that the body must have axial symmetry. Imagine that is not so, i.e. the ring which was described above is not complete - then you can easily see that the free-fall acceleration can be increased by relocating some of the surface material.

For a more detailed solution see PDF format.



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