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If you wait long enough, the ellipse will eventually become planar, so the volume isn't actually a constant - what is true is that the first and second derivatives of the volume are initially zero. (I'm getting that the third derivative is also zero)cygnus2 said:

If you assume that the shape is in fact elliptical, it isn't too hard to prove. It would take some more complicated calculations to confirm that the shape is that of an ellipse.

The volume of an ellipse (actual the surface generated by resolving an ellipse) is

4/3*pi*a*b*c

where a,b,c are the axes of the ellipsoid

The distortion force on the ellipse is just the tidal force. The tidal force has a stretching component in the radial direction, but a compressive component in the transverse directions.

The stretching force per unit distance 2M/r^3 (this is the derivative of G*M/r^2 with respect to r). The compressive forces are both m/r^3. Here M is the mass of a point mass, and r is the distance one is away from the point mass. It takes some vector diagrams to calculate the compressive tidal force, so I'll skip that step unless there is some interest in the details.

If we _assume_ that the figure takes the shape of an ellipse, we can calculate the volume simply by tracking two major axes of the ellipsoid, the one that is elongated, and the two that are shortened, as the area of an ellipsoid is just 4/3*pi*a*b*c, where a,b,c are the major axes

http://www.murderousmaths.co.uk/books/BKMM7xea.htm

The "long" axis is c(t) = c+.5*a*t^2 = c+.5*2*m/r^3*t^2 = c+t^2/r^3

The two shorter axes are a-.5*m/r^3*t^2 = a-.5*t^2/r^3

Initially we know that a=c, so we simplify the expression as

volume = (a+t^2/r^3)*(a-.5*t^2/r^3)*(a-.5*t^2/r^3)

Finding the derivatives, we get that the first derivative is proportional to t^3, so it's zero when t=0

The second derivative is proportioanl to t^2, so it's also zero.

Even the third derivative is zero, as it's proportional to 't'.'

The fourth derivative is non-zero.

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