Q: What should the earth's day length be such that its angular
momentum (relative to the solar system) is 0?
[TB, 29 Nov 1999]
A: 23 milliseconds.
The angular momentum of its orbit around the sun, assuming it's a point mass, is MR2v where M = mass of earth, R = orbital radius, v = angular velocity of yearly orbit.
The angular momentum of its rotation about its axis is (2/5)Mr2w where r = planet radius, w = angular velocity of daily rotation.
Thus we want w = -(5/2)R2v / r2.
Now R = 1.5e11 m, v = 2 pi /(3600*24*365) = 2e-7 radians/s and r = 6.4e6 m, so w = -270 radians/s so the required day length is 23 milliseconds, but in the opposite direction to the actual rotation. Thus the earth would be doing 2600 rpm.
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