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.
Back to puzzles
This page is maintained by Thomas Bending,
and was last modified on 25 October 2017.
Comments, criticisms and suggestions are welcome. Copyright © Thomas Bending 2017.