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 `MR`^{2}`v` 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)`Mr`^{2}`w` where `r` = planet
radius, `w` = angular velocity of daily rotation.

Thus we want `w` =
-(5/2)`R`^{2}`v` /
`r`^{2}.

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.

This page is maintained by Thomas Bending,
and was last modified on 7 March 2017.

Comments, criticisms and suggestions are welcome.
Copyright © Thomas Bending 2017.