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.

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