the angular speed omega withwhich the earth would have to rotate on its axis so that a person on the equator would weight 3/5th of original weight
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True weight at equator is w=mg and observed weight is W' = mg' = 3/5 mg
taking λ= 0.
mg' = mg - mRw^2 cosλ
3/5mg = mg - mRw^2 cos0
= mg - mRw^2
or
mRw^2 = 2/5 mg
w = (2g/5R)^1/2
= 7.8 * 10^-4 rad/sec
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