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Determine the speed with which the Earth would have to rotate on its axis so that a person on the equator would weigh(3/5)th as much as present.take the equatorial radius as 6400 km.
Dear Saurabh True weight at the equator, W = mgObserved weight at the equator,W' = mg' = (3/5)mgat the equator λ = 0 mg' = mg - mRw² cosλ 3mg/5 = mg - mRw². Cos 0 = mg - mRw² mRW² = 2mg/5 w = 7.8 * 10^-4 rad/ sec RegardsArun(askIITians forum expert)
True weight at the equator, W = mg
Observed weight at the equator,
W' = mg' = (3/5)mg
at the equator λ = 0
mg' = mg - mRw² cosλ
3mg/5 = mg - mRw². Cos 0
= mg - mRw²
mRW² = 2mg/5
w = 7.8 * 10^-4 rad/ sec
Regards
Arun(askIITians forum expert)
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