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- ω2R22g
- ω2R2g
- 2ω2R23g
- 2ω2R2g

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Solution

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g′=g−ω2R

So the weight at the equator is mg′=m(g−ω2R)

Gravity above the earth's surface is given by

g′=g(1−2hR)

Implies the weight at the equator above the earth's surface is mg′=mg(1−2hR)

It is given that these two weights are equal,

⇒m(g−ω2R)=mg(1−2hR)

⇒g−ω2R=g−2ghR

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