If radius of the earth is 6347 km, then what will be difference between acceleration of free falls and acceleration due to gravity near the earth's surface?

Options

A. A. 0.34
B. B. 0.034
C. C. 0.0034
D. D. 0.24

Answer

B

Explanation

`g=(GM)/(R^(2))=9.8`
`g_("free fall")=(GM)/(R^(2)-omega^(2)R=9.8-omega^(2)R`
`g-g_("free falls")omega^(2)R=((2pi)/(T))^(2)R`
`=(4pi^(2))/((24xx60xx60)^(2))xx6347xx10^(3)=0.03401`