At what height from the surface of earth will the value of g be reduced by `36%` from the value on the surface? Take radius of earth `R = 6400 km`.

Options

A. 1800 km
B. 1600 km
C. 2000 km
D. 2200 km

Explanation

let at height h the value of g reduces by 36% it becomes 64 % of that at the surface
Thus `g_(h)=64% "of" g =g(64/100)`
`therefore g_(h)=g(R_(e)^(R ))/(R_(e )+h)^(2)`
`rarr 8/10=(R_(e ))/(R_(e )+h) rarr (R_(e )+h)/(R_(e ))=10/8`
`rarr (h)/(R_(e ))+1=10/8 rarr(h)/(R_(e ))=2/8`
`therefore h=(R_(e ))/(4)=1600` km