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Physics Question on Gravitation

Acceleration due to gravity at earth's surface is g ms−2g \,ms ^{-2}. Find the effective value of gravity at a height of 32 km32\, km from sea level :(Re=6400 km):\left(R_{e}=6400 \,km \right)

A

0.5 g ms−20.5\,g\,ms^{-2}

B

0.99 g ms−20.99\,g\,ms^{-2}

C

1.01 g ms−21.01\,g\,ms^{-2}

D

0.90 g ms−20.90\,g\,ms^{-2}

Answer

0.99 g ms−20.99\,g\,ms^{-2}

Explanation

Solution

For height hh above the earth's surface g′=g(1−2hRe)=g(1−646400)g'=g\left(1-\frac{2 h}{R_{e}}\right)=g\left(1-\frac{64}{6400}\right) (∵Re=6400 km)\left(\because R_{e}=6400 \,km \right) =g(1−0.01)=g(1-0.01) =0.99 g ms−2= 0.99\, g\, ms ^{-2}