Acceleration due to gravity is ‘ g ’ on the surface of the earth. The value of acceleration due to gravity at a height of 32 km above earth’s surface is (Radius of the earth = 6400 km )

Options

A. 0.9g
B. 0.99g
C. 1.01g
D. 0.8g

Answer

B

Explanation

`(g_(h))/(g)=((R )/(R+h))^(2)=((1)/(1+(h)/(R )))^(2)=(1+(h)/(R ))^(-2)`
if `h lt lt R`, then `(g_(h))/(g)=1-(2h)/(R )`
`therefore g_(h)=g(1-(2xx32)/(6400))=g(1-(1)/(100))`
`therefore g_(h)=(99)/(100)g=0.99 g`