The acceleration due to gravity at a height `(1//20)^(th)` the radius of the earth above earth s surface is `9m//s^(2)` Find out its approximate value at a point at an equal distance below the surface of the earth .

Options

A. 8.5
B. 9.5
C. 9.8
D. 11.5

Answer

B

Explanation

Given `G_(h)=9=(gR^(2))/(R+R//20)^(2)=(20xx20)/(21xx21)g`
or `g=(9xx21xx21)/(20xx20)`
Now `g_(d)=g(1-(d)/(R ))=(9xx21xx21)/(20xx20)(1-(R//20)/(R ))`
`=9.5 ms^(2)`