Question

At what height from the earth does the acceleration due to gravity falls to 1% that of the surface

a) 9R
b) 10R
d) 100R
c) 99R

Solution

Acceleration due to gravity on the earth surface, g = (GM)/R^2...(1)

Acceleration due to gravity a height h above the Earth's surface
g^' = (GM)/(R+h)^2....(2)

Given that g^' = 0.01 g

From equation (1) and (2), we will get

(GM)/(R+h)^2= (0.01)(GM)/R^2

=> 1/(R+h)^2 = 0.01/R^2

Taking squareroot both the sides,

=> 1/(R+h)= 0.1/R

=> 10R = R+h

=> h = 9R

Therefore, option (a) is correct.