Question

at what height above the surface of the earth will the acceleration due to gravity be 25% of its value on the surface of the earth assume that the radius of the earth is 6400 km

Answer 1

At a height equal to radius of earth
g'=g/4
So 6400+6400=12800Km

Answer 2

Given :

Acceleration due to gravity at height h is 25 % of its value on the surface of the earth .

Radius of the earth surface , R = 6400 km .

To Find :

Value of h .

Solution :

Acceleration to due gravity at earth's surface is given by :

$g=\dfrac{GM}{R^2}$

Now , acceleration due to gravity at height h above from surface of earth is given by :

$g_h=\dfrac{gR^2}{(R+h)^2}$

Now , we have given :

$g_h= \dfrac{g}{4} \\\\\dfrac{gR^2}{(R+h)^2}=\dfrac{g}{4}\\\\\dfrac{R}{R+h}=\dfrac{1}{2}\\\\h=R$

Therefore , value of h is 6400 km .

Hence , this is the required solution .