Question

at which height acceleration due to gravity becomes one forth compare to earth surface

Answer 1

g = G*M/(r)^2

g_prime = G*M/(r+h)^2

g/g_prime = (r+h)^2/(r)^2

sqrt(g/g_prime) = (r+h)/r

substituting the given values in the above equation:

sqrt(g/g/4)=sqrt(4)=2=(r+h)/r

2*r = r+h

h = r

Therefore at a height equal to radius of the earth(6400 km), earth’s acceleration due to gravity will be one fourth of it’s value on the surface of the earth.