Kindly solve the following sum-

1)How far away from the earth's surface does the acceleration due to gravity become 1% of its value on the earth's surface?

    At what height above the earth's surface has acceleration due to gravity diminished by 1/10 th of 1%?

    Assume the earth to be a sphere of radius 6380kms.

Value of g at height h from the earth's surface is given by

 g_h = GM/(R + h)²

 

Where, G= gravitational constant

R = radius of the earth

h = height h from the surface of earth.

 

or  g_h = GM/R²(1 + h/R)²  = g/(1 + h/R)²

where, g = acceleration due to gravity on the earth's surface.

 

so height say 'h' at which g_{h ⁼} 1% of g = g/100 is given by

 

 g/100  = g/(1 + h/R)²

 

 or 100  = (1 + h/R)²

 

now substitute R to find the value of 'h'

 

 

Similarly, to work out for second problem take g_h = g - (g/1000) and proceed.