A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?
Let gh be the acceleration due to gravity at a height equal to half the radius of the Earth (h = R/2) and g its value on Earth’s surface. Let the body have mass m.
We know that
`(g_h)/g = (R/(R+h))^2 or (g_h)/g = (R/(R+R/2))^2 = (2/3)^2 = 4/9`
Let W be the weight of body on the surface oof Earth and `W_h` thge weight of the body at height h.
Then `(W_h)/W = (mg_h)/mg = (g_h)/g = 4/9`
or `W_h = 4/9 W = 4/9 xx 63 N = 28 N`
Weight of the body, W = 63 N
Acceleration due to gravity at height h from the Earth’s surface is given by the relation:
`g = g/((1+h)/R_e)^2`
g = Acceleration due to gravity on the Earth’s surface
Re = Radius of the Earth
For h = `R_e/2`
`g = g/(1+R_e/(2xxR_e))^2 = g/(1+1/2)^2 = 4/9 g`
Weight of a body of mass m at height h is given as:
W' = mg
`= mxx4/9 g = 4/9xx mg`
`= 4/9 W`
`= 4/9 xx 63 = 28 N`