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# 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? - Physics

#### Question

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?

#### Solution 1

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

#### Solution 2

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

Where,

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

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