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?

Asked by Pragya Singh | 1 year ago |  116

##### Solution :-

Given:

Weight of the man, W = 63 N

We know that acceleration due to gravity at height ‘h’ from the Earth’s surface is:

$$g’ = \dfrac{g}{\left [ 1+\left ( \dfrac{h}{R_{e}} \right ) \right ]^{2}}$$

Where, g = Acceleration due to gravity on the Earth’s surface

And, Re = Radius of the Earth

For h = $$\dfrac{R_{e}}{2}$$

g’ = $$\dfrac{g}{\left [ 1+\left ( \dfrac{R_{e}}{2R_{e}} \right ) \right ]^{2}}$$

$$⇒ g’ = \dfrac{g}{\left [ 1+\left (\dfrac{1}{2} \right ) \right ]^{2}}=\dfrac{4}{9}\;g$$

Also , the weight of a body of mass ‘m’ kg at a height of ‘h’ meters can be represented as :

W’ = mg

$$m × \dfrac{4}{9} = \dfrac{4}{9}$$

$$\dfrac{4}{9}$$

$$\dfrac{4}{9} × 63$$  = 28 N.

Answered by Abhisek | 1 year ago

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