A body weighs $72 N$ on the surface of the earth. What is the gravitational force on it at a height equal to half the radius of the earth from the surface?

Question

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Answer 1

Lrt $R_{e}$ be the radius of earth
Given $h = R_{e} / 2$
Since h is comparable to $R_{e}$
So, $g_{h} = \frac{G M}{( R _{e} + h ) ^{2}}$
Putting $h = R_{e} / 2$ ,we get
$\Rightarrow g_{h} = \frac{4}{9} \times \frac{G M}{R _{e}^{2}} = \frac{4 g}{9}$
Where g is acceleration due to gravity at the surface of earth

Therefore weight at a height equal to half of the earth's radius $= \frac{4}{9} \times 72 = 32 N$

Hence correct answer is option $D$

Answer 2

Lrt

R_{e}

be the radius of earth

Given

h = R_{e} / 2

Since h is comparable to

R_{e}

So,

g_{h} = \frac{G M}{( R _{e} + h ) ^{2}}

Putting

h = R_{e} / 2

,we get

\Rightarrow g_{h} = \frac{4}{9} \times \frac{G M}{R _{e}^{2}} = \frac{4 g}{9}

Where g is acceleration due to gravity at the surface of earth

Therefore weight at a height equal to half of the earth's radius

= \frac{4}{9} \times 72 = 32 N

Hence correct answer is option

D