The approximate height from the surface of earth at which the weight of the body becomes ${1 \over 3}$ of its weight on the surface of earth is :
[Radius of earth R = 6400 km and $\sqrt 3$ = 1.732]
Solution
<p>According to the given information</p>
<p>${{GM} \over {{{(R + h)}^2}}} = {1 \over 3} \times {{GM} \over {{R^2}}}$</p>
<p>$\Rightarrow R + h = \sqrt 3 R$</p>
<p>$\Rightarrow h = (\sqrt 3 - 1)R \simeq 4685$ km</p>
About this question
Subject: Physics · Chapter: Gravitation · Topic: Newton's Law of Gravitation
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