The approximate height from the surface of ea | vedclass

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(B) The weight of a body at height $h$ is given by $W' = m g'$,where $g' = g \left( \frac{R}{R+h} \right)^2$.Given that the weight at height $h$ is $\

Vedclass · Accepted Answer

$4685$

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(B) The weight of a body at height $h$ is given by $W' = m g'$,where $g' = g \left( \frac{R}{R+h} ight)^2$.Given that the weight at height $h$ is $\frac{1}{3}$ of the weight on the surface,we have $m g' = \frac{1}{3} m g$,which implies $g' = \frac{g}{3}$.Substituting the expression for $g'$,we get $g \left( \frac{R}{R+h} ight)^2 = \frac{g}{3}$.Taking the square root on both sides,we get $\frac{R}{R+h} = \frac{1}{\sqrt{3}}$.Rearranging the equation,$R+h = R\sqrt{3}$,so $h = R(\sqrt{3} - 1)$.Substituting the values $R = 6400 \, km$ and $\sqrt{3} = 1.732$,we get $h = 6400 	imes (1.732 - 1) = 6400 	imes 0.732$.$h = 4684.8 \, km \approx 4685 \, km$.

The approximate height from the surface of earth at which the weight of the body becomes $\frac{1}{3}$ of its weight on the surface of earth is $.......... \, km$ : [Radius of earth $R = 6400 \, km$ and $\sqrt{3} = 1.732$]

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