The weight of a body on the surface of the ea | vedclass

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Question

Acceleration due to gravity at altitude $h$ is given by

$g^{\prime}=\frac{g}{\left(1+\frac{h}{R_{e}}\right)^{2}}=\frac{g}{\left(1+\frac{R_{e} / 2}{

Vedclass · Accepted Answer

$28$

Vedclass · Answer

Acceleration due to gravity at altitude $h$ is given by

$g^{\prime}=\frac{g}{\left(1+\frac{h}{R_{e}}ight)^{2}}=\frac{g}{\left(1+\frac{R_{e} / 2}{R_{e}}ight)^{2}}$

$=\frac{g}{\left(1+\frac{1}{2}ight)^{2}}=\frac{g}{\left(\frac{3}{2}ight)^{2}}=\frac{4 g}{9} \ldots(i)$

Weight of the body at the earth's surface,

$w=m g=63 N-$....... $(ii)$

Weight of the body at altitude, $h=\frac{R_{e}}{2}$

$w^{\prime}=m g^{\prime}=\frac{4}{9} m g$ $...(iii)$

Using equation $(ii),$ we get

$w^{\prime}=\frac{4}{9} 	imes 63=28 N$

The weight of a body on the surface of the earth is $63\, N$. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth ? (in $N$)

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