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

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(B) Acceleration due to gravity at altitude $h$ is given by the formula:$g^{\prime} = \frac{g}{(1 + \frac{h}{R_{e}})^2}$Given $h = \frac{R_{e}}{2}$,we

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$28$

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(B) Acceleration due to gravity at altitude $h$ is given by the formula:$g^{\prime} = \frac{g}{(1 + \frac{h}{R_{e}})^2}$Given $h = \frac{R_{e}}{2}$,we substitute this into the equation:$g^{\prime} = \frac{g}{(1 + \frac{R_{e}/2}{R_{e}})^2} = \frac{g}{(1 + \frac{1}{2})^2} = \frac{g}{(\frac{3}{2})^2} = \frac{4g}{9} \dots(i)$Weight of the body at the earth's surface is $w = mg = 63 \ N \dots(ii)$Weight of the body at altitude $h = \frac{R_{e}}{2}$ is $w^{\prime} = mg^{\prime} = m(\frac{4g}{9}) = \frac{4}{9}mg \dots(iii)$Substituting the value from equation $(ii)$ into equation $(iii)$:$w^{\prime} = \frac{4}{9} 	imes 63 = 4 	imes 7 = 28 \ N$

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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