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(A) The acceleration due to gravity at a distance $r$ from the center of the Earth is given by $g' = \frac{GM}{r^2}$.Given the distance from the cente

Vedclass · Accepted Answer

$36$

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(A) The acceleration due to gravity at a distance $r$ from the center of the Earth is given by $g' = \frac{GM}{r^2}$.Given the distance from the center is $r = \frac{5}{4}R$,the acceleration due to gravity at this height is:$g' = \frac{GM}{(\frac{5}{4}R)^2} = \frac{GM}{\frac{25}{16}R^2} = \frac{16}{25} \left( \frac{GM}{R^2} ight) = \frac{16}{25}g$.Since weight $W = mg$,the new weight $W' = m g' = \frac{16}{25}mg = \frac{16}{25}W$.The decrease in weight is $\Delta W = W - W' = W - \frac{16}{25}W = \frac{9}{25}W$.The percentage decrease in weight is $\frac{\Delta W}{W} 	imes 100 = \frac{9/25 W}{W} 	imes 100 = \frac{9}{25} 	imes 100 = 36\%$.Thus,the percentage decrease in the weight of the object is $36\%$.

An object is taken to a height above the surface of the Earth at a distance of $\frac{5}{4} R$ from the center of the Earth,where the radius of the Earth is $R = 6400 \, km$. The percentage decrease in the weight of the object will be $....\%$.

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