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

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(B) The weight of a body on the surface of the earth is given by $W = mg = 100\,N$.The acceleration due to gravity at a height $h$ above the surface o

Vedclass · Accepted Answer

$64$

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(B) The weight of a body on the surface of the earth is given by $W = mg = 100\,N$.The acceleration due to gravity at a height $h$ above the surface of the earth is given by $g' = g \left( \frac{R}{R+h} ight)^2$.Given $h = \frac{R}{4}$,we substitute this into the formula:$g' = g \left( \frac{R}{R + R/4} ight)^2 = g \left( \frac{R}{5R/4} ight)^2 = g \left( \frac{4}{5} ight)^2 = \frac{16}{25}g$.The weight at height $h$ is $W' = mg' = m \left( \frac{16}{25}g ight) = \frac{16}{25} 	imes W$.Substituting $W = 100\,N$:$W' = \frac{16}{25} 	imes 100 = 16 	imes 4 = 64\,N$.

The weight of a body on the surface of the earth is $100\,N$. The gravitational force on it when taken at a height,from the surface of earth,equal to one-fourth the radius of the earth is $..........\,N$.

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