A body weighs 63 , N on the surface of the ea | vedclass

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(A) $1$. First,calculate the mass of the body $(m)$. Given weight $W = mg = 63 \, N$. Assuming $g = 10 \, m/s^2$,$m = 63 / 10 = 6.3 \, kg$.$2$. The di

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$27.35$

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(A) $1$. First,calculate the mass of the body $(m)$. Given weight $W = mg = 63 \, N$. Assuming $g = 10 \, m/s^2$,$m = 63 / 10 = 6.3 \, kg$.$2$. The distance of the body from the center of the earth $(r)$ at a height $h = R/2$ is $r = R + h = R + R/2 = 1.5R$.$3$. Given $R = 6.4 	imes 10^6 \, m$,so $r = 1.5 	imes 6.4 	imes 10^6 = 9.6 	imes 10^6 \, m$.$4$. The gravitational force is given by $F = \frac{GMm}{r^2}$.$5$. Substituting the values: $F = \frac{6.67 	imes 10^{-11} 	imes 6 	imes 10^{24} 	imes 6.3}{(9.6 	imes 10^6)^2}$.$6$. $F = \frac{252.186 	imes 10^{13}}{92.16 	imes 10^{12}} = \frac{2521.86}{92.16} \approx 27.35 \, N$.

$A$ body weighs $63 \, N$ on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth (in $, N$)? (Given: Mass of the earth $= 6 \times 10^{24} \, kg$ and radius of the earth $= 6.4 \times 10^{6} \, m$)

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