The acceleration due to gravity at a height o | vedclass

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(A) Given: $g_1 = 2.5 \,ms^{-2}$ at height $h_1 = 6400 \,km$. Radius of earth $R = 6400 \,km$.The acceleration due to gravity at height $h$ is given b

Vedclass · Accepted Answer

$1.11$

Vedclass · Answer

(A) Given: $g_1 = 2.5 \,ms^{-2}$ at height $h_1 = 6400 \,km$. Radius of earth $R = 6400 \,km$.The acceleration due to gravity at height $h$ is given by $g = \frac{GM}{(R+h)^2}$.This implies $g \propto \frac{1}{(R+h)^2}$.For height $h_1 = 6400 \,km$, distance from center is $r_1 = R + h_1 = 6400 + 6400 = 12800 \,km = 2R$.For height $h_2 = 12800 \,km$, distance from center is $r_2 = R + h_2 = 6400 + 12800 = 19200 \,km = 3R$.Taking the ratio: $\frac{g_2}{g_1} = \left( \frac{r_1}{r_2} ight)^2 = \left( \frac{2R}{3R} ight)^2 = \left( \frac{2}{3} ight)^2 = \frac{4}{9}$.Therefore, $g_2 = \frac{4}{9} 	imes g_1 = \frac{4}{9} 	imes 2.5 = \frac{10}{9} \approx 1.11 \,ms^{-2}$.

The acceleration due to gravity at a height of $6400 \,km$ from the surface of the earth is $2.5 \,ms^{-2}$. The acceleration due to gravity at a height of $12800 \,km$ from the surface of the earth is (Radius of the earth $= 6400 \,km$) (in $\,ms^{-2}$)

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