The ratio of the accelerations due to gravity | vedclass

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Question

(A) The acceleration due to gravity at a height $h$ above the surface of the earth is given by the formula: $g' = g \left( \frac{R}{R+h} \right)^2$,wh

Vedclass · Accepted Answer

$25: 16$

Vedclass · Answer

(A) The acceleration due to gravity at a height $h$ above the surface of the earth is given by the formula: $g' = g \left( \frac{R}{R+h} ight)^2$,where $g$ is the acceleration due to gravity at the surface and $R$ is the radius of the earth.Given $R = 6400 \ km$.For $h_1 = 1280 \ km$:$g_1 = g \left( \frac{6400}{6400 + 1280} ight)^2 = g \left( \frac{6400}{7680} ight)^2 = g \left( \frac{5}{6} ight)^2 = \frac{25}{36} g$.For $h_2 = 3200 \ km$:$g_2 = g \left( \frac{6400}{6400 + 3200} ight)^2 = g \left( \frac{6400}{9600} ight)^2 = g \left( \frac{2}{3} ight)^2 = \frac{4}{9} g$.Now,the ratio $\frac{g_1}{g_2}$ is:$\frac{g_1}{g_2} = \frac{25/36 g}{4/9 g} = \frac{25}{36} 	imes \frac{9}{4} = \frac{25}{16}$.Thus,the ratio is $25:16$.

The ratio of the accelerations due to gravity at heights $1280 \ km$ and $3200 \ km$ above the surface of the earth is (Radius of the earth $= 6400 \ km$)

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