At what depth below the surface of the earth, | vedclass

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(D) Let $R = 6400 \, km$ be the radius of the earth.Acceleration due to gravity at height $h = 1600 \, km$ is given by $g_h = g \left( \frac{R}{R+h} \

Vedclass · Accepted Answer

None of these

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(D) Let $R = 6400 \, km$ be the radius of the earth.Acceleration due to gravity at height $h = 1600 \, km$ is given by $g_h = g \left( \frac{R}{R+h} ight)^2$.Substituting $h = 1600 \, km$ and $R = 6400 \, km$,we get $h = R/4$.$g_h = g \left( \frac{R}{R + R/4} ight)^2 = g \left( \frac{R}{5R/4} ight)^2 = g \left( \frac{4}{5} ight)^2 = \frac{16}{25} g$.At depth $d$,the acceleration due to gravity is $g_d = g \left( 1 - \frac{d}{R} ight)$.According to the problem,$g_d = \frac{1}{2} g_h$.$g \left( 1 - \frac{d}{R} ight) = \frac{1}{2} \left( \frac{16}{25} g ight) = \frac{8}{25} g$.$1 - \frac{d}{R} = \frac{8}{25} \Rightarrow \frac{d}{R} = 1 - \frac{8}{25} = \frac{17}{25}$.$d = \frac{17}{25} 	imes 6400 \, km = 17 	imes 256 \, km = 4352 \, km = 4.352 	imes 10^6 \, m$.Since this value is not among the options,the correct choice is $D$.

At what depth below the surface of the earth,will the acceleration due to gravity $g$ be half of its value at $1600 \, km$ above the surface of the earth?

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