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(D) The acceleration due to gravity at a depth $d$ is given by $g_d = g_0 \left(1 - \frac{d}{R}\right)$,where $g_0 = \frac{GM}{R^2}$.The acceleration

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$4800$

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(D) The acceleration due to gravity at a depth $d$ is given by $g_d = g_0 \left(1 - \frac{d}{R}ight)$,where $g_0 = \frac{GM}{R^2}$.The acceleration due to gravity at a height $h = 3R$ is given by $g_h = \frac{GM}{(R+h)^2} = \frac{GM}{(R+3R)^2} = \frac{GM}{(4R)^2} = \frac{GM}{16R^2} = \frac{g_0}{16}$.According to the problem,$g_d = 4 	imes g_h$.Substituting the expressions: $g_0 \left(1 - \frac{d}{R}ight) = 4 	imes \left(\frac{g_0}{16}ight)$.$1 - \frac{d}{R} = \frac{1}{4}$.$\frac{d}{R} = 1 - \frac{1}{4} = \frac{3}{4}$.$d = \frac{3}{4} 	imes R = \frac{3}{4} 	imes 6400 \ km = 4800 \ km$.

At a certain depth $d$ below the surface of the earth,the value of acceleration due to gravity becomes four times its value at a height $3R$ above the earth's surface. Where $R$ is the radius of the earth (Take $R = 6400 \ km$). The depth $d$ is equal to $............ \ km$.

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