At what depth below the surface of the earth | vedclass

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(A) The acceleration due to gravity at height $h$ is given by $g_h = g \left( \frac{R}{R+h} \right)^2$. Given $h = 1600 \ km$ and $R = 6400 \ km$,we h

Vedclass · Accepted Answer

$4.8 	imes 10^6 \ m$

Vedclass · Answer

(A) The acceleration due to gravity at height $h$ is given by $g_h = g \left( \frac{R}{R+h} ight)^2$. Given $h = 1600 \ km$ and $R = 6400 \ km$,we have $g_h = g \left( \frac{6400}{6400+1600} ight)^2 = g \left( \frac{6400}{8000} ight)^2 = g \left( \frac{4}{5} ight)^2 = 0.64g$. The acceleration due to gravity at depth $d$ is given by $g_d = g \left( 1 - \frac{d}{R} ight)$. According to the problem,$g_d = \frac{1}{2} g_h = \frac{1}{2} (0.64g) = 0.32g$. Substituting this into the depth formula: $0.32g = g \left( 1 - \frac{d}{R} ight)$. $0.32 = 1 - \frac{d}{R} \Rightarrow \frac{d}{R} = 1 - 0.32 = 0.68$. $d = 0.68 	imes 6400 \ km = 4352 \ km = 4.352 	imes 10^6 \ m$. Note: If using the approximation $g_h \approx g(1 - 2h/R)$,then $g_h = g(1 - 3200/6400) = 0.5g$. Then $g_d = 0.5 g_h = 0.25g$. $0.25g = g(1 - d/R) \Rightarrow d/R = 0.75 \Rightarrow d = 0.75 	imes 6400 \ km = 4800 \ km = 4.8 	imes 10^6 \ m$. Given the options,the approximation method is intended.

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

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