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(D) The acceleration due to gravity at a depth $d$ is given by $g_d = g(1 - d/R)$.The acceleration due to gravity at a height $h$ is given by $g_h = g

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(D) The acceleration due to gravity at a depth $d$ is given by $g_d = g(1 - d/R)$.The acceleration due to gravity at a height $h$ is given by $g_h = g(1 - 2h/R)$ (for $h \ll R$).Equating the two values for the same change in $g$:$g(1 - d/R) = g(1 - 2h/R)$$d/R = 2h/R$$d = 2h$Given $d = 10 \, km$,we have $10 = 2h$,which implies $h = 5 \, km$ is incorrect based on the standard approximation. Wait,let's re-evaluate: $g_d = g(1 - d/R)$ and $g_h = g(1 - 2h/R)$. If $g_d = g_h$,then $d = 2h$. Thus $h = d/2 = 10/2 = 5 \, km$. However,the provided solution suggests $h = 2x = 20 \, km$. Let's check the exact formula: $g_h = gR^2 / (R+h)^2 \approx g(1 - 2h/R)$. The depth formula is exact: $g_d = g(1 - d/R)$. Equating $1 - d/R = 1 - 2h/R$ gives $h = d/2 = 5 \, km$. If the question implies the same value,the answer is $5 \, km$.

At what height from the ground will the value of $g$ be the same as that in a $10 \, km$ deep mine below the surface of the Earth?

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