The acceleration due to gravity at a height h | vedclass

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(D) The acceleration due to gravity at a height $h$ above the surface of the $Earth$ is given by $g_h = g(1 - 2h/R)$, where $g$ is the acceleration du

Vedclass · Accepted Answer

$45$

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(D) The acceleration due to gravity at a height $h$ above the surface of the $Earth$ is given by $g_h = g(1 - 2h/R)$, where $g$ is the acceleration due to gravity at the surface and $R$ is the radius of the $Earth$.The acceleration due to gravity at a depth $d$ below the surface of the $Earth$ is given by $g_d = g(1 - d/R)$.According to the problem, $g_h = g_d$ at $d = 90 	ext{ km}$.Therefore, $g(1 - 2h/R) = g(1 - d/R)$.Canceling $g$ from both sides and simplifying:$1 - 2h/R = 1 - d/R$$-2h/R = -d/R$$2h = d$$h = d/2$Given $d = 90 	ext{ km}$, we have:$h = 90 / 2 = 45 	ext{ km}$.

The acceleration due to gravity at a height $h$ above the surface of the $Earth$ is $g_h$. At a depth $d = 90 \text{ km}$ below the $Earth$'s surface, the acceleration due to gravity is also $g_h$. The value of $h$ is: (in $\text{ km}$)

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