At what height above the earth's surface is t | vedclass

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(B) The acceleration due to gravity at a height $h$ above the earth's surface is given by $g_h = g(1 - \frac{2h}{R})$.The acceleration due to gravity

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$100$

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(B) The acceleration due to gravity at a height $h$ above the earth's surface is given by $g_h = g(1 - \frac{2h}{R})$.The acceleration due to gravity at a depth $d$ below the earth's surface is given by $g_d = g(1 - \frac{d}{R})$.Given that $g_h = g_d$ for $d = 200 \ km$,we equate the two expressions:$g(1 - \frac{2h}{R}) = g(1 - \frac{d}{R})$Canceling $g$ and $1$ from both sides:$-\frac{2h}{R} = -\frac{d}{R}$This simplifies to:$2h = d$$h = \frac{d}{2}$Substituting $d = 200 \ km$:$h = \frac{200}{2} = 100 \ km$.

At what height above the earth's surface is the value of $g$ the same as in a $200 \ km$ deep mine?

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