Acceleration due to gravity at a height h is | vedclass

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Question

(C) The acceleration due to gravity at a height $h$ above the Earth's surface is given by $g_h = g(1 - \frac{2h}{R_e})$,where $R_e$ is the radius of t

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$d=2 h$

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(C) The acceleration due to gravity at a height $h$ above the Earth's surface is given by $g_h = g(1 - \frac{2h}{R_e})$,where $R_e$ is the radius of the Earth.The acceleration due to gravity at a depth $d$ below the Earth's surface is given by $g_d = g(1 - \frac{d}{R_e})$.According to the problem,$g_h = g_d$.Equating the two expressions:$g(1 - \frac{2h}{R_e}) = g(1 - \frac{d}{R_e})$Canceling $g$ from both sides:$1 - \frac{2h}{R_e} = 1 - \frac{d}{R_e}$Subtracting $1$ from both sides:$-\frac{2h}{R_e} = -\frac{d}{R_e}$Multiplying by $-R_e$:$2h = d$ or $d = 2h$.

Acceleration due to gravity at a height $h$ is equal to that at a depth $d$ below the surface of the earth,if

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