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(C) The acceleration due to gravity at a height $h$ above the Earth's surface is given by $g_h = g(1 - \frac{2h}{R_e})$ for $h \ll R_e$.The accelerati

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$2$

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(C) The acceleration due to gravity at a height $h$ above the Earth's surface is given by $g_h = g(1 - \frac{2h}{R_e})$ for $h \ll R_e$.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,$g_h = g_d$,so:$g(1 - \frac{2h}{R_e}) = g(1 - \frac{d}{R_e})$Canceling $g$ from both sides and simplifying:$1 - \frac{2h}{R_e} = 1 - \frac{d}{R_e}$$\frac{2h}{R_e} = \frac{d}{R_e}$$d = 2h$Given that $d = \alpha h$,we compare the two expressions:$\alpha h = 2h$$\alpha = 2$.

If the acceleration due to gravity experienced by a point mass at a height $h$ above the surface of the Earth is the same as the acceleration due to gravity at a depth $d = \alpha h$ $(h \ll R_{e})$ from the Earth's surface,then the value of $\alpha$ will be: (use $R_{e} = 6400 \ km$)

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