At what height above earth's surface, value of g is same as in a mine 100km deep?

we know that for variation of g with height,

$g={g}_{0}\left(1-\frac{2h}{R}\right)----\left(1\right)$

and for variation in g with depth

$g={g}_{0}\left(1-\frac{d}{R}\right)-------\left(2\right)$

where d is the depth.

according to the question

${g}_{0}\left(1-\frac{2h}{R}\right)={g}_{0}\left(1-\frac{d}{R}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2h}{R}=\frac{d}{R}\phantom{\rule{0ex}{0ex}}\Rightarrow h=\frac{d}{2}=\frac{100}{2}=50km$

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