Question

At what distance above the surface of the earth is the acceleration due to the earth’s gravity 0.980 𝑚/𝑠 2 if the acceleration due to gravity at the surface has a magnitude of 9.80 𝑚/𝑠 2 ?

Answer 1

$g=F/m=GM/r^2\to r=\sqrt{GM/g}=$

$=\sqrt{6.67\cdot 10^{-11}\cdot 5.97\cdot10^{24}/0.98}=$

$=20157518\ (m)$

$h=20157518-6371000=13786518\ (m)$