Q.Show that acceleration due to gravity at height h above the Earth's surface is $g_h=\dfrac{gR^2}{(R+h)^2}$.

The acceleration due to gravity decreases continuously as height h above a planet's surface increases, following the exact inverse-square relation $g_h=\dfrac{gR^2}{(R+h)^2}$, obtained by simply replacing the surface distance R with the larger distance $(R+h)$ in $g=GM/r^2$. For small altitudes where $h\ll R$, this simplifies (via the binomial approximation) to the widely used linear formula $g_h\approx g\left(1-\dfrac{2h}{R}\right)$, accurate for heights genuinely small compared to the roughly 6400 km radius of the Earth, but increasingly inaccurate at satellite altitudes where the full inverse-square formula must be used instead.

Unlike the depth case (which reaches exactly zero at the centre), g due to altitude only ever approaches zero as h tends to infinity -- there is no finite height at which gravity vanishes entirely, only ever weaker attraction, which is exactly why 'escaping' gravity (section on escape velocity) is fundamentally about reaching infinite distance, not some finite height.