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

[!TLDR] Since $g=GM/R^2$ at the surface and $g_h=GM/(R+h)^2$ at height h, dividing the two directly gives $g_h=gR^2/(R+h)^2$. [!ANSWER] $g_h=\dfrac{gR^2}{(R+h)^2}$

At the Earth's surface (distance R from the centre), the acceleration due to gravity is $g=\dfrac{GM}{R^2}.$ At height h above the surface (distance $R+h$ from the centre), it becomes $g_h=\dfrac{GM}{(R+h)^2}.$ Dividing the second equation by the first: $\dfrac{g_h}{g}=\dfrac{GM/(R+h)^2}{GM/R^2}=\dfrac{R^2}{(R+h)^2}.$ The common factor GM cancels exactly, since it is the same Earth in both cases. Rearranging gives the required result: $g_h=g\cdot\dfrac{R^2}{(R+h)^2}=\dfrac{gR^2}{(R+h)^2}.$ This is the EXACT relation, valid for any height h, showing g decreases as an inverse-square function of the distance $(R+h)$ from the Earth's centre. [!ANSWER] $g_h=\dfrac{gR^2}{(R+h)^2}$, obtained by dividing $g_h=GM/(R+h)^2$ by $g=GM/R^2$ to eliminate GM.

Write down g at the surface and at height h separately from $g=GM/r^2$, then divide the two expressions to eliminate the unknown GM and isolate the ratio.

Forgetting that r must be measured from the Earth's CENTRE (so the height-h distance is $R+h$, not just h).