Q.Draw a graph showing the variation of gravitational acceleration due to the depth and altitude from the Earth's surface.

[!TLDR] g rises linearly from zero at the centre to a maximum at the surface, then falls off as an inverse-square curve above the surface. [!ANSWER] A straight line for $r<R$ (through the origin, …

Plotting g as a function of r, the distance from the Earth's centre, with the Earth's radius R marked on the horizontal axis: for $r<R$ (below the surface, i.e. at depth $d=R-r$), the depth formula $g_d=g(1-d/R)$ rewrites, using $r=R-d$, as $g(r)=\dfrac{g}{R}r$ -- a STRAIGHT LINE through the origin (g=0 at the centre, $r=0$) with slope $g/R$, rising linearly up to $g(R)=g$ at the surface. For $r>R$ (above the surface, altitude $h=r-R$), the altitude formula $g_h=gR^2/(R+h)^2$ rewrites, using $r=R+h$, as $g(r)=\dfrac{gR^2}{r^2}$ -- a smoothly DECREASING inverse-square curve, starting at $g(R)=g$ and falling off (never reaching exactly zero) as r increases further. Both pieces of the graph meet at exactly $r=R$, at the same value g -- the overall MAXIMUM of the entire graph -- with the curve rising linearly to that peak from …

Rewrite both the depth formula ($g_d=g(1-d/R)$) and the altitude formula ($g_h=gR^2/(R+h)^2$) in terms of a single variable r (distance f …

Drawing the below-surface part as also an inverse-square curve (it is genuinely LINEAR for a uniform-density Ea …