Gravitational Potential Energy

Initial potential energy at the earth's surface: $U_i=-\frac{GMm}{R}$ Final potential energy at height $h=R$: $U_f=-\frac{GMm}{2R}$ Work done = change in PE: $W=U_f-U_i=\frac{GMm}{2R}$ Using $GM=gR^2$: $W=\frac{gR^2m}{2R}=\frac{1}{2}mgR$

$\Delta U=-GMm\left[\frac{1}{R+h}-\frac{1}{R}\right]=\frac{GMmh}{R(R+h)}$.

Change in PE $= -\frac{GMm}{3R} - \left(-\frac{GMm}{R}\right) = \frac{2}{3}\frac{GMm}{R} = \frac{2}{3}mgR$

Gravitational potential energy at distance r from the earth's centre: $U = -\dfrac{GMm}{r}$
Initial (on the surface, r = R): $U_i = -\dfrac{GMm}{R}$
Final (h = 3R, so r = R + 3R = 4R): $U_f = -\dfrac{GMm}{4R}$
Change in GPE $= U_f - U_i = -\dfrac{GMm}{4R} + \dfrac{GMm}{R} = \dfrac{GMm}{R}\left(1 - \dfrac{1}{4}\right)$
$= \dfrac{3}{4}\dfrac{GMm}{R} = \dfrac{3}{4}\dfrac{GM}{R^2}mR = \dfrac{3}{4}mgR$ (since $g = \dfrac{GM}{R^2}$)
(The formula mgh cannot be used, because h is not small compared with R.)

Apply energy conservation between the surface (r = R) and the highest point (r = 2R, where the speed is zero):
$-\dfrac{GMm}{R} + \dfrac{1}{2}mv^2 = -\dfrac{GMm}{2R}$
$\dfrac{1}{2}mv^2 = \dfrac{GMm}{R} - \dfrac{GMm}{2R} = \dfrac{GMm}{2R}$
$v = \sqrt{\dfrac{GM}{R}}$
So the particle must be projected with velocity $\left(\dfrac{GM}{R}\right)^{1/2}$.