Question

A particle is projected upward from the surface of earth(radius R) with a K.E. equal to half the minimum value needed for it to escape from earth's gravity. To what height does it rises above the surface of earth?

Answer 1

KE needed to escape from the surface:
$$\begin{gathered} E=\frac{1}{2}m{v_e}^2 \\ =\frac{1}{2}m{\left(\sqrt{\frac{2Gm}{r}}\right)}^2 \\ =\frac{1}{2}m\left(\frac{2Gm}{r}\right) \\ =\frac{GMm}{r} \end{gathered}$$

PE at the surface of Earth + $\frac{1}{2}E$= PE at the maximum height
$$\begin{gathered} -\frac{GMm}{r}+\frac{1}{2}\frac{GMm}{r}=-\frac{GMm}{r+h} \\ -\frac{1}{2}\frac{GMm}{r}=-\frac{GMm}{r+h} \\ \frac{1}{2r}=\frac{1}{r+h} \\ 2r=r+h \\ h=r \end{gathered}$$

Answer: h = r