Calculus (3rd Edition)

$626 m/s$
The work needed to move the rocket a distance $r$ from the surface of the earth can be calculated as: $Work \ done =G M_e \ m (\dfrac{1}{R_e}-\dfrac{1}{r+R_e})$ The rocket will attain its maximum height when its kinetic energy reduces to $0$. That is, $\dfrac{1}{2} m V_0^2 =G M_e \ m (\dfrac{1}{R_e}-\dfrac{1}{r+R_e})$ Now, the initial velocity will be $V_0=\sqrt {\dfrac{2G M_e \ m }{m} (\dfrac{1}{R_e}-\dfrac{1}{r+R_e})}$ or, $V_0=\sqrt { 2G M_e(\dfrac{1}{R_e}-\dfrac{1}{r+R_e})}$ Plug in the given data: $V_0=\sqrt { 2 (6.67 \times 10^{-11}) (5.98 \times 10^{24}) (\dfrac{1}{6.37 \times 10^{6}}-\dfrac{1}{20000+6.37 \times 10^{6}})}$ $=626~m/s$