Answer to Question #121145 in Classical Mechanics for Lizwi

Question

What is the ratio of the escape speed of a rocket launched from sea level to the escape speed of one launched from Mt. Everest (an altitude of {\bf \scriptsize{8.85}}8.85{\bf \scriptsize{8.85}} km)? The radius of the earth is {\bf\scriptsize{6.38 \times 10^6}}6.38×106{\bf\scriptsize{6.38 \times 10^6}} m. (The escape speed is the minimum speed that is required to escape from the influence of the gravitational potential.)
1 : 1.0007
2 : 1.0014
3 : 1.0001
4 : 0.9986
5 : 0.9993

Answer 1

Gravitational potential energy is given by $\frac{-GMm}{r}$

escape velocity is $\sqrt{\frac{2GMm}{r}}$

so the only entity that differs here is r

in case of Mt. Everest r= $6.38\times10^6+8.48\times10^3$ m

in case of sea level r=$6.38\times10^6$ m

$ratio:\frac{v_1}{v_2}=\sqrt{\frac{6.38\times10^6}{6.38848\times10^6}}=0.9993$