Question

(this is from a thermoscience class) In the space program and in astronomy in general the concept of escape velocity is sited as a bit of information regarding the earth or other planets. The escape velocity is considered the minimum velocity required at the earth’s surface so that its kinetic energy there directed vertically upward can overcome the gravitational pull on a rocket’s mass that is trying to escape to outer space. If the earth’s radius is 6378 km and the earth’s mass can be taken as 5.983 x 10²⁴ kg, estimate the escape velocity from sea level to outer space. (hint: the kinetic energy is effectively the work required to leave the earth’s gravitational field.)

Answer 1

$\frac {1}{2}mv^2- \frac {GMm}{R}= 0$

$v =\sqrt \frac {2GM}{R}$

$v =\sqrt \frac {2\times 6.67\times 10^{-11}\times 5.983 \times 10^{24}}{6378000} \: m/s = 11.18 \: km/s$