7.3.3 Escape Velocity

“What goes up must come down.” Any object launched vertically upward, must slow down, eventually come to a rest, and fall back down. So says the adage. But wait. The Earth’s gravitational field strength is not constant. Its strength decreases with distance (from Earth). Because of this, it is possible to “jump” out of the Earth and never fall back down if the launch speed is fast enough. This minimum launch speed (from Earth’s surface) is called the escape velocity.

Example 1

Ignore the effects of air resistance. Calculate the escape velocity from the Earth’s surface.

Earth’s radius $=6370\text{ km}$

Earth’s mass $=5.97\times {{10}^{{24}}}\text{ kg}$

Solution

By PCOE,

$\displaystyle \begin{aligned}(KE+GPE)\text{ at launch}&=(KE+GPE)\text{ at infinity}\\\frac{1}{2}m{{v}_{e}}^{2}+(-\frac{{GMm}}{{{{R}_{E}}}})&=0+0\\{{v}_{e}}&=\sqrt{{\frac{{2GM}}{{{{R}_{E}}}}}}\\&=\sqrt{{\frac{{2(6.67\times {{{10}}^{{-11}}})(5.97\times {{{10}}^{{24}}})}}{{6370\times {{{10}}^{3}}}}}}\\&=11.2\text{ km }{{\text{s}}^{{-1}}}\end{aligned}$

–

Example 2

A projectile m, launched vertically with initial kinetic energy KE_i from the surface of a planet of mass M and radius R, eventually returns to the planet.

Sketch the graphs of KE, GPE and TE of the projectile at distance r from the planet’s centre.

Solution

Things to note:

The initial GPE, $\displaystyle GP{{E}_{i}}=-\frac{{GMm}}{R}$
$\displaystyle TE=GPE+KE=-\frac{{GMm}}{R}+K{{E}_{i}}$ is constant. (Effects of air resistance is ignored)
As the projectile moves away from Earth, it gains GPE at the expense of losing KE. So GPE graph is mirror image of KE graph.
At $r={{r}_{{\max }}}$ , KE drops to zero. This is the furthest point reached by the projectile before it reverses direction and embarks on its return journey.
To escape from Earth’s gravitational field, the projectile must be launched with large enough KE_i such that $\displaystyle TE=K{{E}_{i}}-\frac{{GMm}}{R}\ge 0$