Escape Velocity, Escape Energy

As we noted in the previous section, a particle has ``escape energy'' if and only if its total energy is greater than or equal to zero. We define the escape velocity (a misnomer!) of the particle as the minimum speed (!) that it must have to escape from its current gravitational field - typically that of a moon, or planet, or sun. Thus:

$$E_{\rm tot} = 0 = \frac{1}{2}mv_{\rm escape}^2 - \frac{GMm}{r}$$ (26)

so that

$$v_{\rm escape} = \sqrt{\frac{2GM}{r}} = \sqrt{2gr}$$ (27)

where in the last form $g = \frac{GM}{r^2}$ (the magnitude of the gravitational field - see next item). For earth:

$$v_{\rm escape} = \sqrt{\frac{2GM_E}{R_E}} = \sqrt{2gR_E} = 11.2 \quad {\rm km/sec}$$ (28)

(Note: Recall the form derived by equating Newton's Law of Gravitation and $mv^2/r$ in an earlier section for the velocity of a mass $m$ in a circular orbit around a larger mass $M$:

$$v_{\rm circ}^2 = \frac{GM}{r}$$ (29)

from which we see that $v_{\rm escape} = \sqrt{2}v_{\rm circ}$.)

It is often interesting to contemplate this reasoning in reverse. If we drop a rock onto the earth from a state of rest ``far away'' (much farther than the radius of the earth, far enough away to be considered ``infinity''), it will REACH the earth with escape (kinetic) energy. Since the earth is likely to be much larger than the rock, it will undergo an inelastic collision and release nearly all its kinetic energy as heat. If the rock is small, this is not a problem. If it is large (say, 1 km and up) it releases a lot of energy.

$$M = \frac{4 \pi \rho}{3} r^3$$ (30)

is a reasonable equation for the mass of a spherical rock. $\rho$ can be estimated at $10^4$ kg/m$^3$, so for $r \approx 1000$ meters, this is roughly $M \approx 4\times 10^{13}$ kg, or around 10 billion metric tons of rock, about the mass of a small mountain.

This mass will land on earth with escape velocity, 11.2 km/sec, if it falls in from far away. Or more, of course - it may have started with velocity and energy from some other source - this is pretty much a minimum. As an exercise, compute the number of Joules this collision would release to toast the dinosaurs - or us!