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:

(26) |

(27) |

(28) |

(Note: Recall the form derived by equating Newton's Law of Gravitation
and in an earlier section for the velocity of a mass in a
circular orbit around a larger mass :

(29) |

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.

(30) |

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!