system
Given the following post, respond as if you're commenting on Reddit:
human
For how long do you need to maintain 'escape velocity' in order to escape Earth's gravitational pull?
gpt
Some clarification first. Suppose there is some central gravitating mass *M* and we want to calculate the speed *v* needed for some negligible test mass *m* to escape to infinity. That is, *v* is the speed required of a *free* particle (no force other than gravity) for that particle to get arbitrarily far away from the central mass.

The central mass may not be symmetric, but it has some gravitational potential *U* associated with it, which depends, in principle, on (*x*, *y*, *z*), or whatever your favorite coordinate system is. The total energy of the test mass is

> E = mv^(2)/2 + mU

and this energy is conserved. If we use the convention that U = 0 at infinity, then for a free particle to just barely escape to infinity, it would have to be at rest at infinity. In other words, at infinity, both U = 0 and v = 0, which means E = 0. But since energy is conserved, we must have

> 0 = v^(2)/2 + U

at all points along the particle's trajectory. The speed is thus given by

> v = √(-2U)

(Remember that U < 0, so this makes sense.) This is the equation for escape speed. Since *U* depends on (x, y, z), so does *v*. That should make sense to you. The farther away the particle begins, the smaller *v* has to be to guarantee escape. So *v* depends on position as well.

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Okay, now let's make this more specific. Suppose the gravitating mass *M* is spherically symmetric with radius *R*. Then the potential exterior to this mass is U = -GM/r where r measures distance from the center of the body. When we talk about *the* escape speed of this planet (or star or whatever), we mean the escape speed at the surface of the planet. So we mean *v**_esc_* at r = R. From the formula above, we get

> v*_esc_*(R) = √(-2U(R)) = √(2GM/R)

So if the particle is on the surface of the planet and has velocity v = √(2GM/R), it escapes to infinity. (Assuming, of course, it's not initially pointed toward the planet itself.) That's it. There's no maintenance required; we are neglecting any effects of a possible atmosphere. If the particle has that speed on the surface for whatever reason, it escapes; it doesn't have to do anything else.

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I should make this point very explicit though: the escape speed is the speed necessary only for a *free* particle. A rocket ship, for instance, need never reach the local escape speed at all. As long as its rockets give it enough thrust to just barely counteract the gravitational force (fully counteracting gravity is not even necessary), the rocket will escape. It can do so at an arbitrarily slow speed if it wanted to.