# Escape Velocity Derivation

1. Jul 18, 2010

### realitybugll

I tried finding the equation for terminal velocity in a broader attempt to understand planetary orbits better. According to wikipedia's two derivations (one of which i don't understand very well) I'm off by a factor of $$\sqrt 2$$. I would greatly appreciate help in finding my thinking/math mistake. Thanks in advance!

I was reading another thread on the same topic, and I think what I have calculated is the "velocity needed to keep the test particle in circular orbit of radius r" which is apparently different than "escape velocity".

I started off assuming that the planet is questions velocity vector is perpendicular to the force one, attracting it to the larger body whose movement in relation to the planet is negligible. So we should be talking about a circular orbit.

I thought that for every miniscule displacement towards the larger body there has to be a simultaneous one perpendicular such that the distance between the objects remains constant.

Mathematically, $${r - d_1}^{2} + d_2^{2} = r^{2}$$

where $$d_1$$ is the displacement towards the larger body, $$d_2$$ is the displacement perpendicular to that resulting from the planets inherent velocity, and $$r$$ is the radius or displacement between the objects.

$$d_1$$ can be substituted with $$\frac {at^2} {2}$$ where a is the acceleration of the planet towards the larger body and t the time.

similarly $$d_1$$ can be substituted with $$vt$$ where v is the planets escape velocity and t the time.

So rearranging and solving for v we get:

$$d_1 = \sqrt {2rd_2 - d_2^{2}}$$

$$v = \frac {\sqrt {rat^{2} - {\frac {a^{2}t^{4}} {4}}}} {t}$$

and taking the limit of t, the time, as it tends to zero and thus minimizes the two theoretical displacements of the planet (that from the acceleration due the gravity of the larger planet and the perpendicular motion from its own velocity) making the curvature of the orbit more continuous and closer to circular, results in I think (I haven't really learned about limits)

$$v = \sqrt {ra}$$ whereas it really is $$v = \sqrt {2ra}$$

note: I was reading another thread on the same topic, and I think what I have calculated is the "velocity needed to keep the test particle in circular orbit of radius r" which is apparently different than "escape velocity". I had been thinking they were the same, but I guess escape velocity refers to the velocity when elliptical orbits can't form (?).

Last edited: Jul 18, 2010