Q: What are the escape velocities for the earth, solar system and
galaxy?

[29 Nov 1999]

A: If we're a unit mass at distance `r` from an object of
mass `M` and want to get to infinity then the potential energy
required is the integral of `MG` / `x`^{2}
between `x` = `r` and infinity, which is `MG`
/ `r`. To escape, our kinetic energy `v`^{2} /
2 must be equal to this.

If we're on the surface of the object, and it has surface gravity
`g`, then `g` = `MG` /
`r`^{2} so the potential energy required is
`gr` and the escape velocity is √(2`gr`).

If we're in a circular orbit around the object, then `MG` /
`r`^{2} = `rw`^{2}, where `w`
is the angular velocity, so the potential energy required is
`r`^{2}`w`^{2} and the escape velocity
is √(2`r`^{2}`w`^{2}).

Thus we have

- Escape from earth:
`g`= 9.81 m/s^{2},`r`= 6.4e6 m so`v`= 1.1e4 m/s = 2.5e4 mph. - Escape from solar system:
`r`= 1.5e11 m,`w`= 2 π / year = 2e-7 radians/s so`v`= 4.2e4 m/s = 9.4e4 mph. - Escape from galaxy:
`r`= 30000 light years = 2.8e20m,`w`= 2 π / (2e8 years) = 1e-15 radians/s so`v`= 4e5 m/s = 9e5 mph.

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