Escape velocity derives from Ve = sqrt[2µ/r] where Ve = escape velocity in ft/sec., µ = the gravitational constant of the earth and r = the surface radius.
µ can also be stated as GM where G = the gravitational and M = the mass of the planet.
Therefore, Ve can be simply expressed by
Ve = sqrt[2(13)µ/(r/13)] = sqrt[169(2µ/r] =
13 times earth's escape velocity of Ve = 36,747fps = 25,054mph or 325,712mph.
V(new)/V(earth) = 13
Suppose a planet is discovered orbiting a distant star with 13 times the mass of the Earth and 1/13 its radius. How does the escape speed on this planet compare with that of the Earth?
Ve new
------- =
ve earth
1 answer
1 answer