## Comment Re:600 light years... (Score 1) 257

Remember, though, that the mass itself is not really the important part. The question is what is the surface gravity. Using your assumption of identical density (maybe not likely, but I don't know offhand what density is likely), the mass of this new planet in terms of the density rho will be M = rho*4/3*pi*R^3. The gravitational acceleration at the surface will be

a = (G M) / R^2

Combining that with the expression for the mass gives

a = G*rho*4/3*pi*R^3 / R^2 = (G*rho*4/3*pi) R.

All the stuff inside the parenthesis is assumed to be the same for both planets, so if we want to write it in terms of the surface gravity of Earth, g, and radius of Earth, r, then we'd have

a = g*(R/r). Thus, the surface gravity (under the assumption of identical density) is only 2.4 times greater.

Of course, if you're going to visit this place and plan to leave again (maybe not so useful without warp drive, given) then you might also be concerned with the energetic depth of the gravity well. For an object of mass m to escape to interstellar space from the surface will require an amount of energy

E = G*M*m/R

and in terms of the earth value E_earth this would be

E = (E_earth)*(R/r)^2

meaning it will take about 6 times as much energy as getting off Earth.

I was actually just thinking about this issue the other day while playing Mass Effect, because I was wondering if they'd done their numbers right on the planet properties (they had).