The surface gravity of a planet is given by g=GM/r
2. For Earth, that's (6.67x10
-8erg*cm*g
-2(5.97x10
27g)/(6.37x10
8cm), which gives us 981cm/s
2.
We don't know the mass of Kepler 22 b, though we could find out by using the
Radial Velocity method. However, if we assume that the planet is composed of roughly the same materials as Earth, and thus has a similar average mass density ρ as Earth (5.51g/cm
3), then its mass is just ρ*4/3*п*(2.4R
E)
3 = 8.25x10
28g, or about 13.8x more massive than Earth.
Then the surface gravity of Kepler 22 b would be (G*ρ*4/3*п*(2.4R
E)
3)/(2.4R
E)
2. The (2.4R
E)
2 downstairs cancels two of the same upstairs, and we get a result of ~2353cm/s
2, or about 2.4 Earth gravity. Wait, holy ****, was that a coincidence that it scaled exactly with the planet radius? Let's see... duh, of course! Replace M
E with volume times density, and everything but the 2.4 cancels out. So it turns out the surface gravity of a planet scales exactly with the planet radius, if the average mass density is held constant. You learn something new every day.
