Kepler 186f is a very recently (April 2014) announced planet orbiting another st
ID: 1305574 • Letter: K
Question
Kepler 186f is a very recently (April 2014) announced planet orbiting another star, within the star's "habitable zone" (which means that if the planet were to have water on its surface, under some plausible but unproven circumstances that water would be in the liquid phase). The orbital period of the planet is 129.95 days. Using assumed values for the stellar mass (0.48 solar masses) and radius (0.47 solar radii) and timings of transits across the star, the planet appears to have a radius about 1.1 times that of Earth. Assuming a circular orbit, compute the orbital radius (in AU) and velocity (in km/s) for the planet.
Explanation / Answer
We know that v = (GM/r)1/2 and that v = 2?r/T so the period T = 2?r3/2/(GM)1/2 and therefore r = (T/2?)2/3(GM)1/3
= ((129.95 d)(86400 s/d)/2?)2/3 (6.67384 * 10-11 Nm2/kg2)*(0.48 * 1.9885 * 1030 kg)1/3
= 5.88 * 1010 m = 0.393 AU
the velocity is 2pi times the radius divided by the period = 3.29 * 104 m/s
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