Question

Imagine two planets orbiting a star with orbits edge-on to the Earth. The peak Doppler shift for each 75 m/s, but one has a period of 7 days and the other has a period of 700 days. The star has a mass of one solar mass. Assume 1 solar mass equals 2∗10^30.

1.) Calculate the mass of the shorter period planet. (Hint: See Mathematical Insight Finding Masses of Extrasolar Planets)

2.) Calculate the mass of the longer period planet.

Answer 1

The Centre of mass of the star and planet system does not move. Hence, the planet and the star must have equal magnitudes of momentum, and can be set equal to each other. Let us consider the mass of star and the planet are Ms and Mp. The velocity of star and the planet are Vs and Vp. Therefore, you can write

The system as a whole has no momentum relative to this center of mass. The velocity change gives us the star's speed, which tells us the planet's mass from the periodic Doppler shift graph. Here it is 75 m/s.

Each time the planet completes an orbit, it must travel a distance of where r_p is the planet’s average orbital distance. (considering the orbit as circular) The orbital period is T_p. Thus, the planet’s average orbital velocity must be:

The orbital distance (r_p) can be found from Kepler's third law as

Here, G = 6.67 x 10^-11 N.m².Kg^-2, M_s = 2 x 10³⁰ Kg, Therefore, the velocity of planet 1 will be

Similarly,

(1) The mass of the shorter period planet is

(2) The mass of the shorter period planet is