Question

Imagine two planets orbiting a star with orbits edge-on to the Earth. The peak Doppler shift for each is 90 m/s , but one has a period of 6 days and the other has a period of 600 days. The star has a mass of one solar mass. Assume 1 solar mass equals 2*10^30 kg. What is: A. the mass of the planet with the shorter period, and B. the mass of the planet with the longer period.

Answer 1

Equations to consider:

mass of star: 2.0*10^30
Velocity: 90 m/s
periods(convert to seconds): 6 days = 518400 s; 600 days = 51840000

1. mass = (Mstar*Vstar*period)/(2*pi*a) = (2.0*10^30*90*518400)/(2*pi*a)
a=cubed root(((6.67*10^-11)(Mstar)(period^2))/(4*pi^2)
a= cubed root[((6.67*10^-11)(2.0*10^30)(518400^2)/(4*pi^2)] = 7.7*10^9

mass = (2.0*10^30*90*518400)/(2*pi* 7.7*10^9) = 1.93*10^27 kg

2) mass = (Mstar*Vstar*period)/(2*pi*a) = (2.0*10^30*90*51840000)/(2*pi*a)

a=cubed root(((6.67*10^-11)(Mstar)(period^2))/(4*pi^2)
a= cubed root[((6.67*10^-11)(2.0*10^30)(51840000^2)/(4*pi^2)] = 7.7*10^13

mass = (2.0*10^30*90*51840000)/(2*pi* 7.7*10^13) = 1.93*10^25 kg