Imagine two planets orbiting a star with orbits edge-on to the Earth.

Imagine two planets orbiting a star with orbits edge-on to the Earth. The peak Doppler shift for each 70, but one has a period of 9 days and the other has a period of 900 days.

1.Calculate the mass of the shorter period planet.
2.Calculate the mass of the longer period planet.

Alright there are a lot of calculations here so I'll try to lay it all out clearly


Equations to consider:

mass of star: 2.1*10^30
Velocity: 70
periods(convert to seconds): 9 days = 777,600; 900 days = 77,760,000

1. (Mstar*Vstar*period)/(2*pi*a)


(2.1*10^30*70*777,600)/(2*pi*a)

a=cubed root(((6.67*10^-11)(Mstar)(period^2))/(4...

a= cubed root[((6.67*10^-11)(2.1*10^30)(777,600^2...


I got 1.14307*10^38 for the top. I got a = 1.28973*10^10. I got 2pi*a = 8.10362*10^10 (the bottom)

1.14307*10^38/8.10362*10^10 = 1.41057*10^27 1.4*10^27 (maybe 1.3*10^27)

As you can see, there are many places where calculator errors can be made. However, this should be right; if it isn't working, it should be because of significant figures (it should say how many to input if your homework is on masteringastronomy).

For 2., do the same equations but with the period being 77,760,000

Imagine two planets orbiting a star with orbits edge-on to the Earth. The peak Doppler shift for each 60 m/s, but one has a period of 7 days and the other has a period of 700 days.

Also, what is the mass of the longer period planet?

For both planets we have the conservation of momentum for the movement of star and planets: MV = mv, where the capitals refer to the star.

We have the following relations for each planet, (the capitals refer to the star)
- MV = mv consevation of momentum
- mv^2/r = GmM/r^2, which reduces to GM = r v^2 . balance of forces
- P^2 = 4 pi^2 r^2 /GM Kepler's third law.

We know V=60 m/s and P = 7 days or 700 days.

The equations allow us to eliminate the unknowns.