Mass of invisible binary companion

[Kinetic]

Hey guys, got this question that's been giving me some trouble.

The spectral lines in a low-mass main sequence star are observed to show sinusoidal velocity variations with an amplitude of 500km/s and a period of ten hours. Calculate the lower mass limit of the unseen binary companion.

Here's how far I've managed to get:

I started with Newton's adaptation of kepler's third law

1. M₁+M₂=(4(PI)²a³)/(GP²)

where 'a' is the separation of the two masses

2. a=r₁+r₂

I don't know the radius of either orbit but i do know the velocity of one and the period.

3. V₁=(2(PI)r₁)/P
4. V₂=(2(PI)r₂)/P

I then rearranged 3 and 4 to get r in terms of P and V and substituted into 2 to get

5. a=(P/2(PI))*(V₁+V₂)

Then substituting 5 into 1 gives

6. M₁+M₂=(P/2(PI)G)*(V₁+V₂)³

I now want to get rid of the V₂.

V₂/V₁=r₂/r₁=M₁/M₂

So i make V₂= V₁*(M₁/M₂)

Substitute this into 6 and with a bit of fiddling i get

7. M₂³/(M₁+M₂)²=(PV₁³)/(2(PI)G)

I've been told the visible star is a 'low-mass main sequence star' so i can make a rough estimate of M₁. Now my only unknown is M₂...
However! I've been playing with 7 for ages and simply cannot isolate M₂

So I've either messed up somewhere along the way to 7 or my algebra is failing me.

Any suggestions or help would be great guys!

Kinetic

 

[NateD]

energy = P*V^2/(4*pi^2)Using the kinetic energy equation, you can solve for M2. You know the values for P and V, so you just need to solve for M2. You can then use this value to calculate the lower mass limit of the unseen binary companion.