Astrophysics: Finding the mass of a hidden star in a binary system

[knowlewj01]

Homework Statement

The spectral lines in a low mass main sequence star show sinusodal velocity variations with an amplitude of 500 km/s and a time period of 10 hours

calculate a lower limit to the mass of the unseen binary companion

Homework Equations

M₁ + M₂ = $$\frac{4\pi^2a^3}{GP^2}$$

M₁r₁ = M₂r₂

a = r₁ + r₂

The Attempt at a Solution

The redshifted and blueshifted spectral lines show that the star is traveling at 500km/s

in a time period of 10 hoiurs (= 36 000 seconds)

Distance traveled in 1 orbit = 18 000 000 km

radius of orbit = $$\frac{18 000 000}{2\pi}$$ = 28274334 km

assume that the hidden object is much more massive.

so a = radius of this orbit = 28274334000 m

M₁ >> M₂ therefore r₁ << r₂

M₁ = $$\frac{4\pi^2r^3}{GP^2}$$

M₁ = 1.30x10^13 kg

this is wrong,

think i may have made a mistake when i said that the hidden object is much more massive, can this be solved without making an assumption?

 

[Borek]

I think you can calculate ratio of both masses, then use information about mass of the observed star ("low mass main sequence star") to estimate mass of the other one. But that's just intuition, I can be easily wrong.

 

[kerimek]

Yes, this can be solved without making the assumption that the hidden object is much more massive. The key is to use the second equation listed, which relates the masses and radii of the two objects in a binary system. Since we know the distance traveled by the visible star in one orbit (18,000,000 km), we can use this to calculate the radius of the visible star's orbit (r1) using the formula a = r1 + r2. Then, we can use this value for r1 in the equation M1r1 = M2r2 to solve for the mass of the hidden object (M2). This will give us a lower limit for the mass of the hidden object, since we are assuming that M1 >> M2.