Binary System in circular orbit: Separation distance between stars

AI Thread Summary
The discussion centers on calculating the orbital period of a binary star system with a separation distance of 14.3 parsecs. The Doppler shift indicates a velocity of 23.4 km/s for both stars, as they share the same mass. Based on a separation of 0.04 nm every 26 years, the estimated full orbital period is 52 years. The inclination of the orbital plane is noted to be 90 degrees, suggesting an edge-on view of the system. This information is crucial for completing the calculations related to the binary system's dynamics.
Thomas Smith
Messages
12
Reaction score
0
Homework Statement
The binary system has a parallax of 0.07 arc seconds. This system consists of two identical stars in circular orbits.

An absorption line at a wavelength of 512nm is seen in the spectra of both stars. This line is split into two components by the orbital motion and has a maximum separation of 0.04nm, which occurs every 26 years.

What's the separation distance between both stars?
Relevant Equations
Distance to Star System from parallax d=1/p

Doppler Shift Δλ = λ - λ_0

Radical Velocity v_r=Δλ/λ_0 c
Distance is d=1/0.07 = 14.3 parsecs

The Doppler shift of one star is, Δλ = 512 - 512.04 = -0.04

So the radical of the velocity of the star is = (-0.04/512) x (3.00 x 10^5 km/s) = 23.4km/s which is the same for both stars because they have the same mass.
This is as far as I've got.
 
Last edited:
Physics news on Phys.org
From the information provided, can you determine the orbital period of the binary system?
 
jbriggs444 said:
From the information provided, can you determine the orbital period of the binary system?
Well considering the separation of 0.04nm every 26 years, I believe the full orbit period is 52 years.
 
If you make an assumption about orientation, you have tangential velocity and period...
 
jbriggs444 said:
If you make an assumption about orientation, you have tangential velocity and period...
Ah sorry, i missed out the inclination. It says the orbital plane is inclined to the plane of the sky by 90 degrees
 
Thomas Smith said:
Ah sorry, i missed out the inclination. It says the orbital plane is inclined to the plane of the sky by 90 degrees
That was the assumption I'd needed for the problem to make sense. So we are looking at the binary system, "edge on". Does that give you everything you need to finish the exercise?
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top