Astronomy - Determine the sum of the stellar masses

Click For Summary

Homework Help Overview

The problem involves determining the sum of the stellar masses of a double star system located 10 light years away, with a maximum angular separation of 2 arcseconds and a rotation period of 4 years. The context includes concepts from gravitational physics and orbital mechanics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using geometry to find the actual distance between the stars and the implications of their orbits around a common center of mass. There are attempts to express gravitational and centripetal forces symbolically, leading to an expression for the sum of the masses. Questions arise regarding the calculation of the distance (R+r) and its relation to the distance from Earth and the maximum angle.

Discussion Status

The discussion is active, with participants exploring various approaches to relate the geometry of the system to the forces involved. Some guidance has been offered regarding the use of trigonometry to find the necessary distances, but no consensus has been reached on the next steps.

Contextual Notes

Participants are working under the assumption that the stellar orbit is circular and are considering the implications of the given angular separation and rotation period. There is an emphasis on the need for geometric relationships to progress further.

tosv
Messages
11
Reaction score
0

Homework Statement


A double star located at a distance of 10 light years from us. The maximum angle between the stars, as seen from Earth, is 2 arcseconds. (1 arcsecond = 1 / 3600 degrees), we can assume that the stellar orbit is circular and that this angle gives us the real distance between the stars. We observe that the rotation period of the stars is 4 years. What is the sum of stellar masses?

Homework Equations


Newton's law of universal gravitation:
[tex]F_{G}=G\frac{m_{1}m_{2}}{r^{2}}[/tex]
Centripetal force:
[tex]F_c=\frac{mv^2}{r}[/tex]
Kepler's third law:
[tex]P^{2} \propto a^{3}[/tex]

The Attempt at a Solution


The stellar orbit is circular, so the gravitational force should be equal to the centripetalforce, but how do I proceed?
 
Physics news on Phys.org
You can work out the actual distance between the two bodies by simple geometry. You should also realize that the both bodies will orbit a common center of mass, so you can work out the individual radii of their orbits (symbolically) as fractions of that distance which depend upon their relative masses. Their individual velocities can similarly be determined (symbolically) from the resulting circumferences of their orbits and the given period of the orbit.

These, along with your gravitational force and centripetal force expressions, should allow you to find an expression for the sum of the two masses.
 
I rewrite the centripetal force as:
[tex]F_{c}=m\cdot \omega^{2}\cdot r[/tex]

So now I can express the equality between the gravitational force and centripetal force for respective star to be
[tex]M\omega^{2}R=G\frac{Mm}{(R+r)^{2}}[/tex]
[tex]m\omega^{2}r=G\frac{Mm}{(R+r)^{2}}[/tex]

I simply the expression and then and I add the left-hand side of the equations together and the same procedure to the right-hand side and I find this expression for the sum of the masses:
[tex]M+m=\frac{\omega^{2}\cdot (R+r)^{3}}{G}[/tex]

The angular velocity can be expressed as
[tex]\omega=\frac{2\pi}{T}[/tex]

I finally found this expression for the sum of the masses:
[tex]M+m=\frac{(\frac{2\pi}{T})^{2}\cdot (R+r)^{3}}{G}[/tex]
But I’m not sure how I’ll proceed and calculate the distance (R+r), should I relate the distance from Earth and the maximum angle between the stars?
 
tosv said:
But I’m not sure how I’ll proceed and calculate the distance (R+r), should I relate the distance from Earth and the maximum angle between the stars?

Yes. Basic trig.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K