Triple Star System: Find Period of Revolution in Years

In summary: So you can disregard the second term and just use the first one.Now, this is the force on star A. What we need is the acceleration. This is given by F=ma. So we have\frac{G m_B}{R^2} = ma or a = \frac{G m_B}{R^2 m_A} Now, you also have a formula for the acceleration a in terms of the period T, the mass m_A and the radius R of the orbit. So you should be able to solve for T.In summary, the period of revolution for the stars in this system can be calculated using the formula T=2*pi*r^3/2 / sqrt (GM), where
  • #1
NAkid
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Homework Statement


A certain triple-star system consists of two stars, each of mass m = 7.00×1024kg, revolving about a central star of mass M = 1.69×1030kg in the same circular orbit of radius r = 2.00×1011m(see the figure). The two stars are always at opposite ends of a diameter of the circular orbit. What is the period of revolution of the stars in years(yr)?

https://cats.lsa.umich.edu/wiley-HRW6/Graphics/14-39.jpg

Homework Equations


The Attempt at a Solution



I know that the equation for period of motion is

T=2*pi*r^3/2 / sqrt (GM)

I'm just not sure if I can plug in values where M is the aggregate of masses M1 and M2..
 
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  • #2
I'm just a high school physics student, (so I could be wrong), but technically, if the stars are always at opposite ends, aren't the periods of the stars equal to each other? Which would mean using that equation and plugging in the 7E1024 and 1.69E1030 be correct?

Like I said, I could be very very wrong, but this is just my thought on the problem.
 
  • #3
NAkid said:

Homework Statement


A certain triple-star system consists of two stars, each of mass m = 7.00×1024kg, revolving about a central star of mass M = 1.69×1030kg in the same circular orbit of radius r = 2.00×1011m(see the figure). The two stars are always at opposite ends of a diameter of the circular orbit. What is the period of revolution of the stars in years(yr)?

https://cats.lsa.umich.edu/wiley-HRW6/Graphics/14-39.jpg

Homework Equations





The Attempt at a Solution



I know that the equation for period of motion is

T=2*pi*r^3/2 / sqrt (GM)

I'm just not sure if I can plug in values where M is the aggregate of masses M1 and M2..

Go back to the basics:

[tex] F = m \frac{v^2}{R} = m \frac{4 \pi^2 R}{T^2} [/tex]
where here m is the mass of the star that you are focusing on and R is the radius of its orbit.

Now, the force will be the total force acting on that star so it will be the sum of teh force of gravity due to the other two stars. Can you go on from there?
 
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  • #4
ok so the force of gravity is Fg = (GM1M2)/R^2

If total force is what you have above, should I set that equal to 2*(GM1M2)/R^2 -- or the sum of Fg of star1 on central star and star 2 on central star?
 
  • #5
NAkid said:
ok so the force of gravity is Fg = (GM1M2)/R^2

If total force is what you have above, should I set that equal to 2*(GM1M2)/R^2 -- or the sum of Fg of star1 on central star and star 2 on central star?

The equation I wrote is to be applied to one of the star in motion. So the force is the total force due to the other two stars. Note that the other two stars are AT DIFFERENT DISTANCES! So be careful with that
 
  • #6
but their orbital radius is the same though?
 
  • #7
i'm still confused, the force is the total force due to the other two stars - what does that mean? and how does gravitational force work into this equation?
 
  • #8
NAkid said:
i'm still confused, the force is the total force due to the other two stars - what does that mean? and how does gravitational force work into this equation?

Let's say we have stars A, B and C in that order. Star B is at rest. Stars A nd C are moving in a circle. The radius of the circle is the distance between star A and star B, right? (which is the same as between star B and star C) and is equal to 2 x 10^11 m.

Calculate the total force on star A. It's the sum of the force of star B on star A PLUS the force of star C and star A.

But I just noticed that the mass of stars A and C is really tiny compared to the central star (thos enumbers don't make much sense since the masses of stars A and C are comparable to the mass of a planet not of a star!)

Anyway... because the mass of stars A and C is so small compared to mass B, you can neglect the effect of the gravitational pull of star C on star A. In that case, the force of gravity on star A is simply the force of star B on star A only.
 
  • #9
i think its supposed to be 10^24 and 10^30
 
  • #10
the total force on star A would be G(mass starB) / (R)^2 + G(mass starC) / (2R)^2 ?
 
  • #11
NAkid said:
the total force on star A would be G(mass starB) / (R)^2 + G(mass starC) / (2R)^2 ?


Almost. You still have to multiply by the mass of star A. So it is

[tex]\frac{G m_A m_B}{R^2} + \frac{G m_A m_C}{(2R)^2} [/tex]

as I said, it turns out that the second term will be negligible compared to the first since they used such a small mass for "stars" A and C (the masses are actually close to the mass of our planet!).
 

FAQ: Triple Star System: Find Period of Revolution in Years

What is a triple star system?

A triple star system is a system in which three stars are gravitationally bound to each other and orbit around a common center of mass.

How do scientists determine the period of revolution in years for a triple star system?

Scientists use the orbital parameters of the three stars, such as their masses, distances, and velocities, to calculate the period of revolution using Kepler's third law of planetary motion.

Is the period of revolution the same for all three stars in a triple star system?

No, the period of revolution may differ for each star in a triple star system depending on their masses and distances from each other. This can result in one star having a longer or shorter period of revolution than the other two.

Can the period of revolution in a triple star system change over time?

Yes, the period of revolution in a triple star system can change over time due to gravitational interactions between the stars, as well as external influences such as the presence of other nearby stars or planets.

Why is knowing the period of revolution in a triple star system important?

Knowing the period of revolution in a triple star system can provide valuable information about the dynamics and stability of the system, as well as help scientists better understand the formation and evolution of multiple star systems. It can also aid in the search for exoplanets and the study of their potential habitability.

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