# Homework Help: Mass of orbiting stars

1. Mar 7, 2006

### Eivind

Can anybody help me whit this?

"Two stars which have the same mass are orbiting an object between the two stars. The velocity of the stars is 80 km/s and the time used on one orbit is 864000 s. Find the mass of the two stars."

This is what I have done so far:
d(distance)=6.912e10 m
M(sentral object mass)=1.05e30 kg
g(gravitational force at the distance which the stars are)=0.58 m/s^2

Im lost trying to find the mass of the two stars, somebody?

Last edited: Mar 7, 2006
2. Mar 7, 2006

### topsquark

I'm lost, too. How did you get those numbers?

-Dan

3. Mar 7, 2006

### Eivind

d=vt
d=80000 m/s*864000 s
d=6.912e10 m

d=2pi*r
d/2pi=r
r=6.912e10 m/2pi
r=1.1e10 m

yMm/r^2=m4pi^2r/t^2
M=4pi^2r^3/yt^2
M=(4pi^2*1.1e10m^3)/(6.67e-11*864000^2 s)
M=1.05e30 kg

g=yM/r^2
g=(6.67e-11*1.05e30 kg)/1.1e10^2 m
g=0.58 N

4. Mar 7, 2006

### BobG

I think there's some missing info. You can't figure out the mass of the stars unless you know the mass of the central object. I have a feeling that the central object is fictitous - i.e. it's the center of mass for the two stars combined (and the two stars have equal masses).

I used the formula for orbital period and orbital velocity to find the radius:

$$\tau^2=\frac{4 \pi^2 a^3}{GM}$$

$$v^2 = \frac{GM}{a}$$

If you rearrange each to solve for GM, then you'll have two equations that are equal to each other and can solve for the radius.

You can then substitute the radius into one of the equations to solve for mass and then solve for the gravitational force.

Edit: Doh! The number they used for the orbital period was so similar to the number of seconds in a day that I overlooked the extra zero they tossed in.

Last edited: Mar 7, 2006
5. Mar 7, 2006

### Hootenanny

Staff Emeritus
If you assume that the stars are exactly opposite each other in a circular orbit, you can just consider one star. I would start by calculating the required centripital force, which is given by;
$$F = mr\omega^2$$
Where $\omega$ is angular velocity.

6. Mar 7, 2006

### BobG

This is all correct except for the last. You found gravitational acceleration instead of gravitational force.

7. Mar 7, 2006

### Eivind

Oh, sorry
. Ill try to work it out.

8. Mar 7, 2006

### Eivind

I have found the mass of the sentral object, but dont have a clue on the rest(correct answer:4.2e30 kg).

9. Mar 7, 2006

### Hootenanny

Staff Emeritus
As I said before, find out the required centripetal acceleration

10. Mar 7, 2006

### Eivind

Ok, Hootenanny, I`ll try that.

11. Mar 7, 2006

### BobG

I think we crossed posts.

It was me that dropped out a zero. Your calculations are correct except for the force of gravity. You calculated gravitational acceleration instead of gravitational force.

I also don't think there really is a central mass. There is a center of mass. Both stars orbit their combined center of mass. For orbital motion, you can treat all of the mass being located at the center of mass. Since both stars are of equal mass, each star is half of the combined mass.

12. Mar 7, 2006

### tony873004

Just curious, why are you solving for g when the question asks you to solve for M?

I agree with Bob, I think you may have worded the question wrong. It wouldn't make sense to have a central mass. It would have to be much heavier than the stars to keep from being ejected by the stars. And if it were much heavier, you could compute it, but not the mass of the 2 stars.

I don't get 1.05 e30.

Try:

$$M = \frac{{v^3 P}}{{4G\pi }}$$
for the mass of a single star. Multiply by 2 for the combined mass.

Though you won't get an answer that agrees with
which makes me further suspicious that you worded the question wrong.

Last edited: Mar 7, 2006
13. Mar 7, 2006

### lightgrav

Eivind, your post #3 was correct until you wrote:
yMm/r^2=m4pi^2r/t^2

(the only thing in the center is the center-of-mass)

The distance between masses is R = 2r, so
yMm/R^2 = m (2 pi/t)^2 r (= the "m w^2 r" that Hootenany told you to use)
M = (2 pi/t)^2 (4 r^3)/y
(which can be written (2 pi r/t)^3 (4 t)/(2 pi y) , 8x Tony's.

14. Mar 8, 2006

### tony873004

Actually, 4x my answer, since my equation gives the mass of one star and yours seems to give their combined masses.
My formula is assuming r is the distance between the 2 stars, not twice their distance as it should be.

Last edited: Mar 8, 2006
15. Mar 8, 2006

### Eivind

Thanks everyone, I got it right now! I should have realized there was no sentral object, thanks again!

16. Mar 8, 2006

### tony873004

Eivind, what answer did you come up with? Was 4.2e30 given as the correct answer by the teacher or in the back of the book? I don't think it is correct. I think the answer given by the formula in my post 12 gives you the correct answer if by 80 km/s they mean relative to eachother rather than relative to their barycenter, which is a strange way to express velocity.

As a double check, if we use the period formula as supplied by Bob:

$$\tau^2=\frac{4 \pi^2 a^3}{GM}$$

and slightly re-write it as

$$\tau=\sqrt{\frac{4 \pi^2 a^3}{GM}}$$

Then plug in 4.2e30 for mass, you get a period of 433093, roughly half of the period the one given in the question.

Using the formula I provide, you get a mass of 5.27774e29 kg per star or 1.055548e30 kg for the stars combined. Plugging this in for mass into the period formula gives 864000 seconds.

Last edited: Mar 8, 2006
17. Mar 8, 2006

### Eivind

The answer was in the back of the book, and I got 4.2e30 kg, which, according to the book, is correct.

18. Mar 8, 2006

### tony873004

I'm pretty sure the back of the book is wrong. For the sake of getting credit you can turn it in as-is. Since you got the book's answer you probably did it the way they wanted.

But I think the mistake here is that the distance between the 2 stars is 2 * r. In circular orbits, r will represent the semi-major axis. I just looked up data on JPL's web service for the Pluto Charon system. This system has its barycenter above the surface of Pluto, so if distance and semi-major axis were really different, you would expect the Charon/Pluto distance to differ from the Charon semi-major axis by over 1000 km. But it is not. The difference is only about 240 km, which can be explained simply by the small amount of eccentricity in its orbit.

As a further test, I set up this system in Gravity Simulator. 2 stars with a semi-major axis of 1.1e10m and a mass of 2.1e30 each completed their orbits in 5 days, not 10. 2 stars with a mass of 5.27774e29 each completed their orbits in exactly 10 days (864000s) as the question asks.

Sorry for over discussing this since this is the homework section and not the Astrophysics section, but I get obsessed with these kind of problems. I'd be interested in anyones comments.