# Homework Help: Gravitation problem -- Binary star system

1. Aug 4, 2015

### PhysicStud01

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
the solution says to equate the moments about C or equate the centripetal forces
but the moments, they used M1R1 = M2R2
how does the above represent the moment, and why is the moment even equal?

similarly, why is the centripetal force equal?

2. Aug 4, 2015

### SteamKing

Staff Emeritus
The language here is a tad imprecise. Instead of "moment" in the sense of force times distance, the text is using "moment" to represent mass times distance.

In other words, point C represents the center of gravity of the binary system, if that means anything to you.

What would happen to the stars if the net force acting on them was not zero?

3. Aug 5, 2015

### PhysicStud01

there orbit would no longer be circular, i think

but then again, they are in different orbits (of different radius) - so, how is it equal.

also, why is the centripetal force equal?

4. Aug 5, 2015

### haruspex

Well, centre of mass.
No, it's not related to that. There are no external forces on the system. What does that tell you about how the centre of mass of the system moves?

5. Aug 5, 2015

### PhysicStud01

it also moves circular orbit? but how is this related?

6. Aug 5, 2015

### haruspex

If there are no forces acting on a mass, how does it move?

7. Aug 5, 2015

### PhysicStud01

if it's moving at constnat speed, it continues.
if it's at rest, it remains at rest.

but I'm lost here. how is this related to moment. why is the centripetal force equal in both cases?

8. Aug 5, 2015

### haruspex

A centripetal force is that resultant force required to produce the centripetal acceleration. It is not an applied force, but is the resultant of applied forces. What is the applied force that produces it?

9. Aug 5, 2015

### PhysicStud01

it's the gravitational force between the 2 stars, right.

10. Aug 5, 2015

### Bandersnatch

Alright. Write it down for each of the stars. Is it the same both? Is it different for each?

For the stars to be moving in circles, this force, acting on each of the stars separately, has to be equal to the centripetal force. Write down the centripetal force equation for each star. Are they the same? Are they different?

Next, for each star, equate the actual (applied) force that's acting on it (gravity), with the force (resultant) required to be moving in circles. You'll end up with two sets of equations. See if you can do any algebraic magic on them.

Show us your work.

11. Aug 5, 2015

### PhysicStud01

the gravitational force is the same. i understood the thing about centriptal force. thanks.
The solution is already available, i just wanted to understand it.

but one thing i have not yet understood - why are the moment equal? why did they use to equation i stated at first? is this a simplification of something else, or is it a formula itself?

12. Aug 5, 2015

### haruspex

Where is the mass centre of the system in relation to C?

13. Aug 6, 2015

### tms

Do what was suggested above: write down $F = ma$ for each star, and compare the equations. You will then see the answer to your question. Be careful with the radii.

14. Aug 6, 2015

### haruspex

I believe PS01 understands why the forces are the same. The question is why the "mass moments" (mass times distance from C) can be immediately written down as being the same for each without having to think about forces.

15. Aug 6, 2015

### PhysicStud01

sorry, i don't know about this one. i assume it is at a radius r from C, but i think that this centre also moves.

but what is F and a here? m is the mass of the planet into consideration, right.

thanks. yeah, that's why i really want to know. but understanding what the others are saying will be a plus, i believe.

16. Aug 6, 2015

### haruspex

If there are no external forces acting on a system then the mass centre of that system does not accelerate.

17. Aug 6, 2015

### PhysicStud01

but the stars are moving, right. this is a change direction of velocity and thus there is an acceleration. anyway, when the positions of the stars are changing, shouldn't C be changing too?

18. Aug 6, 2015

### tms

$F$ is the only force the stars exert on each other. You get $a$ from the kinematics of uniform circular motion.

19. Aug 6, 2015

### PhysicStud01

so F is the gravitational force between them. but could you give me the formula for a

20. Aug 6, 2015

### haruspex

"The system" is both stars taken together. There is no external force on that system, so the mass centre of that system cannot accelerate. $\Sigma F_{ext} = ma$.
C in the diagram is defined as a fixed point. I'm endeavouring to prove to you that it is the mass centre of the system. M1R1 = M2R2 would follow immediately from that by definition of mass centre.

21. Aug 6, 2015

### PhysicStud01

so, when the centre is fixed at C, we can write M1R1 = M2R2. how does this formula come? is it an actual formula, for example momentum = mv?

22. Aug 6, 2015

### Bandersnatch

If you follow the steps I outlined earlier, you should arrive at the equation provided.

23. Aug 6, 2015

### tms

Yes, $F$ is gravity. For $a$, go back and look at uniform circular motion; there is an equation that relates the angular speed to the radius and mass.

24. Aug 6, 2015

### tms

It comes from the definition of the center of mass.

25. Aug 6, 2015

### PhysicStud01

so, this is called moments?