Gravitation problem -- Binary star system

In summary: The equation you are mentioning is used to represent the moment of a force. It is simply a way of representing the mass (m) and distance (d) of the point of application of the force. It is not a simplification of anything.
  • #1
PhysicStud01
174
0

Homework Statement


problem.png


Homework Equations

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?

thanks in advance
 
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  • #2
PhysicStud01 said:

Homework Statement


View attachment 86837

Homework Equations

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?
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.

similarly, why is the centripetal force equal?

thanks in advance
What would happen to the stars if the net force acting on them was not zero?
 
  • #3
SteamKing said:
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?
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
SteamKing said:
In other words, point C represents the center of gravity of the binary system, if that means anything to
Well, centre of mass.
PhysicStud01 said:
there orbit would no longer be circular, i think
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
haruspex said:
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?
it also moves circular orbit? but how is this related?
 
  • #6
PhysicStud01 said:
it also moves circular orbit? but how is this related?
If there are no forces acting on a mass, how does it move?
 
  • #7
haruspex said:
If there are no forces acting on a mass, how does it move?
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
PhysicStud01 said:
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?
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
haruspex said:
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?
it's the gravitational force between the 2 stars, right.
 
  • #10
PhysicStud01 said:
it's the gravitational force between the 2 stars, right.
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
Bandersnatch said:
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.
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
PhysicStud01 said:
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?
Where is the mass centre of the system in relation to C?
 
  • #13
PhysicStud01 said:
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?

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
tms said:
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.
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
haruspex said:
Where is the mass centre of the system in relation to C?
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.

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

haruspex said:
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.
thanks. yeah, that's why i really want to know. but understanding what the others are saying will be a plus, i believe.
 
  • #16
PhysicStud01 said:
i think that this centre also moves.
If there are no external forces acting on a system then the mass centre of that system does not accelerate.
 
  • #17
haruspex said:
If there are no external forces acting on a system then the mass centre of that system does not accelerate.
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
PhysicStud01 said:
but what is F and a here? m is the mass of the planet into consideration, right.

##F## is the only force the stars exert on each other. You get ##a## from the kinematics of uniform circular motion.
 
  • #19
tms said:
##F## is the only force the stars exert on each other. You get ##a## from the kinematics of uniform circular motion.
so F is the gravitational force between them. but could you give me the formula for a
 
  • #20
PhysicStud01 said:
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?
"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
haruspex said:
"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.
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
If you follow the steps I outlined earlier, you should arrive at the equation provided.
 
  • #23
PhysicStud01 said:
so F is the gravitational force between them. but could you give me the formula for a

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
PhysicStud01 said:
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?

It comes from the definition of the center of mass.
 
  • #25
tms said:
It comes from the definition of the center of mass.
so, this is called moments?
 
  • #26
PhysicStud01 said:
so, this is called moments?
Google moment of mass.
 
  • #27
haruspex said:
Google moment of mass.
OK. i tried and obtained moment of inetia. is this the same as moment of mass. but the formulae there seems complicated for this level of question though
 
  • #28
PhysicStud01 said:
OK. i tried and obtained moment of inetia. is this the same as moment of mass. but the formulae there seems complicated for this level of question though
Moment of inertia is also known as the second moment of mass. That's because it involves an integral of the form ##\int r^2.dm##. For finding mass centre use the first moment of mass, which has the form ##\int r.dm##.
 
  • #29
I don't still quite understand how the centripetal forces are equal? Aren't the radii and the masses different?
 
  • #30
Anyone?
 
  • #31
Angie Tom said:
I don't still quite understand how the centripetal forces are equal? Aren't the radii and the masses different?
In this system the centripetal forces are provided by gravity. No other forces are acting. Can the force due to gravity acting on the two bodies be different?
 
  • #32
Yes, if the masses and radii are different
 
  • #33
gneill said:
In this system the centripetal forces are provided by gravity. No other forces are acting. Can the force due to gravity acting on the two bodies be different?
yes, if the masses and radii are different
 
  • #34
Angie Tom said:
yes, if the masses and radii are different
No. Review Newtonian gravity. There is only one distance that matters, and that is the separation between the centers of the two spherical bodies.
 
  • #35
gneill said:
No. Review Newtonian gravity. There is only one distance that matters, and that is the separation between the centers of the two spherical bodies.
I know the law! attractive force between any 2 point masses is directly proportional to the product of the masses and inversely proportional to the distance squared between them right?
Can u please clarify what exactly is point c
 

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