Newton's law and double stars moving in circles

In summary: You have a misunderstanding here. When you have motion in a plane angular velocity can be taken to be simply a scalar quantity, usually positive if anticlockwise and negative if clockwise.In this case, you don't need to deal with vectors at all.This is similar to the case of 1D linear motion, where velocity is taken to be a signed scalar and not a 2D or 3D vector.The more general case, where motion is not necessarily in a plane, must use the full vector definition of angular velocity and acceleration.My advice in this case is to draw a diagram of the motion, showing the planets and their common center of mass, then use this to
  • #1

Felipe Lincoln

Gold Member
99
11

Homework Statement


The two components of a double star are observed to move in circles of radii ##r_1## and ##r_2##. What is the ratio of their masses? (Hint: Write down their accelerations in terms of the angular velocity of rotation ##\omega##)

Homework Equations


##m\ddot{\vec{x}}= \vec{F}_i##
Being ##\vec{F}_i## all the interaction forces acting on body ##i##
##v=\omega r##

The Attempt at a Solution


##m_i\ddot{\vec{x}_i}= \vec{F}_{ij}##
##m_j\ddot{\vec{x}_j}= \vec{F}_{ji}##
##\vec{F}_{ij}=-\vec{F}_{ji}\implies m_i\ddot{\vec{x}_i}+m_j\ddot{\vec{x}_j}=0##
##m_i\dot{\vec{\omega_i}}r_i+m_j\dot{\vec{\omega_j}}r_j=0##
I guess ##\dot{\vec{\omega_j}} = \dot{\vec{\omega_i}}## but can't argue properly why. It just feels that since their movement depend on each other and there's no reason to have different angular acceleration, I don't know..
But this way we conclude that ##\dfrac{m_i}{m_j}=-\dfrac{r_j}{r_i}##
I don't think these scalars ##r_i## and ##r_j## can be negative. What's wrong with my resolution?
 
Physics news on Phys.org
  • #2
What is the only kind of acceleration present when a body is moving in circles?
 
  • #3
centripetal acceleration, given by ##v^2/r## so it turns to ##\omega^2r## but again, I'll cut the ##\omega^2## and will get the same result
 
  • #4
Felipe Lincoln said:
##\vec{F}_{ij}=-\vec{F}_{ji}\implies m_i\ddot{\vec{x}_i}+m_j\ddot{\vec{x}_j}=0##
##m_i\dot{\vec{\omega_i}}r_i+m_j\dot{\vec{\omega_j}}r_j=0##
The second equation above does not follow from the first. In vectors, ##\ddot {\vec x}=\vec r\times\vec{\alpha}##.
Since the two r vectors are in opposite directions, that will lead to opposite signs when you switch to scalars.
 
  • Like
Likes Felipe Lincoln
  • #5
Oh I can see, you're right.
But what argument can I use to state that ##\vec{\alpha}_i=\vec{\alpha}_j?##
Other question is: how you came to this relation ##\ddot {\vec x}=\vec r\times\vec{\alpha}##. I though the relation was ##\ddot {\vec x}=r\cdot\vec{\alpha}## and ##r## is any real number. Is it right to think of it as a scalar product? How you came to the cross product relation?
Thank you!
 
  • #6
If we can assume that each of the two stars does uniform circular motion, with angular velocities ##\omega_1## and ##\omega_2## then ##\omega_1=\omega_2## because if they were different, their center of mass would change through time, as they rotate, which means that there would be an external force acting on the system.
 
  • Like
Likes Felipe Lincoln
  • #7
Felipe Lincoln said:
Oh I can see, you're right.
But what argument can I use to state that ##\vec{\alpha}_i=\vec{\alpha}_j?##
Other question is: how you came to this relation ##\ddot {\vec x}=\vec r\times\vec{\alpha}##. I though the relation was ##\ddot {\vec x}=r\cdot\vec{\alpha}## and ##r## is any real number. Is it right to think of it as a scalar product? How you came to the cross product relation?
Thank you!
Actually, what I posted before is not right either... more later.
 
  • Like
Likes Delta2
  • #8
Felipe Lincoln said:
Oh I can see, you're right.
But what argument can I use to state that ##\vec{\alpha}_i=\vec{\alpha}_j?##
Other question is: how you came to this relation ##\ddot {\vec x}=\vec r\times\vec{\alpha}##. I though the relation was ##\ddot {\vec x}=r\cdot\vec{\alpha}## and ##r## is any real number. Is it right to think of it as a scalar product? How you came to the cross product relation?
Thank you!

You have a misunderstanding here. When you have motion in a plane angular velocity can be taken to be simply a scalar quantity, usually positive if anticlockwise and negative if clockwise.

In this case, you don't need to deal with vectors at all.

This is similar to the case of 1D linear motion, where velocity is taken to be a signed scalar and not a 2D or 3D vector.

The more general case, where motion is not necessarily in a plane, must use the full vector definition of angular velocity and acceleration.

My advice in this case is to draw a diagram of the motion, showing the planets and their common centre of mass, then use this to simplify the mathematics as much as possible.
 
  • Like
Likes Felipe Lincoln
  • #9
Delta² said:
If we can assume that each of the two stars does uniform circular motion, with angular velocities ##\omega_1## and ##\omega_2## then ##\omega_1=\omega_2## because if they were different, their center of mass would change through time, as they rotate, which means that there would be an external force acting on the system.
Why can't their center of mass remain the same but their angular velocities be different? Is it only dependent of the center of mass? I still don't get it.

PeroK said:
You have a misunderstanding here. When you have motion in a plane angular velocity can be taken to be simply a scalar quantity, usually positive if anticlockwise and negative if clockwise.

In this case, you don't need to deal with vectors at all.

This is similar to the case of 1D linear motion, where velocity is taken to be a signed scalar and not a 2D or 3D vector.

The more general case, where motion is not necessarily in a plane, must use the full vector definition of angular velocity and acceleration.

My advice in this case is to draw a diagram of the motion, showing the planets and their common centre of mass, then use this to simplify the mathematics as much as possible.
Alright!
So to make the scalar treatment should I work only with modules? Starting by ##||m_i\ddot{\vec{x}_i}|| = ||m_j\ddot{\vec{x}_j}||##
Another question about the dynamics of the system. Are these two bodies always in opposite side?
For me it's being though to understand how the bodies behaves, I just know that they attract with other by gravitational forces but it's not easy to visualize how is their movements, I first thought in working with vector wondering if it would let me understand the problem better.
 
  • #10
The two stars are rotating around the centre of mass of the system, in such a way that the two stars and the centre of mass , which is also the centre of rotation are always in the same line (three points that are always in the same line, so the distance between the two stars is always ##r_1+r_2##).

If the angular velocities were different, then the way i see it , the stars would get closer at some times and farther at some other times (their distance wouldn't be always ##r_1+r_2## ), hence their c.o.m would change position. IF the c.o.m of a system changes position it usually means that there are external forces acting on the system. But in this case we assume that the only forces are the internal gravitational forces between the two stars.
 
Last edited:
  • Like
Likes CWatters and Felipe Lincoln
  • #11
And yes work, with modules and with centripetal accelerations, the tangential accelerations are zero anyway cause we assume that they do uniform circular motion.
 
  • #12
Felipe Lincoln said:
Why can't their center of mass remain the same but their angular velocities be different? Is it only dependent of the center of mass? I still don't get it.

This is the purpose of the diagram. Draw the stars in their starting position, and the centre of mass. Then draw their positions some time later. If the centre of mass doesn't move, then where are the stars in relation to each other?

This should help you understand the motion.
 
  • Like
Likes Felipe Lincoln
  • #13
Now that I think of it, this problem proves to be more challenging. Suppose we are given that:

The two stars are rotating in circles of radius ##r_1## and ##r_2## and the only forces acting in the system are the internal gravitational forces.

and we are asked to prove that

1)Their centre of rotation is common and it is their centre of mass and
2)Their motion is uniform circular motion
3)Their angular velocities are equal,##\omega_1=\omega_2 ##
4)The radii ##r_1## and ##r_2## are inversely proportional to their masses ##m_1## and ##m_2##.

I am not sure if this exercise asks only for 4) or for 1 to 3 as well.
 
  • Like
Likes Felipe Lincoln
  • #14
Delta² said:
Now that I think of it, this problem proves to be more challenging. Suppose we are given that:

The two stars are rotating in circles of radius ##r_1## and ##r_2## and the only forces acting in the system are the internal gravitational forces.

and we are asked to prove that

1)Their centre of rotation is common and it is their centre of mass and
2)Their motion is uniform circular motion
3)Their angular velocities are equal,##\omega_1=\omega_2 ##
4)The radii ##r_1## and ##r_2## are inversely proportional to their masses ##m_1## and ##m_2##.

I am not sure if this exercise asks only for 4) or for 1 to 3 as well.

I think it's safe to assume 1), that in general bodies orbit the common centre of mass.

We are given circular orbits. Uniform circular motion follows simply from this.
 
  • Like
Likes Delta2
  • #15
PeroK said:
I think it's safe to assume 1), that in general bodies orbit the common centre of mass.

We are given circular orbits. Uniform circular motion follows simply from this.
Ok well, if we assume 1 , I can see now that 2 and 3 follow easily (for example for 3, if at the same time t they swept different angles ##\phi_1\neq\phi_2##, then their center of mass wouldn't be their common centre of rotation, that is their centre of rotation wouldn't lie at the line segment that connects the two bodies). BUT what is a proof for 1 using Newton's laws and/or conservation of energy/momentum?
 
  • #16
Delta² said:
Ok well, if we assume 1 , I can see now that 2 and 3 follow easily (for example for 3, if at the same time t they swept different angles ##\phi_1\neq\phi_2##, then their center of mass wouldn't be their common centre of rotation, that is their centre of rotation wouldn't lie at the line segment that connects the two bodies). BUT what is a proof for 1 using Newton's laws and/or conservation of energy/momentum?

That's a general problem, which can be solved by using the concept of reduced mass.

I can't see that we would need to prove that for this problem. It looks natural to assume we know this general result.
 

1. What is Newton's Law of Universal Gravitation?

Newton's Law of Universal Gravitation states that every object in the universe attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

2. How does Newton's Law apply to double stars moving in circles?

In the case of double stars, each star exerts a gravitational force on the other, causing them to orbit around their center of mass. This movement is governed by Newton's Law of Universal Gravitation.

3. Can Newton's Law explain the circular motion of double stars?

Yes, Newton's Law explains the circular motion of double stars by stating that the gravitational force between the stars is always directed towards their center of mass, causing them to move in a circular path.

4. What other factors affect the motion of double stars in circles?

Besides gravitational force, the mass and distance between the two stars also affect the motion of double stars. The more massive the stars and the closer they are to each other, the faster they will orbit.

5. Is Newton's Law applicable to all celestial bodies?

Yes, Newton's Law applies to all celestial bodies, including double stars. It is a fundamental law of physics that governs the motion of objects in the universe.

Suggested for: Newton's law and double stars moving in circles

Replies
5
Views
245
Replies
3
Views
610
Replies
15
Views
1K
Replies
15
Views
1K
Replies
8
Views
1K
Replies
3
Views
592
Back
Top