Angular Velocity for Constant Separation Distance

Click For Summary

Discussion Overview

The discussion centers on determining the required angular velocity $\omega$ for two equal masses to maintain a constant separation distance $r_0$. The context involves concepts from gravitational physics and rotational dynamics, with implications for binary systems and orbital mechanics.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant suggests balancing potential energy due to gravity with rotational kinetic energy to find the angular velocity for two equal masses rotating around each other.
  • Another participant emphasizes the need for the masses to rotate fast enough to avoid being pulled together by gravity while not rotating so fast that they fly apart.
  • A participant proposes using the gravitational force equation and centripetal acceleration to derive the necessary conditions for maintaining constant separation.
  • There is a question regarding whether the position vector should be defined with respect to the center of gravity of the two bodies, with a suggestion that the bodies orbit this center in a uniform circle.
  • Another participant notes that the choice of reference frame, such as the center of gravity or one of the bodies, should not affect the outcome of the angular velocity calculation.

Areas of Agreement / Disagreement

Participants express varying perspectives on the reference frame and the approach to calculating angular velocity, indicating that multiple competing views remain without a consensus on the best method or interpretation.

Contextual Notes

There are unresolved assumptions regarding the definitions of forces and reference frames, as well as the specific conditions under which the angular velocity is to be calculated.

Dustinsfl
Messages
2,217
Reaction score
5
Determine the required angular velocity $\omega$ of the relative position vector in order to maintain a constant separation distance $r_0$ for two equal masses.

What equations are needed for this?
 
Mathematics news on Phys.org
To me it seems as if you have to balance the potential energy due to gravity against the rotational kinetic energy. That is, it seems as if you're in outer space with two equal masses, and they are rotating around each other like a binary star system. If the distance between them is constant, how fast are they rotating around each other?
 
Ackbach said:
To me it seems as if you have to balance the potential energy due to gravity against the rotational kinetic energy. That is, it seems as if you're in outer space with two equal masses, and they are rotating around each other like a binary star system. If the distance between them is constant, how fast are they rotating around each other?

Fast enough to not be pulled in by the others gravity and a slow enough that they don't fly off.
 
dwsmith said:
Fast enough to not be pulled in by the others gravity and a slow enough that they don't fly off.

Exactly. So I think you'd have the force due to gravity being

$$|F_{g}|=\frac{m^{2}G}{r_{0}^{2}},$$

and the centripetal acceleration being

$$a_{c}=\frac{v_{\tan}^{2}}{r_{0}}=\frac{F_{c}}{m}.$$

Also note that $v_{\tan}=\omega \,r_{0}$.
 
Is this question talking about the position vector for the center of gravity of the two bodies?

If we set the origin to be the center of gravity, the bodies orbit it in a uniform circle with radius 1/2r_0.
I can just find the angular velocity on this circle?
 
Last edited:
dwsmith said:
Is this question talking about the position vector for the center of gravity of the two bodies?

If we set the origin to be the center of gravity, the bodies orbit it in a uniform circle with radius 1/2r_0.
I can just find the angular velocity on this circle?

It shouldn't matter. You could choose your frame of reference to be at the center of one of the bodies just as well.
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
26
Views
2K
  • · Replies 20 ·
Replies
20
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
3
Views
877
Replies
1
Views
839
  • · Replies 39 ·
2
Replies
39
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K