# What is their common angular velocity?

Sir,
In a double star system, 2 stars of masses m and M separated by a distance d rotate about their common centre of mass. What is their common angular velocity?

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siddharth
Homework Helper
Gold Member
Amith, you've been here long enough to know that you need to show your work before you get some help. What are your thoughts on this problem? Any ideas on how to solve it?

Amith2006 said:
Sir,
In a double star system, 2 stars of masses m and M separated by a distance d rotate about their common centre of mass. What is their common angular velocity?
I solved it in the following way:
The gravitational force of attraction between the masses = (GMm)/d^2
This force is balanced by the centrifugal force of (M + m)dw^2 acting on the combined system.
i.e. GMm/d^2 = (M + m)dw^2 [ here w is the angular velocity of the system]
By solving we get,
w = {(GMm)/((M + m)d^3)}^(1/2)
But the answer given in my book is {(GMm)/d^3}^(1/2).
Here the symbol ^ represents power.

Hootenanny
Staff Emeritus
Gold Member
Amith2006 said:
I solved it in the following way:
The gravitational force of attraction between the masses = (GMm)/d^2
This force is balanced by the centrifugal force of (M + m)dw^2 acting on the combined system.
i.e. GMm/d^2 = (M + m)dw^2 [ here w is the angular velocity of the system]
By solving we get,
w = {(GMm)/((M + m)d^3)}^(1/2)
But the answer given in my book is {(GMm)/d^3}^(1/2).
Here the symbol ^ represents power.
Centrifugal force! Sorry thats my pet hate. There is no such force as the centrifugal force. If the forces are balanced, why would the system be undergoing circular motion?

You were correct in stating that the force between them is;

$$F = \frac{GMm}{d^2}$$

Now what is the equation for the centripetal acceleration?

~H

Last edited:
Curious3141
Homework Helper
Hootenanny said:
Centrifugal force! Sorry thats my pet hate. There is no such force as the centrifugal force.
I myself sometimes invoke the centrifugal force in explanations. but I am always careful to clarify that these are non-Newtonian inertial "forces" which would vanish if the problem is considered from a proper inertial perspective. In any case, I understand what I'm doing - I appreciate the D'Alembert formulation connecting the inertial/non-inertial perspectives.

Hootenanny is absolutely right, at a student level, there is no place for centrifugal force or other fictitious forces in answers to exam questions.

The answer in your textbook is definitely incorrect as the dimensions do not match.

mukundpa
Homework Helper
First of all tell me about the center of the circular path(s).