# Finding the period of rotation of this system....

• Vitani11
In summary: I'm so sorry!In summary, the period of rotation for the three stars is different due to the different masses.
Vitani11

## Homework Statement

Three identical stars, each with mass m, form the verticies of an equilateral triangle with side length d and rotate in a circular orbit due to their mutual gravitation. What is the period τ of their rotation?

I set up the FBD for each star and am now trying to figure out what to do from there. I know how to get period into the equation from circular motion, but is that all there is to it? The FBD gives different equations for each mass so when I solve I get different periods for each star. What should I be doing here?

Force equations

Star at bottom left point:

x: Fcosθ+F = max
y: Fsinθ = may

Star at bottom right:

x: -(Fcosθ+F) = max
y: Fsinθ = may

Star at top:

y: -2F = may (or should this be -2Fcos(θ/2)?)
x: forces cancel to 0

If you draw an equilateral triangle you will see what I'm talking about and then how I oriented my axes around the stars.

When I solve for period my units are seconds - so I know my method is correct. Anyway I would solve each of the force equations above for period and compare because they should be the same (or at least what I have been thinking) and they are off by a factor due to the cosθ or sinθ

Instead of working with x and y components of force, think about centripetal components of forces. Use what you know about circular motion.

Vitani11
I should neglect the force that each is exerting on each other in x or y and only focus on the forces going towards the center of the system (triangle)? I thought about that but I was afraid to try it.

Okay I think this might be right. Using the centripetal components I was able to solve for period using the star at the top which my answer was T = (2πd3/2)/(√(Gm))(31/4). Units check out.

(2Gm2cos(θ/2))/d2 = mv2/d this was my equation from F = ma.

Is the orbital radius equal to d?

Vitani11
Vitani11 said:
Okay I think this might be right. Using the centripetal components I was able to solve for period using the star at the top which my answer was T = (2πd3/2)/(√(Gm))(31/4). Units check out.

(2Gm2cos(θ/2))/d2 = mv2/d this was my equation from F = ma.

Vitani11
dsinθ-dsinθ/2 is the distance? Here is the work. Sorry for the blur.

#### Attachments

• IMG_20170112_165357.jpg
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OH MY GOD I DID IT THANK YOU haruspex as usual and Tsny . No dsinθ -dsinθ/2 was definitely not it

## What is the period of rotation?

The period of rotation is the time it takes for an object to complete one full rotation on its axis.

## How is the period of rotation measured?

The period of rotation is typically measured in units of time, such as seconds, minutes, or hours.

## What factors can affect the period of rotation?

The period of rotation can be affected by the mass, shape, and distance from other objects in the system.

## Can the period of rotation change over time?

Yes, the period of rotation can change due to external forces, such as gravity or collisions with other objects.

## Why is finding the period of rotation important?

Finding the period of rotation can help us understand the motion and behavior of objects in a system, and can also provide valuable information for predicting future movements and events.

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