What is the period of rotation for three stars forming an equilateral triangle?

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Homework Help Overview

The discussion revolves around calculating the period of rotation for three stars of solar mass forming an equilateral triangle with given side lengths. The problem involves gravitational forces and rotational dynamics.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the gravitational forces acting on the stars and how to account for the mass of all three stars in the calculations. There are questions about the setup and whether the radius of rotation differs from the triangle's side length.

Discussion Status

Participants are actively engaging with the problem, sharing calculations and questioning assumptions about the geometry involved. Some guidance has been offered regarding the need to consider all masses and the correct interpretation of the radius of rotation.

Contextual Notes

There is a specific length given for the sides of the triangle, and participants are working under the constraints of the problem's parameters, including the assumption of an equilateral triangle and the need for rotational dynamics to prevent the stars from collapsing into one another.

DRC12
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Homework Statement



Three stars, each with the mass of our sun (1.9891E31 kg), form an equilateral triangle with sides long. The triangle has to rotate, because otherwise the stars would crash together in the center. What is the period of rotation?

Homework Equations


F1=GM2/r2
ƩF=0
F2=ma
a=4π2r/T2

The Attempt at a Solution


F1=2.639E26
Because the sum of the forces is equal to zero F1=F2
2.639E26=(1.9891E31)*a
a=.000133m/s2=4π2/T2
T=5.45E8 s

Do I have to do something else because there's three stars instead of two?
 
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What is the length of the sides of the triangle?

It's difficult to judge what calculations you're performing when you only show results; show more of your work.

Yes, you need to take all the masses into account. Think summation of vectors.
 
The length of the sides of the triangle is 1E12m.
F1=(6.67E-11)(1.9891E30kg)/(1E12)2=2.639E26
Because it's a equilateral triangle all angles are 60 so I when summing the vectors I made one side at 0o and the other 60o above it so the coordinates of the first side is Fx=2.639E26 Fy=0 and the coordinates of the other side is Fx=2.639E26 cos(60)=1.32E26 Fy=2.639E26 sin(60)=2.29E26.

the total Fx=3.959E26 and Fy=2.29E26 F=(Fx2+Fy2).5=4.57E26

a=F/m so a=4.57E26/1.9891E30=2.3E-4
a=4π2r/T2
2.3E-4=4π2*1E12/T2
T=4.14E8

The correct answer: 3.15E8
 
DRC12 said:
The length of the sides of the triangle is 1E12m.
F1=(6.67E-11)(1.9891E30kg)/(1E12)2=2.639E26
Because it's a equilateral triangle all angles are 60 so I when summing the vectors I made one side at 0o and the other 60o above it so the coordinates of the first side is Fx=2.639E26 Fy=0 and the coordinates of the other side is Fx=2.639E26 cos(60)=1.32E26 Fy=2.639E26 sin(60)=2.29E26.

the total Fx=3.959E26 and Fy=2.29E26 F=(Fx2+Fy2).5=4.57E26

a=F/m so a=4.57E26/1.9891E30=2.3E-4
a=4π2r/T2
2.3E-4=4π2*1E12/T2
T=4.14E8

The correct answer: 3.15E8
Ah. Well the radius of the circle around which the stars are rotating is not the same as the length of the triangle sides.
 
Yes that was the problem thanks for the help
 

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