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

AI Thread Summary
The discussion revolves around calculating the period of rotation for three stars, each with the mass of the sun, arranged in an equilateral triangle with sides of 1E12 meters. The initial calculations yielded a period of 5.45E8 seconds, but further analysis revealed the need to consider the correct radius for rotation, which is not the same as the triangle's side length. After summing the forces and adjusting for vector components, the correct acceleration was determined to be 2.3E-4 m/s². This led to a revised period of rotation of approximately 3.15E8 seconds. The importance of accurately accounting for the geometry and dynamics of the system was emphasized in the calculations.
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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