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

In summary, the conversation discusses a problem involving three stars with the mass of our sun forming an equilateral triangle with sides 1E12m long. The objective is to calculate the period of rotation for the triangle. Using the equations F1=GM2/r2, ƩF=0, F2=ma, and a=4π2r/T2, the solution is found to be 3.15E8 s. However, some errors were made due to not considering all three masses and using the incorrect radius for the circle of rotation. After correcting for these errors, the correct answer is found to be 3.15E8 s.
  • #1
DRC12
41
0

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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  • #2
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.
 
  • #3
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
 
  • #4
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.
 
  • #5
Yes that was the problem thanks for the help
 

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

1. What is the period of stars rotating?

The period of stars rotating is the time it takes for a star to complete one full rotation on its axis.

2. How is the period of stars rotating measured?

The period of stars rotating is measured using a technique called spectroscopy, which analyzes the changes in a star's light spectrum over time.

3. What factors affect the period of stars rotating?

The period of stars rotating can be affected by the mass, size, and composition of the star, as well as external factors such as gravitational forces from nearby objects.

4. Why is studying the period of stars rotating important?

Studying the period of stars rotating can provide insight into a star's internal structure and evolution. It can also help scientists understand the dynamics of star formation and the formation of planetary systems.

5. Can the period of stars rotating change over time?

Yes, the period of stars rotating can change over time due to various factors such as stellar interactions, changes in the star's mass, and internal processes such as nuclear fusion. However, these changes are typically gradual and may not be noticeable within a human lifespan.

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