Question

Three stars, each with the mass of our sun, form an equilateral triangle with sides 1.0×1012m long. (This triangle would just about fit within the orbit of Jupiter.) The triangle has to rotate, because otherwise the stars would crash together in the center. What is the period of rotation?

Answer 1

Answer:

period of rotation is 9.9843 years

Explanation:

Given data

sides  = 1.0×10^12 m

to find out

period of rotation

solution

first we use the gravititional formula i.e

F(g) = 2 F cos 30

here F = G  × mass(sun)² / radius²

F = 6.67 × $10^{-11}$ × (1.99× $10^{30}$)² / (1× $10^{30}$)²

so

F(g) = 2  ×  6.67 × $10^{-11}$ × (1.99× $10^{30}$)² / (1× $10^{30}$)²  ×  (1.73/2)

F(g) = 4.57 ×  $10^{26}$ N

and

by the law of sines

r/sin30 = s/sin120

so r will be

r = 1.0×10^12  (0.5/.87)

r = 5.75 ×  $10^{11}$ m

we know that gravititional force = centripetal force

so  here

centripetal force = mv² /r

centripetal force = 1.99× $10^{30}$ v² / 5.75 ×  $10^{11}$

so

4.57 ×  $10^{26}$  = mv² /r

4.57 ×  $10^{26}$  = 1.99× $10^{30}$ v² / 5.75 ×  $10^{11}$

v² = 4.57 ×  $10^{26}$ × 5.75 ×  $10^{11} / 1.99× [tex]10^{30}$

and

v = 11480 m/sec

and we know period of rotation formula

t = d /v

t = 2 $\pi$ r /v

t = 2 × $\pi$ ×  5.75 ×  [tex]10^{11} / 11480

t = 314 × [tex]10^{5} / 31536000 year

t = 9.9843 year

so period of rotation is 9.9843 years