Question

Three stars, each with the mass of our sun, form an equilateral triangle with sides 1.0 x 10¹² m 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?

Short answer

The period of rotation of stars is 8.6 years

Step-by-step solution

Given information

Three stars are at vertex of an equilateral triangle at a distance of 1x10¹²m.
This equilateral triangle fits the orbit of Jupiter.
Mass of each star = mass of sun, m=1.99 x 10³⁰ kg

Explanation

Lets draw a diagram to understand and solve the problem as below

Lets first find r from the figure using simple trigonometric ratio

$$\begin{gathered} r=\frac{L}{2}\cos 30° \\ r=\frac{1.0\times {10}^{12}m}{2}\cos 30° \\ r=4.3\times {10}^{11}\mathrm{m} \end{gathered}$$

Find the gravitational force between two stars as

$$\begin{gathered} F_g=\frac{G{m}^2}{L^2} \\ {F}_g=\frac{\left(6.67\times {10}^{-11}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2\right)\times {\left(1.99\times {10}^{30}\mathrm{kg}\right)}^2}{{\left(1\times {10}^{12}\mathrm{m}\right)}^2} \\ {F}_g=2.64\times {10}^{26}\mathrm{N} \end{gathered}$$

Find the component towards center using simple trigonometry

$$\begin{gathered} F_{\text{center}}=2F_g\cos 30° \\ {F}_{\text{center}}=2\times \left(2.64\times {10}^{26}\mathrm{N}\right)\cos 30° \\ {F}_{\text{center}}=4.57\times {10}^{26}\mathrm{N}.............................\left(1\right) \end{gathered}$$

This force is equal to centripetal force

Centripetal force is given by

$F_c=mr{\omega }^2=m\times r\times {\left(\frac{2\pi }{T}\right)}^2..........................\left(2\right)$

Equate (1) and (2) and Solve for T, we get

$$\begin{gathered} F_{\text{center}}=F_c \\ T=\sqrt{\frac{4{\pi }^{2}mR}{4.57\times {10}^{26}}} \\ T=\sqrt{\frac{4{\pi }^2\times \left(1.99\times {10}^{30}\right)\times \left(4.3\times {10}^{11}\right)}{4.57\times {10}^{26}}} \\ T=2.72\times {10}^8\mathrm{s}\times \frac{1\mathrm{hr}}{3600\mathrm{s}}\times \frac{1\text{day}}{24\mathrm{hr}}\times \frac{1\text{year}}{365\text{days}} \\ T=8.6\text{years} \end{gathered}$$

So period is 8.6 years