Question

Three identical stars of massMform an equilateral triangle that rotates around the triangle’s center as the stars move in a common circle about that center. The triangle has edge lengthL. What is the speed of the stars?

Short answer

The speed of the star is $\mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{L}}}$.

Step-by-step solution

Step 1: Given

Three identical stars of mass M form an equilateral triangle that rotates around the triangle’s center as the stars move in a common circle about the center.

Determining the concept

Find the speed of the planet by equating the gravitational force of attraction and centripetal force in a three-star system and putting the radius of orbit using the concept of center of mass. Gravitational force is the force exerted by the Earth towards it.

Formulae are as follows:

${\mathrm{F}}_{\mathrm{c}}=\frac{{\mathrm{Mv}}^2}{\mathrm{r}}$

${\mathrm{F}}_{\mathrm{g}}=\frac{\mathrm{GMm}}{{\mathrm{r}}^2}$

where F is force, G is gravitational constant, M and m are masses, v is velocity and r is the radius.

Determining the speed of the star

The gravitational force acting on each star due to the other two stars is,

$\begin{array}{c}\mathrm{F}=\frac{\mathrm{GMm}}{{\mathrm{L}}^2}\cos 30°+\frac{\mathrm{GMm}}{{\mathrm{L}}^2}\cos 30°\\ =\frac{2\mathrm{GMm}}{{\mathrm{L}}^2}\cos 30°\end{array}$

All stars rotate about the center of mass of the system. From the figure, write the coordinates of the center of mass as,

$\begin{array}{c}{\mathrm{x}}_{\mathrm{c}}=\frac{0+\mathrm{L}+\frac{\mathrm{L}}{2}}{3}\\ =\frac{\mathrm{L}}{2}\end{array}$

$\begin{array}{c}{\mathrm{y}}_{\mathrm{c}}=\frac{0+0+\frac{\sqrt{3\mathrm{L}}}{2}}{3}\\ =\frac{\mathrm{L}}{2\surd 3}\end{array}$

The distance between the star and the center of mass is,

$\begin{array}{l}\mathrm{R}=\sqrt{{{\mathrm{x}}_{\mathrm{c}}}^2+{{\mathrm{y}}_{\mathrm{c}}}^2}\\ \mathrm{R}=\sqrt{{\frac{\mathrm{L}}{2}}^2+{\left(\frac{\mathrm{L}}{2\sqrt{3}}\right)}^2}\\ \mathrm{R}=\frac{\mathrm{L}}{\sqrt{3}}\end{array}$

This gravitational force provides centripetal acceleration to the stars to orbit in the circle. So,

$\begin{array}{c}\frac{2{\mathrm{GM}}^2}{{\mathrm{L}}^2}\cos 30°=\frac{{\mathrm{Mv}}^2}{\mathrm{R}}\\ \frac{2{\mathrm{GM}}^2}{{\mathrm{L}}^2}\left(\frac{\sqrt{3}}{2}\right)=\frac{{\mathrm{Mv}}^2}{\frac{\mathrm{L}}{\sqrt{3}}}\\ \frac{\mathrm{GM}}{{\mathrm{L}}^2}=\frac{{\mathrm{v}}^2}{\mathrm{L}}\\ \mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{L}}}\end{array}$

Hence, the speed of the star is $\mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{L}}}$.

Therefore, using the formulae for the gravitational force of attraction between two objects and the centripetal force, the velocity of one of the objects can be found.