91Ó°ÊÓ

Figure shows three stars, each of mass m. which move in the plane of the page along a circle of radius r.

It's easier to label some of the lengths involved in the geometry of this three-body gravitational system by drawing a figure. Due to the gravitational attraction of the other two masses, each mass is moving in a circular motion. The mass at the lower right will be considered, and the Centre of mass of the other two masses (point P) will impose a gravitational pull force (F) on the mass at the lower right.

In order to obtain distance d as a function of radius of path r, employ the law of cosines (considering the triangle formed by the upper mass, the mass at the lower right, and the Centre of mass of the whole system, which happens to be the same as the geometric Centre of the figure).

$\begin{array}{l}d=\sqrt{2r^2-2\cos \left({120}^{∘}\right)r^2}\\ d=\sqrt{2-2\left(-\frac{1}{22}\right)rn}\\ d=\sqrt{3rn}\end{array}$

After getting the value of d, apply the Pythagorean theorem to get I as a function of r . This time, look at the triangle created by the point p, the higher mass, and the bottom right mass.

$$\begin{gathered} I=\sqrt{d^2-{\left(\frac{d}{2}\right)}^2} \\ I=\sqrt{3r^2-\frac{3}{4}{r}^2} \\ I=\frac{3}{2}r \end{gathered}$$

It is to note that d expression is used in terms of r.

Write Newton's second law, in which use Newton's rule of universal gravity to substitute F for the relevant expression.

In this statement, the upper mass's Centre of mass and the mass at the lower left, on the one hand, and the mass at the right, on the other hand, are taken into account (as if we were considering two objects that attract each other).

In addition, the angular speed(omega)has been given in terms of the period (T)

Finally, solve for (T) in this equation.

Equating force exerted on mass and centripetal force

$$\begin{gathered} F=m{a}_c \\ \frac{G\left(2m\right)\left(m\right)}{I^2}=m\left({\mathrm{Ó\neg }}^{2}r\right)n\frac{2Gm}{I^2}{\mathrm{Ó\neg }}^{2}r \\ \mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{T} \\ \frac{2Gm}{I^2}=\frac{4{\mathrm{Ï€}}^2\mathrm{r}}{T^2} \\ {T}^2=\frac{2{\mathrm{Ï€}}^2{\mathrm{I}}^2\mathrm{r}}{Gm} \\ I=\frac{3}{2}r \end{gathered}$$

localid="1668602466972" $$\begin{gathered} T=\sqrt{\frac{2{\mathrm{Ï€}}^2{\left({\displaystyle \frac{3}{2}}\mathrm{r}\right)}^2\mathrm{r}}{2Gm}} \\ T=\sqrt{\frac{9{\mathrm{Ï€}}^2{\mathrm{r}}^3}{2Gm}} \\ T=3\mathrm{Ï€}\sqrt{\frac{{\mathrm{r}}^3}{2Gm}} \end{gathered}$$

Therefore, the expression obtained in the preceding step has been substituted byI.

So, the duration that the system takes to make one complete revolution is $3\mathrm{Ï€}\sqrt{\frac{{\mathrm{r}}^3}{2\mathrm{Gm}}}$.