Question

Plaskett’s binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This statement implies that the masses of the two stars are equal (Fig. P13.21). Assume the orbital speed of each star is $\left|\overline{v}\right|\mathbf{=}\mathbf{220}\mathit{k}\mathit{m}\mathbf{/}\mathit{s}$ and the orbital period of each is 14.4 days. Find the mass M of each star. (For comparison, the mass of our Sun is 1.9931030 kg.)

Short answer

Thus the required mass of each star is$M=3.16\times {10}^{-9}kg$

Step-by-step solution

Step 1:

It is important to know the gravitational force because it is the force that allows orbiting to exist. A central body exerts a gravitational force on the orbiting body to keep in it orbit. Centripetal force is also important, as this is the force responsible for circular motion.

Step 2:

LetorbitalspeedofstarisVandTimeperiodisT,

It is important to know the gravitational force because it is the force that allows orbiting to exist. A central body exerts a gravitational force on the orbiting body to keep in it orbit. Centripetal force is also important, as this is the force responsible for circular motion.

We know that, for circular orbit net force toward the center will be equals to centripetal force,

$$\begin{gathered} \frac{GMm}{r^2}=\frac{m{v}^2}{r} \\ v=\sqrt{\frac{GM}{4R}} \\ {v}^3=\frac{{\left(GM\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{8R^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}.....................\left(1\right) \end{gathered}$$

Now we know that time period is given by,

$$\begin{gathered} T=\frac{2\mathrm{\pi R}}{v} \\ T=\frac{2\mathrm{\pi }\times 2\times {\mathrm{R}}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{\sqrt{GM}} \\ T=\frac{4{\mathrm{\pi R}}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{\sqrt{GM}}........\left(2\right) \end{gathered}$$

From (1) and (2), we get

$$\begin{gathered} T=\frac{4\pi GM}{8v^3} \\ M=\frac{8v^{3}T}{4\pi G} \\ M=\frac{2v^{3}T}{\pi G} \end{gathered}$$

Now we substitute the values of V and T, we get

$$\begin{gathered} M=\frac{2\times {\left(220\times {10}^3\right)}^3\times 14.4\times 24\times 3600}{3.14\times 6.67\times {10}^{-11}} \\ M=3.16\times {10}^{9}kg \end{gathered}$$