Star Waltz

Let,
$r_1 =$ The distance of $3M$ from the Center of Mass
$r_2 =$ The distance of $M$ from the Center of Mass

It is clear that,
$r_1+r_2=d$
$3Mr_1=Mr_2$

$\Rightarrow r_1=\frac{d}{4}$ and $r_2=\frac{3d}{4}$

Also, writing the Force equation for $3M$ ,
$\frac{G(M)(3M)}{d^2} = \frac{(3M)v_1^2}{r_1}$

Using the derived equations,
$v_1=\sqrt{\frac{GM}{4d}}$

Thus,
$w = \frac{v_1}{r_1} = \sqrt{\frac{4GM}{d^3}}$

We know,
$T=\frac{2\pi}{w} = \pi\sqrt{\frac{d^3}{GM}}$

First of all, we need to find the centre of mass of the given system.

CENTER OF MASS

Assume the stars to be point objects. Let the mass $M$ be at origin i.e. $(0,0)$ . Since the distance between the stars is $d$ , the star with mass $3M$ will be at $(d,0)$ . The center of mass $(x_c ,y_c)$ can now be computed as:

$(x_c ,y_c)=(\frac{M\times 0 + 3M\times d}{M+3M},\frac{M\times 0 + 3M\times 0}{M+3M})=(\frac{3d}{4},0)$ .

TIME PERIOD

We know that for an object revolving around a point at a distance $R$ with a velocity $v$ , the time period can be computed using:

$T=\frac{2\pi R}{v}$ ...........................(1)

So now, we have to find $v$ .

We know that in rotational motion, $a=\frac{v^2}{R}$ . Also, $a=\frac{F}{mass}\Rightarrow \frac{v^2}{R}=\frac{F}{mass}$ .

Substituting $R=\frac{d}{4}$ , $F=\frac{G(M)(3M)}{d^2}$ , we get

$\frac{v^2}{\frac{d}{4}}=\frac{3GM^2}{d^23M}\Rightarrow v=\frac{1}{2}\sqrt{\frac{GM}{d}}$ .

Substituting this in equation (1), we get:

$T=\frac{2\pi \frac{d}{4}}{\frac{1}{2}\sqrt{\frac{GM}{d}}}$

or $\boxed{T=\pi \sqrt{\frac{d^3}{GM}}}$

$$ Two stars in a binary system are bound by gravity and revolve around a common center of mass. In this case, we may be treated as if all its mass the system is concentrated at its center of mass, which is $M'=M+3M=4M$ . Therefore, by using Newton's Law of gravity we obtain $$ $\begin{aligned} F&=\frac{GM'm}{R^2}\\ ma&= \frac{G(4M)m}{d^2}\\ a&=\frac{4GM}{d^2}\\ \omega^2d&=\frac{4GM}{d^2}\\ \frac{2\pi}{T}&=\sqrt{\frac{4GM}{d^3}}\\ T&=\pi\sqrt{\frac{d^3}{GM}} \end{aligned}$ $$ For extra information, this problem can also be solved by using Kepler's $3^{rd}$ Law. $$ $\text{\# }\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}$

Let,

F =gravitational force between the two stars (=G $\frac{(m_{1})(m_{2})}{d^{2}}$ )

$\omega_{a}$ & $\omega_{b}$ =angular velocities of larger(3M) and smaller(M) stars respectively

$r_{a}$ & $r_{b}$ =distance of larger and smaller stars from the common centre of mass

Stars are in an orbit about their common center of mass. So we have

gravitational force(inwards) = centrifugal force(outwards)

Using this,

For larger star

G $\frac{(3M)(M)}{d^{2}}$ =3M $\omega_{a}^{2}$ $r_{a}$

=> $\omega_{a}$ = $\sqrt{\frac{GM}{d^{2}r_{a}}}$ -----(1)

For smaller star

G $\frac{(3M)(M)}{d^{2}}$ =M $\omega_{b}^{2}$ $r_{b}$

=> $\omega_{b}$ = $\sqrt{\frac{3GM}{d^{2}r_{b}}}$

Now,

Since the common centre of mass does not change its position, we have that, time taken to complete one orbit for both stars is equal.

i.e, $t_{a}$ = $t_{b}$

=>(2 $\pi$ / $\omega_{a}$ )=(2 $\pi$ / $\omega_{b}$ )

=> $\omega_{a}$ = $\omega_{b}$

=> $\sqrt{\frac{GM}{d^{2}r_{a}}}$ = $\sqrt{\frac{3GM}{d^{2}r_{b}}}$

=>3 $r_{a}$ = $r_{b}$

Also we have $r_{a}$ + $r_{b}$ =d

=> $r_{a}$ = $\frac{d}{4}$

Substituting above value of $r_{a}$ in eqn (1), we have

$\omega_{a}$ = $\sqrt{\frac{GM}{d^{2}r_{a}}}$ = $\sqrt{\frac{4GM}{d^{3}}}$

Hence,

$t_{a}$ =(2 $\pi$ / $\omega_{a}$ )= $\pi\sqrt{\frac{d^{3}}{GM}}$ ANS