Question: In a double-star system, two stars of mass$\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{30}}\mathit{k}\mathit{g}$ each rotate about the system’s center of mass at radius $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{11}}\mathit{m}$. (a) What is their common angular speed? (b) If a meteoroid passes through the system’s center of mass perpendicular to their orbital plane, what minimum speed must it have at the center of mass if it is to escape to “infinity” from the two-star system?

The mass of the star is $M=3.0\times {10}^{30}kg$

The orbital radius of the system is $r=1.0\times {10}^{11}m$

Every particle in the universe is attracted to every other particle with a force that is directly proportional to the product of the masses and inversely proportional to the square of the distance, according to Newton's Law of Universal Gravitation.

We can use the concept that the gravitational force on the two stars provides the centripetal force on them. From this, we can find the orbital speed of the system of stars, the angular speed, and radius of its circumference of circle of system. Then equating the total energy of the system to zero, we can find the minimum speed ofthemeteoroid at the center of mass if it is to escape to infinity from the two-star system.

Formula:

$F=\frac{GMm}{r^2}$
$$\begin{gathered} F_c=\frac{m{v}^2}{r} \\ v=r\omega \end{gathered}$$

Where,

F is the gravitational force

F_c is the centripetal force

G is the gravitational constant ( $6.67\times {10}^{-11} {\text{m}}^{\text{3}}\text{/kg} ·{\text{s}}^{\text{2}}$)

M is the mass of earth

V is the speed of object

m is the mass of object

w is the angular speed of object

r is the orbital radius of object

The angular speed of the two stars:

According to the Newton’s law of gravitation, the gravitational force acting on the spherical planet is

$F=\frac{G{M}^2}{{\left(2r\right)}^2}$

The gravitational force on the star provides the centripetal force on the star. Hence, the expression for centripetal force is

$F_c=\frac{M{v}^2}{r}$

The gravitational force can be balanced by the centripetal force. Therefore,

$\frac{G{M}^2}{{\left(2r\right)}^2}=\frac{M{v}^2}{r}$ (i)

According to the expression of the relation between orbital speed, angular speed, and radius of its circumference of circle of system,

$v=r\omega$

Equation (i) becomes

$$\begin{gathered} \frac{G{M}^2}{{\left(2r\right)}^2}=\frac{M{\left(r\omega \right)}^2}{r} \\ {\omega }^2=\frac{GM}{4r^3} \\ \omega =\sqrt{\frac{GM}{4r^3}} \\ \omega =\sqrt{\frac{\left(6.67\times {10}^{-11}\frac{{\text{Nm}}^{\text{2}}}{{\text{kg}}^{\text{2}}}\right)\times \left(3.0\times {10}^{30}\text{kg}\right)}{4{\left(1.0\times {10}^{11}\text{m}\right)}^3}} \\ \omega =2.2\times {10}^{-7}\frac{\text{rad}}{\text{s}} \end{gathered}$$

Therefore, the common angular speed of stars is $2.2\times {10}^{-7}\frac{rad}{s}.$

A meteoroid passes through the system’s center of mass, and if it escapes to infinity from the two star systems, it means it has total energy equal to zero. If mass of the meteoroid is m

$$\begin{gathered} \frac{1}{2}m{v}^2-\frac{GMm}{r}-\frac{GMm}{r}=0 \\ \frac{1}{2}m{v}^2=\frac{GMm}{r}+\frac{GMm}{r} \\ \frac{1}{2}m{v}^2=\frac{2GMm}{r} \\ {v}^2=\frac{4GM}{r} \end{gathered}$$

$$\begin{gathered} dv=\sqrt{\frac{4GM}{r}} \\ =\sqrt{\frac{4\left(6.67\times {10}^{-11}\frac{{\text{Nm}}^{\text{2}}}{{\text{kg}}^{\text{2}}}\right)\times \left(3.0\times {10}^{30}\text{kg}\right)}{1.0\times {10}^{11}\text{m}}} \\ =8.9\times {10}^4\frac{\text{m}}{\text{s}} \end{gathered}$$

Therefore, the minimum speed of the meteoroid at the center of mass if it is to escape to infinity from the two-star system is $8.9\times {10}^4\frac{\text{m}}{\text{s}}$