Question

Two identical stars with mass M orbit around their center of mass. Each orbit is circular and has radius

Answer 1

Answers: (A)$F=G\frac{M^2}{4R^2}$ (B) $V=\sqrt{\frac{GM}{4R}}$ (C)$T=4\pi R\sqrt{\frac{R}{GM}}$ (D)

$E=-\frac{GM^{2}}{4R}$

Explanation:

(A) Gravitational force of one star on the other

According to the law of universal gravitation:

$F=G\frac{m_{1}m_{2}}{r^2}$   (1)

Where:

$F$ is the module of the gravitational force exerted between both bodies  

$G$ is the universal gravitation constant.

$m_{1}$ and $m_{2}$ are the masses of both bodies.

$r$ is the distance between both bodies

In the case of this binary system with two stars with the same mass $M$ and separated each other by a distance $2R$, the gravitational force is:

$F=G\frac{(M)(M)}{(2R)^2}$   (2)

$F=G\frac{M^2}{4R^2}$   (3) This is the gravitational force between the two stars.

(B) Orbital speed of each star

Taking into account both stars describe a circular orbit and the fact this is a symmetrical system, the orbital speed $V$ of each star is the same. In addition, if we assume this system is in equilibrium, gravitational force must be equal to the centripetal force  $F_{C}$ (remembering we are talking about a circular orbit):

So: $F=F_{C}$   (4)

Where $F_{C}=Ma_{C}$  (5) Being $a_{C}$ the centripetal acceleration

On the other hand, we know there is a relation between $a_{C}$ and the velocity $V$:

$a_{C}=\frac{V^{2}}{R}$  (6)

Substituting (6) in (5):

$F_{C}=M\frac{V^{2}}{R}$ (7)

Substituting (3) and (7) in (4):

$G\frac{M^2}{4R^2}=M\frac{V^{2}}{R}$   (8)

Finding $V$:

$V=\sqrt{\frac{GM}{4R}}$ (9) This is the orbital speed of each star

(C) Period of the orbit of each star

The period $T$ of each star is given by:

$T=\frac{2\pi R}{V}$  (10)

Substituting (9) in (10):

$T=\frac{2\pi R}{\sqrt{\frac{GM}{4R}}}$  (11)

Solving and simplifying:

$T=4\pi R\sqrt{\frac{R}{GM}}$  (12) This is the orbital period of each star.

(D) Energy required to separate the two stars to infinity

The gravitational potential energy $U_{g}$ is given by:

$U_{g}=-\frac{Gm_{1}m_{2}}{r}$  (13)

Taking into account this energy is always negative, which means the maximum value it can take is 0 (this happens when the masses are infinitely far away); the variation in the potential energy $\Delta U_{g}$ for this case is:

$\Delta U_{g}=U-U_{\infty}$ (14)

Knowing $U_{\infty}=0$ the total potential energy is $U$ and in the case of this binary system is:

$U=-\frac{G(M)(M)}{2R}=-\frac{GM^{2}}{2R}$  (15)

Now, we already have the potential energy, but we need to know the kinetic energy $K$ in order to obtain the total Mechanical Energy $E$ required to separate the two stars to infinity.

In this sense:

$E=U+K$ (16)

Where the kinetic energy of both stars is:

$K=\frac{1}{2}MV^{2}+\frac{1}{2}MV^{2}=MV^{2}$ (17)

Substituting the value of $V$ found in (9):

$K=M(\sqrt{\frac{GM}{4R}})^{2}$ (17)

$K=\frac{1}{4}\frac{GM^{2}}{R}$ (18)

Substituting (15) and (18) in (16):

$E=-\frac{GM^{2}}{2R}+\frac{1}{4}\frac{GM^{2}}{R}$ (19)

$E=-\frac{GM^{2}}{4R}$ (20) This is the energy required to separate the two stars to infinity.