Physics, 10.07.2019 17:20 mia1518

Two identical stars with mass m orbit around their center of mass. each orbit is circular and has radius
r, so that the two stars are always on opposite sides of the circle.
(a) find the gravitational force of one star on the other.
(b) find the orbital speed of each star.
(c) find the period of the orbit of each star.
(d) how much energy would be required to separate the two stars to infinity?

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Answer from: kataldaine

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:

is the module of the gravitational force exerted between both bodies

is the universal gravitation constant.

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

is the distance between both bodies

In the case of this binary system with two stars with the same mass and separated each other by a distance , 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 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 :

$a_{C}=\frac{V^{2}}{R}$ (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 :

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

The period 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 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 in order to obtain the total Mechanical Energy 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 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.

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