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IOAA 2007, Теория — Задача 3. BINARY SYSTEM

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Всички материали от IOAA 2007

Theoretical Competition — 1st IOAA — December 5, 2007 · 5 декември 2007 г. · 10 т.

Условие

Two stars M1 and M2 (black dots) joined by a line, the centre of mass CM on the line closer to M1 with a curved arrow showing rotation; below, a double arrow between dotted vertical lines marks their separation D.

A binary star system consists of $M_{1}$ and $M_{2}$ separated by a distance $D$ . $M_{1}$ and $M_{2}$ are revolving with an angular velocity $\omega$ in circular orbits about their common centre of mass. Mass is continuously being transferred from one star to the other. This transfer of mass causes their orbital period and their separation to change slowly with time.

In order to simplify the analysis, we will assume that the stars are like point particles and that the effects of the rotation about their own axes are negligible.

a) What is the total angular momentum and kinetic energy of the system? [2 т.]

b) Find the relation between the angular velocity $\omega$ and the distance $D$ between the stars. [2 т.]

c) In a time duration $\Delta t$ , a mass transfer between the two stars results in a change of mass $\Delta M_{1}$ in star $M_{1}$ , find the quantity $\Delta\omega$ in terms of $\omega$ , $M_{1}$ , $M_{2}$ and $\Delta M_{1}$ . [3 т.]

d) In a certain binary system, $M_{1} = 2.9\ \mathrm{M}_{\odot}$ , $M_{2} = 1.4\ \mathrm{M}_{\odot}$ and the orbital period, $T = 2.49$ days. After 100 years, the period $T$ has increased by 20 s. Find the value of $\dfrac{\Delta M_{1}}{M_{1}\Delta t}$ (in the unit “per year”). [1,5 т.]

e) In which direction is mass flowing, from $M_{1}$ to $M_{2}$ , or $M_{2}$ to $M_{1}$ ? [0,5 т.]

f) Find also the value of $\dfrac{\Delta D}{D\Delta t}$ (in the unit “per year”). [1 т.]

You may use these approximations:

$(1+x)^{n} \sim 1+nx$ , when $x \ll 1$ ;

$(1+x)\left(1+y\right) \sim 1+x+y$ , when $x \ll 1$ , $y \ll 1$ .

Отговори

Решение

QUESTION 3 BINARY SYSTEM

Solution

a) The total angular momentum of the system is

L = I\omega = (M_{1}r_{1}^{2} + M_{2}r_{2}^{2})\omega

(0.5 points)

We have also $M_{1}r_{1} = M_{2}r_{2}$ and $D = r_{1} + r_{2}$

which yield

L = \frac{M_{1}M_{2}}{M_{1} + M_{2}}D^{2}\omega \qquad \text{……………..(1)}

(0.5 points)

The kinetic energy of the system is

K.E. = \frac{1}{2}M_{1}\left(r_{1}\omega\right)^{2} + \frac{1}{2}M_{2}\left(r_{2}\omega\right)^{2} = \frac{1}{2}\left(M_{1}r_{1}^{2} + M_{2}r_{2}^{2}\right)\omega^{2}

(0.5 points)

= \frac{1}{2}\frac{M_{1}M_{2}}{(M_{1} + M_{2})}D^{2}\omega^{2} \qquad \text{…………..(2)}

(0.5 points)

b) From Newton’s laws of motion we have

M_{1}\omega^{2}r_{1} = M_{2}\omega^{2}r_{2} = \frac{GM_{1}M_{2}}{D^{2}}

(1 point)

These equations together with those in a) yield

\omega^{2} = \frac{G\left(M_{1} + M_{2}\right)}{D^{3}} \qquad \text{……………….(3)}

(1 point)

c) In order to find the quantity $\Delta\omega$ , we must also realize that

M_{1} + M_{2} = \text{constant} \qquad \text{…………….(4)}

(0.5 points)

And that, since there is no external torque acting on the system, the total angular momentum must be conserved.

L = \frac{M_{1}M_{2}}{M_{1} + M_{2}}D^{2}\omega = \text{constant}

that is,

M_{1}M_{2}D^{2}\omega = \text{constant}

(0.5 points)

Now, after the mass transfer,

\omega \rightarrow \omega + \Delta\omega

M_{1} \rightarrow M_{1} + \Delta M_{1}

M_{2} \rightarrow M_{2} - \Delta M_{1}

D \rightarrow D + \Delta D

Hence, $M_{1}M_{2}D^{2}\omega = \left(M_{1} + \Delta M_{1}\right)\left(M_{2} - \Delta M_{1}\right)\left(D + \Delta D\right)^{2}\left(\omega + \Delta\omega\right)$

After using the approximation $\left(1 + x\right)^{n} \sim 1 + nx$ and rearranging, we get

\frac{\Delta\omega}{\omega} + 2\frac{\Delta D}{D} = \left(\frac{M_{1} - M_{2}}{M_{1}M_{2}}\right)\Delta M_{1} \qquad \text{………..(5)}

(1 point)

From equation (3), $\omega^{2}D^{3}$ is also constant. That is,

\omega^{2}D^{3} = \left(\omega + \Delta\omega\right)^{2}\left(D + \Delta D\right)^{3}

This gives,

\frac{\Delta D}{D} = -\frac{2}{3}\frac{\Delta\omega}{\omega} \qquad \text{……………….(6)}

(0.5 points)

Hence

\Delta\omega = -\frac{3\left(M_{1} - M_{2}\right)}{M_{1}M_{2}}\omega\,\Delta M_{1} \qquad \text{………..(7)}

(0.5 points)

d) Given that

$M_{1} = 2.9\ \mathrm{M}_{\odot}$ , $M_{2} = 1.4\ \mathrm{M}_{\odot}$

orbital period, $T = 2.49$ days

and that $T$ has increased by 20 s in the past 100 years,

we have

\omega = \frac{2\pi}{T} \text{ and } \Delta\omega = -\frac{\omega}{T}\Delta T \qquad \text{…………(8)}

(0.5 points)

\Delta M_{1} = +\frac{1}{3}\left(\frac{M_{1}M_{2}}{M_{1} - M_{2}}\right)\frac{\Delta T}{T}

(0.5 points)

\frac{\Delta M_{1}}{M_{1}\Delta t} = \frac{1}{3}\left(\frac{1.4}{(2.9 - 1.4)}\right)\left(\frac{20}{2.49 \times 24 \times 3600}\right)\left(\frac{1}{100}\right) = 2.89 \times 10^{-7}\ \text{per year}

(0.5 points)

e) Mass is flowing from $M_{2}$ to $M_{1}$ . (0.5 points)

f) From equations (6) and (8):

\frac{\Delta D}{D\Delta t} = -\frac{2}{3}\frac{\Delta\omega}{\omega} = +\frac{2}{3}\frac{\Delta T}{T} = 6.20 \times 10^{-7}\ \text{per year}

(1.0 points)

Оригинал в Архива: exam_theory.pdf · официални решения: exam_theory_solution.pdf