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Nuclear Fission

Consider a bi...

Question

Consider a binary star system of star $A$ and star $B$ with masses $m_{A}$ and $m_{B}$ revolving in a circular orbit of radii $r_{A}$ and $r_{B}$ , respectively. If $T_{A}$ and $T_{B}$ are the time period of star $A$ and star $B$ , respectively, then :

A

T_{A} > T_{B}

(if

m_{A} > m_{B}

)

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B

\frac{T_{A}}{T_{B}} = {(\frac{r_{A}}{r_{B}})}^{\frac{3}{2}}

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C

T_{A} = T_{B}

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D

T_{A} > T_{B}

(if

r_{A} > r_{B}

)

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Solution

The correct option is C $T_{A} = T_{B}$

In the diagram, $C$ is the centre of mass of given binary star system.

Since, $C$ is the centre of mass, so

$m_{A} r_{A} = m_{B} r_{B} . . . . . (1)$

Two forces, gravitational force and centripetal force are acting on each star, and these forces are responsible for rotation of the star.

$∴ F_{g r a v} = F_{c e n t}$

For stars $A & B :$

$\frac{G m_{A} m_{B}}{(r_{A} + r_{B})^{2}} = \frac{m_{B} v_{B}^{2}}{r_{B}} = \frac{m_{A} v_{A}^{2}}{r_{A}}$

Putting the value of $m_{B}$ from eq. $(1)$ ,

$\frac{m_{A} r_{A}}{r_{B}} \frac{v_{B}^{2}}{r_{B}} = \frac{m_{A} v_{A}^{2}}{r_{A}}$

$\Rightarrow \frac{v_{B}}{r_{B}} = \frac{v_{A}}{r_{A}} . . . . . . . . . (2)$

Now time periods of stars,

$\frac{T_{A}}{T_{B}} = \frac{(2 π r_{A} / v_{A})}{(2 π r_{B} / v_{B})} = 1 [from eq. (2)]$

$∴ T_{A} = T_{B}$

Hence, option $(B)$ is correct.

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