A system of binary stars of masses $\mathrm{m}_{\mathrm{A}}$ and $\mathrm{m}_{\mathrm{B}}$ are moving in circular orbits of radii $\mathrm{r}_{\mathrm{A}}$ and $\mathrm{r}_{\mathrm{B}}$ respectively. If $\mathrm{T}_{\mathrm{A}}$ and $\mathrm{T}_{\mathrm{B}}$ are the time periods of masses $\mathrm{m}_{\mathrm{A}}$ and $\mathrm{m}_{\mathrm{B}}$ respectively, then

(D) $\mathrm{T}_{\mathrm{A}}=\mathrm{T}_{\mathrm{B}}$

(A) $\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\left(\frac{\mathrm{r}_{\mathrm{A}}}{\mathrm{r}_{\mathrm{B}}}\right)^{3 / 2}$

(B) $\mathrm{T}_{\mathrm{A}}>\mathrm{T}_{\mathrm{B}}$ (if $\mathrm{r}_{\mathrm{A}}>\mathrm{r}_{\mathrm{B}}$)

(C) $\mathrm{T}_{\mathrm{A}}>\mathrm{T}_{\mathrm{B}}$ (if $\mathrm{m}_{\mathrm{A}}>\mathrm{m}_{\mathrm{B}}$)

JEE Advanced · 2006 · 09 Apr 2006 · Gravitation

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A system of binary stars of masses

m_{A}

and

m_{B}

are moving in circular orbits of radii

r_{A}

and

r_{B}

respectively. If

T_{A}

and

T_{B}

are the time periods of masses

m_{A}

and

m_{B}

respectively, then

(A) $\frac{T _{A}}{T _{B}} = (\frac{r _{A}}{r _{B}})^{3/2}$
(B) $T_{A} > T_{B}$ (if $r_{A} > r_{B}$ )
(C) $T_{A} > T_{B}$ (if $m_{A} > m_{B}$ )
(D) $T_{A} = T_{B}$ ✓ Correct

Answer: (D)

Solution

In a binary star system, both stars orbit their common center of mass with the same angular velocity (and hence the same period).

The gravitational force provides the centripetal force for each star:

\frac{G m _{A} m _{B}}{( r _{A} + r _{B} ) ^{2}} = \frac{m _{A} r _{A} 4 π ^{2}}{T _{A}^{2}} = \frac{m _{B} r _{B} 4 π ^{2}}{T _{B}^{2}}

From the equality of the centripetal forces:

\frac{m _{A} r _{A}}{T _{A}^{2}} = \frac{m _{B} r _{B}}{T _{B}^{2}}

Also, for the center of mass to remain fixed:

m_{A} r_{A} = m_{B} r_{B}

Substituting this into the force equation shows that

T_{A} = T_{B}

.

Therefore, both stars have the same period, which is option (D).

JEE Advanced · 2006 · 09 Apr 2006 · Single correct option

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JEE Advanced · 2006 · 09 Apr 2006 · Gravitation

Solution

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