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JEE Advanced 2010 · System of Particles and Rotational Motion

A binary star consists of two stars A (mass 2.2Ms)2.2M_{s}) and B (mass 11Ms),11M_{s}), where MsM_{s} is the mass of the sun. They are separated by distance d and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star B about the centre of mass is

Answer: 66

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Step 1: Identify the given parameters.

The masses of the two stars and their separation distance are given.

mA=2.2Msm_A = 2.2 M_s mB=11Msm_B = 11 M_s d=rA+rBd = r_A + r_B

Step 2: Relate the orbital radii to the center of mass.

Since the stars rotate about their stationary center of mass, the position of the center of mass satisfies the balancing condition.

mArA=mBrBm_A r_A = m_B r_B rArB=mBmA\frac{r_A}{r_B} = \frac{m_B}{m_A}

Step 3: Express the orbital radii in terms of the separation distance.

Using the total distance d=rA+rBd = r_A + r_B, we find the individual radii.

rA=mBmA+mBd=1113.2d=56dr_A = \frac{m_B}{m_A + m_B} d = \frac{11}{13.2} d = \frac{5}{6} d rB=mAmA+mBd=2.213.2d=16dr_B = \frac{m_A}{m_A + m_B} d = \frac{2.2}{13.2} d = \frac{1}{6} d

Step 4: Write the expressions for the angular momenta.

Both stars rotate with the same angular velocity ω\omega about the center of mass.

LB=mBrB2ωL_B = m_B r_B^2 \omega LA=mArA2ωL_A = m_A r_A^2 \omega Ltotal=LA+LB=(mArA2+mBrB2)ωL_{\text{total}} = L_A + L_B = \left(m_A r_A^2 + m_B r_B^2\right) \omega

Step 5: Calculate the ratio of the total angular momentum to the angular momentum of star B.

Divide the total angular momentum by the angular momentum of star B.

LtotalLB=1+mArA2mBrB2\frac{L_{\text{total}}}{L_B} = 1 + \frac{m_A r_A^2}{m_B r_B^2}

Substitute the ratio rArB=mBmA\frac{r_A}{r_B} = \frac{m_B}{m_A} into the expression.

mArA2mBrB2=mAmB(mBmA)2=mBmA\frac{m_A r_A^2}{m_B r_B^2} = \frac{m_A}{m_B} \left(\frac{m_B}{m_A}\right)^2 = \frac{m_B}{m_A} LtotalLB=1+mBmA\frac{L_{\text{total}}}{L_B} = 1 + \frac{m_B}{m_A}

Step 6: Substitute the given mass values to find the final ratio.

Substitute mA=2.2Msm_A = 2.2 M_s and mB=11Msm_B = 11 M_s.

LtotalLB=1+112.2\frac{L_{\text{total}}}{L_B} = 1 + \frac{11}{2.2} LtotalLB=1+5=6\frac{L_{\text{total}}}{L_B} = 1 + 5 = 6

Answer: The ratio of the total angular momentum of the binary star to the angular momentum of star B about the centre of mass is 6.

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