Consider a star of mass kg revolving in a circular orbit around another star of mass kg with . The heavier star slowly acquires mass from the lighter star at a constant rate of kg/s. In this transfer process, there is no other loss of mass. If the sepa

Question

Consider a star of mass $m_{2}$ kg revolving in a circular orbit around another star of mass $m_{1}$ kg with $m_{1} ≫ m_{2}$ . The heavier star slowly acquires mass from the lighter star at a constant rate of $γ$ kg/s. In this transfer process, there is no other loss of mass. If the separation between the centers of the stars is $r$ , then its relative rate of change $\frac{1}{r} \frac{d r}{d t}$ (in $s^{- 1}$ ) is given by: [2025]

Smart Achievers · Accepted Answer

(2)

$Here, \frac{m_{2} v^{2}}{r} = \frac{G m_{1} m_{2}}{r^{2}}, or, m_{2} ω^{2} r = \frac{G m_{1} m_{2}}{r^{2}}$

$ω = \sqrt{\frac{G m_{1}}{r^{3}}}$

$L = I ω = m_{2} ω r^{2} = m_{2} \sqrt{\frac{G m_{1}}{r^{3}}} r^{2}$

$L = m_{2} \sqrt{G m_{1} r} = constant$

$l n L = l n m_{2} + l n G + \frac{1}{2} l n m_{1} + \frac{1}{2} l n r$

$0 = \frac{d m_{2}}{m_{2}} + \frac{1}{2} \frac{d m_{1}}{m_{1}} + \frac{1}{2} \frac{d r}{r}$

$\frac{d r}{r d t} = - \frac{2}{m_{2}} \frac{d m_{2}}{d t} - \frac{1}{m_{1}} \frac{d m_{1}}{d t} \approx - \frac{2 γ}{m_{2}}$ $(∵ \frac{d m}{d t} = γ)$

Solution

1 $- \frac{3 γ}{2 m_{2}}$

2 $- \frac{2 γ}{m_{2}}$

3 $- \frac{2 γ}{m_{1}}$

4 $- \frac{3 γ}{2 m_{1}}$

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Solution

1 -3γ2m2

2 -2γm2

3 -2γm1

4 -3γ2m1

Ans. (2) Here, m2v2r=Gm1m2r2, or, m2ω2r=Gm1m2r2 ω=Gm1r3 L=Iω=m2ωr2=m2Gm1r3r2 L=m2Gm1r=constant lnL=lnm2+lnG+12lnm1+12lnr 0=dm2m2+12dm1m1+12drr drr dt=-2m2dm2dt-1m1dm1dt≈-2γm2 (∵dmdt=γ)

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1 $- \frac{3 γ}{2 m_{2}}$

2 $- \frac{2 γ}{m_{2}}$

3 $- \frac{2 γ}{m_{1}}$

4 $- \frac{3 γ}{2 m_{1}}$