Binary stars of comparable masses rotate under the influence of each others gravity at a distance - 8G- 2-1-3- where - is the angular velocity of each of the system- If difference between the masses of two star is 6 units- Find the ratio of the masses of smaller to bigger star-

The correct option is B 1:7 REF.Image. m_2 m_1 = 6u r = ( 8Gω^2 )^1/3 a_1 = (F_net)m_1 = Gm_1m_2r^2 m_1 = Gm_2r^2 as; ω^2 r_1 = Gm_2r ...

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Standard XII

Physics

Center of Mass as an Average Point

Binary stars ...

Question

Binary stars of comparable masses rotate under the influence of each others gravity at a distance ${[\frac{8 G}{ω^{2}}]}^{1 / 3}$ , where $ω$ is the angular velocity of each of the system. If difference between the masses of two star is 6 units. Find the ratio of the masses of smaller to bigger star.

4 : 10

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1 : 7

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2 : 8

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3 : 9

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Solution

The correct option is B $1 : 7$
REF.Image.
$m_{2} - m_{1} = 6 u$

$r = {(\frac{8 G}{ω^{2}})}^{1 / 3}$

$a_{1} = \frac{(F_{n e t})}{m_{1}} = \frac{G m_{1} m_{2}}{r^{2} m_{1}} = \frac{G m_{2}}{r^{2}}$

as; $ω^{2} r_{1} = \frac{G m_{2}}{r^{2}} . . . (1)$

and similarly
$ω^{2} r_{2} = \frac{G m_{1}}{r^{2}} . . . (i i)$

dividing (1) and (ii)
$\frac{r_{1}}{r_{2}} = \frac{m_{2}}{m_{1}}$
adding (i) and (ii)

$ω^{2} (r_{1} + r_{2}) = \frac{G}{r^{2}} (m_{1} + m_{2})$

$\Rightarrow ω^{2} r^{3} = G (m_{1} + m_{2}) = 8 G$

$\Rightarrow 8 G = G (m_{1} + m_{2})$ (On putting the value of $r$ from question)

$m_{1} + m_{2} = 8$

$m_{2} - m_{1} = 6 u$

$m_{1} + m_{2} = 8 u$

$m_{1} = 1 u$

$m_{2} = 7 u$

$m_{1} : m_{2} = 1 : 7$ Ans
Option (B)

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Binary stars of comparable masses rotate under the influence of each others gravity at a distance [8Gω2]1/3, where ω is the angular velocity of each of the system. If difference between the masses of two star is 6 units. Find the ratio of the masses of smaller to bigger star.