Question

A triple star system consists of two stars each of mass m, in the same circular orbit about the control star of mass M. The two outer stars always lie at opposite ends of a diameter of their common circular orbit. The radius of the circular orbit of each outer star is r and the orbital period of each outer star is T. The mass of each star is:-
A) (16 pi^2 r^3)/(GT^2)
B) [ (16 pi^2 r^3)/(GT^2) ] - M
C) [ (16 pi^2 r^3)/(GT^2) ] - 2M
D) [ (16 pi^2 r^3)/(GT^2) ] - 4M

Plz answer with steps and proper explanation. Don't spam pls.

Answer 1

Given:

A triple star system consists of two stars each of mass m, in the same circular orbit about the control star of mass M. The two outer stars always lie at opposite ends of a diameter of their common circular orbit. The radius of the circular orbit of each outer star is r and the orbital period of each outer star is T.

To find:

Value of mass m ?

Calculation:

The gravitational force will be equal to the centripetal force of each of the mass:

$\therefore \: \dfrac{m {v}^{2} }{r} = \dfrac{GMm}{ {r}^{2} } + \dfrac{G {m}^{2} }{ {(2r)}^{2} }$

$\implies\: \dfrac{m {v}^{2} }{r} = \dfrac{Gm}{ {r}^{2} } \bigg( M + \dfrac{m}{4} \bigg)$

$\implies\: {v}^{2} = \dfrac{G}{ r } \bigg( M + \dfrac{m}{4} \bigg)$

$\implies\: v = \sqrt{\dfrac{G}{ r } \bigg( M + \dfrac{m}{4} \bigg)}$

Now, time period of the mass be T :

$\therefore \: T = \dfrac{2\pi r}{v}$

$\implies \: T = \dfrac{2\pi r}{ \sqrt{ \dfrac{G}{r} (M + \dfrac{m}{4} } )}$

$\implies \: {T}^{2} = \dfrac{4 {\pi}^{2} {r}^{2} }{ \dfrac{G}{r} (M + \dfrac{m}{4})}$

$\implies \: M + \dfrac{m}{4} = \dfrac{4 {\pi}^{2} {r}^{3} }{G {T}^{2} }$

$\implies \: \dfrac{m}{4} = \dfrac{4 {\pi}^{2} {r}^{3} }{G {T}^{2} } - M$

$\implies \: m = \dfrac{16 {\pi}^{2} {r}^{3} }{G {T}^{2} } - 4M$

So , final answer is:

$\boxed{ \bf m = \dfrac{16 {\pi}^{2} {r}^{3} }{G {T}^{2} } - 4M}$