A triple star system consists of two stars each of mass m in the same circular orbit about central star with mass `M=2xx10^(33)kg`. The two outer stars always lie at opposite ends of a diameter of their common circular orbit the radius of the circular orbit is `r=10^(11)` m and the orbital period each star is `1.6xx10^(7)s` [take `pi^(2)=10` and `G=(20)/(3)xx10^(-11)Mn^(2)kg^(-2)`]
Q. The mass m of the outer stars is:

Options

A. `16/15xx10^(30) kg`
B. `11/8xx10^(30) kg`
C. `15/16xx10^(30) kg`
D. `8/11xx10^(30) kg`

Answer

B

Explanation

`F_(m m)`=Gravitational force between two outer stars `=(Gm^(2))/(4r)`.
`F_(mM)`=Gravitational force between central star and outer star=`(GmM)/(r^(2))` for circular motion of outer star, `(mv^(2))/r=F_(m m)+F_(m M)`
`:. V^(2)=(G(m+4M))/(4r)`
`T`=period of orbtal motion `=(2pir)/v`
`m=(16pi^(2)r^(2))/(GT^(2))-4M=(150/16-8)10^(30)=11/8xx10^(30)kg`