In a double star, two stars one of mass `m_(1)` and another of mass `m_(2)`, with a separation d, rotate about their common centre of mass. Find
(a) an expression for their time period of revolution.
(b) the ratio of their kinetic energies.
(c) the ratio of their angular momenta about the centre of mass.
(d) the total angular momentum of the system.
(e) the kinetic energy of the system.

Answer

A, B, C, D

Explanation

`r_(1) + r_(2) = d` ..(i)
`m_(1)r_(1) = m_(2)r_(2)` ..(ii)

Solving these two equations we get,
`r_(1) = ((m_(2))/(m_(1) + m_(2))) d` or `r_(2) = ((m_(1))/(m_(1) + m_(2))) d`
The centripetal force is provided by gravitational firce,
`m_(1) r_(1) omega^(2) = m_(2) r_(2) omega^(2) = (Gm_(1)m_(2))/(d^(2))`
Solving these equations, we get
`omega = sqrt((G(m_(1) + m_(2)))/(d^(3))`
`:. T = (2pi)/(omega) = 2pi sqrt((d^(3))/(G(m_(1) + m_(2)))`
(b) `(K_(1))/(K_(2))= ((1)/(2) I_(1) omega^(2))/((1)/(2) I_(2) omega^(2)) = (I_(1))/(I_(2)) = (m_(1)r_(1)^(2))/(m_(2)r_(2)^(2))`
`= ((m_(1))/(m_(2))) ((r_(1))/(r_(2)))^(2) = ((m_(1))/(m_(2))) ((m_(2))/(m_(1)))^(2) = (m_(2))/(m_(1))`
(c ) `(L_(1))/(L_(2)) = (L_(1) omega)/(I_(2) omega) = (I_(1))/(I_(2)) = (m_(2))/(m_(1))`
(d) `L = L_(1) + L_(2) = (I_(1) + L_(2)) omega`
`= (m_(1)r_(1)^(2) + m_(2)r_(2)^(2)) omega `
`= [(m_(1)m_(2)^(2)d^(2))/((m_(1) + m_(2))^(2)) + (m_(2)m_(1)^(2)d^(2))/((m_(1) + m_(2))^(2))]omega`
`= mu omega d^(2)`
where, `mu = (m_(1)m_(2))/(m_(1) + m_(2)) =` reduced mass
(e) `K = (1)/(2) (I_(1) + I_(2)) omega^(2) = (1)/(2) mu omega^(2) d^(2)`