A binary star consists of two stars A (mass 2.2Ms) and B (mass 11Ms), where Ms is the mass of the sun. They are separated by distance d and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star B about the centre of mass is

To solve this problem, we need to understand the concept of angular momentum. The angular momentum (L) of a rotating object is given by the product of its mass (m), its velocity (v), and the distance (r) from the point about which it is rotating. In mathematical terms, L = mvr.

In a binary star system, the two stars rotate about their common center of mass. The total angular momentum of the system is the sum of the angular momenta of the two stars.

Let's denote the distance of star A from the center of mass as rA and the distance of star B from the center of mass as rB. The total angular momentum of the system (Ltotal) is then given by:

Ltotal = LA + LB = mAvArA + mBvBrB

The angular momentum of star B about the center of mass (LB) is given by:

LB = mBvBrB

We are asked to find the ratio of the total angular momentum of the binary star to the angular momentum of star B about the center of mass. This is given by:

Ltotal/LB = (mAvArA + mBvBrB) / (mBvBrB)

Since the stars are rotating about their common center of mass, their velocities are related by the equation vArA = vBrB. Substituting this into the equation for the ratio gives:

Ltotal/LB = (mAvBrB + mBvBrB) / (mBvBrB) = (mA + mB) / mB

Substituting the given values for the masses of the stars gives:

Ltotal/LB = (2.2Ms + 11Ms) / 11Ms = 13.2/11 = 1.2

So, the ratio of the total angular momentum of the binary star to the angular momentum of star B about the center of mass is 1.2.