Question 455: Gravitation

Question

A binary star consists of two stars A (mass 2.2M
s
) and B (mass 11M
s
), where M
s
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

Accepted Answer

Let stars A and B are rotating about their centre of mass with angular velocity $$\omega$$.
Let distance of stars A and B from the centre of mass be r
A
and r
B
respectively as shown in the figure.
Total angular momentum of the binary stars about the centre of mass is
$$L = {M_A}r_A^2\omega  + {M_B}r_B^2\omega $$
Angular momentum of the star B about centre of mass is
$${L_B} = {M_B}r_B^2\omega $$
$$	herefore$$ $${L \over {{L_B}}} = {{({M_A}r_A^2 + {M_B}r_B^2)\omega } \over {{M_B}r_B^2\omega }} = \left( {{{{M_A}} \over {{M_B}}}} ight){\left( {{{{r_A}} \over {{r_B}}}} ight)^2} + 1$$
Since $${M_A}{r_A} = {M_B}{r_B}$$
or, $${{{r_A}} \over {{r_B}}} = {{{M_B}} \over {{M_A}}}$$
$$	herefore$$ $${L \over {{L_B}}} = {{{M_B}} \over {{M_A}}} + 1 = {{11{M_S}} \over {2.2{M_S}}} + 1 = {{11 + 2.2} \over {2.2}} = 6$$

Gravitation

Integer / Numeric Answer Type