To carry out this exercise, it is necessary to use the equations made to Centripetal Force and Gravitational Energy Conservation.
By definition we know that the Centripetal Force is estimated as
[tex]F_c = M\omega^2R[/tex]
Where,
M = mass
[tex]\omega =[/tex] Angular velocity
R = Radius
From the 'linear' point of view the centripetal force can also be defined as
[tex]F_c = \frac{GM^2}{R^2}[/tex]
PART A ) Equating both equations we have,
[tex]\frac{M}{\omega^2R}=\frac{GM^2}{R^2}[/tex]
Re-arrange to find \omega
[tex]\omega = \sqrt{\frac{Gm}{r^3}}[/tex]
Replacing with our values
[tex]\omega = \sqrt{\frac{(6.67*10^{-11})(4.6*10^{30})}{(1.9*10^{11})^3}}[/tex]
[tex]\omega = 2.115*!0^{-7}rad/s[/tex]
Therefore the angular speed is [tex]\omega = 2.115*!0^{-7}rad/s[/tex]
PART B) For energy conservation we have to
[tex]KE_{min} = PE_{cm}[/tex]
Where,
[tex]KE_{min} =[/tex] Minimus Kinetic Energy
[tex]PE_{cm} =[/tex] Gravitational potential energy at the center of mass
Then,
[tex]\frac{1}{2} mv^2_{min} = \frac{2GMm}{R}[/tex]
Re-arrange to find v,
[tex]v_min = \sqrt{\frac{4GM}{R}}[/tex]
[tex]v_min = \sqrt{\frac{4(6.67*10^{-11})(2.2*10^{30})}{(1.9*10^{11})}}[/tex]
[tex]v_min = 5.55*10^4m/s[/tex]
Therefore the minimum speed must it have at the center of mass if it is to escape to "infinity" from the two-star system is [tex]v_min = 5.55*10^4m/s[/tex]
