Determine the approximate mass m2 of the dark star.?

Question

1. Observations of the light from a certain star indicate that it is a part of a binary (two-star) system. The visible star has orbital speed v= 270 km/s, orbital period T = 1.70 days, and approximate mass m1 = 6*mass of sun (6*1.99*10^30 kg. Assuming that the visible star and its companion star, which is dark and unseen, are both in circular orbits, determine the approximate mass m2 of the dark star.

Answer 1

more data required.

Answer 2

♠ thus m1*r1=m2*r2 is equation for their mass center, where r1 is radius of rotation of visible m1, r2 is radius of rotation of invisible m2, radii being measured from their common mass center – the origin of the system; thus r2=r1*(m1/m2);
♣ also circumference of orbit of m1 is 2pi*r1=v1*T, hence r1=v1*T/2pi;
♦ the force of their mutual gravitation F=G*m1*m2/(r1+r2)^2
is also their mutual centripetal force F= m1*v1^2/r1; thus
G*m2/(r1+r2)^2=v1^2/r1; or; G*m2= (v1^2/r1)*(r1+r2)^2;
plugging r2 of (♠) we get G*m2= v1^2*r1 *(1+m1/m2)^2;
plugging r1 of (♣) we get G*m2= v1^2* (v1*T/2pi) *(1+m1/m2)^2;
or; m2^3=v1^3*T/(2pi*G) *(m2+m1)^2; or;
m^3= v1^3*T/(2pi*G*m1) *(m+1)^2, where m=m2/m1, G= 6.6742x10^-11,
v1=0.270·10^6 m/s, T=1.7*24*3600s, m1=6* 6*1.99·10^30 kg;
♥ m^3=0.096232* (m+1)^2, hence m=0.6364, m2= 3.8184 Suns;
now Rajjy, take your hands out of your pockets and check me!