Chapter 13, Problem 054 Hunting a black hole. Observations of the light from a c

Question

Chapter 13, Problem 054 Hunting a black hole. Observations of the light from a certain star indicate that it is part of a binary (two-star) system. This visible star has orbital speed v-280 km/s, orbital period T 22.2 days, and approximate mass m- 6.1Ms, where Ms is the Sun's mass, 1.99 x 1030 kg. Assume that the visible star and its companion star, which is dark and unseen, are both in circular orbits (see the figure). Find the ratio of the approximate mass m2 of the dark star to Ms Number Units

Explanation / Answer

given speed of visible star, v = 280,000 m/s
T = 22.2 days
m1 = 6.1Ms
Ms = 1.99*10^30 kg

now, from the kepler's laws for binaries
T = 2*pi/sqroot(G(m1 + m2)/a^3)
a = r1 + r2

and
Tv = 2*pi*r1
also
m1*r1 = M2r2
hence
a = Tv/2*pi(1 + m1/m2)

so,
G(m1 + m2)T^2 = 4*pi^2*T^3*v^3*m2^3/8*pi^3(m1 + m2)^3
2*pi*G(m1 + m2)^4 = T*v^3*m2^3
(m1 + m2)^4 = 1.004697*10^32*m2^3
(m1^2 + m2^2 + 2m1m2)^2 = m1^4 + m2^4 + 6m1^2m2^2 + 4m2^3m1 + 4m1^3m2
(m1 + m2)^4 = 2.17135*10^124 + m2^4 + 8.84131*10^62*m2^2 + 4.8556*10^31*m2^3 + 7.15498*10^93*m2 = 1.004697*10^32*m2^3
2.17135*10^124 + m2^4 + 8.84131*10^62*m2^2 - 5.19137*10^31*m2^3 + 7.15498*10^93*m2 = 0
solving for m2, we get solution
we can then find m2/Ms = answer

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