Hunting a black hole. Observations of the light from a certain star indicate tha

Question

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 = 270 km/s, orbital period T = 17.0 days, and approximate mass m1 = 5.5Ms, 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.

Explanation / Answer

The force on each star is the same. M = mass of second star (unknown) m = mass of first star = 6 time the mass of the sun G = gravitational constant R = distance between the centers of the stars F = G*M*m/R^2 The standard form for force is just mass times acceleration. And for a circular orbit we have the usual centripetal acceleration stuff. So for star 1: F = m*a And we have a = V^2/R Put these together: G*M*m/R^2 = m*[V^2/R] M = V^2*R/G Now for the period of the first star. P = 2*pi*SQRT[R^3/(G*M)] ................ period for star 1 R = [P^2*G*M/(4*pi^2)]^(1/3) .............. solving for R Put the expression for R in the one for M. M = (V^2/G) [P^2*G*M/(4*pi^2)]^(1/3) M^(2/3) = V^2 [P^2/(4*pi^2*G^2)]^(1/3) Just a quick check on units to make sure this at least gives the right units. G -> kg^-1 m^3 s^-2 P -> s V -> m/s (m/s)^2[s^2 kg^2 s^4 m^-6]^(1/3) = (m/s)^2[s^2 m^-2 kg^(2/3)] = kg^(2/3) So the units are good kg^(2/3) on each side In the expression for M above. take the 3/2 power for each side to get an expression for M. M = V^3 [P^2/(4*pi^2*G^2)]^(1/2) M = V^3*P/(2*pi*G) P = 1.70 days = 1.4688x10^5 s ....... period for star 1 V = 270 km/s = 2.7x10^5 m/s ......... orbital velocity for star 1 G = 6.673x10^(-11) m^3 kg^-1 s^-2 M = 2.04057x10^30 kg Mass of the sun = 1.98892x10^30 kg So the mass of the unknown star is approximately 1 solar mass. I checked this equation using the earth-sun and the moon-earth and got 1.962x10^30 kg for the mass of the sun and 6.11x10^24 kg for the mass of the earth. Mass of the sun is given above and mass of the earth is 5.974x10^24 kg so these answers look pretty good so the equation looks good. Note. Yes center of mass, etc. I considered that but also wanted to get some sort of answer since the question is asking for only an approximate mass. On further reflection I can see an error: I used a = V^2/R when it should have been a = V^2/r Where r = distance of first star from the center of mass And R = r + s if I call s the distance of 2nd star from CM. This is true since the stars are in circular orbit and always opposite each other with respect to the CM (which is what makes the problem solvable). P = period which is the same for both stars since the stars are in circular orbits V = velocity of first star U = velocity of second star (unknown) m*a = F from gravity m*V^2/r = G*M*m/(r + s)^2 I can write P*V/(2*pi) = r from P = 2*pi*r/V 2*pi(V/P) = G*M/(r + s)^2 We will now use: U = V(m/M) 2*pi*r/V = P ..... r = V*P/(2*pi) 2*pi*s/U = P .... s = U*P/(2*pi) = V*P(m/M)/(2*pi) r + s = [V*P/(2*pi)][1 + m/M] 2*pi(V/P) = G*M/{ [V*P/(2*pi)][1 + m/M]}^2 [2*pi(V/P)][V*P/(2*pi)]^2 = G*M/[1 + m/M]^2 V^3P/(2*pi*G) = M^3/(M + m)^2 V^3P/(2*pi*G) = 2.87E+29 = C M^3 = C*(M^2 + 2Mm + m^2) Scale by solar mass C = C = 0.144 M^3 - C*M^2 - 12*C*M - 36*C = 0 M = 2.11 solar masses So my corrected answer is 2.11 solar masses. Further note of clarification. I used U = V(m/M) and want to make sure you know where this came from. m*r = M*s r/s = M/m Circular orbits are a special case and the periods of both stars will be the same (as I used above). 2*pi*r/V = P = 2*pi*s/U r/V = s/U r/s = V/U Combine the r/s statements: V/U = M/m And U = V(m/M) I suggest you consult the book "Astronomy: a physical perspective" by Marc Leslie Kutner and the sections starting around page 88. My derivation differs from his in some ways but in principle is the same and, if you do check this source, you will see that the problem can indeed by solved as stated because of the special circumstances of circular orbits. I suggest you rethink the problem. The key is the fact that they will have the same periods and that they will be directly opposite each other with respect to the center of mass. ANOTHER NOTE. By the way Jack, I suggest you take a look at the wikipedia entry and pay particular attention to the statement "The Keplerian problem assumes an elliptical orbit". It is very clearly stated in the problem that the orbits are NOT elliptical but are circular and this is done precisely so a solution can be found. Check the second source. I had the page number wrong it should have been 87. The section is entitled "Binary Stars and Circular Orbits". Books are very good things. You can learn a lot from them.

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