# Mass Transfer in a Binary Star System

Homework Statement:
A binary star system consists of M1 and M2 separated by a distance D. M1 and M2 are revolving with an angular velocity w in circular orbits about their common center of mass. Mass is continuously being transferred from one star to the other. This transfer of mass causes their orbital period and their separation to change slowly with time. Assume the stars are like point particles and that the effects of the rotation about their own axes are negligible. In a time duration dt, a mass transfer between the two stars results in a change of mass dM1 in star M1. Find the quantity dw in terms of w, M1, M2, and dM1.
Relevant Equations:
Total angular momentum of system: L = (wM1M2D^2) / (M1+M2)
Relation between angular velocity and distance between stars: w^2 = G(M1+M2) / D^3
Homework Statement: A binary star system consists of M1 and M2 separated by a distance D. M1 and M2 are revolving with an angular velocity w in circular orbits about their common center of mass. Mass is continuously being transferred from one star to the other. This transfer of mass causes their orbital period and their separation to change slowly with time. Assume the stars are like point particles and that the effects of the rotation about their own axes are negligible. In a time duration dt, a mass transfer between the two stars results in a change of mass dM1 in star M1. Find the quantity dw in terms of w, M1, M2, and dM1.
Homework Equations: Total angular momentum of system: L = (wM1M2D^2) / (M1+M2)
Relation between angular velocity and distance between stars: w^2 = G(M1+M2) / D^3

My strategy for solving this problem was to consider that angular momentum remains constant and thus to take ##dL/dt = 0##. Plugging in the expression for L, I obtained $$\frac{d}{dt} \frac {wM_1M_2D^2} {M_1+M_2} = 0$$
I plugged in the given expression for D in terms of w, M1, and M2 and used the quotient rule to take the derivative of this expression with respect to time, assuming that ##dM_2/dt = -dM_1/dt##. I got $$0 = \frac {\frac {dM_1}{dt} (M_2-M_1)(M_2+M_1)w - \frac{1}{3} (M_1+M_2) \frac {dw}{dt}}{(w(M_1+M_2))^{4/3}}$$
Thus, solving for dw, I obtained for my final answer $$dw = 3w(M_2-M_1) dM_1$$
However, this answer is obviously wrong - the units don't match up, and the answer given in the text is ##dw = 3w(M_2-M_1) dM_1 / M_1M_2##. The authors of the text provide a step-by-step solution which I understand and I can provide to anyone who wishes to see it, but it uses a completely different method which does not involve taking the derivative at all, and I am not sure why my method did not work. Was it a simple math error, a fault in my initial assumptions, etc? Any help would be much appreciated! (It is also my first time using LaTeX or posting on this site, so if I have done anything wrong please notify me.)

Delta2

mfb
Mentor
Something went wrong with the derivative, it has mismatching units already.

Instead of assuming the angular momentum stays the same it might be easier to assume the square of the angular momentum stays the same. That is equivalent, of course, but then you get ##\omega^2## which you can replace using the relation between ##\omega## and D, that way you avoid fractional exponents.

jim mcnamara
Something went wrong with the derivative, it has mismatching units already.

Instead of assuming the angular momentum stays the same it might be easier to assume the square of the angular momentum stays the same. That is equivalent, of course, but then you get ##\omega^2## which you can replace using the relation between ##\omega## and D, that way you avoid fractional exponents.

Thank you so much! I cubed it instead so that I could replace D with the expression for D in terms of w and it worked out. I'll keep this in mind in the future!

mfb