Question

Consider two stars in orbit about a mutual center of mass. If a1 is the semi major axis of the orbit of the star of mass m1 and a2 is the semi major axis of the orbit of the star of m2, prove that the semi major axis of the orbit of the reduced mass is given by a=a1+a2 hint: recall that r=r2-r1

Answer 1

Recall from the definition of the center of mass that m1m1r1 = −m2r2, where r1 and r2 are measured from the center of mass This gives us |r2| = m1 m2 |r1| as a result when star 1 is at its furthest distance from the center of mass star 2 is also at its furthest distance from the center of mass Remember that the furthest distance from focus 1 to a point on an ellipse occurs when an object is on its major axis near focus 2 at which point its distance from focus 1 is a+ae, where a and e are the semimajor axis and eccentricity of the ellipse in addition the center of mass is a shared focus of the two elliptical orbits which tells us that star 1 is at a distance a1(1 + e) from the center of mass when star 2 is at a distance a2(1+e) from the center of mass (note that we have assumed here that both stars orbit in ellipses with the same eccentricity it’s true and if we wanted to prove it we could use our expressions for the stars’ positions relative to the center of mass) At this moment, since the stars are always on opposite sides of the center of mass we have r1 = −a1nˆ and r2 = a2nˆ where ˆn is a unit vector that points from the center of mass towards the apoastron of star 1 Thus the corresponding position of the reduced mass in the associated one-body problem is r = r2 − r1 = (a1 + a2)(1 + e)ˆn This is the maximum distance of the reduced mass from the origin in the associated one-body problem because it is the maximum value of r (the maximum separation between stars 1 and 2) The distance is thus equal to a(1 + e), where a is the semimajor axis of the reduced mass’s orbit Setting a(1 + e) = (a1 + a2)(1 + e) gives a = a1 + a2