Consider an eclipsing binary whose orbital plane is parallel...

Consider an eclipsing binary whose orbital plane is parallel to our line of sight. Doppler measurements of the radial velocity of each component of the binary are shown in Figure $13-36 .$ Assume that the mass $m_{1}>m_{2}$ and that the orbits of each component about the center of mass are circular. ( $a$ ) What is the period $T$ and the angular frequency $\omega$ of the binary? ( $b$ ) Show that in this case $\left(m_{1}+m_{2}\right)=\left(\omega^{2} r^{3}\right) / G,$ where $r=$ separation of the binary. ( $c$ ) Compute the values of $m_{1}, m_{2}$, and $r$ from the data in the $v$ versus $t$ graph.

Key Concepts

Doppler Effect in Astronomical Measurements

The Doppler Effect involves the change in the observed wavelength of light due to the relative motion between a source and an observer. In the study of eclipsing binary systems, measuring the Doppler shifts in the spectral lines of each star provides estimates of their radial velocities, which can then be used to infer orbital parameters such as period, separation, and masses. This technique is fundamental in astrophysics for probing the dynamics of systems not directly spatially resolved.

Orbital Period and Angular Frequency

In orbital mechanics, the orbital period (T) is the time required for an object to complete one full orbit, and the angular frequency (?) is defined as the rate at which the object sweeps through its orbit (? = 2?/T). This concept is fundamental for characterizing the timing and geometry of orbital motion in systems ranging from planets to binary stars.

Kepler's Third Law and Newtonian Gravitation

Kepler's Third Law, when combined with Newton's law of gravitation, relates the sum of the masses of two orbiting bodies to their orbital separation (r) and angular frequency (?) via the formula (m1+m2) = (?²r³)/G. This relationship is key for determining stellar masses from observable orbital dynamics, as it arises from equating gravitational forces with the centripetal force required for circular motion.

Center of Mass and Orbital Dynamics in Binary Systems

In a binary star system, the center of mass is the balance point about which both stars orbit. The distances of the stars from this center depend inversely on their masses, meaning the more massive star moves on a smaller orbit compared to its companion. Understanding this concept is essential for deciphering how variations in radial velocity, as seen through Doppler measurements, relate to the individual masses and orbital characteristics of the system.

Transcript

00:01 Okay, in this question, i'm talking about the orbital dynamics of binary star system.

00:07 So we're given this graph, which shows the velocity of the two stars seen from a direction perpendicular to the plane of the orbit.

00:20 And you know that the mass of star 1 is created in the mass of star 2.

00:28 All right, so question a starts by asking us, what is the period? and also the angle frequency of this star system.

00:40 The period is just the amount of time it takes to complete one cycle, which is the same for both stars.

00:46 They both complete one cycle at this point, which is at 12 days, which in seconds is 1 .037 times 10 to the 6 seconds.

01:01 Okay, and the angle of frequency is equal to 2 pi over town.

01:11 That's how it's fine.

01:13 So this is equal to 6 .006 times 10 negative 6 radians per 7.

01:23 All right.

01:26 And in question b, they would like us to prove that the following equation is true.

01:35 M1 plus m2 is equal to omega squared r cubed over g.

01:44 All right, so one thing we know about things in orbit is that the centripetal force is balanced out by the gravitational force.

01:56 So if we just substitute in the definitions of these two forces, we should get an equivalent expression.

02:04 All right, so let's start with centrifugal force.

02:07 Centripetal acceleration is equal to omega squared r.

02:13 And so since force is equal to mass times the acceleration, the centripetal force is equal to the reduced mass times the acceleration.

02:25 And reduced mass, mu is equal to m1 times m2 over m1 plus m2.

02:33 So we'll plug that in and we'll plug in this.

02:37 So we find that centripetal force is equal to m1, m2 over m1 plus m2 times omega squared r.

02:52 All right, now the gravitational force, this one's a little bit simpler.

02:56 It's our normal formula, g, m1, m2, over r squared.

03:03 All right, so now we just want to set these two things equal to each other.

03:07 So i'll do that now.

03:09 And then we can immediately cancel this top term, because it appears in both sides.

03:16 And then if we multiply the r squared to this side, multiply both sides by m1 plus m2 and divide both sides by g.

03:24 We get exactly the expression we're looking for, which is m1 plus m2 is equal to omega squared r cubed over g.

03:37 All right.

03:37 Now, question c has a few more parts, but we're going to get through it, so let's hold on.

03:43 So this question asks us to find the radius that represents the distance between the two stars and also each planet's mass.

03:58 Okay, so we're going to need several equations since we have several unknowns.

04:05 All right, so let's start with finding the radius...

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