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An extrasolar planet can be detected by observing the wobble it produces on the star around which it revolves Suppose an extrasolar planet of mass $m_{B}$ revolves around its star of mass $m_{A}$ . If no external force acts on this simple two-object system, then its $\mathrm{CM}$ is stationary. Assume $m_{\mathrm{A}}$ and $m_{\mathrm{B}}$ are in circular orbits with radii $r_{\mathrm{A}}$ and $r_{\mathrm{B}}$ about the system's $\mathrm{CM} .(a)$ Show that$r_{\mathrm{A}}=\frac{m_{\mathrm{B}}}{m_{\mathrm{A}}} r_{\mathrm{B}}$(b) Now consider a Sun-like star and a single planet with the same characteristics as Jupiter. That is, $m_{B}=1.0 \times 10^{-3} m_{A}$ and the planet has an orbital radius of $8.0 \times 10^{11} \mathrm{m} .$ Determine the radius $r_{A}$ of the star's orbit about the system's CM.(c) When viewed from Earth, the distant system appears to wobble over a distance of 2$r_{A} .$ If astronomers are able to detect angular displacements $\theta$ of about 1 milliarcsec $\left(1$ arcsec $=\frac{1}{3600}$ of a degree), from what \right. distance $d$ (in light-years) can the star's wobble be detected $\left(1 \mathrm{ly}=9.46 \times 10^{15} \mathrm{m}\right) ?(d)$ The star nearest to our Sun is about 4 $\mathrm{ly}$ away. Assuming stars are uniformly distributed throughout our region of the Milky Way Galaxy, about how many stars can this technique be applied to in the search for extrasolar planetary systems?

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Physics 101 Mechanics

Chapter 9

Linear Momentum

Motion Along a Straight Line

Kinetic Energy

Potential Energy

Energy Conservation

Moment, Impulse, and Collisions

Cornell University

University of Michigan - Ann Arbor

Hope College

University of Winnipeg

Lectures

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Okay, This is the system that we're dealing with. Your, uh, this point over here is a center of mass. Of the whole system from radius are a from that is the star and bang from radius R B from that is the planet R R B. And that's on the other side, uh, of the star. So decided the orbit a star on. So in part A, we use the definition off center of mass coordinates. So that's ah s abi sum Of all the masses times, senator, last coordinates over the total less so in this case, it will be a m A times are a plus n b times minus r be upset side over m a plus. All right, and we set our center of mass to be a zero points of that zero. Uh, therefore, uh, lips are will be equal to M B over mm times R b er and part B. We figure out what our rays are. A is, um, one times 10 to the negative three a over m a times eight tons. 10 of the 11 meters for R B. So the Emmys cancel and you get eight times 10 to the eight meters are in part. See, our situation is something like this. Um, two are twice the radius of the diameter of the star stars orbit is well, we can see, uh, the distance d is the distance to it. And data is the angle that we, uh that it's a tense in the sky. And so, uh, been so Tam Fada is a cz two r a over duty, therefore, and and and so since you know that this angle is really, really small, it's a fraction of an arc second, which is a small fraction of a degree, which is a small fraction, which is a fraction of a radiant. Anyway, you get the point data. It's small, it's tiny. So, uh, for a small Seita lets astronomers use will be what I like to call what we call the small angle approximation. Uh, Zeta is approximately 10 data. That's the approximation. So in this, so too are a over D is approximately Saito. Therefore d will be to our air over Seita. Ah, Now, data is the whole beast itself. It's ah, one mill yard second, so one thousands of an arc second, except times which is on art. Second is one thes 3600 off a degree. Saw 3600 art seconds in one degree. That cancels out the art seconds. But we want it in terms of radiance. So remember, 1 80 degrees has pie. Radiance of it. Okay, so and so this gives us gets us 4.84 eight for eight times 10 to the negative nine radiance that stayed up. All right. And so this means that D is equal to two times eight times 10 to the eight meters over 4.8448 times 10 to the negative nine radiance. Um and so that's saying it. We want it in. And so what we really have is a unit of meat meters Ah, and meters. And so this is 3.3 times 10 to the 17 meters. But we wanted in terms of light years. So one later is 9.46 times 10 to the five ah meters or attend to the 15 meters. Excuse me. So this gives us 35 light years for the distance to the star, okay? And part D finally is Ah, probably the most interesting part of the problem. You want to figure out how many of these stars were gonna detect assuming a uniform distribution of stars in the galaxy relive in, which is a Milky Way. How many of these stars care how many of these wobbles? How many of these stars can be detected? Um, in this method by this method. So the number of starts. So we're assuming we can assume spherical volumes throughout this part. Right? So we're gonna so well, this means is that first we have volume of wobble, so volume off stars that are detectable by this wobble method. So that's four pi over three. Volume of a sphere is four pi over three times r Cubed R is the distance to it. So that's 35 light years cubed on. And then we have the volume of a start itself which, uh, we which we will assume to be four again. Four pi over two times 1/2 the distance to the Mirror Star, which is two light years cubed. Therefore therefore, the number of stars number of stars, eyes equal to vault volume detectable from the wobble method are, uh, wrecked over the Volume ave off the nurse of a single star. That's four pi over three times 35 light years cubed huge sorry over a four pi over to e times. Two light years cubed. And so this gives us about 5400 stars, and that's it.

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