Obviously our first step here is to define $a_p$ and $a_*$. Key to understanding this is realizing that planets don't orbit stars per se--in reality, both objects orbit the center of mass of the system. As it happens, though, stars are so enormous compared to planets that this point tends to be well inside the body of the star itself. However, this does mean that the star is "wobbling" as it orbits a point that isn't quite its center. There are some great animations of this happening in different setups here.
The barycenter (center of mass) of Sun and Jupiter is just outside the Sun's surface. Note that this isn't quite the barycenter of the Solar System as a whole--considering all planets and other objects complicates the problem enormously. However, Jupiter is so massive compared to the other planets that our Solar System can be thought of in this regard as being the Sun-Jupiter system (source).
We found in an earlier problem that the offsets of each object from the system's barycenter proportionate to mass--in other words, $a_p \times M_p = a_* \times M_*$. To figure out the velocity of each object, we'll consider the period of each, which we can use in a distance-rate-time equation: the circumference of each's orbit must be equal to its orbital velocity multiplied by its period (note that the period for both objects must be the same). In mathematical terms, we can solve for period and set up the following inequality:\[\frac{2\pi a_*}{v_*} = \frac{2\pi a_p}{v_p}\] Dividing out the $2\pi$ from both sides, we have\[\frac{a_*}{v_*} = \frac{a_p}{v_p}\]
b) Express the speed of the star, $K$, in terms of the orbital period $P$, the mass of the star $M_*$, and the mass of the planet $M_p << M_*$.
We know right off the bat that $KP = 2\pi a_*$--again, this is just distance equals rate times time. From there, all we need to do is substitute in for $a_*$. Since we know that $a_p \times M_p = a_* \times M_*$, we can solve for $a_*$ to find that $a_* = a_p \times M_p / M_*$. We can then substitute this back into our original equation and solve for $K$ to find that\[K = \frac{2\pi M_p a_p}{PM_*}\]
c/d) We can measure the velocity of a star along the line of sight using a technique similar to the way in which you measured the speed of the Sun's limb due to rotation. Specifically, we can measure the Doppler shift of stellar absorption lines to measure the velocity of the star in the radial direction, towards or away from the Earth, also known as the "radial velocity." What is the time variation of the line-of-sight velocity of the star as a planet orbits? Sketch the velocity of a star orbited by a planet as a function of time.
This problem is worded somewhat confusingly--it's important to understand for it that we're discussing the radial velocity of a distant star, being tugged about by a large companion planet. This is also assuming that we're sitting in or close to the plane of that planet's orbit, so it won't be getting tugged up and down, just around in a circle. Knowing all of this, it should be fairly intuitive to realize that the star's radial velocity would follow a sine wave--when the planet is approaching us, the star will be moving away, and when the planet is moving away from us, the star will be moving closer. When the planet is between us and the star, and vise-versa, the star would appear to be sitting still.
Velocity of the star, plotted against time. The amplitude of the star's velocity is $K$ (when it moves directly toward or away from us, we see it moving at its true orbital velocity of $K$ or $-K$), and the length of one period is (surprise!) $P$.
e) What is the velocity amplitude, $K$, of an Earth-mass planet in a 1-year orbit around a Sun-like star?
We have the equation for $K$ that we came up with in part b), so this is merely a matter of plugging in $M_\oplus$ for $M_p$, 1 AU for $a_p$, 1 year ($3.16 \times 10^7$ seconds) for $P$, and $M_\odot = 2\times10^{33}$ g for $M_*$. The result: a whopping 9 cm/s, which WolframAlpha helpfully lets us know is, among other things, 10% faster than the top speed of a sloth.
Credit and appreciation goes out to Jennifer Shi, Louise Decoppet, and Scott Zhuge for working through this one with me.
Very good! So if we don't see sloths red- or blue-shifting, it's pretty clear why detecting Earth-like planets is such a challenge... 4/4.
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