Worksheet 13.1, Problem 2: Finding Exoplanets from Stellar Wobble

This problem discusses one of the ways in which we can detect exoplanets. This method involves detecting a planet's gravitational effect on a star's motion, known as wobble. In the system of the planet and the star, the two objects have a center of mass around which they both orbit, called a barycenter, as shown in the image below. 

Two masses orbiting around a common barycenter.
Image Credit: http://upload.wikimedia.org/wikipedia/commons/5/5a/Orbit4.gif

By measuring the Doppler shifts in the spectrum of the star, we can detect the "wobble" of the star as it orbits its barycenter. Such an observation would indicate the existence of an exoplanet orbiting the star. 

a) Start with the relationship between $a_p$ and $a_\star$ to find the relationship between speed of the planet and the star, $v_p$ and $v_\star$.

Let $a_p$ be the distance of the planet away from the barycenter, and $a_\star$ be the distance of the star from the barycenter. Also let $m_p$ be the mass of the planet and $M_\star$ be the mass of the star. Setting the point of the center of mass to be zero, from the center of mass equation, we get the relation:

$0=\dfrac{(M_\star )(a_\star ) - (m_p)(a_p)}{M_\star + m_p}$

Which gives us:

$a_\star M_\star = a_p m_p$

Now, multiplying both sides of the equation by $\frac{2\pi}{P}$, where $P$ is the mutual orbital period of the objects around the barycenter, we can use the relation, $P=\frac{2\pi r}{v}$ to solve for velocity:

$\left(\dfrac{2\pi a_\star }{P}\right) M_\star = \left(\dfrac{2\pi a_p}{P}\right) m_p$

Notice that $v=\frac{2\pi r}{P}$, so we can rewrite our equation:

$v_\star M_\star = v_p m_p$

From this relationship and the relationship we derived from the center of mass equation, we found that:

$\dfrac{a_p}{a_\star}=\dfrac{m_p}{M_\star}=\dfrac{v_p}{v_\star}$

b) Express the speed of the star, $v_\star$, in terms of the orbital period $P$, the mass of the star $M_\star$ and the mass of the planet $m_p \ll M_\star$. Convert your units so that your expression is given as a speed in meters per second, with $P$ measured in years, $M_\star$ in solar masses ($M_\odot$) and planet mass in Jupiter masses $M_{Jup}$.

From part (a), we found that the velocity can be written as $v=\frac{2\pi r}{P}$, so the speed of the star would be:

$K=\dfrac{2\pi a_\star}{P}$

To find $a_\star$, making the approximation $m_p +M_\star\approx M_\star$ since $m_p\ll M_\star$, we can use Kepler's Law to get:

$a = \left( \dfrac{P^2 GM_\star}{4\pi^2}\right)^{1/3}$

Where $a$ is the total distance between the planet and the star. We can then use the relationships $a=a_\star +a_p$ and $a_p=\frac{a_\star M_\star}{m_p}$ from the center of mass equation to solve for $a_\star$:

$a_\star + \dfrac{a_\star M_\star}{m_p} = \left( \dfrac{P^2 GM_\star}{4\pi^2}\right)^{1/3}$

Which gives us:

$a_\star=\dfrac{\left( \frac{P^2 GM_\star}{4\pi^2}\right)^{1/3}}{1+\frac{M_\star}{m_p}}$

Now plugging this back into our equation for Kepler's Law, we find that the speed of the star is:

$K=\dfrac{2\pi \left( \frac{P^2 GM_\star}{4\pi^2}\right)^{1/3}}{P\left( 1+\frac{M_\star}{m_p}\right)}$

Simplifying, we can multiply by $\frac{m_p}{m_p}$ and approximate $M_\star +m_p\approx M_\star$ to get:

$K=\dfrac{2\pi m_p\left( \frac{P^2 GM_\star}{4\pi^2}\right)^{1/3}}{P\left( m_p+M_\star\right)}$

$K=\dfrac{m_p\left( 2\pi G\right)^{1/3}}{P^{1/3}M_\star^{2/3}}$

Now converting our units into meters per second, years, Solar masses, and Jupiter masses, we assume that we are changing our units from cgs units:

$K=\dfrac{\left( m_p\cdot 2\times 10^{30}\frac{g}{M_{Jup}}\right)\left( 2\pi G\cdot 10^{-6}\frac{m^3}{cm^3}\right)^{1/3}}{\left( P\cdot \pi \times 10^7 \frac{s}{yr}\right)^{1/3}\left(M_\star\cdot 2\times 10^{33} \right)^{2/3}}$

c) 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?

As the star orbits its barycenter, we are able to observe a Doppler shift of the star's spectrum while it is moving away and moving towards the observer, as shown in the image below:

As the star moves away, it is redshifted, and as it moves towards us, it is blueshifted.
Image Credit: http://obswww.unige.ch/~udry/planet/Images/doppler.jpg

We only care about the line-of-sight direction, or the radial direction, since this will be the only direction which contributes to the Doppler shift. Since the star is rotating in a circular orbit around the center of mass, we can express its motion as a sine function. We know that the star is orbiting with a velocity of $K$, so its amplitude will be $K$ since this is the maximum radial velocity we will see, and we know that it is orbiting with a period $P$. So this gives us the expression:

$v_{rad}(r)=\sin\left(\frac{2\pi}{P}t\right)$

d) Sketch the velocity of a star orbited by a planet as a function of time. Denote the maximum velocity as $K$, and express the x-axis in terms of the number of periods, in intervals of $P/4$.

The following is a graph of the radial velocity of a star's wobble:

An example of a curve for the velocity amplitude of a star, 51 Pegasi, is shown in the image below:

Velocity of Wobble for 51 Pegasi
Image Credit:
https://exoplanetmusings.files.wordpress.com/2011/10/51_peg_rv.png?w=600

e) What is the velocity amplitude, $K$, caused by an Earth-mass planet in a 1-year orbit around a Sun-like star?

Using our equation from part (b), we can plug in known values:

$K=\dfrac{\left( M_\oplus\cdot 2\times 10^{30}\frac{g}{M_{Jup}}\right)\left( 2\pi G\cdot 10^{-6}\frac{m^3}{cm^3}\right)^{1/3}}{\left( P\cdot \pi \times 10^7 \frac{s}{yr}\right)^{1/3}\left(M_\odot\cdot 2\times 10^{33} \right)^{2/3}}$

The mass of the Sun is $M_\star =M_\odot$. We know that the mass of Earth is $M_\oplus\approx \frac{1}{318}M_{Jup}$, and the period of Earth's orbit around the sun is $P=1$ year. Plugging these values in, we get:

$K=\dfrac{\left( \frac{1}{318}M_{Jup}\cdot 2\times 10^{30}\frac{g}{M_{Jup}}\right)\left( 2\pi 6.67\times 10^{-8}\text{ cm}^3\text{ g}^{-1}\text{ s}^{-2}\cdot 10^{-6}\frac{m^3}{cm^3}\right)^{1/3}}{\left(1\text{ yr}\cdot \pi \times 10^7 \frac{s}{yr}\right)^{1/3}\left(M_\odot\cdot 2\times 10^{33} \right)^{2/3}}$

So we have a velocity amplitude of:

$K\approx 0.09\text{ m/s}\approx 9$ cm/s

This is a very small velocity, which makes sense when comparing the sizes of the Earth and the Sun.

I would like to thank Charles Law, Daniel Chen, and Andy Mayo for their help in solving this problem.