Exoplanets and Their Orbits

 

Planets do not orbit stars. Rather, the star and planet both orbit the center of mass of the system. The stars therefore "wobble" due to the tug of gravity from their planet. This wobble allows astronomers to detect planets orbiting stars other than the sun using the doppler technique. The light coming from the planet and star shifts when the star's line of sight velocity changes.

1. (a) The center of mass of a planet star system in general is given by $x_{com }= \sum_{i} m_i x_i/  \sum_{i} m_i$ Set up the problem by drawing an x-axis with the star at $x = -́a_*$ with mass $M_*$, and the planet at $x = a_p$ and mass $m_p$. Also, set $x_{com}= 0$. How do $a_p$ and $a_*$ depend on the masses of the star and planet?

$x_{com }= \sum_{i} m_i x_i/  \sum_{i} m_i$

$m_pv_p=M_*v_*$
$\frac{m_pv_p^2}{2}= \frac{m}{2}- \frac{M_*^2v_*^2}{m_p^2}=\frac{M_*^2v_*^2}{2m_p}$

$x_{com }= \sum_{i} m_i x_i/  \sum_{i} m_i$

$0=\frac{m_pa_p-M_*a_*}{m_p+M_*}$

$M_*a_*=m_pa_p$

$\frac{M_*}{m_p}=\frac{a_p}{a_*}$

(b) In a two-body orbital system the variable a is the mean semimajor axis, or the sum of the
planet’s and star’s distances away from their mutual center of mass: a “ ap ` a‹. Label this
on your diagram. Now derive the relationship between the total mass M‹ ` mp « M‹, orbital
period P , and the mean semimajor axis a, starting with the Virial Theorem for a two-body
orbit (assume circular orbits from here on).

$K=-\frac{U_g}{2}$

$1/2(M_*v_*^2+m_pv_p^2)=-1/2(\frac{-GM_*m_p}{a})$
$1/2(M_*^2+ \frac{M_*^2v_*^2}{m_p})=\frac{GM_*m_p}{2a}$
$\frac{M_*v_*^2}{2}(1+ \frac{M_*}{m_p})= \frac{GM_8m_p}{2a}$

$\frac{M_*v_*^2}{m_p}=\frac{Gm_p}{a}$

$aM_*(\frac{2 \pi}{P}a_*)=Gm_p^2$

$aM_* \frac{4 \pi ^2}{P^2}a^2 \frac{m_p^2}{M_*^2}=Gm_p^2$

$\frac{4 \pi ^2}{G}= \frac{P^2M_*}{a^3}$

(c) By how much is the Sun displaced from the Solar System’s center of mass (a.k.a. the Solar
System “barycenter”) as a result of Jupiter’s orbit? Express this displacement in a useful
unit: Solar radii (Rd). Here are some useful numbers: Md «1000 MJup, PJup « 12 years,
Rd « 7 ˆ 1010 cm.

$\frac{4 \pi ^2}{G}= \frac{P^2M_*}{a^3}$

$a_j=(\frac{P_j^2M_*G}{4 \pi^2})^{1/3}$
$a_o=a_j\frac{M_j}{M_o}$
$a_o=(\frac{P_j^2M_oG}{4 \pi^2})^{1/3}\frac{M_j}{M_o}$

Now we plug in our values

$a_o=(\frac{(12\times 365 \times 24 \times 3600)^2 \times 2 \times 10^{30} \times 6.7 \times 10^{-11}}{4 \pi^2})^{1/3}\times 10^{-3}$

$a_o= 3.76 \times 10^8 cm$

$$a_o= 0.5 R_o$

(d) Start with the relationship between ap and a‹ to find the relationship between speed of the
planet and the star, v_p and v_*.

$\frac{2 \pi a_*}{v_*}=P=\frac{2 \pi a_p}{v_p}$

$\frac{a_*}{a_p}=\frac{v_*}{v_p}$

(e) For this part, we’ll take a bit of a diversion to learn a new way of expressing equations. Start
with the expression of the surface gravity of a panet, g, as a function of the planet’s mass, $m_p$,
and radius $R_p$, and re-expess it as a power law of the form
g =[some numbers] cm\s^2 (\frac{M_p}{M_o})^{\alpha}\times (\frac{R_p}{R_o})^β
The grouping of constants you pull out front should give you some
number of $cm/s^2$.

$mg=\frac{Gmm_p}{R_p^2}$
$g=G\frac{m_p}{R_p^2}=G\frac{m_p/M_o}{(R_p/R_o)^2}\times \frac{M_o}{R_o^2}$
$g= \frac{GM_o}{R_o^2} \times \frac{m_p}{M_o} \times (\frac{R_p}{R_o)^{-2}}$

$\alpha=1, \beta= -2$
$10^2=\frac{6.7 \times 10^{-11} \times 6 \times 10^{24}}{6.4 \times 10^12}cm/s^2=10^3 cm\s^2$

(f) Express the speed of the star, $v_*$, in terms of the orbital period P , the mass of the star $M_*$
and the mass of the planet mp, where $m_p  < M_*$. Give your answer in the form of a power law,
with units of meters per second (for some reason astronomers switch to m.k.s. for the radial
velocity technique, but nothing else).
$v= [some number] m/s $(\frac{M_*}{M_p})^{\alpha}(\frac{P}{1 year})^β (\frac{m_p}$ ${M_{Jup}})^{\gamma}$

We know that $v_p = \frac{2 \pi a_p}{P}$
We can solve for $a_p$

$a=a_pc\frac{m_p}{M_*}$

$a=a_* +a_p= a* (a+\frac{M_*}{M_p})$
$a_*=a \frac{m_p}{M_*+m_p}approximately a \frac{m_p}{M_*}$ since $m_p  < M_*$
$\frac{4 \pi ^2}{G}= \frac{P^2GM_*}{a^3}$
This gives us $a=(\frac{P^2GM_*}{4 \pi^2})^{1/3}$

We can substitute a back into $v_p = \frac{2 \pi a_p}{P}$

$v_p= \frac{2 \pi}{P} \frac{m_p}{M_*} \frac{P^{2/3}G^{1/3}M^{1/3}}{2 \pi^{2/3}}= [some number] m/s $(\frac{M_*}{M_p})^{\alpha}(\frac{P}{1 year})^β (\frac{m_p}$

$v_p=(\frac{Gm_p^3}{2 \pi PM_* ^2})^{1/3}$

$\alpha = -2/3 \beta = -1/3 \gamma = 1$

$some number= G^{1/3}M^{-2/3}1yr^{-1/3}M_j 2 \pi ^{-1/3}$

Substituting in our constants and given values we get $ 30 m/s$

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.”

3. Let’s look at some actual data. The following plots are from Prof. Johnson’s PhD thesis. Each data
point is a radial velocity measured from an observation of the star’s spectrum, and the dashed line
is the best-fitting orbit model. Prof. Johnson found the planet around HD 167042 when he was a
grad student, and each data point represents a trip from Berkeley, CA to Mt. Hamilton and a long
night at the telescope.

(a) What are the periods, velocity amplitudes and planet masses corresponding to the two radial

velocity time series below? The star 18 Del has $M_D= 2.3 M_o$, and HD 167042 has $M_H=1.5 M_o$.

We can get the velocities from the radial velocity curves

$v_H=35m/s$

$v_D=120 m/s$

We can also determine the periods from the radial velocity curves by measuring the distance between peaks.

$P_H=1.2 years$

$P_D=2.7 years$

we use the following formula to solve for the planet masses.

$v_*= 30 m/s $(\frac{M_*}{M_o})^{-2/3}(\frac{P}{1 year})^{-1/3} \frac{m_p}{M_{Jup}}$

$m_p= \frac{v_*}{30m/s}\frac{M_*}{M_o})^{-2/3}\frac{P}{1 year})^{-1/3}M_{Jup}$

$m_H=2.4 \times 10^{27}kg$

$m_D=1.5 \times 10^{28}kg$

(b) For the figure showing HD 167042, “trend removed” means that in addition to the sinusoidal
variations, there was also a linear change in the star’s velocity. What would cause such a
“trend?”

 Such a trend could be caused by another large mass in the star planet system. This would cause acceleration of the center of mass which would lead to a change in the sinusoidal curve of the radial velocity graph.

Worksheet 13

 

Some exoplanetary systems are aligned just right, such that the planet is seen to eclipse its host star.
When this happens, astronomers can observe a periodic dimming of the star’s light. This worksheet will
guide you through some of the basics of the transit method of finding exoplanets

Now draw the star projected on the sky, with a dark planet passing in front of the star along the
star’s equator.

(a) How does the (fractional) depth of the transit depend on the stellar and planetary physical
properties?

$\frac{F_p}{F_*} \sim \frac{A_p}{A_*}=\frac{\pi R_p^2}{\pi R_*^2}$

transit depth = $\delta =( \frac{R_p}{R_*})^2$

(b) What is the depth of a Jupiter-sized planet transiting a Sun-like star?

 $\delta _J =( \frac{0.1R_o}{R_o})^2
 $\delta _J =( \frac{R_J}{R_*})^2$= 0.1^2= 0.01$

(c) Compare the transit depth of an Earth-sized planet to the depth of a Jupiter-sized planet.

 $\delta _E =( \frac{0.1R_E}{R_o})^2=( \frac{6400}{690000})^2= 1 \times 10^{_-4}$

$\frac{\delta _J}{\delta _E}= \frac{1 \times 10^{-2}}{1 \times 10^{-4}}= 1 \times 10^2$

(d) In terms of the physical properties of the planetary system, what is the transit duration, T,
defined as the time for the planet’s center to pass from one limb of the star to the other?

$\frac{(2R_*+2R_p)}{2 \pi a}=\frac{T}{P}$

$T=P \frac{R_*+R_p}{\pi a}$

(e) What is the duration of “ingress” and “egress”, τ , in terms of the physical parameters of the
planetary system?

$\tau = \frac{PR_p}{\pi a}$

3. For the transit light curve and radial velocity time series on the next page, and assuming a 3.1 day
orbit,

(a) What is the qualitative brightness distribution of the star’s surface as viewed from the Earth
(HINT: think about the bottom portion of the transit light curve)
The star's surface is brightest in the middle and becomes dimmer around the edges.

(b) What is the planet’s radius compared to the star’s radius ($R_p /R_*$)? (HINT: Use the average
of the portion of the light curve during which the planet is completely in front of the star)

$\delta = (\frac{R_p}{R_*})^2$

$(\delta )^{1/2} = \frac{R_p}{R_*} $

$0.027^{1/2}= 0.17$

The radius of the planet is 0.17 times the radius of the star.

(c) What is the planet’s semimajor axis compared to the star’s radius ($a/R_*$)?

$T=\frac{P}{\pi a} \times R_*(1+ \frac{R_p}{R_*})$
$T= \frac{P}{\pi } \frac{R_*}{a} (a+ \delta^{1/2}$
$\frac{R_*}{a}= \frac{\pi T}{P(1+ \delta ^{1/2}}= 2/24$
$\frac{a}{R_*}=12$

(d) What is the size of the planet compared to Jupiter if the star has a radius of $0.8 R_o$?

$\delta = (\frac{R_p}{R_*})^2$

$\frac{R_p}{0.8R_o}=0.17$

$\frac{R_p}{R_J}=\frac{0.14R_o}{R_J}$

$0.14/0.1= 1.4 $

The planet's radius is 1.4 times the radius of Jupiter

(e) Show that the scaled semimajor axis, $a/R_*$, is related to the average stellar density, $ρ_*$

$\frac{a}{R_*}= \frac{P(1+\frac{R_p}{R_*})}{\pi T}$

$\frac{4 \pi ^2}{G}=\frac{P^2 M_*}{a^3}$

$P^2= \frac{4 \pi ^2 a^3}{GM_*}$

Now we can substitute our above equation in for P

$\frac{a}{R_*}=\frac{1+\frac{R_p}{R_*}}{\pi T} \times \frac{2 \pi}{G^{1/2}} \times \frac{a^{3/2}}{M_*^{1/2}}=\frac{1+ \frac{R_p}{R_*}}{\frac{P(R_p+R_*)}{\pi a}} \times \frac{a^{3/2}}{G^{1/2} \rho _*^{1/2}R_*^{3/2}} $

$(\frac{a}{R_*})^2= (\frac{1+ \frac{R_p}{R_*}}{\frac{P(R_p+R_*)}{\pi a}} \times \frac{a^{3/2}}{G^{1/2} \rho _*^{1/2}R_*^{3/2}})^2$

$\frac{a}{R_*}=( \frac{P^2G \rho _*}{\pi ^2})^{1/3}$