Worksheet 13.2, Problem 3
This problem is about the WASP-10 planetary system, which has a transiting hot Jupiter with a 3.1 day orbit. We are given the transit light curve and the radial velocity curve, shown below.
 |
| Transit Light Curve |
 |
| Radial Velocity Plot |
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 hint points us towards the bottom of the transit light curve, which is valley-shaped. That means that the most brightness is lost when the planet transits in the middle of the star, while less brightness is lost when it transits in the edges. Therefore, we can conclude that the star is brightest in the middle, and darkens as we move away from the center. This is known as "limb darkening".
b)
What is the planet's radius compared to the star's radius ($R_p/R_{\star}$)? (HINT: Use the average
of the portion of the light curve during which the planet is completely in front of the star)
We use the equation for transit depth, $\delta$ to compare the radius of the planet to that of the star:
$$\delta = \left(\frac{R_p}{R_{\star}}\right)^2$$
$$\frac{R_p}{R_{\star}} = (\delta)^{\frac{1}{2}}$$
From the transit light curve, we can see that the transit depth $\delta = 0.03$. Plug in to solve:
$$\frac{R_p}{R_{\star}} = (0.03)^{\frac{1}{2}} = 0.173$$
c)
What is the planet's semimajor axis compared to the star's radius ($a/R_{\star}$)?
From Problem 2 on this worksheet, we solved for transit duration $T$. We can rearrange to isolate $a/R_{\star}$.
$$T = \frac{PR_{\star}}{\pi a}(1-b^2)^{\frac{1}{2}}$$
$$a/R_{\star} = \frac{P}{\pi T}(1-b^2)^{\frac{1}{2}}$$
We know that period $P = 3.1 \text{ days}$. From the transit light curve, we know that the transit duration $T = 2 \text{ hr}$, and the duration of "ingress" and "egress", $\tau = 1 \text{ hr}$.
Using the equations for $T$ and $\tau$, we can solve for b:
$$T = \frac{PR_{\star}}{\pi a}(1-b^2)^{\frac{1}{2}}$$
$$\tau = \frac{PR_{p}}{\pi a}\frac{1}{(1-b^2)^{\frac{1}{2}}}$$
Rearrange:
$$\frac{\tau (1-b^2)^{\frac{1}{2}}}{R_p} = \frac{T}{R_{\star}(1-b^2)^{\frac{1}{2}}}$$
$$b^2 = (1-\frac{T}{\tau}\frac{R_p}{R_{\star}})$$
$$b^2 = (1-\frac{2 \text{ hr}}{1 \text{ hr}}(0.03)^{\frac{1}{2}}) = 0.65$$
Now we can plug in to solve for the ratio $a/R_{\star}$:
$$a/R_{\star} = \frac{3.1 \text{ days} \times \frac{24 \text{ hr}}{\text{day}}}{\pi \times 2 \text{ hr}}(1-0.65)^{\frac{1}{2}} = 7.0$$
d)
What was the transit probability for this system (how lucky were we to find it transiting)?
$$p_{tr} = \frac{R_{\star}}{a} = \frac{1}{7.0} = 0.14$$
e)
What is the size of the planet compared to Jupiter if the star has a radius of 0.8 $R_{\odot}$?
$$\delta = \left(\frac{R_P}{R_{\star}}\right)^2$$
$$\delta = \left(\frac{R_P}{0.8 R_{\odot}}\right)^2$$
$$R_P = 0.03^{\frac{1}{2}}\times 0.8 R_{\odot} = 0.139 R_{\odot}$$
$$R_P = 0.139 \frac{R_{\odot}}{R_{Jup}} R_{Jup}$$
$$R_P = 0.139 \frac{69.551 \times 10^9 \text{ cm}}{6.9911 \times 10^9 \text{ cm}} R_{Jup}$$
$$R_P = 1.38 R_{Jup}$$
f)
Show that the scaled semimajor axis, $a/R_{\star}$, is related to the stellar density, $\rho_{\star}$.
We start with Kepler's 3rd Law:
$$P^2 = \frac{4\pi^2 a^3}{GM_{\star}}$$
Now, we can express $M_{\star}$ in terms of $\rho \times V$:
$$M_{\star} = \rho_{\star} \times \frac{4}{3}\pi R_{\star}^3$$
Let's plug in this expression for mass in Kepler's 3rd Law:
$$P^2 = \frac{4\pi^2 a^3}{G\rho_{\star} \frac{4}{3}\pi R_{\star}^3}$$
$$\left(\frac{a}{R_{\star}}\right)^3 = \frac{P^2 G \rho_{\star}}{3\pi}$$
$$\frac{a}{R_{\star}} = \left(\frac{P^2 G \rho_{\star}}{3\pi}\right)^{\frac{1}{3}}$$
g)
What is the density of the star compared to the density of the Sun?
Let's solve for $\rho_{\star}$ and then compute it in terms of the density of the Sun, $\rho_{\odot}$. For reference, $\rho_{\odot} = 1410 \text{ kg/m}^3$.
$$\left(\frac{a}{R_{\star}}\right)^3 = \frac{P^2 G \rho_{\star}}{3\pi}$$
$$\rho_{\star} = \frac{3\pi}{P^2 G}\left(\frac{a}{R_{\star}}\right)^3$$
$$\rho_{\star} = \frac{3\pi}{P^2 G}\left(\frac{a}{R_{\star}}\right)^3\frac{1}{\rho_{\odot}}\rho_{\odot}$$
$$\rho_{\star} = \frac{3\pi}{(3.1 \text{ days}\cdot \frac{24 \text{ hr}}{\text{ day}}\cdot\frac{3600 \text{ s}}{\text{hr}})^2 \cdot 6.67\times 10^{-11}\text{ m}^3\text{ kg}^{-1}\text{ s}^{-2}}(7.0)^3\frac{1}{1410 \text{ kg/m}^3}\rho_{\odot}$$
$$\rho_{\star} = 0.48 \rho_{\odot}$$
The star is about half as dense as the Sun.
h)
What is the density of the planet compared to the density of Jupiter?
Let's express the density of the planet:
$$\rho_p = m_p/V_p$$
$$\rho_p = \frac{m_p}{\frac{4}{3}\pi R_p^3}$$
From Worksheet 12, we know that the peak radial velocity of the star is:
$K = \frac{2\pi a_{\star}}{P}$
We already know $P$, so let's solve for $a_{\star}$:
From the center of mass equation, we have:
$$m_p a_p = M_{\star}a_{\star}$$
$$a_p = \frac{M_{\star}}{m_p}a_{\star}$$
$$a = a_p + a_{\star} = a_{\star}(1 + \frac{M_{\star}}{m_p})$$
Let's plug this into the equation for Kepler's Third Law:
$$P^2 = \frac{4\pi^2 a^3}{GM}$$
$$P^2 = \frac{4\pi^2 \left(a_{\star}(1 + \frac{M_{\star}}{m_p})\right)^3}{G(M_{\star} + m_p)}$$
$$a_{\star} = \left(\frac{P^2G(M_{\star}+m_p)}{4\pi^2}(1 + \frac{M_{\star}}{m_p})^{-3}\right)^{\frac{1}{3}}$$
Then, plug back into the equation for $K$:
$$K = \frac{2\pi}{P}\left(\frac{P^2G(M_{\star}+m_p)}{4\pi^2}(1 + \frac{M_{\star}}{m_p})^{-3}\right)^{\frac{1}{3}}$$
$$K = \frac{2\pi}{P}\left(\frac{P^2Gm_p^3}{4\pi^2}\frac{1}{(M_{\star}+m_p)^2}\right)^{\frac{1}{3}}$$
$$K = \frac{2\pi m_p}{P}\left(\frac{P^2G}{4\pi^2(M_{\star} + m_p)^2}\right)^{\frac{1}{3}}$$
We can approximate $M_{\star} + m_p \approx M_{\star}$
$$m_p \approx \frac{KP}{2\pi}\left(\frac{P^2G}{4\pi^2M_{\star}^2}\right)^{-\frac{1}{3}} = \frac{KP^{\frac{1}{3}}M_{\star}^{\frac{2}{3}}}{2^{\frac{1}{3}}\pi^{\frac{1}{3}}G^{\frac{1}{3}}}$$
$$m_p = \frac{500 \text{ m/s}(3.1 \text{ days}\cdot \frac{24 \text{ hr}}{\text{ day}}\cdot\frac{3600 \text{ s}}{\text{hr}})^{\frac{1}{3}}M_{\star}^{\frac{2}{3}}}{2^{\frac{1}{3}}\pi^{\frac{1}{3}}G^{\frac{1}{3}}}$$
Now, we need to solve for $M_{\star}$:
$$M_{\star} = \rho_{\star}V_{\star} = \rho_{\star}\frac{4}{3}\pi R_{\star}^3$$
$$M_{\star} = 0.48 \rho_{\odot} \frac{4}{3}\pi (\frac{1.38}{0.173} R_{Jup})^3$$
$$M_{\star} = 0.48 \times 1410 \text{ kg/m}^3 \cdot \frac{4}{3}\pi (\frac{1.38}{0.173} \cdot 71492 \text{ km})^3 = 5.26 \times 10^{20} \text{ kg}$$
Now, solve for $\rho_p$:
$$\rho_p = \frac{500 \text{ m/s}\cdot(3.1 \text{ days}\cdot \frac{24 \text{ hr}}{\text{ day}}\cdot\frac{3600 \text{ s}}{\text{hr}})^{\frac{1}{3}}(5.26 \times 10^{20} \text{ kg})^{\frac{2}{3}}}{2^{\frac{1}{3}}\pi^{\frac{1}{3}}(6.67\times 10^{-11}\text{ m}^3\text{ kg}^{-1}\text{ s}^{-2})^{\frac{1}{3}}}\cdot \frac{1}{\frac{4}{3}\pi (1.38 \cdot 71492 \text{ km})^3}$$
$$\rho_p = 6.97 \times 10^5 \text{ kg/m}^3$$
This is $\frac{6.97 \times 10^5 \text{ kg/m}^3}{1330 \text{ kg/m}^3} = 524.6$ times the density of Jupiter.
Acknowledgement: collaborated with Allison Tsay and Daniel Gatega
General Post
In a new paper published in Nature, astrophysicists Szabolcs Marka at Columbia University and Imre Bartos at the University of Florida describe the collision of two neutron stars 4.6 billion years ago as the source of some of Earth's most precious elements. They found that this event created about 0.3 percent of the heaviest elements found on earth, such as uranium, platinum, and gold. Bartos makes a human connection to this finding, noting that about 10 milligrams of any wedding ring would have been formed 4.6 billion years ago.
Meteorites formed in the early solar system contain trace radioactive isotopes, whose decay can be used to date the meteorites. The researchers made their findings by comparing the composition of meteorites to simulations of the Milky Way galaxy. They found a single collision of neutron stars could have happened right before Earth was formed, predating it by 100 million years, about 1000 light years away from the gas cloud that would eventually form our Solar System.
