Exoplanet Activity - Part 1

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ASTR 3000 — Exoplanet Properties, Part I 1

Introduction

How do astronomers learn so much about a planet just by watching how a star wiggles around? I could lecture a couple of days and explain it, but I am not sure it would really sink in, so instead we will find out what we can about an exoplanet by following the same steps astronomers take. Here is our basic scenario: The star HD4308 is a K0 star with an apparent magnitude of 6.54. It is located in the constellation Tucana at a distance of 21.9 pc. Observations taken from the European Southern Observatory’s 3.6 meter telescope at La Sillia Observatory with the HARPS eschelle spectrograph over between September 7, 2003 and July 28, 2005 indicate that this star has a slight wobble, most likely due to a planetary companion. The radial velocity of the star for each observation is shown in figure 1. These data indicate that the orbital period of the star-planet system is 15.56 days.

Figure 1: Intermediate season of HARPS radial velocities for HD4308. The best fit of the data gives an orbital period of 15.56 days for the planet (from Udry et al. 2006. “The HARPS search for southern extra-solar planets V.” Astronomy & Astrophysics 447: 361–367)

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Orbital Distance

The first priority is to find the average distance between the star and the planet. We can use a generalization of Kepler’s law to find this information. This general Kepler’s law states that if two objects, masses MA and MB are orbiting eachother than their orbital period (P ), measured in years and average distance (a) between the star and planet, measured in astronomical units, will follow the equation MA + MB =

a3 P2

.

(1)

Unfortunately it is at this point where you can get stuck. After all we don’t know the mass of the star plus the planet yet, and the period is given in days. We can get around these little problems though. 1. Convert the period to years, there are conversion factors in the appendix.

2. Find the mass of HD4308 using its spectral class. A table of properties of Main sequence stars is in the appendix.

3. Assume that the mass of the planet is negligible compared with the star. Solve for the average distance (a) between the star and planet using the equation above. We will check that this assumption is valid later.

Assuming the mass of the planet is small we have found how far apart the star and planet are. Now we can start to explore some interesting properties.

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Habitability

If we assume that life on other planets will depend on the presence of liquid water located on the surface of a planet as it does on Earth, then there is a limited zone around a star that can support life. This zone is called the habitable zone. Since this depends only on the amount of energy provided by the star, it varies depending on the spectral class, or mass of the star. Figure 2 shows the range of distances from different Main sequence stars in which a planet can have liquid surface water.

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Figure 2: Habitable zone for Main sequence stars (from NASA’s Kepler mission http://kepler.nasa.gov/sci/ basis/nature.html). 1. What is the range of the habitable zone for the star HD4308? .

2. Is the planet too close to the star to support liquid surface water? within the habitable zone? too far from the star to support liquid surface water?

Oh well, we still don’t even know what kind of planet it is. Let’s delve deeper.

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Planetary Mass

Now we are ready to estimate the mass of this planet. We will use the principle of conservation of momentum. Basically the star and its planet exert constant tugs of equal force on each other. Since the star is large it is moved

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at low speeds, but the planet is small, so it gets pulled around at far greater speeds. This is encapsulated in the equation M? M p 2πa M? v? = , (2) M? + M p P where M? is the mass of the star, v? is the average speed of the star in its orbit, M p is the mass of the planet, a is the average separation between the star and planet, and P is the period of the orbit. It looks a little complicated, but we will simplify it a little. We will still assume the planet is much less massive than the star. In that case the equation simplifies to become 2πa M? v? = M p . (3) P This is more manageable, but we now run into a little complication. In order to find the mass of the planet (M p ) we need to know the speed of the star, and we can’t find this exactly. Unfortunately we can only measure the line of sight speed of the star from Doppler shift. The star may also be moving in the plane of the sky. In essence we can only measure v? sin i where i is the angle by which the orbit is inclined relative to the plane of the sky. The inclination, i, is not 0 or we wouldn’t see the Doppler shift of the star at all, but it lies somewhere between 0 and 90 degrees. If we add the inclination to our equation we get ” — 2πa M? [v? sin i] = M p sin i . P

(4)

This means we cannot uniquely find the mass of the planet, but we find M p sin i which depends on an inclination which is between 0 and 90 degrees. The equation above, solved for M p sin i becomes ”

— M? [v? sin i] P M p sin i = 2πa

(5)

1. Find [v? sin i] using figure 1. v? sin i is just the difference between the maximum and minimum velocity divided by 2.

2. Convert the average distance between the star and planet (a) to kilometers using the conversion factor in the appendix.

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3. Convert the period to seconds using the conversion factor in the appendix.

” — 4. Solve for M p sin i using equation 5. Be sure to use the distance in kilometers and period in seconds you found above. The masses should be in solar masses.

5. If we assume the inclination is 30 degrees, what is the mass of the planet M p ?

” — 6. Is M p sin i the minimum or maximum mass of the planet?

7. Is the planet much less massive than the star as we have assumed?

We will learn more about this planet during our next class.

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Main sequence star properties Class O2 O5 B0 B5 A0 A5 F0 F5 G0 G2 G5 K0 K5 M0 M5 M8

Radius R/R 16 14 5.7 3.7 2.3 1.8 1.5 1.2 1.05 1.0 0.98 0.89 0.75 0.64 0.36 0.15

Mass M /M 158 58 16 5.4 2.6 1.9 1.6 1.35 1.08 1.0 0.95 0.83 0.62 0.47 0.25 0.10

Luminosity L/L 2,000,000 800,000 16,000 750 63 24 9.0 4.0 1.45 1.0 0.70 0.36 0.18 0.075 0.013 0.0008

Temperature K 54,000 46,000 29,000 15,200 9,600 8,700 7,200 6,400 6,000 5,700 5,500 5,150 4,450 3,850 3,200 2,500

Example Sanduleak -71 51 Sanduleak -66 41 Phi1 Orionis Pi Andromedae A Vega Beta Pictoris Gamma Virginis Eta Arietis Beta Comae Berenices Sun Alpha Mensae 70 Ophiuchi A 61 Cygni A Gliese 185 EZ Aquarii A Van Biesbroeck’s star

Properties of Main sequence stars (from Wikipedia http://en.wikipedia.org/wiki/Main_sequence.

B

Solar System Data Planet Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune

Radius km 2,440 6,052 6,378 3,394 71,492 60,268 25,559 24,766

Mass kg 3.30 × 1023 4.87 × 1024 5.97 × 1024 6.42 × 1023 1.90 × 1027 5.68 × 1026 8.68 × 1025 1.02 × 1026

Mean Density kg m−3 5,430 5,240 5,520 3,930 1,330 690 1,270 1,640

Escape Velocity km s−1 4.2 10.4 11.2 5.0 60 36 21 24

Taken from Eric Chaisson & Steve McMillan. Astronomy Today Upper Saddle River, NJ: Pearson Prentice Hall, 2005.

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Conversion Factors • 1 year = 365.25 days = 3.16 × 107 seconds. • 1 AU = 1.496 × 108 km. • 1 M = 1.99 × 1030 kg = 1,050 M Jupiter = 333,000 MEarth . • 1 M Jupiter = 1.90 × 1027 kg.

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• 1 MEarth = 5.98 × 1024 kg. • 1 km = 1000 m.

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Other Equations • The mass M of an object with volume V and density ρ is given by the expression

• The volume V of a sphere of radius r is

M = ρV .

(6)

4 V = πr 3 . 3

(7)