By Mo Liu, IV Form
Smozaturn D3 — A Home in Alpha Centauri
In this Advanced Physics independent study unit, I decided to create a planet that is hypothetically habitable by humans in the nearest star system from our very own Solar System — Alpha Centauri. More specifically, the planet, which I have given the name Smozaturn D3, is rotating in set orbit around Proxima Centauri, the dimmest star among all three stars in the Alpha Centauri system, approximately 4.22 light years away. There are many factors that determine the habitability of a planet, including: the chemicals present on the planet, the construction of its atmosphere, and most importantly its distance from the star. Due to this complexity, there are theories like the Rare Earth Hypothesis that argues that complicated and biological life is a very improbable phenomenon and is likely to be extremely rare. However, there is an alternative view known as the principle of mediocrity that argues that the universe is friendly to complex life, since Earth is a tropical rocky planet in a common planetary system.
I personally believe in the existence of earth-like planets in and outside the Milky Way, and with the recent discovery of planets by the Kepler space telescope, astronomers are looking forward to finding potentially habitable worlds from about three-dozen candidates that Kepler has detected.
When we are searching for habitable worlds, one of the most basic criteria is that liquid water must be present, as we currently believe that liquid water is essential to all life. Considering this necessity, the range of the Circumstellar Habitable Zone has to be within the bounds where the planet receives enough radiant energy from its central star. That’s why we also call the CHZ the Goldilocks Zone — it has to be right in the middle, between the two extremes. Although there are many restrictions in identifying CHZs, and they are usually fractions of the size of their star systems, we are able to define the boundaries of a typical CHZ by considering the essential life-supporting elements. The inner edge of the zone is the distance from the star where the runaway greenhouse effect vaporizes all water in liquid form. These are the conditions on the surface of Venus. The outer edge of the zone would be the distance where the planet is so far away from the star that it could not be above the freezing point, no matter how much carbon dioxide is emitted in order to strengthen the greenhouse effect.
Smozaturn D3, like some of the other planets in space, is in thermal equilibrium with its surroundings, meaning that it is absorbing and radiating away energy at the same rate. Smozaturn D3 is also an ideal black body — it absorbs all incident radiation, and it has a functional structured atmosphere that allows it to stay in thermal equilibrium. After setting the initial conditions for my planet, I apply the equation of surface temperature using planet luminosity:
where L is the luminosity of the star and D the distance between the planet and the star. The luminosity of Proxima Centauri is 0.0017L⊙(solar luminosity), and the standard solar luminosity defined by the International Astronomical Union is 3.828×1026 W, therefore the luminosity of Proxima Centauri would be Lpc=6.5076×1023 W. Because the temperature has to be restricted to allow liquid water to exist, the inner bound temperature would be TI=373K and the outer bound temperature would be TO=273K. Plugging the numbers in, the inner bound distance from Proxima Centauri would be
(373K)4= (6.5076×1023 W)/DI2
DI=5.7982×106 km=3.8759×10-2 AU
(1AU=1.496×108 km, the estimated distance between the earth and the sun)
Similarly, the outer bound distance from Proxima Centauri would be
(273K)4= (6.5076×1023 W)/DO2
DO=1.0824×107 km=7.2354×10-2 AU
The distance between the two edges of the CHZ is 3.3595×10-2 AU, which tells us that the range of habitable zone in the Alpha Centauri star system is very small. According to the Vladilo et al. estimation in 2013, the Goldilocks zone in the solar system is within about 0.87 and1.18 AU from the sun, with Earth between the two extremes. Compared to the solar system data, Alpha Centauri’s CHZ is significantly closer to the central star, mainly because Proxima Centauri itself is only a fraction of the size of the sun. The radius of Proxima Centauri is 100,900 km, which is 0.145 R⊙(solar radius), and its mass is only 2.446×1029 kg, only 0.123 M⊙(solar mass). As I mentioned before, the luminosity of Proxima Centauri — the most important factor that determines thermal radiation — is only a small fraction of that of the sun. Smozaturn D3 would be right between the edges at the distance of 5.5557×10-2AU or 8.3112×106km away from Proxima Centauri.
Before I determine the atmosphere structure of Smozaturn D3, I set the radius of my planet to be 2.700×103 km and the gravitational acceleration to 13.5 m/s2. Besides having liquid water, the atmosphere of Smozaturn D3 must contain the essential volatiles that are necessary for life, such as carbon dioxide, oxygen, methane, and ammonia. I use the thermal escape limits of nitrogen and hydrogen to set the boundaries of the mass of Smozaturn D3 with my set radius and gravitational acceleration. If the atmosphere is able to hold gaseous nitrogen it would also be able to hold the heavier volatiles like carbon dioxide and oxygen. Using the same logic, if the planet is able to hold hydrogen, which is the most abundant and also the lightest element in the universe, the pressure would be so high that carbon dioxide and water would only exist in solid form at the surface level, and the planet would therefore be inhabitable. To calculate the mass of the planet using the equation
r = mMw/zTeq
We need to first calculate the equilibrium temperature of Smozaturn D3. The planet has a cross-sectional area of πRp2, and a surface area of 4πd2. When we divide the cross-sectional area by the surface area, we get the ratio of energy that the sphere receives.
πRp2/4πd2 = (Rp/2d)2
Because the energy received by the planet is the luminosity of the star, and because the star is an ideal emitter that radiates like a blackbody, in this particular case the luminosity of Proxima Centauri would be
Lpc = 4πRpc2σTpc4
The total amount of energy received by the planet would therefore be the product of luminosity and the ratio
E = 4πRpc2σTpc4 × (Rp/2d)2
But this equation predicts that the planet absorbs all the incident energy it receives, which is not usually not the case because of albedo expressed as A. Albedo is the ratio of reflected radiation upon the planet’s surface. Different materials have different albedo, and the Earth has an average albedo of 0.3. Smozaturn D3 has the same land structure as the Earth, so it would also have an albedo of about 0.3. If we take the albedo into account, the actual percentage of radiation received would then be 1- A. The energy that the planet absorbs would also be equal to the luminosity of the planet.
Lp = (1-A) × 4πRpc2σTpc4 × (Rp/2d)2 = 4πRp2σTp4
Tp4 = (1-A)Tpc4 (Rpc/2d)2
Tp= Tpc(1-A)1/4 (Rpc/2d)1/2
The surface temperature of Proxima Centauri, Tpc, is approximately 3042K; the albedo of Smozaturn D3 is 0.3, making (1-A) equal to 0.7; the equatorial radius of Proxima Centauri is equal to 1.009×105 km; and the distance from Smozaturn D3 to Proxima Centauri is 8.3112×106 km.
Tp= Tpc(1-A)1/4 (Rpc/2d)1/2
Tp = (3042)(.7)1/4(1.009×105/2(8.3112×106))1/2
Tp = 216.79K
The freezing point of water is at 273.15K, so if this was the mean surface temperature on Smozaturn D3, liquid water would not exist. There has to be something else to ensure the temperature is above the freezing point. The greenhouse effect traps the radiation from the planet’s surface and raises the temperature on Smozaturn D3 to around 300K. The Earth, having an actual equilibrium temperature of 260K, is habitable only because of the existence of greenhouse gases that raise the temperature.
Knowing the radius and the equilibrium temperature gives me the ability to calculate the mass of Smozaturn D3. Using the escape velocity of nitrogen and hydrogen, I can set the boundaries of a planets mass that would be theoretically habitable.
r = mMw/zTeq
where m and r are the mass and radius in Earth units(🜨); Mw is the atomic weight of nitrogen(14.0067g/mol) or hydrogen(1.008g/mol); z is a constant with the value of 2.000×10-2 mol/(gK); and Teq is the equilibrium temperature of Smozaturn D3 — 216.79K.
The smallest value would occur when the planet is just massive enough to hold nitrogen in its atmosphere. The radius of Smozaturn D3 is equal to 2.700×103 km, which is .423795R🜨(Earth radius).
r = m1MNw/zTeq
.423795R🜨 = m1(14.0067)/(2.000×10-2 × 216.79)
m1 = .423795R🜨/3.230476
m1 = .1312494M🜨
m1 = 7.8382×1023 kg
Similarly, the greatest value of mass would occur when the planet is massive enough and the gravitational force is so strong that it would hold gaseous hydrogen in its atmosphere, which means that the high pressure would make heavier volatiles like carbon dioxide solid.
r = m2MHw/zTeq
.423795R🜨 = m2(1.008)/(2.000×10-2 × 216.79)
m2 = .423795R🜨/.232483
m2 = 1.822907M🜨
m2 = 1.08864×1025 kg
Now that I have a range of masses, the atmospheric composition is the key to finding the exact value. If we look at the Earth’s atmosphere, nitrogen accounts for 78.09% of dry air, oxygen 20.95%, argon 0.93%, and many other gases like carbon dioxide, nitrous oxide, and methane are also present. Looking back at the Earth’s thermal equilibrium temperature, if it weren’t because of the greenhouse effect, there wouldn’t be any liquid water. In order for the temperature to rise, the Earth’s atmosphere must hold all essential chemicals involved in the greenhouse process. Some of the most crucial elements involved are water vapor, carbon dioxide, nitrous oxide, and methane, which are also present in the Earth’s atmosphere. Therefore, in order to make Smozaturn D3 an inhabitable planet, the essential elements of the greenhouse effect must be present. I set the atmosphere composition to be
4% carbon dioxide, nitrous oxide, methane
Then I computed a weighted average to determine the mass. Since about 80% of the chemicals are going to be heavy volatiles, the mass would be
m = 1/2(.8m1+.2m2)
m = 1/2(.8(7.8382×1023 kg)+.2(1.08864×1025 kg))
m = 1.402168×1024 kg
With the recent discovery of more than 1,200 exoplanets that Kepler telescope discovered just weeks earlier, we are constantly expanding our outermost reach of the cosmos to a greater limit. Among those 1,200 newly inducted planets, about three-dozen of them are in their own respective Goldilocks zones. Natalie Batalha, the mission scientist, said “calculation suggests tens of billions potentially habitable worlds”(The Guardian). It is hopeful that Kepler is going to add more exoplanets onto our list as it continues its journey through the cosmos, and we will soon discover a world that is habitable for complex life.
Smozaturn D3 Orbital and Physical Characteristics
Equatorial radius: 2.700×103 km
mass: 1.402168×1024 kg
surface area: 1.6965×104 km2
volume: 2.2902×107 km3
mean density: 6.1224×1016 kg/km3
surface gravity: 13.5m/s2
thermal equilibrium temperature: 216.79K
mean surface temperature: 279K
Atmosphere type: CO2-H2O-N2
structure: terrestrial, iron-rock
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