|MadSci Network: Astronomy|
The article you mentioned has a number of facts about the planet found orbiting Gl 581. It also has a lot of speculation.
The most important fact is that the planet has relatively low mass, as extrasolar planets go. Its mass is about 5 times the mass of the Earth. Normally, astronomers quote the mass of extrasolar planets in terms of the mass of Jupiter. Let MJ mean the mass of Jupiter. Obviously M(Jupiter) = 1 MJ. In these units, M(Earth) = 0.003 MJ, and the mass of that planet is M = 0.016 MJ.
As pointed out in the article, most extrasolar planets have M about 1 MJ. This is one of the least massive extrasolar planets ever found, which is one reason to call it "earthlike".
Now of course it really is Earthlike if it has the same temperature as the Earth does. There is no direct measurement of this planet's temperature, but it is possible to estimate it using the calculations that you have done. Indeed the Wikipedia article on Gliese 581c has a calculation of the average temperature of the planet.
Let me present a calculation which will show us how the temperature of a planet depends on the distance to its star.
All objects of any temperature emit energy at a certain rate, called the
luminosity (L). The luminosity is expressed in Watts; a 100W lightbulb
emits energy at a rate of 100 Watts (not all in the form of light). The
luminosity L is related to the temperature T and the radius R via
L = 4 pi R2 sigma T4,
where pi is the mathematical constant and sigma is a physical constant called the Stefan-Boltzmann constant. We can estimate the temperature T of the planet by requiring that its luminosity L be equal to the rate of energy that it receives from its star. The Earth, for example, receives a large amount of energy every second from the Sun, which we will call L(input). If we know that, we set L = L(input) in the above equation, and then solve for T.
Now let L(star) be the total energy emitted by the star. It is related to
R(star) and T(star) using the above equation. The planet will capture some
of this light, depending on its distance D from the star. Working through
all the math yields at estimated temperature T for the planet:
T = sqrt(R(star) / 2D) T(star).
The equation is pretty simple because a lot of the factors cancel out. Note that the estimated temperature does not depend on the radius R of the planet, but only on the star's temperature and the distance D between the star and the planet.
NOTE that I am ignoring an important factor, namely the fraction of light that the planet reflects. Some of the light striking the Earth is immediately reflected into space by the clouds, water, and land, and does not go into heating the Earth. This calculation, however, is useful for computing what would happen for a planet which varies in distance from its star, as Gliese 581c does.
If you insert values appropriate for the Earth, you derive T = 278 K, or +5 Celsius. The actual temperature is higher because of the greenhouse effect. The link takes you to an article from Wikipedia, which quotes that the equililbrium temperature of the Earth, computed if we include its reflectivity, would be -18 Celsius.
Now consider Gliese 581c. We can get the relevant information about the planet from the compilation you cited, the Extrasolar Planets Encyclopedia. All we need is the distance D, which is 0.073 astronomical units (= 1/14 the distance from the Earth to the Sun!). The temperature of the star is about T = 3350 K, and its radius would be about 1/4 that of our Sun. (These are numbers I gleaned from various sources in the astronomical literature.)
Putting in the values, we derive T = +25 Celsius, which is Earthlike indeed. For reasonable estimates of the planet's reflectivity, however, the average temperature would be about 0 Celsius (see the Wikipedia article). That's still "earthlike" by most people's reckoning, at least compared to Venus (average temperature = 457 C) or Mars (T = -63 C).
Anyhow, your main concern was with the temperature variations due to the variation in distance between Gliese 581c and its star. The orbital eccentricity is 0.16, less than that of Mercury. Its minimum distance is 0.84 the average distance, and its maximum distance is 1.16 times the average.
Now note that the distance in our equation enters in the square root part of the equation, so it won't vary much. If we calculate the effect of temperature for a 16% variation in distance, that comes out as an 8% variation in the temperature when it expressed in Kelvin. That means the average temperature using our equation varies by plus or minus 23 Celsius during the orbit.
All these estimates are too simple, of course. We don't know what kind of atmosphere this planet has, so we don't know anything about the planetwide circulation of heat. The orbital period is short, only 12 days, so the rise and fall of input energy happens quickly. If the planet is tidally locked to its star, one hemisphere will always be pointed to the star and the other will mostly be in darkness. (The Moon is tidally locked to the Earth, which is why we only see one hemisphere from our vantage point on the Earth.)
Still, even these crude estimates show that "Earthlike" is not such a bad label to apply to this planet.
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