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Astronomers Claim Discovery of Earth-like Planet

Posted by CmdrTaco on Monday July 21, @10:29AM
from the that-would-be-a-rough-civ-map dept.
Raver32 writes "A team of astronomers announced they have discovered the smallest and potentially most Earth-like extrasolar planet yet. Five times as massive as Earth, it orbits a relatively cool star at a distance that would provide earthly temperatures as well, signaling the possibility of liquid water. 'The separation between the planet and its star is just right for having liquid water at its surface,' says astronomer and team spokesperson Stephane Udry of the Observatory of Geneva in Versoix, Switzerland. 'That's why we are a bit excited.' But researchers do not yet know if the planet contains water, if it is truly rocky like Earth, which might make it hospitable to life as we know it, or whether it is blanketed by a thick atmosphere. 'What we have,' Udry says, 'is the minimum mass of the planet and its separation" from its star.'"

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  • TFA (Score:5, Informative)

    by bsDaemon (87307) on Monday July 21, @10:32AM (#24273759)

    TFA is dated 24 April, 2007 -- I'm pretty sure that this is old news.

  • by Fallen Andy (795676) on Monday July 21, @10:53AM (#24274133)
    can be found here [wikipedia.org](Gliese 581) and here [wikipedia.org](Gliese 581c). It's a sad day when wikipedia seems to be more reliable than SciAm, but oh well, the rot set in many years ago...

    Andy

  • by sjonke (457707) on Monday July 21, @10:58AM (#24274229) Journal

    Invade!

    • by CastrTroy (595695) on Monday July 21, @10:32AM (#24273755) Homepage
      That's only if the radius from the centre of the objects is the same. Remember, gravity decreases as a function of the square of the distance.
      • by oldspewey (1303305) on Monday July 21, @10:41AM (#24273927)

        I think that in planetary terms we can safely assume 5x mass will create an environment of roughly 5g ... maybe give or take 20%. Enough to ensure that the simple act of getting out of bed would be a gruelling ordeal.

        Another problem I noticed after actually reading TFA:

        Gliese 581 c, orbits at one fourteenth the distance between Earth and the sun. But the red dwarf is 50 times cooler than the sun. The group estimates that the planet would experience temperatures in the zero-to-40-degree-Celsius (32-to-104-Fahrenheit) range.

        It is my understanding that red dwarfs, while generating reduced heat and light output, produce solar flares that are almost as intense as those produced by a G class star. So if a planet exists in the habitable zone it is also exposed to periodic sterilizing blasts of charged particles.

        Maybe if we're lucky the planet happens to have a really strong magnetic field ... then we just have the crushing g load to contend with.

        • Plus you can take into account all the other advantages life on Earth has had to make it possible:

          - In a solar system with a large gas giant, which helps keep catastrophic impacts with asteroids and comets from happening too often

          - Has a large satellite, which may help stabilize climate

          - Is in a quiet part of the galaxy, and is not too near other stars, avoiding interactions with other stars/gamma ray bursts/etc.

        • by Anonymous Coward on Monday July 21, @11:15AM (#24274595)

          I think that in planetary terms we can safely assume 5x mass will create an environment of roughly 5g ... maybe give or take 20%. Enough to ensure that the simple act of getting out of bed would be a gruelling ordeal.

          I don't know about you, but getting out of bed is a grueling ordeal at any gravity for me!

        • by Grokmoo (1180039) on Monday July 21, @11:26AM (#24274797)

          I think that in planetary terms we can safely assume 5x mass will create an environment of roughly 5g ... maybe give or take 20%. Enough to ensure that the simple act of getting out of bed would be a gruelling ordeal.

          Another problem I noticed after actually reading TFA:

          No, it would not. It would need to be much denser than Earth for that to happen. This is basically impossible for an object of that mass.

          Assuming roughly Earth like density (which is quite plausible), Radius will scale like Mass to the 1/3, while gravity scales like mass / radius squared. This works out to about 1.7 times Earth gravity at the surface.

        • by flabbergasted (518911) on Monday July 21, @11:32AM (#24274935)

          Oh for god's sake, people! Do a little math.

          Let's assume that the average density of the earth-like planet is the same as Earth. (It wouldn't be an earth like planet if it were significantly different.) Then we can use the volume of the sphere to relate the mass and the surface radius. Since M = 4/3 * \pi * R^3 * \rho, where \rho is the density, it is easy to see that the surface radius goes like the cube root of the mass. Putting this into Newton's equation, we can see that a = GM/R^2 means that the surface gravity is also going to go like the cube root of the mass. If the mass is five times that of Earth, then the surface gravity will be the cube root of 5 greater than Earth's or about 1.7 times Earth normal.

          Taking differences in the mean density into account is no more difficult, but I leave that as an exercise for the reader.

        • by AdamHaun (43173) on Monday July 21, @10:56AM (#24274183)

          You're probably thinking of the shell theorem, which says that a uniform sphere of mass is gravitationally equivalent to a point mass located at the center of the sphere. This theorem does imply that a larger radius = less gravity at the surface.

        • by the_other_chewey (1119125) on Monday July 21, @11:00AM (#24274265)
          To calculate the gravitational effect of a massive sphere, its whole mass can be
          considered accumulated in its center as long as you are outside of it.
          So the gravitational acceleration indeed only depends on mass an distance.
          Mathematical fact.

          Neat additional trivia:
          - Inside a hollow sphere, there is no gravitational effect by the sphere's mass - it cancels out exactly.
          That's why
          - Inside a massive sphere, gravitational acceleration increses linearly with the radial distance to the center.
          (the mass increases with r^3 as you get further out, its effect decreases by 1/r^2 - and as it can be considered
          concentrated in the middle, you get an increase by a factor of r^3/r^2 = r
          Gravity is fun :-)

          All this of course only for constant density.
        • by zacronos (937891) on Monday July 21, @11:46AM (#24275179)

          It's a poorly written and shite article, but the box off to the side says:

          One of two newly discovered exoplanets is nearly the size of Earth...

          So, assuming they're talking about the same one, it should be roughly 5 times our gravity.

          Not so. If the planet has twice the diameter of earth, that falls well within the category of "nearly the size of Earth" for astronomers. Since gravity decreases proportionally to the square of the distance, gravity would be only 5/(2^2) times as strong as on Earth, an increase of a mere 20%.

          If it has approximately the same density as earth, then since volume of a sphere increases proportionally to the cube of the radius/diameter, it would have 5^(1/3) times as large a diameter as earth, which is about 1.71 -- even closer to the size of our Earth. It would also wind up with gravity 1.71 times as strong, since 5/((5^(1/3))^2) == 5/(5^(2/3)) == 5^(1/3).

    • It may be earthlike, but it sure wouldn't be a comfortable place to spend any amount of time.

      For you maybe, but anything that evolved in that environment would be really strong on this planet, be able to leap...wait, what was the name of that planet?

      Can you imagine the women, why they'd be build like a brick shi...cue the Commodores.

    • Nope. (Score:5, Informative)

      by Ihlosi (895663) on Monday July 21, @10:38AM (#24273879)
      It may be earthlike, but it sure wouldn't be a comfortable place to spend any amount of time.

      Mass alone says very little about the surface gravity of a planet - you need to know the radius of the object to make any statement about its surface gravity. Earth's moon has slightly over a percent of the mass of Earth, but about 1/6g surface gravity. Mars has only about 10% of the mass of Earth, while having 1/3g surface gravity.

    • Re:Interesting find. (Score:5, Interesting)

      by WibbleOnMars (1129233) on Monday July 21, @10:49AM (#24274049)

      Since Drake's equation needs to know the proportions of stars with planets, it would require us to have known negative results as well as known positives in order for it to give any meaningful results.

      At the moment, we can say there are a few hundred planets, out of maybe a few thousand stars that we've scanned, but for the stars where we haven't found anything, we don't know for sure whether that's because there isn't anything there, or because we just aren't looking hard enough.

      • by edalytical (671270) on Monday July 21, @12:35PM (#24276209) Homepage
        Forget the proportion of stars with planets. Fl is the real unknown. Why assume all planets that can support life will develop life? What if life is actually pretty rare? Try plugging in values less that 1 for Fl (0.1, 0.01, 0.001) and you'll get some disappointing results from this equation. Trying to quantify something with so many unknowns seems pretty silly to me. On the other hand maybe life isn't rare, but that's just me being hopeful.