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Astronomers Claim Discovery of Earth-like Planet
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.'"
TFA is dated 24 April, 2007 -- I'm pretty sure that this is old news.
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.
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.