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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.'"
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.
Anthropic principle: We see the universe the way it is because if it were different we would not be here to see it.
TFA is dated 24 April, 2007 -- I'm pretty sure that this is old news.
5x gravity only if it were the same physical size.
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.
I live in constant fear of the Coming of the Red Spiders.
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.
Acceleration due to gravity scales as mass over Radius squared, whereas Radius will scale as Mass to the one third, assuming relatively constant density. So, the radius would be about 1.7 times that of Earth.
Thus, acceleration due to gravity would be about 5 / (1.7 ^ 2) or about 1.7 times Earth acceleration (10 m / s ^ 2). This is all assuming it is Earth like in composition, which we don't know for sure.
1.7 times Earth gravity would be pretty high, but it might be livable. It is worth noting, however, that this is only a MINIMUM mass estimate - the mass could be higher.
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.
The Earth has a density of about 5.5, so if that planet was 5 times heavier than Earth, it would have a density of 27.5 am I right? Even if it was made of pure gold it couldn't be that dense.
You just got troll'd!
From the blurb itself, it's five time the size of earth, it's revolving around a cooler sun than earth, and it might not have liquid water or a thick atmosphere. Yeah, that's exactly like earth!
You're missing the point. By Earth-like they mean telluric planet, as in, not a gas giant. That's all. And that matters because until now we haven't found so many of them, most of the planets we've found were gas giants orbiting close to their star. But as time goes by we find ever decreasingly large planets that get closer and closer to the Earth in size.
You just got troll'd!
No. While it is hard to measure, gravity drops off with any altitude at all. The gravity you feel standing on top of Mt. Everest is ever so slightly less than that in Death Valley.
Andy
... they really don't know jack squat about this planet and they're just making wild speculations about it given the extremely tiny amount of gravitational aberration data that they've managed to collect with their instruments, which only really suggests anyway that there *might* be a planet of some certain mass and orbital radius there, if their instruments' measurement errors or perhaps some other gravitional phenominon aren't what they actually seeing anyway.
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.
Visit the
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.
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.
Isn't the case, that, at least while you are on the planet, the math works out such that the gravity is consistent at whatever point you are on?
No, we can even measure how surface gravity varies from g=9.78 m/s2 to g=9.82 m/s2 when moving from the equator towards a pole. And this is because Earth is not perfectly round, the people at higher latitudes are closer to the center of Earth and fall faster.
5 times the mass means 5 times the gravity assuming the same volume.
If we assume this planet is truly "earth-like" and has similar density, then it is not a true statement that it will have 5 times the gravity of earth. If the density is similar to that of Earth, then the size of the planet will be larger. The radius will be larger by a factor of the cube root of 5 (the real one, that is), which is about 1.7, which is also almost exactly the square root of 3.
Since gravity is proportional to the inverse of the radius squared, the gravity of this planet at its surface is 1/3 the magnitude it would have if the volume were the same as Earth's.
Comparing, this means that this planet will have 5/3 the gravity of earth, or 1.67 times. 1 kg of mass would weigh about 3.68 lbs, so my meager 75kg frame would weigh 276 Earth lbs. Wouldn't be the first time...
Yeah, that's an unreasonable density, but you should bear in mind that compression occurs. (Earth's uncompressed density is significantly lower than the actual density, for example.)
That said, you can fairly assume that the density is nearly the same as Earth's. In that case, the surface gravity is only about 70% higher than here. It'd still be tough walking around.
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.
When they say "nearly the size of earth" they're speaking in an astronomical scale, which would qualify something 5-times as large as the earth as 'nearly'. It's not composed of gold or other heavy metal.
!#&*
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.
You're missing the point. By Earth-like they mean telluric planet, as in, not a gas giant.
In this case, they don't actually mean even this. From TFA:
"What we have," Udry says, "is the minimum mass of the planet and its separation" from its star.
They don't actually know if it's rocky. All that they know is that the mass is about right, and it's about close enough to a red dwarf for liquid water.
--
$tar -xvf
I think that in planetary terms we can safely assume 5x mass will create an environment of roughly 5g ... maybe give or take 20%.
How do you justify that remark? Mars has a mass 1/9 of Earth's but a surface gravity over 1/3 of Earth's. Mercury has a mass 1/18 that of Earth but has gravity slightly higher than that of Mars.
There's just no way you can have confidence within 20% that the gravity will be proportional to the mass.
Patrick Doyle
I mod down every jackass who puts his moderation policy in his sig. Oh, wait a sec....
If it's less than 14 Earth masses it's rocky. Simple as that ;-)
You just got troll'd!
Mass =/= Density Density = Masss / Volume 5x Mass =/= 5x Density, unless the planet's volume was equal. Which it probably is not.
Uncompressed density = density of material at 1 atm
Compressed density = density of material under given pressure
If we took all the stuff Earth is made of, took it apart and measured the average density of all those rocks at 1 atm, we would get a significantly lower average than what we get by dividing the estimated mass of Earth by its estimated volume.
Those who would give up liberty to obtain working drivers, deserve neither liberty nor working drivers.
indeed... see Mars... it's internal warmth died off quickly, being not only a small, but light-for-its-size, planet... Earth is waning now, but remained internally hot for billions of years longer... and even in spite of that, it once spent a fair period of time as an icecube before the cambrian explosion... just goes to show that location is important, but not enough.
To be pedantic, Jupiter doesn't have a surface. It just has a really thick atmosphere that turns into molten stuff when you get deep enough. It's not a stretch to say there's nothing solid in the entire planet.
The latest results of Messenger's first flyby of Mercury confirms a magnetic field and molten outer core. Conversely, Venus which is Earth's twin in size, seems a lot more dead. A more important factor may have been chemical composition at the time of formation - Mercury had more metal. Elements may have been unevenly distributed as function of distance from the Sun in the original planetary nebula.
point a big-ass telescope at it
The problem is the relative brightness of the parent star. Even if (as in this case) it's a red dwarf, it's still about a bazillion times brighter than the reflected light reaching earth from the planet itself.
If libertarians are so opposed to effective government, why don't they all move to Somalia?