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Rocky Planet Discovered
Fraser Cain writes "Astronomers have discovered a rocky, terrestrial planet orbiting a nearby star, Gliese 876. The planet has approximately 7.5 times the mass of the Earth, double its radius, and orbits its parent star once every two days. This is the most Earthlike extrasolar planet discovered so far." Reader Karthik Narayanaswami points out that "the planet was discovered by the famed Berkeley astronomer Geoff Marcy," and adds a link to the news release from Berkeley.
Here is the link to the Berkeley press release and information on Berkeley astronomer Geoff Marcy.
And oh, looks like Slashdot is continuing to mirror Boing Boing.
g_newplanet = G(7.5M_Earth)/4(r_Earth^2), where g is the gravitational field strength and G is the gravitational constant. This is less than 2g_earth.
Seems close enough for life. Although I'm not sure why a stronger gravitational field would necessarily be a huge constraint on the development of life.
And to preempt the trolls, my little convenient formula is good enough. Relativistic effects can be ignored.
Is life possible at all under such gravity? Any examples?
:) But seriously, check out this article.
And where do you propose we find such examples? In space perhaps?
Want to improve your Karma? Instead of "Post Anonymously", try the "Post Humously" option.
The gravity would be about twice Earth's.
Of course life would be possible with that gravity. Microorganisms don't care particularly about gravity, and any multicellular life that might evolve would adapt to whatever the local conditions are.
What would make life untenable would be a lack of liquid water. This world is very close to the star and would be tidally locked to the star. Unless there are deep ocean basins on the nightside to permit the water to cycle back, the water would eventually freeze out on the nightside. Even if the water didn't freeze out, with a temperature of over 200 Celsius the world would not be very hospitable. Life may be possible but the conditions are not ideal.
The only thing necessary for the triumph of evil is for good men to do nothing. - Edmund Burke
No you idiot, it said extra solar there, replied to myself.
The correct formula is Fg = G((m1*m2)/(r^2))
...which Google says is approximately 45.92 N. A one kilogram mass on Earth should exert a downward force due to gravity of 9.8 N if I remember my physics classes correctly.
Using Google to come up with necessary constants gives me:
((6.67300 × 10E-11) * (7.5 * 5.97200E24)) / ((2 * 12 756 300)^2)
(I used a theoretical 1 kilogram test mass, at the planet's surface, to simplify things.)
So, call it about 4.7 times the gravity of Earth. Life? Possibly - but I sure as hell wouldn't want to move there.
--Ender
Loose things are easy to lose. You're getting your hair cut. They're going there to see their aunt.
No.
A body moving in a circle of radius R at a uniform speed V experiences an acceleration a = (V*V)/R towards the center of the circle. In neither of the cases you mention does any centripetal acceleration come close to the local gravitational acceleration at the surface of the planet.
Case 1: The Earth: orbital speed V = 30 km/s, and R = 150 million km, so (V*V)/R is of order (10^8)/(10^11) m/s^2, or about 10^(-3) m/s^2. The local gravitational acceleration is about 10 m/s^2, of course. If you speak of the Earth's rotational motion at the equator, then very roughly V = 500 m/s and R = 6,400,000 m, so (V*V)/R has magnitude roughly (2.5 x 10^5) / 6.4 x 10^6 = 0.03 m/s^2; again, much less than 10 m/s^2 due to the gravitational pull of the Earth.
Case 2: The new planet. Its orbital radius is about 2 billion meters, so the circumference is about 7 billion meters; if it travels that distance in a period of 2 days = 170,000 seconds, then it speed is about V = 40,000 m/s. The orbital centripetal acceleration is therefore of order (16 x 10^8)/(2 x 10^9) = 0.8 m/s^2. That's much larger than the Earth's orbital centripetal acceleration, but still far less than the likely gravitational acceleration at the surface (or cloudtops) of this planet.
Michael Richmond "This is the heart that broke my finger."
mwrsps@rit.edu http://stupendous.rit.edu
It is, that's why we haven't flung off into the void, or been dragged into the Sun.
Also, we really can't judge what it's semimajor axis is unless we know how massive the star is, but if we know that, then we also have the velocity.
(6.67e-11)*M=(v^2)(r)
where M is the mass of the star.
Also, the centripetal force has a magnitude of v^2/r.
HTH
Fnord.
The wobble method of planet detection is more effective for lower mass stars. A planet of a given mass will move a smaller M star more than a bigger G star like the Sun. Plus on top of that, only 8% of stars in the galaxy are G class or bigger, so as we look outwards the closest stars tend to be M and K which biases the planet finds to lower mass stars. If you use the occulation method (the planet covers part of the star as it passes between us and the star), again in the M star case, the planet would cover a bigger fraction of the star's area since the red dwarf is so small. If some alien astronomer looked at the solar system and the earth passed in front of the Sun, the Earth would only cover 1/10000 th the area of the sun, but if earth orbited a red dwarf it would be more like 1/200 (0.5%) -- much easier to detect the change in light. Mark
From TFA:
Um, no, that's not true - there certainly are bacteria which can survive these temperatures and have adapted to them (those living near hydrothermal vents at the bottom of the ocean, for example). Whether this new planet could (even theoretically) host life is another question entirely, of course, but the statement that we do not know life that can endure such temperatures is simply not true.
quidquid latine dictum sit altum videtur.
... is 6378.5 kilometers, not 12756.3 as you have above. The correct values for the force on a 1kg mass on the surface are:
Earth: 9.785 N
New planet: 18.366 N
So the grandparent poster is correct, the surface gravity would be about 1.9 times that of Earth.
I'm going to guess there is no night and day. The plant is probably tidally locked.
Patrick Doyle
I mod down every jackass who puts his moderation policy in his sig. Oh, wait a sec....
Though, just to be fair to everyone, you *are* assuming that the density of the new planet is the same as Earth's. A perfectly reasonable assumption given this planet is meant to be "Earth-like", but worth flagging up all the same.
An alternative method is to look for eclipses of the planet passing in from its stars. About 5% of the planets have been discovered this way. One estimate is about one in two hundred stars have suitable orientations and plantary systems for this method, if one could observe them long enough. A @hundred megapixel space probe called Kepler might be launched around 2008 to observe light curves of several hundred thousand stars for several years. This might find dozens of eclipsing planets plus understand the abundance of planets.
Except that out of those, only microlensing is likely to detect non-giant planets (orbiting non-pulsars) and it's based on random luck.
In practical terms, if you want to find earthlike planets, you use the doppler method.