Earth sized Goldilock planet found

[dragon]

Only 1.1 times the sive of Earth and its a Goldilocks planet previous was 1.4 times no clue how far away though.

The search for a new Earth outside the solar system seems to be nearing its end. NASA's Ames Research Center astronomer Thomas Barclay has found a planet nearly the size of Earth in the habitable zone of a star in the Milky Way.

Barclay's announcement at the Search for Life Beyond the Solar System conference hasn't been officially published yet, so the details are scarce. We know that:

1. It's an M1 red dwarf star (maybe we should call it Krypton.)

2. It's a goldilocks planet, orbiting within the zone where liquid water (and life) can exist.

3. It's radius is only 1.1 times the size of Earth. Until now the minimum size for a new Earth candidate was 1.4 times—Kepler-62f, which orbits a star about 1,200 light years away from us.

4. At least five other planets are orbiting this red dwarf.

I can't wait for that new telescope starshade that will let us take actual photos of these new worlds.

link

[StarSword]

Only 1.1 times the sive of Earth and its a Goldilocks planet previous was 1.4 times no clue how far away though.

Follow the citations. The second link in the article gives this article, which identifies the star as Kepler-62, 1200 LY from us.

[Kitsune]

It orbits a red dwarf which means it is tide locked in that orbit

[Eternal_Freedom]

It orbits a red dwarf which means it is tide locked in that orbit

Do you have a source for this? Not this specific case but in general, since I've heard this before but no one had any sources for it.

[Borgholio]

Even if it's tidal locked it could still very well be habitable. It's always possible that it's far enough away from the star that the side facing the sun is pleasantly warm.

[GrandMasterTerwynn]

Only 1.1 times the sive of Earth and its a Goldilocks planet previous was 1.4 times no clue how far away though.

Follow the citations. The second link in the article gives this article, which identifies the star as Kepler-62, 1200 LY from us.

Err, no. Kepler 62f was the previous recordholder for smallest planet orbiting a star in its habitable zone.

It orbits a red dwarf which means it is tide locked in that orbit

So? A planet in the habitable zone of a red dwarf with thick enough atmosphere and well-placed oceans ought to maintain sufficient heat transport from hemisphere to hemisphere that everything doesn't end up freezing out on the dark side of the planet.

Do you have a source for this? Not this specific case but in general, since I've heard this before but no one had any sources for it.

It's simple orbital mechanics. The radius of a red dwarf's habitable zone is usually such that a planet in this region will orbit its parent star in time periods on the order of a week or two. To put that into perspective, Titan orbits Saturn in a bit over two weeks, and it's tidally locked.

The time it takes for a body to become tidally locked to its parent can be approximated thusly:
6 * ((semi-major axis6 * planet's radius * 3x1010)/(mass of star * mass of planet2)), and the result is in tens of billions of years.

For the largest red dwarves (~0.6 solar mass,) the habitable zone is at 0.3 AU. For a red dwarf like Proxima Centauri (~0.12 solar mass,) the habitable zone is at roughly 0.03 AU.

So, for an Earth-sized planet orbiting a massive red dwarf (0.6 solar mass) with a habitable zone of 0.3 AU, the time to tidally lock is all of 4.1 billion years.

The average red dwarf, however, has about a third of a solar mass, and a habitable zone of about 0.15 AU. The time to tidally lock an Earth-sized planet around the typical red dwarf? 110 million years. Ergo, the average planet around the average habitable zone of the average red dwarf is highly likely to be tidally locked.

[Eternal_Freedom]

Ah thank you. Like I said, I've often seen that stated but with no actual explanation.

[Broomstick]

Very nicely put, by the way. That's one of the most concise and coherent explanations for that which I've heard.

[Kitsune]

It orbits a red dwarf which means it is tide locked in that orbit

Do you have a source for this? Not this specific case but in general, since I've heard this before but no one had any sources for it.

http://en.wikipedia.org/wiki/Habitabili ... rf_systems
At the close distances that red dwarf planets would have to maintain to their stars in order to maintain liquid water at their surfaces, tidal locking to the host star is likely, causing the planet to rotate around its axis once for every revolution around the star; as a result, one side of the planet would eternally face the star and another side would perpetually face away, creating great extremes of temperature. For many years, it was believed that life on such planets would be limited to a ring-like region known as the terminator, where the star would always appear on the horizon.

The article points out other problems as well

[Guardsman Bass]

The bigger problem would be the one that Sky and Telescope pointed out about the planets of red dwarf stars. These stars apparently tend to be quite volatile for the first few billion years, which could leave planets close-in uninhabitable.

I still think Kepler-62f is a much better candidate. At 1.4 times the radius of Earth, its mass with an Earth-like density would be less than 3 Earth masses, and it's below a threshold for size that some models indicate as a dividing line between "rocky planet" and "rocky planet with thick, hot hydrogen-helium atmosphere". It only gets about 42% of the solar insolation that Earth gets, but it's a larger planet with (probably) a thicker atmosphere.

[Irbis]

It's simple orbital mechanics. The radius of a red dwarf's habitable zone is usually such that a planet in this region will orbit its parent star in time periods on the order of a week or two. To put that into perspective, Titan orbits Saturn in a bit over two weeks, and it's tidally locked.

Out of curiosity, would it always be the case? For example, IIRC astronomers always thought Mercury would be tidally locked due to this, but its eccentricity makes the orbit locked 3:2 instead of 1:1. Or would distance be so small to render eccentricity meaningless in this case?

[Borgholio]

Mercury is not close enough to our sun's gravity well to be fully tidally locked. A red dwarf is so much smaller and dimmer, that a planet in the habitable zone has to be proportionally closer to the dwarf than a planet is to our sun. Thus it's far more likely to be tidally locked.