Habitable zone planet discovered

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[PLAIN]http://fastcache.gawkerassets.com/assets/images/8/2011/09/xlarge_hd85512btop.jpg [Broken]
HD 85512 b
Wikipedia said:
HD 85512 b is an extrasolar planet orbiting the star HD 85512 approximately 36 light-years away in the constellation of Vela.

HD 85512 b is considered to be the best candidate for habitability as of August 25, 2011.

The planet has a minimum Earth mass of 3.6 ± 0.5 and an estimated temperature of 298 K (25 C or 77 F) at the top of its atmosphere.

It orbits the parent star at a distance of about 0.26 AU, with an orbital period of about 54 days.

For the temperature to be below 270 K, for a circular orbit, the planetary albedo should be 0.48 ± 0.05 and for an eccentricity of 0.11, the planetary albedo should be 0.52. If the planet has 50% cloud cover, water may exist in liquid form in the planet provided its atmosphere is similar to our own and the planet would be considered habitable. Also, if the albedo of the planet is increased due to cloud cover in the planet, water could be present in its liquid form in the planet, which would mean that, the planet is on the edge of habitability.

Reference:
http://en.wikipedia.org/wiki/HD_85512_b" [Broken]
 
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  • #2
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Alternatively if it's a desert planet with open water restricted to near the poles, then it could be stable against higher insolation levels than a planet that's mostly ocean covered.
 
  • #3
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Really cool when they do speculations about how it would or should look like from what information we know about this. What do they exactly mean by it's on the "edge for habitability"? Is it the edge of the habitability zone or rather because it has a bigger mass than Earth that's making it doubtful? A planet like this, with that mass (3.6 ± 0.5) of Earths, I'm quite sure it has to have, at least one orbital moon. And maybe the size of earth? Spectacular with the picture, and I hope that solar system has additional planets. It looks appropriate to have more. These smaller stars (red/orange) are very surprising when it comes to planets, don't you agree?
 
  • #4
Drakkith
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That's pretty cool! According to my calculations, if this planet had a similar density to Earth, it would have an acceleration due to gravity of 14-15 m/s^2 at it's surface.(At minimum of about 3.1 Earth masses) Earth has 9.8 m/s^2 for comparison. Hope that's all correct...:uhh:
 
  • #5
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Yeah and if everything is appropriate. Then, according to some certain tests and researches, a regular human can survive in 1.5-2 g in a longer period without elementary injuries. Drakkith, what eqvation did you use to calculate the gravity? :)
 
  • #6
Drakkith
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Yeah and if everything is appropriate. Then, according to some certain tests and researches, a regular human can survive in 1.5-2 g in a longer period without elementary injuries. Drakkith, what eqvation did you use to calculate the gravity? :)
Searched for gravitational force calculator on google. Then had to get the numbers for the Earth, such as density and mass and such, and then figured out through trial and error using that calculator plus another calculator for density and radius and volume or something to find what radius the planet would be to keep the same density as the Earth. I think the radius was just over 9,000 km. So it would be about 50% larger in radius and gravity than the Earth.
 
  • #7
Really cool when they do speculations about how it would or should look like from what information we know about this. What do they exactly mean by it's on the "edge for habitability"? Is it the edge of the habitability zone or rather because it has a bigger mass than Earth that's making it doubtful? A planet like this, with that mass (3.6 ± 0.5) of Earths, I'm quite sure it has to have, at least one orbital moon. And maybe the size of earth? Spectacular with the picture, and I hope that solar system has additional planets. It looks appropriate to have more. These smaller stars (red/orange) are very surprising when it comes to planets, don't you agree?
What they mean by edge of habitable zone is the planet's temperature remains between 0 and 100°C so that there's always liquid water. Necessary for all life we've yet discovered. They estimate it's temperature based on it's tiny orbit of it's red dwarf star and it's size. Good luck living on a planet where a supermodel weighs in at 360lbs.
 
  • #8
Drakkith
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What they mean by edge of habitable zone is the planet's temperature remains between 0 and 100°C so that there's always liquid water. Necessary for all life we've yet discovered. They estimate it's temperature based on it's tiny orbit of it's red dwarf star and it's size. Good luck living on a planet where a supermodel weighs in at 360lbs.
I'm not sure they know the force of gravity on this planet. That will depend on the density and size of the planet. The bigger and less dense it is, the less the force of gravity will be.
 
  • #9
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I'm not sure they know the force of gravity on this planet. That will depend on the density and size of the planet. The bigger and less dense it is, the less the force of gravity will be.
Extrasolar planet's sizes can be determined by the amount of light they block out from the host star. The mass can be determined by several ways, I've done a little bit of research work in this area (pre-masters, summer research stuff), and the way we determined the planets mass was by finding the host stars mass, then we could take the spectra of the star and determine the planets mass using the radial velocity method.

Using these two variables you can determine the force of gravity on the planet.
 
  • #10
Drakkith
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Extrasolar planet's sizes can be determined by the amount of light they block out from the host star. The mass can be determined by several ways, I've done a little bit of research work in this area (pre-masters, summer research stuff), and the way we determined the planets mass was by finding the host stars mass, then we could take the spectra of the star and determine the planets mass using the radial velocity method.

Using these two variables you can determine the force of gravity on the planet.
Are you finding the radial velocity of the star?
 
  • #12
Drakkith
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Yes.
Kk. Didn't know what radial velocity was, so I had to go look it up real quick. That's why I was asking.
 
  • #13
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Kk. Didn't know what radial velocity was, so I had to go look it up real quick. That's why I was asking.
No problem. It's all about learning :smile:
 
  • #14
DaveC426913
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Extrasolar planet's sizes can be determined by the amount of light they block out from the host star.
Only if their orbit is edge-on. Rare. This one was not discovered by transiting.
 
  • #15
Drakkith
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Only if their orbit is edge-on. Rare. This one was not discovered by transiting.
That's too bad! We might have been able to get a size estimate.
 
  • #16
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Only if their orbit is edge-on. Rare. This one was not discovered by transiting.
It doesn't have to be completely edge on, just edge on enough for it to pass in front of our sight line to the star. :redface:

I'm trying to find out if there are other methods of determining the size of the planet if it doesn't transit across our line of sight...
 
  • #17
DaveC426913
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It doesn't have to be completely edge on, just edge on enough for it to pass in front of our sight line to the star. :redface:
"1 part in 100" is edge-on in my books. (eg. Sol = 800K/93M = 8.6 parts in 1000)

Simple probability dictates that less than 1% of systems will have this config.
 
  • #18
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Yeah and if everything is appropriate. Then, according to some certain tests and researches, a regular human can survive in 1.5-2 g in a longer period without elementary injuries. Drakkith, what eqvation did you use to calculate the gravity? :)
I realize you asked Drakkith, it is my wish to be helpful.


Assuming planet x is spherical with the same density as earth, if the Mass of planet x is stated as being 3.6 times greater than the Mass of earth, then:

a = acceleration "in m/s^2"
G = Gravitational constant, 6.67e-11
M = earth’s Mass, 5.976e+24 kg
r = earth’s radius, 6.378e+6 meters

(3.6)^1/3 * r = radius of planet x

Therefore,

a = (GM * 3.6) / [(3.6)^1/3 * r]^2

Applying the numerical values:

(6.67e-11 * 5.976e+24 kg * 3.6) / [(3.6)^1/3 * 6.378e+6 meters]^2 = 15.0 m/s^2
 
  • #19
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I realize you asked Drakkith, it is my wish to be helpful.


Assuming planet x is spherical with the same density as earth, if the Mass of planet x is stated as being 3.6 times greater than the Mass of earth, then:

a = acceleration "in m/s^2"
G = Gravitational constant, 6.67e-11
M = earth’s Mass, 5.976e+24 kg
r = earth’s radius, 6.378e+6 meters

(3.6)^1/3 * r = radius of planet x

Therefore,

a = (GM * 3.6) / [(3.6)^1/3 * r]^2

Applying the numerical values:

(6.67e-11 * 5.976e+24 kg * 3.6) / [(3.6)^1/3 * 6.378e+6 meters]^2 = 15.0 m/s^2
A bit of algebra would tell you that, for a constant density, the surface gravity increases with the cube-root of the mass. However for realistic planets of Earth mass and larger, the density increases with the mass, meaning a higher surface gravity for a given mass.

R ~ M^b where R is in Earth radii and M is in Earth masses, and b is in the range 0.28-0.26. Work the algebra and it means the surface gravity goes up with the 0.48-0.44 power of the mass. Thus 3.6 Earth masses means a surface gravity of 1.85-1.76 gees.
 

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