New 'super-Earth' found in space

[Casual Flashlight User]

(love the BBC title...where else would you find a "super Earth? in your pocket? down the back of the setee?)

Um, Anyways....

Astronomers have found the most Earth-like planet outside our Solar System to date, a world which could have water running on its surface.

The planet orbits the faint star Gliese 581, which is 20.5 light-years away in the constellation Libra.

Scientists made the discovery using the Eso 3.6m Telescope in Chile.

They say the benign temperatures on the planet mean any water there could exist in liquid form, and this raises the chances it could also harbour life.

"We have estimated that the mean temperature of this 'super-Earth' lies between 0 and 40 degrees Celsius, and water would thus be liquid," explained Stephane Udry of the Geneva Observatory, lead author of the scientific paper reporting the result.

"Moreover, its radius should be only 1.5 times the Earth's radius, and models predict that the planet should be either rocky - like our Earth - or covered with oceans."

http://news.bbc.co.uk/1/hi/sci/tech/6589157.stm


CFU

 

[LEDcandle]

Wonder if there are any flashaholics over there

 

[Casual Flashlight User]

Well, this place revolves around a "dim", "cold" star...so I think a few decent flashlights are in order.

CFU

 

[DM51]

The star may be dim, but I doubt it is cold if the planet's temperature is in the range 0-40degC. Gravity on the planet is ~4x earth's though, so everything would feel heaaavvyyy and you'd probably want to consider small lights rather than big ones.

 

[Casual Flashlight User]

The star may be dim, but I doubt it is cold if the planet's temperature is in the range 0-40degC. Gravity on the planet is ~4x earth's though, so everything would feel heaaavvyyy and you'd probably want to consider small lights rather than big ones.

Cold as in "a lot cooler than our own Sun"...this planet is very close to it's parent star (it's "year" is only 2 weeks long)...I suggest the new Novatac Bumble-Bee flashlight for use on this new world...light, solid construction and it should have a decent output being of the HDS stable.

CFU

 

[DM51]

The star must be a red / infra-red source with not much white, so I don't think I'll take my A2 Aviatrix with the red McNuggets - I'll take the UV or Y/G version.

With a year @ only 2 weeks, you'd age quite fast. Also, with all that gravity, you wouldn't get much throw as the lumens would be a lot heavier and they would fall to the ground quite quickly.

 

[ViReN]

It's 20 light-years away, what ever we see today is actually has happened atleast 20 years before. actually we are viewing the past of that planet. to view the present occurring, we have to travel 20 years to the future.

Oh, one more thing, the star is dim, close to it's end, a stage will be reaching when every thing is going to collapse in a black hole (one of the reason, planets are revolving so fast and so close).

 

[elgarak]

The star may be dim, but I doubt it is cold if the planet's temperature is in the range 0-40degC. Gravity on the planet is ~4x earth's though, so everything would feel heaaavvyyy and you'd probably want to consider small lights rather than big ones.

Ummm... It's radius is 1.5 x Earth's radius, it's mass is 5 x Earth's mass. That means it's gravity on the surface is about 2.2 g (5/1.5^2), not 4.

 

[DM51]

I defer to your greater knowledge! The 4g was a figure I heard on the radio, and it sounded about right to me as I thought the g figure was directly proportional to mass/radius, not mass/radius^2.

Being so close to the star, and spinning so fast (1 rev / 2 wks) would there be any centrifugal force affecting gravity on the planet? And would the the gravity due to the star itself come into play, similar to the way the moon affects tides here?

 

[Diesel_Bomber]

Where do I buy a ticket? Starship Earth is just tooooooooo crowded for my tastes.


:buddies:

 

[Mike Painter]

I defer to your greater knowledge! The 4g was a figure I heard on the radio, and it sounded about right to me as I thought the g figure was directly proportional to mass/radius, not mass/radius^2.

Being so close to the star, and spinning so fast (1 rev / 2 wks) would there be any centrifugal force affecting gravity on the planet? And would the the gravity due to the star itself come into play, similar to the way the moon affects tides here?

The gravity is a function of density as well as radius and without knowing the density you can't use just the radius. (A lead planet with the same diameter of earth would have a higher gravity.)
The planet can't be seen and the only way we know it is there is from the interaction of the planet and the sun. This causes a "wobble" for lack of a better word. This wobble is a function of the distance and mass of the planet. Assuming the radio got it right, the 4x would probably have been correct.

 

[elgarak]

The gravity is a function of density as well as radius and without knowing the density you can't use just the radius. (A lead planet with the same diameter of earth would have a higher gravity.)
The planet can't be seen and the only way we know it is there is from the interaction of the planet and the sun. This causes a "wobble" for lack of a better word. This wobble is a function of the distance and mass of the planet. Assuming the radio got it right, the 4x would probably have been correct.

Nope. As long as the mass distribution is spherical (no 'blobs' of high density on one side), you can assume that all the gravity is primarily going to the center of the planet. This approximation works pretty well for 'large' planets; it's pretty difficult to 'make' a large stable planet when there's too much variation from a spherical distribution. It would blow itself up before it could solidify. With this mass and radius estimates as large as they are (5x m_Earth, 1.5x r_Earth), it's pretty sure that this thing is a sphere, with a spherical density distribution.

 

[elgarak]

I defer to your greater knowledge! The 4g was a figure I heard on the radio, and it sounded about right to me as I thought the g figure was directly proportional to mass/radius, not mass/radius^2.

Being so close to the star, and spinning so fast (1 rev / 2 wks) would there be any centrifugal force affecting gravity on the planet? And would the the gravity due to the star itself come into play, similar to the way the moon affects tides here?

Hmmm. Let's see... 11million km orbit, 13 days period, that means about 0.0009g centrifugal acceleration... guess that's too small to have a influence on everyday activity (compared to the 2.2g).

However, with the orbit parameters as they appear right now, there's a good chance this planet is tidally locked... means is faces the same side towards Gliese 581 all the time, like the Moon does to Earth.

 

[highorder]

Also, with all that gravity, you wouldn't get much throw as the lumens would be a lot heavier and they would fall to the ground quite quickly.

you do understand that photons have no mass, right?

 

[kakster]

Photons *do* have mass, as light obsverved from distance stars is affected by gravity.
It's one of the ways astronomers hunt black holes.

 

[Data]

Hmmm. Let's see... 11million km orbit, 13 days period, that means about 0.0009g centrifugal acceleration... guess that's too small to have a influence on everyday activity (compared to the 2.2g).

However, with the orbit parameters as they appear right now, there's a good chance this planet is tidally locked... means is faces the same side towards Gliese 581 all the time, like the Moon does to Earth.

In the low order approximation of forces on any two objects (that are not spinning about their own axis) in orbit about each other, the "centrifugal acceleration" force that you refer to is always exactly equaled by the gravitational attraction between the two bodies. That is what creates an orbit. If you temporarily zero their motion relative to each other, they will accelerate toward each other and collide.

The major forces our bodies experience while standing on the surface of the earth come from the earth's gravity, not from the son. Some force comes from the earths rotation and that also creates a force that causes the planet to bulge at the equator.

Cheers
Dave

 

[The_LED_Museum]

If a plant's rotation is tidally locked, one side will always be very hot and the other side will be very cold; but there would be a habitable region right at the planet's terminator (the point on the horizon where the sun appears to rise or set).

 

[elgarak]

If a plant's rotation is tidally locked, one side will always be very hot and the other side will be very cold; but there would be a habitable region right at the planet's terminator (the point on the horizon where the sun appears to rise or set).

If it's just a rock, yes.

Things vary when there's an atmosphere or an ocean covering the surface, which could carry heat around the planet.

From The Bad Astronomer:

If so, how does this affect the atmosphere? Models indicate that the air should carry the warmth of the star around the planet, so the temperatures should actually be fairly moderate on both the day and night sides of such a world. But if it's covered by an ocean, how does having one side of the planet eternally locked into daylight affect it?

Criminy, what would life be like on a tidally-locked ocean world?

 

[Mike Painter]

Nope. As long as the mass distribution is spherical (no 'blobs' of high density on one side), you can assume that all the gravity is primarily going to the center of the planet. This approximation works pretty well for 'large' planets; it's pretty difficult to 'make' a large stable planet when there's too much variation from a spherical distribution. It would blow itself up before it could solidify. With this mass and radius estimates as large as they are (5x m_Earth, 1.5x r_Earth), it's pretty sure that this thing is a sphere, with a spherical density distribution.

I'm not sure what you are saying nope to. My comment had nothing to do with a non (nearly) spherical planet. The wobble I mentioned is the interaction between planet and star that let them make these calculations.
This was before I read that the mass was 5x so your 2.2 figure is correct.
(Based on what was reported.)

 

[elgarak]

I'm not sure what you are saying nope to. My comment had nothing to do with a non (nearly) spherical planet. The wobble I mentioned is the interaction between planet and star that let them make these calculations.
This was before I read that the mass was 5x so your 2.2 figure is correct.
(Based on what was reported.)

Ah. I misunderstood. Sorry. Maybe it's just that I'm so used to ignore the density of planets... for most calculations, only the mass and the position of the planet is important, and you can safely ignore the radius and density.