Astronomers Find What May Be the Closest Exoplanet So Far

The Bad Astronomer writes: Astronomers have found a 5.4 Earth-mass planet orbiting the star Gliese 15A, a red dwarf in a binary system just 11.7 light years away (PDF). Other exoplanets candidates have been found that are closer, but they are as yet unconfirmed. This is more evidence that alien planets are common in the galaxy.

Comments

Re:OK Another one

[i kan reed]

5.4 earth masses puts it at about 1/3rd of a Neptune or a tiny fraction of a Jupiter or a Saturn.

It might even have a thin enough atmosphere to not completely crush a human.

Time to travel 11 light years

[goombah99]

traveling with a 1G acceleration:
1/2g t^2 = 1/2*11*3E8

so t = 3.3 years to half way. 6.6 years to go all the way and thus 13.2 years for the round trip.

Thus you could easily go there and come back in your lifetime.

Note that this is also Faster than light can make the round trip. However that is not any violation of relativity. THe people on earth would have aged a lot more than 13.3 years during your trip. But you would only have aged 13.3 years.

Re:Time to travel 11 light years

[i kan reed]

Teah, they call that "thust to weight ratio" you're referring to specific impulse

Re:Time to travel 11 light years

[goombah99]

Let's see if I can work this out correctly;
First assume the spaceships weight negligibly different than the mass of the fuel. The thrust needed to push the weight at a steady 1g will be proportional to the mass of the ship at each interval of time. SO the rate of mass burn is proportional to the mass which means the mass is a decaying exponential.

M = Mo * exp( -g * time / thrust_to_weight )

If you think about this for a moment it becomes clear that any amount of mass would do since as the mass gets lighter it takes less fuel so the ship could go indefinitely at 1g. The problem is the assumption that the ship weighs nothing. so let's fix that.

dM/dt = -g*(M+Ms)/thrust_to_weight.

where Ms = mass of ship and M = mass of fuel.

I'm spacing on how to solve that equation so I'll approximate it by saying that until M = Ms we can mostly ignore the ship mass. therfore for a 6.6 year flight time the fuel required is about:

Mfuel = Ms * exp( g* (6.6 years)/thrust_to_weight )

Mfule = Ms * exp( +303,800,000/thrust_to_weight).

So you need a rather high thrust to weight ratio due to the coefficient in the exponetial.

Let the pillory for my "obvious" math errors begin!

Re:OK Another one

[bill_mcgonigle]

It might even have a thin enough atmosphere to not completely crush a human.

If the gravity isn't too high, we can engineer around all the rest. Ought to be just fine for bots if the solder doesn't flow at its temps. A giant pot of natural resources at 11LY is very exciting for colonials!

Re:OK Another one

[mark-t]

As a larger planet, however, since force of gravity is inversely proportional to the square of the distance, the surface gravity of a world otherwise equivalent in density to another ends up rises linearly with the diameter of the planet. If it is of similar composition to earth, then 5.4 earth masses would make it cbrt(5.4) times the size of earth, or roughly 1.75g at the planet's surface. Assuming that the atmospheric density is comparable to earth's (possible, even with greater gravity if the atmosphere itself is proportionally thinner), then this is theoretically survivable by human beings for short periods, or even prolonged ones if they were able to acclimate to the increased gravitation pull gradually, over a span of several years, giving time for skeletal tissue to build up and strengthen the body's structure to survive the increased tension.

ok, so, what now

[confused one]

So close... and yet still a freeking impossible distance away.

Oh.. it's just 11 light years away. That's a small number, right? As much as I'd like to be able to say we have a "warp drive" or "jump drive" or something like that... at the moment 11 light years might as well be 11 million light years. it makes no difference to our ability to get there.

Re:OK Another one

[ShanghaiBill]

But, of course, we don't know that the density of the planet is comparable to earth.

It is probably less. Of all the planets and spherical moons is our solar system, no other has a density as high as Earth. Earth's density is 5.5 gm/cc. The moon is 3.3. Mars is 3.9. If this planet has a density similar to the moon, its surface gravity would be about the same as Earth's.

Re:alien planets

[Jason Levine]

At one point, the prevailing scientific theory was that planets were a rarity. Then we found the first exoplanet and astronomers started wondering if they might be more common. By now, with the thousands of exoplanets found, we know that planets are plentiful. We don't know how many Earth-like ones are out there, but many astronomers think that this is more of a deficiency in our planetary detection methods than a rarity of Earth-like worlds. (Bigger planets are easier to detect.)

Venus

[luis_a_espinal]

If the surface gravity were about the same as the Earth's, wouldn't that mean that its atmospheric pressure at the surface would be about the same also. After all, it's gravity holding the gas down, and technically the atmospheric pressure is the weight of the gas above that point. Assuming the gas is trapped to the planet by the gravity, then you might have about the same amount of gas trapped above a point by a similar amount of gravity.

I'm just speculating though.

No. Atmospheric pressure is not simply a function of gravity. It is more a function of how much stuff there is in the atmosphere.

Consider that Venus' surface gravity is 0.904g wrt to Earth's (1g). And yet Venus's atmospheric pressure at the surface is 9.2 Megapascals whereas Earth's atmospheric pressure is 101.325 kilo-pascals (or 0.101325 Megapascals).

That is, even though Venus gravity is 90.4% that of Earth, its atmospheric pressure is 92 times that of Earth.