Terxpahseyton
Nobody
- Joined
- Sep 9, 2006
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- 10,759
Makes sense.
Life on another Planet? Gliese 581d
- Thread starter Kal'thzar
- Start date
LucyDuke
staring at the clock
Cool, I guess, but it'll be ages before we can actually investigate whether there might be life there.
r_rolo1
King of myself
I have some doubts regarding this, especially the part of the atmosphere ( spectrometric analysis of a planetary atmosphere out of our solar system is not exactly easy and knowing the atmosphere pressure is even harder ) ... most likely it is theorical speculation ( that it is not as reliable as that ).
It is interesting , but I'm not seeing this news bringing anything of substance to the space exploration programs and to the almost defunct SETI
It is interesting , but I'm not seeing this news bringing anything of substance to the space exploration programs and to the almost defunct SETI
It's a bit of a silly restriction, since thermoelectric power from radioactive decay is allowed. A 60 kg brick of Pu-238 exploding in the upper atmosphere is going to be a pretty big problem.
That would be pretty bad, but a nuclear reactor going overcritical in the atmosphere would be even worse.
Well you wouldn't send the reactor up all at once! You'd send small amounts of fuel up and put the reactor together in space.
How far are we away from being able to analyze the spectra of extrasolar planets' atmospheres for telltale signs of life, like high concentrations of oxygen? I feel like that's the most plausible way to get evidence for ET life.
I seem to recall that when Galieo (or was it Cassini?) did a fly-by of earth they pointed one of its sensors at earth to see if they could detect life here. I don't think the results were encouraging.
Rather far, perhaps practically impossible. If we're considering rocky, earth-like bodies, the largest hurdle would be actually seeing the planet.
There's a limit to how finely you can resolve* detail, dependant on distance from the object, size of the object, size of your viewing equipment and the wavelength of the light you're using. For a rocky planet light years away, you're trying to see something rather tiny when compared to the distance. It's for this reason that you'll rarely see an optical light microscope go beyond 1000x.
Back-of-the-envalope calculations for light with a wave length of 562 nanometres (visible).
Let's start a real life example, the Hubble telescope and its 2.5m mirror. It should resolve down to .04 arc seconds at the wave length we're using. An arc second is 1/60th of an arc minute, which is 1/60th of a degree. Thus an arc second is 1,296,000th of a circle (60*60*360). Let's put Hubble into a low earth orbit of 500km (real orbit is ~569km) and take pictures of the ground. A circle with Hubble in the centre of it has a a circumference of 3,141.5 km or 314,150,000cm and can resolve down to 9.68cm (314,150,000 / 1,294,000 * .04)
So let's move this out to an exoplanet.
A 10m telescope should resolve down to .0116 arc seconds. Let's extend a circle to Gliese 581, 20 light years away and we get a circumference of ~125.66ly (20*2*3.14).
Convert that to kilometres, divide that by 1296000 for arc seconds and then multiply by .01 and we get a resolving power for a 10m telescope of... 9,166,833.93 kilometres, or ~15 times the diameter of our sun -- needless to say, few planets are that large.
Now, if you were to use different wave lengths of light, your resolution would be different. Further, without taking a moment to google and confirm, I assume spectroscopy would work at any wave length of light, thus it might be possible to resolve a rocky planet in another wave length, and still attempt to do spectral analysis of it. Well, except for the fact you run head first into perhaps the biggest problem.
Planets are dark. As far as I'm aware, all the planets currently being imaged are not only very large, rather distant from their parent star, but are also quite warm (giving off a lot of light in the infrared). A rocky planet is unlikely to do this. If it's far enough from its parent star to be easily seen, it will be cold. If it's warm enough to be detected in infrared, it's too close to its star to be seen.
*Disclaimer! I'm not a physicist nor am I by any means great at math. I could be really confused here and using the wrong terms. As I understood in my physics classes, resolution in this sense would be the size an object must be to be seen. If that resolution were say, 20cm, you would be able to see a cylinder on my desk where a can of Diet Coke currently sits, but not the word "coke"
edit, was layin' in bed thinking about this post and realized I'd only done the math to how large an arc second is at those distances -- I'd forgot to multiply by the fraction of resolution! Needless to see I sat up like a bolt and ran to correct myself
r_rolo1
King of myself
^^But with a terran planet, getting enough atmosphere in the way of our spectrometer to make a band detectable looks a little dificult. And even then it is hard to correlate those bands with partial pressures in the said planet even if we identify the coumponds ( especially if the atmosphere has carbon coumpounds, like it is stated in the link )... and you can't know temperatures without it ( just for kicks, Martian and Venusian atmospheres are quite similar in composition ( barring maybe the high atmoshere sulfuric acid clould in Venus ) ... but the pressure makes things a little diferent on both )
IMHO there is a lot of wishful thinking around this planet
IMHO there is a lot of wishful thinking around this planet
... yes, but that wasn't the question. He wanted to know about resolving down to the point of directly imaging a rocky planet and how far that technology would be off.
wapamingo
Prince
- Joined
- Sep 19, 2006
- Messages
- 399
I always wondered, and maybe some of you wise people can provide an answer.
We are looking at these stars and planets, and we are looking at light spectra, radiation and so forth. I know that it takes time for light and radiation and such to travel, so when we are making observations the information is time sensitive. So are we looking at this planet as it looked X million of years in the past, or more? In other words, is there a significant lag in what the planet is right now and what we see it as?
We are looking at these stars and planets, and we are looking at light spectra, radiation and so forth. I know that it takes time for light and radiation and such to travel, so when we are making observations the information is time sensitive. So are we looking at this planet as it looked X million of years in the past, or more? In other words, is there a significant lag in what the planet is right now and what we see it as?
duckstab
Child of Noble Family
As far as I know, all the extrasolar planets discovered so far are within a few hundred light years. So, yes, there's a lag, but not on a geologic time scale.
Yes, you're seeing the star or the planet as it was X years ago, where X = distance in light years.
In the same sense, you're seeing the sun as it was 8 minutes ago, and the moon as it was 1 second ago. If you were to turn the sun "off," we wouldn't know for 8 minutes, as the sun would continue to look like it's "on" for that time.
C is the speed limit of the universe. Nothing goes faster than it, not even information. There's a weird consequence of this that, out side a sphere with a radius of your age (if you're 25, a sphere with a radius of 25 light years) you don't exist as far as the universe is concerned.
I followed most of your post, but didn't understand this one part that i've set in bold.
Why is it necessary to establish a circle with the detector at the center, and what has that got to do with the size of Earth?
It has nothing to do with the size of the earth, but rather the size of the telescope's resolution. An arc minute / arc second are fractions of the circumference of a circle centred on the observer.
I'm still not following where the circle's (circumference of 3141.5 km) dimensions are coming from. But that number is suspiciously similar to pi...
I understand that the circle is centered on the observer, but I don't understand how the radius to the edge of the circle is determined.
I understand that the circle is centered on the observer, but I don't understand how the radius to the edge of the circle is determined.
We were trying to see how finely Hubble could resolve if we pointed Hubble towards the surface of Earth instead of at stars. For a bit of simplicity, I moved Hubble's orbit from 569km above the surface of the earth to 500km up.
A circle's circumfrence is diameter*pi or 2radius*pi. So if we're 500km up in space and want to draw a circle that touches the surface of the earth, and has us in the middle, we'd say 500km*2*pi and get a circle with a circumference of 3141.5km. If we then wanted to know how finely we could resolve a length of that circumference, we'd run it through the math I did earlier in this thread. Hubble's 2.5m mirror can resolve .04 arc seconds, or 1/32,400,000 of a circle (1,296,000th is an arc second divide that by .04)
1/32,400,000'th of a circle with a circumference of 3,141.5km is 9.6cm.
*facepalm*
Completely obvious now. Thanks for taking me [baby] step by [baby] step through it.
Completely obvious now. Thanks for taking me [baby] step by [baby] step through it.
Abaddon
Deity
I hope we have some massive techno-jump so either I can survive until such exploration takes place, or that such exploration suddenly becomes a lot more viable.
There would be a massive worry of sending our own bacteria there by accident. How could we be sure not to "spoil" the samples/planet when we visit it?
There would be a massive worry of sending our own bacteria there by accident. How could we be sure not to "spoil" the samples/planet when we visit it?
