Astronomers have identified exoplanets from which potential life forms are likely to be able to observe a transit of one of our solar system's planets:
"The detection of thousands of extrasolar planets by the transit method naturally raises the question of whether potential extrasolar observers could detect the transits of the Solar System planets," they wrote in a paper published [open, DOI: 10.1093/mnras/stx2077] [DX] last month in the Monthly Notices of the Royal Astronomical Society.
[...] The transit method only works if a planet is aligned in a way that it crosses the star. In the Solar System, the terrestrial planets – Mercury, Mars, Earth and Venus – are more likely to be spotted in this way than the gas and ice giants – Jupiter, Saturn, Uranus and Neptune. Up to three planets in various combinations can be seen at any one time, the researchers found. The next step is to find which boundaries are located in the best positions to observe more than one of the terrestrial planets crossing the Sun, and count up the number of exoplanets inside these "transit zones."
Katja Poppenhaeger, co‑author of the study and assistant professor at Queen's University Belfast, estimated that "a randomly positioned observer would have roughly a 1 in 40 chance of observing at least one planet. The probability of detecting at least two planets would be about ten times lower, and to detect three would be a further ten times smaller than this." A full sweep shows there are currently 68 known exoplanets that are in a good spot to catch a planet transiting the Sun. From this list, nine of them are temperate and have sizes similar to Earth, but none are considered to be habitable. That doesn't mean the chances of aliens potentially spying on Earth are completely zero. The researchers estimate that there are ten other unconfirmed exoplanets that have more favorable conditions of sustaining life, and are within the transit zones.
(Score: -1, Offtopic) by Anonymous Coward on Tuesday September 12 2017, @12:32AM (2 children)
So 68 civilizations know about Trump's hair.
(Score: -1, Offtopic) by Anonymous Coward on Tuesday September 12 2017, @01:07AM
No, lots more know about it. Somebody from the α Draconis system has been posting all over the galactic net about it, complaining about being stationed in BFE. It's right there next to the pics showing that he tapes his ties. There's been an uproar in the All Thing about how anybody could possibly be considering admitting a species to the federation that elected an idiot who tapes his ties to the planet's highest political office!
It's too bad, too, because that fellow who opened the global athletic competition on the first audio/video broadcast received was so well dressed! WTF happened???
(Score: 2) by bob_super on Tuesday September 12 2017, @06:30PM
Gladly, the speed of light cuts that number way down.
Unless they are looking at our solar system with their gravity-sensing mohawks, and telepathing each other about why they would need occlusions, or yellow non-mohawkish head implants.
(Score: 4, Insightful) by Arik on Tuesday September 12 2017, @12:43AM
This certainly makes it sound like we actually have some solid information regarding e.g. the climate of these planets. Nine out of 68, we are informed, are 'temperate' though none are habitable. And it sounds like we're pretty darn sure we know which ones are 'temperate' and which ones are habitable, otherwise they wouldn't need to invoke unknown or unconfirmed exoplanets in order to arrive at a positive chance of ((aliens)) - obviously.
Trouble is, correct me if I'm wrong, but I'm pretty sure our state of knowledge is nowhere near the level implied. We have estimates of total mass and orbital distance, and could make educated guesses based on those estimates, but that's nowhere near enough to justify the sort of precision that's being claimed in this report.
If laughter is the best medicine, who are the best doctors?
(Score: 2, Interesting) by pTamok on Tuesday September 12 2017, @08:35AM (1 child)
Hmm. I wonder if it is technologically feasible to build a sufficiently powerful laser to mess with the transit luminosity profile enough to send a (possibly slow) message? That is, deliberately do something that looks something like KIC 8462852 [wikipedia.org] to an observer on an exoplanet looking in Earth's direction.
An observer on an exoplanet watching a transit of Earth would see a predictable change in the sun's luminosity. If you fire up a sufficiently large light source on earth during the transit, pointing directly at the exoplanet, the luminosity profile the observer sees would change. By modulating the light source intensity, you could probably transmit quite a few bits of information. If you wanted to communicate, the round-trip-delay would be non-negligible.
While I have a lot of blank envelope-backs, I don't have the smarts to work out what power of light source might be required.
There's a few tricks you could do to optimise things:
1) Use a laser. It's directional, so you minimise how much power you throw away in useless directions.
2) Use a laser - it's pretty monochromatic, so you could generate output of about the order of the level of solar luminosity for only a small part of the solar spectrum. This assumes observers on an exoplanet would monitor the spectrum and see the abnormality.
3) Turn it on and off in an easily identifiable pattern. This reduces the overall power requirement and/or allows higher peak power, and makes the signal more obviously artificial.
4) Build the laser to operate at a wavelength that corresponds to a relatively low-radiance emission wavelength for the sun - i.e. your background noise is lower, so you get more signal. I think somewhere near Hydrogen Lyman-alpha might be right, but I'm not sure.
Pointing lasers at celestial objects:
XKCD What-if "Laser Pointer" [xkcd.com]
Discussion about the above on the XKCD forums. [xkcd.com]
(Score: 2, Interesting) by pTamok on Tuesday September 12 2017, @12:01PM
Hmm.
Equatorial radius of Sun: 695,700 km -> Area of solar disc: 4,371,000 km2 (4.371x1012 m,sup>2)
Equatorial radius of Earth: 6,378 km -> Area of Earth's disk: 40,074 km2 (40.074x109 m,sup>2)
Percentage of total solar disc covered by earth during full transit (i.e. not at beginning or end): 0.9168 %
If an observer can detect the sun dimming by 1%, then assuming that if we can output light at 1% of the sun's output, it will be detectable during the transit at a blip in luminosity. That's rather a lot. Especially as the laser aperture is probably not the size of the earth's disc. If the laser aperture is 1 square meter, then it would need to be 40.074x109 times brighter than the sun (possibly 'only' at a particular frequency) to be detectable. Full spectrum full sunlight is about 1366 W/m2, so for a 1 m2 laser aperture, we would need need 1366 x 40.074x109 watts, or 5.474x1013 watts of power. That's 5.474 Terawatts. I don't think that is easily achievable continuously for any reasonable. However - if we 'only' need to achieve that for a small subset of the spectrum - say 0.1 %, that reduces the power requirement to 'only' 5.4 Gigawatts and we use a group of 1 metre aperture lasers - say 100 - then each laser 'only' needs 54 Megawatts of power.
By comparison, the lasers used at the USA's Argonne National Laboratory to investigate Inertial Confinement Fusion deliver 1.9 Megajoules in just over a nanosecond - about 500 Terawatts. Hmm. I doubt that observers would be set up to see nanosecond blinks in luminosity during a transit, but I'm beginning to think it might actually be do-able.
(Score: 3, Interesting) by FatPhil on Tuesday September 12 2017, @08:48AM (7 children)
? Mrad/Maxis
4.1 E-5
? Vrad/Vaxis
5.5 E-5
? Erad/Eaxis
4.3 E-5
? Jrad/Jaxis
9.1 E-5
Pah, let's ignore mars and generously call all other planets Jupiter-ish, because it sweeps out more area than either the inner planets (numerically as demonstrated above, in contradiction of Katja the Inexplicable's statement in TFS), or the outer planets (as it's bigger and closer).
Then lets pretend that none of the swept out areas overlap, so we get the maximum possible coverage.
=> Total area covered by the planets < 0.0005, or 1 part in 2000.
Where did the 1 in 40 come from?
Great minds discuss ideas; average minds discuss events; small minds discuss people; the smallest discuss themselves
(Score: 2) by dx3bydt3 on Tuesday September 12 2017, @11:13AM (6 children)
I would guess that isn't about the area subtended by the planet, rather it is the observable angles where at least one planet would cross the disc of the sun. 1 in 40 implies that 4.5° to either side of the plane of the solar system. Since all the planets have orbital planes that are inclined relative to one another that figure sounds plausible.
(Score: 2) by FatPhil on Tuesday September 12 2017, @12:37PM (5 children)
If she's not done the hard maths required to come up with a believable number, then I would still reserve the right to call fake. Simply being lined up with the ecliptic does *not* mean there'll be a possible transit; if spacey (and timey-wimey) stuff worked like that, there'd be a solar eclipse every 29 days, and the moon's *way* bigger (to the earth) than any planet (is to the sun).
Great minds discuss ideas; average minds discuss events; small minds discuss people; the smallest discuss themselves
(Score: 0) by Anonymous Coward on Tuesday September 12 2017, @04:53PM (1 child)
The reason we don't have a solar eclipse every month (or alternatively, never) is that the earth orbits the sun. Other stars don't orbit the sun.
(Score: 2) by FatPhil on Tuesday September 12 2017, @09:26PM
What utter, utter, codswallop.
The moon occludes millions of stars all the freaking time, and we don't rotate around any of those suns.
Great minds discuss ideas; average minds discuss events; small minds discuss people; the smallest discuss themselves
(Score: 3, Interesting) by dx3bydt3 on Tuesday September 12 2017, @05:46PM (2 children)
That sentence would then be more accurate to say, "a randomly positioned observer would have roughly a 1 in 40 chance of observing at least one planet... eventually". Venus and Mercury have the greatest inclination of their orbits, and Mercury takes over 230,000 years for the orbital ellipse to make one full rotation against the background stars.
(Score: 2) by FatPhil on Tuesday September 12 2017, @09:34PM (1 child)
Great minds discuss ideas; average minds discuss events; small minds discuss people; the smallest discuss themselves
(Score: 3, Interesting) by dx3bydt3 on Tuesday September 12 2017, @11:20PM
Oh, stripes indeed, but it turns out, if my math is right they all overlap handily. The precession of mercury's orbit is 5.5 arcseconds per year, and it is inclined by under 7°. if I didn't mess up a few decimal places, that means that in one earth year, Mercury's perihelion height change relative to some invariable plane would amount to only 1/16 of the planet's diameter. And that's using earth years, so roughly 4 times as many Mercury years, so only 1/66th of Mercury's diameter per orbit.
disclaimer: I have had a few glasses of wine, and my only references are the top hits on google.