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Detecting planets or other cold objects is hard, when they are far from the sun. Illumination from the sun falls with inverse square law (~1/r^2). When the refl
by Qem 3y ago
Detecting planets or other cold objects is hard, when they are far from the sun. Illumination from the sun falls with inverse square law (~1/r^2). When the reflected light does the return trip, it also falls with (~1/r^2). The effect compounds on both legs of the trip, and overall the brightness of a planet falls proportional to ~1/r^4, quickly getting lost amidst all the noise in the night sky as distance increases.
- gray_-_wolf 3y agoCould you build an uber radar for detecting the planet, possibly working together with the radiotelescopes we already have? Given the article, it should be ~150 light hours from sun, which round trip time wise sounds doable.
- pfdietz 3y agoRadar faces the same inverse fourth power effect.
- Qem 3y agoNot sure. But to map polar ice in mercury it was necessary to use the Arecibo radio telescope as radar[1], with the benefit we knew previously where to point the instrument, exactly. Given planet nine is hundreds of times more distant, it would probably take a humungous radio telescope and a lot of power to conduct the search. [1] https://www.nature.com/articles/369213a0 https://www.nature.com/articles/369213a0
- bastawhiz 3y agoMy suspicion is that you'd have a really hard time doing that. For one, if so little of the light from the sun (notably the brightest thing in our solar system) is being reflected, we'd be hard pressed to build something capable of spraying the whole search area and getting back enough particles to detect and count. But even more than that, the power requirements would be immense. We have retro reflectors on the moon, and even with knowing where the moon is and where they are on the moon, we get almost no photons back. From Wikipedia: > Out of a pulse of 3×1017 photons[25] aimed at the reflector, only about 1–5 are received back on Earth, even under good conditions. That's with a dedicated reflector positioned well on the moon's surface, about 0.002au away. You're talking about hitting a rocky lump in front of a bright background 400-800au away. The power requirements wouldn't be five orders of magnitude more, it would be perhaps hundreds of orders of magnitude more.
- catlifeonmars 3y agoIs it fair to say it is easier to detect planets in other solar systems?
- bastawhiz 3y agoYes! We often detect planets in other solar systems because they pass in front of their stars, causing the brightness to dip. You can also look for Doppler shifts in the light you see from stars, which is caused by the planet pulling the star slightly. You can also look for the sorts of brightness changes you'd expect as a result of gravity lensing. In almost all cases, you're looking for changes in something you can already see. The problem with a planet 9 is that so little light is getting to it that it's going to be very very dim.
- greggsy 3y agoWouldn’t it still be passing in front of something, or is it likely to be comparatively ‘stationary’ against a similarly stationary backdrop of stars and other objects?
- bastawhiz 3y agoWhen a star is occluded by a planet, it'll dip in brightness at a pretty regular interval (because of the orbit). So you can look look for a series of dips, which gives you a candidate. That's then looked at more closely. Since a planet 9 wouldn't transit a star at any regular interval, it wouldn't show up as a candidate for that search technique.
- catlifeonmars 3y agoSo we would need to potentially deliver the telescope far outside of our solar system and have it look back at us, kind of like a very expensive selfie stick?
- bastawhiz 3y ago