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Comment by antognini | Hacker News Reader

They do, and at the end of the press release it says that Brown has begun an observing campaign to find the planet. Unfortunately detecting these very distant objects is very difficult. The light we see from them is reflected light from the Sun, so their apparent brightness goes as 1/r^4, not 1/r^2 like most objects. And because they are so close, there is a strong parallax from the motion of the Earth. THe consequence from this is that the object will actually drift across the image as you are taking it, smearing it out. In order to detect it you either have to take very short images so that it doesn't move too much (in which case you don't get very much light), or have some idea as to the direction of the smear so that you can correct for it. Either way it's a very hard observation to make! Fortunately if this planet is as big as they claim it to be it should be relatively easy to detect compared to Sedna and the other Kuiper Belt objects.

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In a previous life, back in the pre-CCD days I worked at an observatory searching for new asteroids. Used to be the way it was done you would point a telescope at the part of the sky directly opposite the sun (min r^4 so best chance of detection), expose a photographic plate, and repeat a few hours later. Develop the plates, mount them in a blink comparator [1] and scan manually, flipping back and forth between the plates for ~3 hours to look for anything that moved. Exclude false positives like plate blemishes (aka Eastman-Kodak Objects) or dust grains. If you see something, and it is smeared indicating motion during the exposure, mark it as a possible candidate. Revisit the same patch of sky the next night, search again. Use the datapoints to extrapolate forward, check again a couple of weeks later. If you haven't lost it (which frequently happened if it was at the limit of visibility), extrapolate back to the previous opposition (~4 years for a main belt asteroid) and search the archives for a match. If you get one, then extrapolate further back and check all previous oppositions. If you get a match on three separate oppositions, congratulations, you have a discovery.

Vastly different now - a single observation can net large numbers of asteroids and sensitivity is so much better. Still the search for this potential new planet is definitely going to be daunting given the distance.

[1] https://en.wikipedia.org/wiki/Blink_comparator

Also, we don't know where in its orbit the planet happens to be right now. If we're lucky, it's relatively close to the sun, and it will be detectable. If we're unlucky, it's relatively far away from the sun, and it could easily be undetectable by current observatories because the orbit is so eccentric.

[Source: I saw Mike Brown give a talk on this topic at the Cal Academy of Sciences, where he hoped he was lucky...]

Note that an eccentric orbit spends most of its time out in the far reaches, since it moves much slower out there. The odds are against the happy case.

Perhaps one of the "stubby hubble" telescopes the NRO donated to NASA would be useful for this task[0]. They are essentially the hubble telescope with a much shorter focal length(hence the nickname), so one of the uses suggested for them is observing faint objects in the solar system like asteroids.

[0]https://en.wikipedia.org/wiki/2012_National_Reconnaissance_O...

I would expect a smear with characteristics of the planet would be as effective to identify it as a single snapshot. Wouldn't the direction be well identified from indirect detection? Is the blur also less effective because of increased background exposure?

Well, if the smear takes the light level below the noise floor of the sensor then you can't even say that there's a smear.

That's why you don't see any cars in long exposures of highways.

That's why you use Goldstone and Arecibo, and bounce your signal!

Are you sure it's r^4, not 4r^2?

Yes. The solid angle subtended by the planet (and hence the amount of reflected light) goes as 1/r^2, and then the apparent brightness of the reflected light to us goes as an additional 1/r^2. The total dependence is then 1/r^2 * 1/r^2 = 1/r^4.

This is a really good physics puzzler.

The key point is that Pluto does not reflect light like a mirror, it scatters light relatively uniformly, causing more light loss to an observer than, say, a mirror that points the reflection directly at an Earth observer.

The light reaching the object from the sun drops off by 1/r^2, and then the light reaching Earth from the object drops again by 1/r^2. Since the latter light is a portion of the former, you multiply them for 1/r^4.

Yes, it's definitely r^4. If we assume that the planet is far from the sun compared to the earth, such that the earth-planet distance and the sun-planet distance are both approximately r, then the amount of light the planet gets from the sun is proportional to 1/r^2. Then the planet reflects that light in all directions, and the fraction of it that comes back to us on earth is also proportional to 1/r^2. 1/r^2 * 1/r^2 = 1/r^4.

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