There are odd things about the orbits of trans-Neptunian objects that suggest ( to some) that there might be an undiscovered super-Earth-sized planet a few hundred AU from the Sun..
We haven’t seen it, but then it would be very hard to see. The underlying reason is simple enough, but I have never seen anyone mention it: the signal from such objects drops as the fourth power of distance from the Sun. Not the second power, as is the case with luminous objects like stars, or faraway objects that are close to a star. We can image close-in planets of other stars that are light-years distant, but it’s very difficult to see a fair-sized planet a few hundred AU out.
West Hunter 
Yup.
Actually, it’s even worse. A 100% reflective body would give you the fourth power. But in practise, you can’t have even that. The approximately spherical shape of such an object would also redirect most of the Sun’s light away from us. So, it’s even much worse.
Nope. It’s true that the light is not 100% reflected, but it does not change how bright the plane appear in function of the distance. If you want to break the 1/d^4 function, you need to make reflectance vary with sun distance… It can, somehow, indirectly through temperature effet on the albedo (rock->water->methane clouds) but that will not happen that far from the sun, and I doubt you had this kind of effect in mind anyway…
We’re talking about how the light intensity decreases with distance. Reducing the albedo reduces the light emitted independent of distance.
According to Wikipedia: “Many small objects in the outer Solar System[19] and asteroid belt have low albedos down to about 0.05.[20] A typical comet nucleus has an albedo of 0.04.[21] ”
https://en.wikipedia.org/wiki/Albedo
Farther out you go, smaller the albedos.
Not really. That’s like talking about the constant factor in an n log n algorithm: doesn’t dominate.
What if Planet Nine isn’t a planet at all, but is a primordial black hole?
If we find it we have to call it Pluto.
Proserpina.
Shouldn’t astronomers be able to determine the approximate position of this hypothetical planet by carefully measuring the orbital perturbations of the trans-Neptunian objects?
I’m guessing that if astronomers can’t do this it’s because even measuring the orbits of those trans-Neptunian objects is hard to do with precision.
That and, there are lots of things of various sizes floating around out there. All of the known large bodies have their effects and eliminating all these effects from so many bodies is very tricky. You’d practically have to observe some body changing it’s orbit unexpectedly: Hey, we predicted this body would be here, but it’s there.
The Planet Nine hypothesis
https://physicstoday.scitation.org/doi/10.1063/PT.3.4172
Good article. Thanks.
Why not just consult the wee chaps in Area 51?
https://news.nationalgeographic.com/2016/01/150119-new-ninth-planet-solar-system-space/
https://iopscience.iop.org/article/10.3847/0004-6256/151/2/22;jsessionid=9DAB98EED9CB30448604A2F4CA0F8752.c5.iopscience.cld.iop.org
Detectable by it occluding a star? More money on telescopes and computing power required rather than the crap we spend it on now maybe.
If it’s Jupiter-sized planet, then it could emit much more heat than it receives from Sun, so fourth power doesn’t apply…
This is very true. Every Jupiter size (even less) object can’t be very cold. At least not yet. So many radioactive atoms in such a big mass and so relatively small surface for that energy produced inside to escape through it — makes this surface relatively hot.
There are other contributors, too. Like compressing under gravity. Every gold atom falling slowly toward the center makes heat. Every hydrogen molecule, falling into a small hole beneath its previous position, makes heat. Every place swap between a lighter and heavier isotope also makes heat.
For big objects, 1/r^2 still applies.
Generally, people haven’t been talking about a Planet X as big as Jupiter.
According to Michael Brown (the infamous Pluto killer) this is the best case for hypothetical Planet Nine.
https://physicstoday.scitation.org/doi/10.1063/PT.3.4172
In a sophisticated comparison of solar-system observations to numerical simulations, we find a best match to be a putative Planet Nine that is approximately six times the mass of Earth, inclined with respect to the ecliptic by a little less than 20 degrees, and in a moderately eccentric orbit about 400 times as distant from the Sun as Earth.
Much of that mass will be ice, lower density than Earth’s iron and silicon. So it might be six times the mass but, oh, eight times the radius. Sixteen times the surface-area as viewed head-on – which it will be, since we’re only going to see the sunny side.
Let’s see if I’m getting this right:
A perfectly flat, perfectly reflective Sun-facing mirror orbiting at a distance x would produce a coherent image of the Sun at an apparent distance of 2x, so its signal strength would go as 1/(2x)^2 = 1/4x^2, i.e. still an inverse-square.
But a planet(oid) isn’t a flat mirror, it gives diffuse reflections. Incident radiation is reflected back in all directions in an expanding (hemi)sphere, so we get another inverse-square factor for the journey back which stacks on top of the original inverse-square, yielding an overall inverse-fourth relationship.
You are getting it mostly right, I guess, but in fact, it’s much more complicated. Assume, that we have quite a perfect sphere in the great distance from the Sun. Assume also, that the Sun is above this sphere’s north pole. At the north pole, the incident light would be reflected back to the Sun (a 180 degrees turn), but at a small distance from the north pole, the incident light would reflect slightly away, let say 179 degrees turn). At every sphere’s point, the angle of incidence rule applies. So at 45 degrees north parallel on that sphere, every ray (photon) from the Sun would be reflected 90 degrees from its previous course. Near the sphere’s equator, those rays would be almost unscattered.
Fortunately, we are very close to the Sun, so only the light reflected from a thin ring around the sphere’s north pole would ever reach us. It’s a bit worse than 1/r^4.
This is purely geometrical consideration. But there are also small albedos for distant objects. Not only small reflections of very diluted sunlight but also less and less reflective stuff when you are going away from the Sun. As some deep space black “mold” sets in when there is not enough light, but a lot of cosmic rays instead.
Doesn’t that logic only work for specular reflections? For diffuse reflections we’d get (a little bit of) light back from every part of the planet that was illuminated.
Random perturbations would give you the same number of photons in your camera, on average.
For analogous reasons, the equation for detection range of radar has an inverse fourth power term in it as well. Ever wonder why the F-22 and F-35 are advertised as having such teeny weeny radar cross sections in the approximately .001M^2 region, or whatever it is? Because it has to be that small to get tactically useful reductions in detection range.
You wanna halve the range at which a given radar system picks up an airplane (or missile or whatever)? RCS has to go down by a factor of 16. You want to cut the range to a third? RCS has to go down by a factor of 81. And so on. This type of engineering is possible, but spendy.