It's too dim to be seen during a normal survey during the majority of its
orbit.
Update: Scientists at the University of Bern have modeled a hypothetical 10 Earth mass planet in the proposed orbit to estimate its detectability with more precision than my attempt below.
The takeaway is that NASAs WISE mission would have probably spotted a planet of at least 50 Earth masses in the proposed orbit and that none of our current surveys would have had a chance to find one below 20 earth masses in most of its orbit. They put the planets temperature at 47K due to residual heat from formation; which would make is 1000x brighter in infrared than it is in visible light reflected from the sun.
It should however be within reach of the LSST once it is completed (first light 2019, normal operations beginning 2022); so the question should be resolved within a few more years even if its far enough from Batygin and Brown's proposed orbit that their search with the Subaru telescope comes out empty.
My original attempt to handwave an estimate of detectability is below.
The paper gives potential orbital parameters of $400-1500~textrm{AU}$ for the semi major axis, and $200-300~textrm{AU}$ for perihelion. Since the paper doesn't give a most-likely case for orbital parameters, I'm going to go with the extreme case that makes it most difficult to find. Taking the most eccentric possible values from that gives an orbit with a $1500~textrm{AU}$ semi-major axis and a $200~textrm{AU}$ perihelion has a $2800~textrm{AU}$ aphelion.
To calculate the brightness of an object shining with reflected light, the proper scaling factor is not a $1/r^2$ falloff as could be naively assumed. That is correct for an object radiating its own light; but not for one shining by reflected light; for that case the same $1/r^4$ scaling as in a radar return is appropriate. That this is the correct scaling factor to use can be sanity checked based on the fact that despite being similar in size, Neptune is $sim 6x$ dimmer than Uranus despite being only $50%$ farther away: $1/r^4$ scaling gives a $5x$ dimmer factor vs $2.25$ for $1/r^2$.
Using that gives a dimming of 2400x at $210~textrm{AU};.$ That puts us down $8.5$ magnitudes down from Neptune at perihelion or $16.5$ magnitude. $500~textrm{AU}$ gets us to $20$th magnitude, while a $2800~textrm{AU}$ aphelion dims reflected light down by nearly $20$ magnitudes to $28$ magnitude. That's equivalent to the faintest stars visible from an 8 meter telescope; making its non-discovery much less surprising.
This is something of a fuzzy boundary in both directions. Residual energy from formation/radioactive material in its core will be giving it some innate luminosity; at extreme distances this might be brighter than reflected light. I don't know how to estimate this. It's also possible that the extreme cold of the Oort Cloud may have frozen its atmosphere out. If that happened, its diameter would be much smaller and the reduction in reflecting surface could dim it another order of magnitude or two.
Not knowing what sort of adjustment to make here, I'm going to assume the two factors cancel out completely and leave the original assumptions that it reflects as much light as Neptune and reflective light is the dominant source of illumination for the remainder of my calculations.
For reference, data from NASA's WISE experiment has ruled out a Saturn-sized body within $10,000~textrm{AU}$ of the sun.
It's also likely too faint to have been detected via proper motion; although if we can pin its orbit down tightly Hubble could confirm its motion.
Orbital eccentricity can be calculated as:
$$e = frac{r_textrm{max} - r_textrm{min}}{2a}$$
Plugging in the numbers gives:
$$e = frac{2800~textrm{AU} - 200~textrm{AU}}{2cdot 1500~textrm{AU}} = 0.867$$
Plugging $200~textrm{AU}$ and $e = 0.867$ into a cometary orbit calculator gives a $58,000$ year orbit.
While that gives an average proper motion of $ 22~textrm{arc-seconds/year};,$ because the orbit is highly eccentric its actual proper motion varies greatly, but it spends a majority of its time far from the sun where its values are at a minimum.
Kepler's laws tell us that the velocity at aphelion is given by:
$$v_a^2 = frac{ 8.871 times 10^8 }{ a } frac{ 1 - e }{ 1 + e }$$
where $v_a$ is the aphelion velocity in $mathrm{m/s};,$ $a$ is the semi-major axis in $mathrm{AU},$ and $e$ is orbital eccentricity.
$$v_a = sqrt{frac{ 8.871 times 10^8 }{ 1500 } cdot frac{ 1 - 0.867 }{ 1 + 0.867 }} = 205~mathrm{m/s};.$$
To calculate the proper motion we first need to convert the velocity into units of $textrm{AU/year}:$
$$205 mathrm{frac{m}{s}}; mathrm{frac{3600 s}{1 h}} cdot mathrm{frac{24 h}{1 d}} cdot mathrm{frac{365 d}{1 y}} cdot mathrm{frac{1; AU}{1.5 times 10^{11}m}} = 0.043~mathrm{frac{AU}{year}}$$
To get proper motion from this, create a triangle with a hypotenuse of $2800~textrm{AU}$ and a short side of $0.043~textrm{AU}$ and then use trigonometry to get the narrow angle.
$$sin theta = frac{0.044}{2800}\ implies theta = {8.799×10^{-4}}^circ = 3.17~textrm{arc seconds};.$$
This is well within Hubble's angular resolution of $0.05~textrm{arc seconds};$ so if we knew exactly where to look we could confirm its orbit even if its near its maximum distance from the sun. However its extreme faintness in most of its orbit means that its unlikely to have been found in any survey. If we're lucky and it's within $sim 500~textrm{AU},$ it would be bright enough to be seen by the ESA's GAIA spacecraft in which case we'll located it within the next few years. Unfortunately, it's more likely that all the GAIA data will do is to constrain its minimum distance slightly.
Its parallax movement would be much larger; however the challenge of actually seeing it in the first place would remain.
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