See you all in a couple weeks

That's a fair question and the answer depends on where it is in its orbit. By Kepler's laws of motion, orbits subtend equal area in equal time. What that means is that it's moving much faster at perihelion and very slowly at aphelion. The consequence is that it spends the vast majority of its time closer to aphelion than close to perihelion, so for any randomly selected point in time (ie, our lifetimes) an object like this in an eccentric orbit that is tens of thousands of years long is much more likely to be closer to aphelion than perihelion. This unfortunately means that it's not likely that a scope like mine can see it at all since it's very likely very far away and probably no brighter than Neptune.

But as to whether it's strictly ruled out, let's take a best-case scenario for being able to see it. Let's say it's only about 200 AU away. The planet it's probably most similar to in size and overall physical characteristics is Neptune, so let's use Neptune as a proxy for its brightness. Of course Neptune is much closer than 200 AU, so how can we do that? Well you can't just look up an answer or load up a program that will tell you, you have to do the math. We can calculate how bright Neptune would be from 200 AU away using the equations for the magnitude of the planets as published in Volume III of "The Solar System" by Gerard Kuiper in 1953 (as always I like to avoid using NASA data to show that you don't have to depend on what they say - they didn't even exist until 1958). This particular section was written by Daniel Harris.

[link to h.dropcanvas.com]

[link to h.dropcanvas.com]

Using these numbers, we can plug in the distance of our "Planet X" for an opposition where it is 200 AU from the sun, 199 AU from earth, as a best case scenario on viewing it.

Apparent magnitude = -6.87 + 5*log(200*(200-1))

The answer is that the apparent magnitude in a best case scenario is about magnitude 16.1. That's within reach of my telescope, but this is a very unlikely situation. It is much more likely to be far from perihelion since it would spend most of its thousands of years orbit closer to aphelion. Let's take a more realistic look at what it would be, close to aphelion. For an orbit with a semi-major axis of about 700 AU and eccentricity of about 0.6, aphelion is equal to 1120 AU. Let's use a 1000 AU distance as a nice round number and see what the situation is there.

Apparent magnitude = -6.87 + 5*log(1000*(1000-1))

That equals a whopping magnitude 23.1. That's beyond my telescope's abilities. It's of the order of magnitude that requires phenomenal effort by amateurs to detect; a magnitude 23.1 star is what amateurs used as a "magnitude signpost" when recording one of the dimmest objects ever recorded by an amateur astronomer.
[link to www.itelescope.net]

I hope that puts things in perspective on how difficult things become at that magnitude. It's a bit challenging even for professionals looking for near earth asteroids; they generally can't detect them until they get close enough to reach about magnitude 21 or so at least, about the same magnitude as Sedna ( [link to en.wikipedia.org (secure)] ).

Of course the aphelion distance could be even greater than this, so the worst case scenarios (which are much more likely than the best case scenario due to how eccentric orbits work) amateurs won't be able to see it at all since it could be even much dimmer than magnitude 23.