9th planet location?

Question

I've seen a number of news reports indicating there is likely a 9th planet in our Solar System, something with an orbital period of between 10k-20k years, that is 10 times Earth's mass. I haven't seen any real indication of where this object might be. If I had access to a sufficient telescope, would I be able to find this planet, and what way would I point a telescope to find it? How far is it likely to be, or is that not well known?

Answer 1

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:

$$e = \frac{2800~\textrm{AU} - 200~\textrm{AU}}{2\cdot 1500~\textrm{AU}} = 0.867$$

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 $\textrm{AU/year}:$

$$205 \mathrm{\frac{m}{s}}\; \cdot \mathrm{\frac{3600 s}{1 h}} \cdot \mathrm{\frac{24 h}{1 d}} \cdot \mathrm{\frac{365 d}{1 y}} \cdot \mathrm{\frac{1\; AU}{1.5 \times 10^{11}m}} = 0.043~\mathrm{\frac{AU}{year}}$$

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.

Answer 2

The position of the hypothetical object is not known with any certainty, so it's hard to know where to point your telescope.

The paper proposes a wide range of orbital distances anywhere from 400 to 1500 AU semi-major axis, with a perihelion (closest approach to the sun) of 200-300AU. This is 8 times as far as Neptune. (I didn't read the article closely enough to determine whether the body would be near perihelion or not at present; it could be over 1000 AU away, 30 times Neptune's distance.)

With a mass of 10 Earths, we would expect the body to be something like 2-5 times Earth's radius -- somewhat smaller than Neptune.

The combination of distance and size suggests the body would be far fainter than Neptune, no brighter than magnitude 16.5 at perihelion, and likely much dimmer.

Answer 3

Citing the original article:

We find that the observed orbital alignment can be maintained by a distant eccentric planet with mass $\geq \approx 10$ m⊕ whose orbit lies in approximately the same plane as those of the distant KBOs, but whose perihelion is 180° away from the perihelia of the minor bodies.

and

As already alluded to above, the precise range of perturber parameters required to satisfactorily reproduce the data is at present difficult to diagnose. Indeed, additional work is required to understand the tradeoffs between the assumed orbital elements and mass, as well as to identify regions of parameter space that are incompatible with the existing data.

So, finding out likely orbital parameters is work in progress.

Answer 4

Batygin and Brown made a website which describes the search for the 9th planet in clear terms. They specifically note the following:

perihelion (its closest approach to the sun) at around a Right Ascension in the sky of 16 hours, which means that the perihelion position is straight overhead in late May. Conversely, the orbit comes to aphelion (the furthest point from the sun) at about 4 hours, or straight overhead in late November.

So to look for it, one should look along the ecliptic, concentrating mostly on the area directly overhead in late November. Note that this is the part of the sky where the galactic center also appears. The inclination is estimated to be 30 degrees, plus or minus 20, so that distance from the ecliptic should be searched as well.

Answer 5

If you had access to a sufficient telescope, you could theoretically see it, if you looked in the right place (although no one knows where the right place might be). But if it's anywhere near aphelion there are only a handful of sufficient telescopes in the world (let's say an 8m mirror or larger), so I think it highly unlikely that you have access to one of them.