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

<p>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.)</p>

<p>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. </p>

<p>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.</p>

CC BY-SA 3.0
2016-01-20T21:45:35.670Z
If the orbital period is $\sim 30,000$ years, then the proper motion on the sky will be a massive $\sim 40$ arcsecs/yr. If it were as bright at $V=16$ I'm surprised this hasn't been picked up already in photographic surveys. It might be tricky because the proper motion is too big! Either way, a candidate could be confirmed in a matter of weeks if identified.
ProfRob
CC BY-SA 3.0
2016-01-20T21:51:30.883Z
I'm not much of an astronomer (I answered this Q when it was in space.sx rather than astronomy.sx). If you want to take a stab at reformulating the last paragraph, please do! I'm also not totally sure of the magnitude computation.
Russell Borogove
CC BY-SA 3.0
2016-01-21T00:00:56.053Z
See my updated answer below.  The problem you and I made was using a 1/r^2 falloff with distance; when since we are talking about reflected light 1/r^4 is the correct term.  As a result, even at its closest it would be far fainter than your estimate.
Dan Is Fiddling By Firelight
CC BY-SA 3.0
2016-01-21T00:02:44.347Z
@DanNeely I had just figured that out myself. This is a serious error in the answer.
ProfRob
CC BY-SA 3.0
2016-01-21T00:10:32.397Z
Ah! Of course. Good catch.
Russell Borogove
CC BY-SA 3.0
2016-01-20T21:14:07.457Z
2016-01-21T00:20:11.000Z
2016-01-21T00:20:11.000Z
Russell Borogove
<p>Citing the <a href="http://iopscience.iop.org/article/10.3847/0004-6256/151/2/22/meta" rel="noreferrer">original article</a>:</p>

<blockquote>
  <p>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.</p>
</blockquote>

<p>and</p>

<blockquote>
  <p>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.</p>
</blockquote>

<p>So, finding out likely orbital parameters is work in progress.</p>

CC BY-SA 3.0
2016-01-20T20:39:10.220Z
2016-01-20T20:39:10.220Z
Gerald
<p><a href="http://www.findplanetnine.com/p/blog-page.html" rel="nofollow">Batygin and Brown made a website</a> which describes the search for the 9th planet in clear terms. They specifically note the following:</p>

<blockquote>
  <p>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.</p>
</blockquote>

<p>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.</p>

CC BY-SA 3.0
2016-01-23T23:20:45.327Z
2016-01-24T00:33:15.307Z
2016-01-24T00:33:15.307Z
PearsonArtPhoto
<p>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.</p>

CC BY-SA 3.0
2016-01-21T16:18:04.897Z
While this is technically an answer to the question, it is sparse on the things that make good answers (citations, detailed explanations, and math).
Donald.McLean
CC BY-SA 3.0
2016-01-21T13:15:03.807Z
2016-01-21T13:15:03.807Z
Mike Scott
<p><strong>It's too dim to be seen during a normal survey during the majority of its
orbit.</strong></p>
<p><strong>Update:</strong>  Scientists at the University of Bern have <a href="http://www.unibe.ch/news/media_news/media_relations_e/media_releases/2016_e/media_releases_2016/planet_9_takes_shape/index_eng.html" rel="noreferrer">modeled</a> a hypothetical 10 Earth mass planet in the proposed orbit to estimate its detectability with more precision than my attempt below.</p>
<p>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.</p>
<p>It should however be within reach of the <a href="https://en.wikipedia.org/wiki/Large_Synoptic_Survey_Telescope" rel="noreferrer">LSST</a> 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.</p>
<p><strong>My original attempt to handwave an estimate of detectability is below.</strong>
The <a href="https://iopscience.iop.org/article/10.3847/0004-6256/151/2/22/meta" rel="noreferrer">paper</a> gives potential orbital parameters of <span class="math-container">$400-1500~\textrm{AU}$</span> for the semi major axis, and <span class="math-container">$200-300~\textrm{AU}$</span> 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 <span class="math-container">$1500~\textrm{AU}$</span> semi-major axis and a <span class="math-container">$200~\textrm{AU}$</span> perihelion has a <span class="math-container">$2800~\textrm{AU}$</span> aphelion.</p>
<p>To calculate the brightness of an object shining with reflected light, the proper scaling factor is not a <span class="math-container">$1/r^2$</span> 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 <span class="math-container">$1/r^4$</span> scaling as in a <a href="https://en.wikipedia.org/wiki/Inverse-square_law" rel="noreferrer">radar return</a> 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 <span class="math-container">$\sim 6x$</span> dimmer than Uranus despite being only <span class="math-container">$50\%$</span> farther away: <span class="math-container">$1/r^4$</span> scaling gives a <span class="math-container">$5x$</span> dimmer factor vs <span class="math-container">$2.25$</span> for <span class="math-container">$1/r^2$</span>.</p>
<p>Using that gives a dimming of 2400x at <span class="math-container">$210~\textrm{AU}\;.$</span> That puts us down <span class="math-container">$8.5$</span> magnitudes down from Neptune at perihelion or <span class="math-container">$16.5$</span> magnitude.  <span class="math-container">$500~\textrm{AU}$</span> gets us to <span class="math-container">$20$</span>th magnitude, while a <span class="math-container">$2800~\textrm{AU}$</span> aphelion dims reflected light down by nearly <span class="math-container">$20$</span> magnitudes to <span class="math-container">$28$</span> magnitude. That's equivalent to the <a href="https://en.wikipedia.org/wiki/Apparent_magnitude" rel="noreferrer">faintest stars visible from an 8 meter telescope</a>; making its non-discovery much less surprising.</p>
<p>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.</p>
<p>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.</p>
<p>For reference, data from NASA's <a href="https://en.wikipedia.org/wiki/Nemesis_%28hypothetical_star%29#Past.2C_current.2C_and_pending_searches_for_Nemesis" rel="noreferrer">WISE</a> experiment has ruled out a Saturn-sized body within <span class="math-container">$10,000~\textrm{AU}$</span> of the sun.</p>
<p><strong>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.</strong></p>
<p>Orbital eccentricity can be calculated as:</p>
<p><span class="math-container">$$e = \frac{r_\textrm{max} - r_\textrm{min}}{2a}$$</span></p>
<p>Plugging in the numbers gives:</p>
<p><span class="math-container">$$e = \frac{2800~\textrm{AU} - 200~\textrm{AU}}{2\cdot 1500~\textrm{AU}} = 0.867$$</span></p>
<p>Plugging <span class="math-container">$200~\textrm{AU}$</span> and <span class="math-container">$e = 0.867$</span> into a <a href="http://www.phas.ubc.ca/%7Enewhouse/p210/orbits/#bottom" rel="noreferrer">cometary orbit calculator</a> gives a <span class="math-container">$58,000$</span> year orbit.</p>
<p>While that gives an average proper motion of <span class="math-container">$ 22~\textrm{arc-seconds/year}\;,$</span> 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.</p>
<p>Kepler's laws tell us that the <a href="http://people.bridgewater.edu/%7Erbowman/ISAW/KeplerCalc.html" rel="noreferrer">velocity at aphelion</a> is given by:</p>
<p><span class="math-container">$$v_a^2 = \frac{ 8.871 \times 10^8 }{ a }  \frac{ 1 - e }{ 1 + e }$$</span></p>
<p>where <span class="math-container">$v_a$</span> is the aphelion velocity in <span class="math-container">$\mathrm{m/s}\;,$</span> <span class="math-container">$a$</span> is the semi-major axis in <span class="math-container">$\mathrm{AU},$</span> and <span class="math-container">$e$</span> is orbital eccentricity.</p>
<p><span class="math-container">$$v_a = \sqrt{\frac{ 8.871 \times 10^8 }{ 1500 } \cdot \frac{ 1 - 0.867 }{ 1 + 0.867 }} = 205~\mathrm{m/s}\;.$$</span></p>
<p>To calculate the proper motion we first need to convert the velocity into units of <span class="math-container">$\textrm{AU/year}:$</span></p>
<p><span class="math-container">$$205 \mathrm{\frac{m}{s}}\; \cdot  \mathrm{\frac{3600 s}{1 h}} \cdot \mathrm{\frac{24 h}{1 d}} \cdot \mathrm{\frac{365 d}{1 y}} \cdot \mathrm{\frac{1\; AU}{1.5 \times 10^{11}m}} = 0.043~\mathrm{\frac{AU}{year}}$$</span></p>
<p>To get proper motion from this, create a triangle with a hypotenuse of <span class="math-container">$2800~\textrm{AU}$</span> and a short side of <span class="math-container">$0.043~\textrm{AU}$</span> and then use trigonometry to get the narrow angle.</p>
<p><span class="math-container">$$\sin \theta = \frac{0.044}{2800}\\ \implies \theta = {8.799×10^{-4}}^\circ   = 3.17~\textrm{arc seconds}\;.$$</span></p>
<p>This is well within Hubble's angular resolution of <span class="math-container">$0.05~\textrm{arc seconds};$</span> 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 <span class="math-container">$\sim 500~\textrm{AU},$</span> it would be bright enough to be seen by the ESA's <a href="https://en.wikipedia.org/wiki/Gaia_%28spacecraft%29" rel="noreferrer">GAIA</a> 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.</p>
<p>Its parallax movement <a href="https://astronomy.stackexchange.com/a/13283/186">would be much larger</a>; however the challenge of actually seeing it in the first place would remain.</p>

CC BY-SA 3.0
2016-01-21T00:04:50.200Z
The proper motion is massive and easily detectable, but you would need JWST (or perhaps just HST) in order to measure it and they have small fields of view, so you'd need to know more-or-less where it was.
ProfRob
CC BY-SA 3.0
2016-01-21T00:26:41.680Z
Actually this is just at the faint end, and of course mabe its brighter in the IR. I read that the Subaru telescope is already looking.
ProfRob
CC BY-SA 3.0
2016-01-21T02:04:02.623Z
For anyone who read earlier versions, I made an ~60x error in calculating proper motion at aphelion; it would be readily observable in Hubble observations; but is probably too faint to have been picked out in any proper motion surveys.
Dan Is Fiddling By Firelight
CC BY-SA 3.0
2016-01-21T06:05:03.057Z
Such a great answer. At aphelion, how bright would the sun appear from Planet Nine? http://astronomy.stackexchange.com/questions/13282/how-bright-would-the-sun-appear-from-the-hypothetical-planet-nine-proposed-by-ca
joseph.hainline
CC BY-SA 3.0
2016-01-21T07:49:33.537Z
Note here http://astronomy.stackexchange.com/questions/13280/how-would-parallax-affect-an-object-at-200-1000-au-ie-the-9th-planet/13283#13283 almost everyone (including me) forgot about the parallax, which is much bigger than the proper motion. It would be clearly identifiable by a large telescope within days. Gaia is all-sky, but limited to about 20th Mag.
ProfRob
CC BY-SA 3.0
2016-01-21T12:57:19.370Z
@RobJeffries: I'm glad I remembered it:-) Still, it might be difficult to see, depending on the orientation, so it might be best to do the search at 3 month increments. Hmmm...
PearsonArtPhoto
CC BY-SA 3.0
2016-01-21T14:56:40.193Z
@RobJeffries yeah, I facepalmed when I realized I'd forgotten that too.  I saw that question last night; but was too brainfried to attempt another answer at that point.
Dan Is Fiddling By Firelight
CC BY-SA 4.0
2016-01-20T23:07:09.203Z
2022-12-11T06:21:24.830Z
2022-12-11T06:21:24.830Z
Dan Is Fiddling By Firelight
<p>I've seen a number of news reports indicating there is likely a <a href="http://news.discovery.com/space/astronomy/9th-planet-may-lurk-in-the-outer-solar-system-160120.htm">9th planet in our Solar System</a>, 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?</p>

<solar-system><planet><9th-planet>
9th planet location?
PearsonArtPhoto
# 9th planet location?

(Asked by: username_0 on 2016-01-20)

I've seen a number of news reports indicating there is likely a [9th planet in our Solar System](http://news.discovery.com/space/astronomy/9th-planet-may-lurk-in-the-outer-solar-system-160120.htm), 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?

**Tags:** solar-system, planet, 9th-planet


---

## 5 answers

### Answer by username_3 on 2016-01-20. Score: 37 | Accepted answer

**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](http://www.unibe.ch/news/media_news/media_relations_e/media_releases/2016_e/media_releases_2016/planet_9_takes_shape/index_eng.html) 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](https://en.wikipedia.org/wiki/Large_Synoptic_Survey_Telescope) 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](https://iopscience.iop.org/article/10.3847/0004-6256/151/2/22/meta) 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](https://en.wikipedia.org/wiki/Inverse-square_law) 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](https://en.wikipedia.org/wiki/Apparent_magnitude); 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](https://en.wikipedia.org/wiki/Nemesis_%28hypothetical_star%29#Past.2C_current.2C_and_pending_searches_for_Nemesis) 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](http://www.phas.ubc.ca/%7Enewhouse/p210/orbits/#bottom) 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](http://people.bridgewater.edu/%7Erbowman/ISAW/KeplerCalc.html) 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](https://en.wikipedia.org/wiki/Gaia_%28spacecraft%29) 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](https://astronomy.stackexchange.com/a/13283/186); however the challenge of actually seeing it in the first place would remain.

> **Comments:**
> **username_2**: The proper motion is massive and easily detectable, but you would need JWST (or perhaps just HST) in order to measure it and they have small fields of view, so you'd need to know more-or-less where it was.  
> **username_2**: Actually this is just at the faint end, and of course mabe its brighter in the IR. I read that the Subaru telescope is already looking.  
> **username_3**: For anyone who read earlier versions, I made an ~60x error in calculating proper motion at aphelion; it would be readily observable in Hubble observations; but is probably too faint to have been picked out in any proper motion surveys.  
> **username_7**: Such a great answer. At aphelion, how bright would the sun appear from Planet Nine? http://astronomy.stackexchange.com/questions/13282/how-bright-would-the-sun-appear-from-the-hypothetical-planet-nine-proposed-by-ca  
> **username_2**: Note here http://astronomy.stackexchange.com/questions/13280/how-would-parallax-affect-an-object-at-200-1000-au-ie-the-9th-planet/13283#13283 almost everyone (including me) forgot about the parallax, which is much bigger than the proper motion. It would be clearly identifiable by a large telescope within days. Gaia is all-sky, but limited to about 20th Mag.  
> **username_0**: @RobJeffries: I'm glad I remembered it:-) Still, it might be difficult to see, depending on the orientation, so it might be best to do the search at 3 month increments. Hmmm...  
> **username_3**: @RobJeffries yeah, I facepalmed when I realized I'd forgotten that too.  I saw that question last night; but was too brainfried to attempt another answer at that point.  

---

### Answer by username_1 on 2016-01-20. Score: 10

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.

> **Comments:**
> **username_2**: If the orbital period is $\sim 30,000$ years, then the proper motion on the sky will be a massive $\sim 40$ arcsecs/yr. If it were as bright at $V=16$ I'm surprised this hasn't been picked up already in photographic surveys. It might be tricky because the proper motion is too big! Either way, a candidate could be confirmed in a matter of weeks if identified.  
> **username_1**: I'm not much of an astronomer (I answered this Q when it was in space.sx rather than astronomy.sx). If you want to take a stab at reformulating the last paragraph, please do! I'm also not totally sure of the magnitude computation.  
> **username_3**: See my updated answer below.  The problem you and I made was using a 1/r^2 falloff with distance; when since we are talking about reflected light 1/r^4 is the correct term.  As a result, even at its closest it would be far fainter than your estimate.  
> **username_2**: @DanNeely I had just figured that out myself. This is a serious error in the answer.  
> **username_1**: Ah! Of course. Good catch.  

---

### Answer by username_4 on 2016-01-20. Score: 5

Citing the [original article](http://iopscience.iop.org/article/10.3847/0004-6256/151/2/22/meta):

> 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.


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### Answer by username_0 on 2016-01-23. Score: 3

[Batygin and Brown made a website](http://www.findplanetnine.com/p/blog-page.html) 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.


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### Answer by username_5 on 2016-01-21. Score: -1

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.

> **Comments:**
> **username_6**: While this is technically an answer to the question, it is sparse on the things that make good answers (citations, detailed explanations, and math).  

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