6
$\begingroup$

This is Yeola~Camay. A double planet orbiting Sharro, an ordinaryish star somewhere in the universe of Far Far Away.

enter image description here

As you can plainly see, Yeola is the planet on the left & Camay is the planet on the right.

Here is a map of Narutanea, the Eastlands of Yeola:

enter image description here

East is to the right, west is to the left. The markings around the edges of the map show degrees north of the equator and east of the prime meridian. The space between each is six degrees.

The land to the south of the Sea of the Moon and to the west of the Ocean of Sunrise is labelled Auntimoany and its chief city is Pycleas. Pycleas is roughly 24 deg north and 36 deg east (close enough for government work!). (This is point of interest no. 1). Directly to the west is the country of Harathalliê, where you can see a lake. In the vicinity of the lake are the West Downs and the old North Tower. (This is point of interest no. 2)

The two locations are about 1200mi distant, as the raven flies.

This is a map of the night sky, centered at the ancient observatory just west of Pycleas:

enter image description here

If you lie on the courtyard of the Observatory with your head towards the east (right side of the map) and tilt your head back, you'll see the red gash of The Chasm down towards your right hand and you'll see a portion of Camay above your head.

This is a map of the eastern sky, showing approximately what an observer would see looking out east over the waters of the Ocean:

enter image description here

In this map, the observer is looking towards the east, making the north poles of both planets off towards the left.

I figure, from this particular vantage point, the observer can see about 25% to 30% of the twin planet. Since they're tidally locked with respect to one another, that aspect of the sky never changes.

The Question!

Given that Camay would be more visible the farther east I travel from Pycleas, because I'm moving towards the line in space where the twin planet's cores line up; about how far west can I go before Camay disappears entirely?

Pertinents: Yeola is 34,851 mi in circumference (11,093 mi in diametre); Camay is 33,124 mi in circ. (10,543 mi diametre); the distance between their poles is 2.6 planetary diameters or about 28,128 mi. (NB: if that makes Camay "look too big" in the sky, then I may opt to place them further apart.)


To make life easier, while there are mountains (none terribly high) and hills and valleys and forests and so forth between Point no. 1 and Point no. 2, we can probably sweep those differences in elevation under the most convenient of metaphorical rugs. I'm looking for a broad range (say within 50 or a hundred miles) rather than a definitive point, brought out to fourteen decimal places.

Also, you can ignore that their miles and our miles are slightly different in length.


The very allerbestest of responses would couple visuals with numbers. A merely acceptable response will simply state a rough distance (in miles) west of Pycleas where the twin planet disappears.

$\endgroup$
12
  • 1
    $\begingroup$ Do they orbit each other parallel to the plane of the ecliptic or at some other angle to it, it matters. $\endgroup$ Commented May 30, 2019 at 19:19
  • $\begingroup$ I think the most important data here are the diameters of the planets and the distance between them. $\endgroup$
    – Cumehtar
    Commented May 30, 2019 at 19:19
  • 1
    $\begingroup$ @Cumehtar I was thinking of optical effects re the sun setting behind the twin, it changes things. $\endgroup$ Commented May 30, 2019 at 19:23
  • 1
    $\begingroup$ @elemtilas, it's probably just me. "Twin planet" in my head means two co-orbital bodies like a counter-earth. "Binary planet" (aka double planet) makes more sense to me. $\endgroup$
    – JBH
    Commented Jun 1, 2019 at 14:26
  • 1
    $\begingroup$ Note that, if the planets' mutual orbits are elliptical and/or Yeola's rotation is oblique to the orbital plane, Camay will wobble in the sky. $\endgroup$ Commented Jun 3, 2019 at 0:57

2 Answers 2

9
$\begingroup$

I'm amazed at the amount of detail here, but most of it is really not necessary. We only need to know 3 things:

  • The diameters of the planets
  • The distance between the planets
  • Which point on Yeola is closest to Camay

To elaborate on the last bullet, we need the point that lies on the axis connecting the two cores, like you mentioned. I'll call this point P. You'll need to find P for yourself because it's not clear from your maps where it is exactly (though you mentioned it was east of Pycleas, so presumably you have some idea).

I'll take your answer of 11093 miles as the diameter of both planets from your comment. Having the same diameter means that the answer to "where on Yeola can you see Camay" becomes really easy. Simply divide Yeola into hemispheres, with one of the hemispheres centered at P. If you're in the hemisphere that contains P, you can see Camay. If you're not, you can't (Camay will be below the horizon). Easy as that!

For a more detailed answer that includes how large in the sky Camay is at every point, we need to know how far apart the two planets are, and how they rotate around each other. All of this information is in your skymap.

Camay is (nearly) exactly in the eastern part of the sky, which seems to me can only happen if they orbit around each other in the same plane as their equators. On Earth, we defined north and south as the directions toward our pole of rotation. It seems logical to me (and supported by the evidence of the skymap) that the people of Yeola did the same. Since Yeola and Camay are tidally-locked, Yeola's rotation axis is parallel to the rotation of Camay around it, so I think I'm safe to assume that Camay and Yeola orbit in the plane of each other's equators. That simplifies things a lot, too.

So how far away is Camay? If the skymap is accurate, ~19353 mi away. To see why, look at the diagram below.

enter image description here

The skymap indicates that Camay takes up $0.1913 \pi$ radians of the sky (I measured the pixels N-S). We know that its cross-section on the sky is roughly its diameter, 11093 miles, so $11093/ (0.1913 \pi) = d = 18458$ miles.

Next, we'll use the fact that 55% of Camay's diameter is visible above the horizon at Pycleas (you mentioned 25-30% of the surface was visible, or about 50-60% of the hemisphere facing Yeola). A bunch of annoying similar triangles later, we find that $D = \sqrt{2(0.55)R^2+d^2} = 19353$ miles.

Which is to say that these planets are frighteningly close to each other. There's a little less than one planet width between the two of them. This will be true as long as the skymap is accurate, even if you eventually change the scales of the planets (at present they're about 40% wider than Earth).

But fair enough, let's actually use this to calculate the size of Camay in the sky as a function of your angle away from P, which we'll call $\theta$.

How do you find $\theta$? Well, if you're directly north or south of P, $\theta$ is just your latitude (N or S), and if you're directly east or west of P, it's the difference between your longitude and that of P. If you're at a different position, you can find it by simply calculating the distance to P along the surface, then dividing by Yeola's radius (this answer is in radians, so multiply by $180/\pi$ afterward to convert to degrees).

Anyway, the angular size $S$, in degrees, of Camay for any theta is

$S = \frac{360}{\pi}[(\frac{D}{R})^2-2\frac{D}{R}cos\theta+1]^{-1/2}$,

where the ratio $\frac{D}{R} = 3.489$.

Important angles:

  • At P, which corresponds to $\theta = 0$, Camay is right in the middle of the sky all the time. Its apparent size is a whopping 46.0 degrees wide, nearly as big as a steering wheel looks when you're driving. For comparison, the moon as seen from Earth is half a degree. If I lived on Yeola, I would be looking up, awe-inspired, like, all the time.
  • At $\theta = arccos(2R/D) = 55.0$ degrees, Camay touches the horizon. At all angles between this and 90 degrees, Camay will be partially hidden behind the curve of Yeola. Here, Camay is 37.8 degrees wide, like a basketball held at arm's length.
  • At $\theta = arccos(R/D) = 73.4$ degrees, Camay is halway below the horizon, and looks like a semicircle. Based on your view of the sea from Pycleas, its $\theta$ value is a little below this. $S$ is 34.3 degrees here, so just a bit smaller than the previous bullet.
  • At $\theta = 90$ degrees, Camay disappears just barely below the horizon, but you can see it if you stand on a tall ladder or a mountain or something. It's 31.6 degrees wide here, about the size of a soccer ball at arms' length.

Here's a handy reference (pun definitely intended) if you want to be able to picture how large in the sky it is. (Source)

enter image description here

Just hold out your hand at arm's length, and these gestures show about how big these angular sizes look.

ANSWERING THE QUESTION

I have enough information to calculate where P is without you telling me, and thus give you an exact answer, but I'm not sure how valid it is to assume that Camay in the star map is perfectly sized. Regardless, I think this answer is close, and certainly $\theta$ for Pycleas is roughly 70 degrees, since this is required in order to have Camay near the horizon as seen from there.

By adding this ~70 degree angle to Pycleas' longitude, subtracting a bit to account for it being 24 degrees north of the equator, I would place P on the equator at around 90 degrees E. Conveniently, that means that Camay disappears right at the prime meridian, so it's even easier to tell when you can see it: if you're in the eastern hemisphere, you can see it. If you're in the western hemisphere, you can't.

Thus, you stop seeing Camay when you cross into the western hemisphere, ~36 degrees (~3,500 miles) west of Pycleas.

$\endgroup$
1
  • $\begingroup$ Minor nitpick: You can see Camay from slightly more than a hemisphere. Looking through the atmosphere at low angle bends the light slightly, you end up being able to look a little bit below the horizon. (Obviously, I'm assuming a decent atmosphere but given the question I think that's reasonable.) $\endgroup$ Commented Jun 3, 2019 at 4:10
2
$\begingroup$

Trace a circle around your globe that passes through both poles. Make the plane of this circle be perpendicular to the line between the cores of Yeola and Camay.

This circle splits the world in to hemispheres.

On the one closest to Camay, Camay is visible. On the farside of the world, Camay is never visible.

Given a circumferance of Yeola, the distances would be simple formulas. Calculating the circumferance given the information you've provided is likely doable, but I am unable to do so at this time.

$\endgroup$
2
  • $\begingroup$ What's the simple formula? $\endgroup$
    – elemtilas
    Commented May 31, 2019 at 1:27
  • 1
    $\begingroup$ @elemtilas Assuming they're tidally locked, Camay is visible from Yeola anywhere within 90 degrees latitude or longitude from the point on the equator directly below Camay. You may get a few additional miles in specific places if the terrain is high enough. $\endgroup$
    – Ryan_L
    Commented May 31, 2019 at 5:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .