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So in my universe, there's only a single planet, contained within a spherical shell. Orbiting around it is a dragon, holding a "sun" for lack of a better word. This sun is not a star as we know it in our universe, but it does provide light and warmth to the planet in the same way.

The universe is magical and not bound by science, but to make it feel more real, I do want to figure out the math regarding size and distance.

How far away and how large would this sun need to be, to be visually indistinguishable from how we experience the sun on Earth?

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    $\begingroup$ The perceived size of something depends both on its size and how far away it is. So when you allow both of those to be variable, there's no single answer. At best you can have an equation. $\endgroup$ Jan 6, 2023 at 17:16
  • $\begingroup$ @basklein Great question! All we need to answer it with solid math is these two details: 1) Is your planet rotating at all; if so, how fast (eg. how many hours per rotation)? 2) How long is day supposed to be, on average (eg. how many hours per day, on the equinox)? $\endgroup$
    – Qami
    Jan 6, 2023 at 17:19
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    $\begingroup$ You could have it be the size of the sun, and at the same distance as the sun. Is that not good enough? $\endgroup$
    – kaya3
    Jan 7, 2023 at 13:09
  • $\begingroup$ @basklein Good question, since even fantasy stories generally have the same geometry as the real world. I upvoted. It is possible that you might find my answer "illuminating". $\endgroup$ Jan 8, 2023 at 6:48

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Part One: A Globular light source emitting light in all directions.

The question says:

How far away and how large would this light source need to be, to be visually indistinguishable from how we experience the sun on Earth?

The light source would have to be rather far away to be visible from about half of the surface area of the planet at any one time, as the Sun is from the Earth.

The question doesn't describe the planet so it could be much larger or smaller than the Earth.

It is impossible for a globular light source spreading its light in all directions illuminate exactly one half of a spherical object, because the rays from the light source would have to be parallel to illuminate exactly half the spherical object, and the spherical light source would have to be infinitely far away for its beams to be parallel when they strike the spherical object.

In the case of Earth and the Sun, the Sun is about a hundred times as wide as the Earth and it's distance is over ten thousand times the diameter of the Earth.

And you can make the light source much bigger and father away, relative to the planet, than the Sun is to Earth, or make it much smaller and closer relative to the planet than the Sun is to the Earth. If you want the light source to have the same angular diameter as seen from the planet as the Sun from the Earth, you have to keep the ratio between the diameter and the distance of the light source the same as for the Sun and the Earth.

If you want a relatively small and nearby light source but want about an entire hemisphere of the planet to be illuminated at any one time there will be a limit to how small and close the light sources can be. If it get too close it will illuminate only a part of one hemisphere at a time and many parts of the planet will be in eternal darkness, which is not eh case on Earth.

So now I will have to draw a picture with words since I don't have a drawing program, and you have to draw it yourself or picture it in your mind's eye.

Picture a circle, representing your planet, with a long line though the center of the planet. Put a dot on the line representing the light source. Draw a line from the light source to graze the circumference of the circle representing the planet.

Draw a line from the point where the line from the light source grazes the circle to the center of the circle. There should be a right angle of 90 degrees at the the pint where the angle from the light sources intersects the lien frm the circumference to the center of the circle. You have now made a right angled triangle. The angle at the center of the circle representing the planet, and the angle at the light source should add up to 90 degrees, but they don't have to be equal to each other.

In order for the light source to illuminate exactly one hemisphere of the planet the angle at the center of the planet has to be exactly 90 degrees, the angle where the lien from the light source grazes the surface of the circle/planet has to be exactly 90 degrees, and the angle at the light source has to be exactly zero degrees, which is impossible in a triangle.

So the light source will be unable to illuminate exactly one hemisphere of he planet, because there will be a part of the near hemisphere in shadow where the curve of the planet blocks the light from the light source. Since you want approximately half of the planet illuminated by the light source at any one time, you need to figure out how close to a full hemisphere you want.

If, for example, you want the angle at the center of the planet to be 75 degrees, the angle at the light source will be 15 degrees.

If, for example, you want the angle at the center of the planet to be 80 degrees, the angle at the light source will be 10 degrees.

If, for example, you want the angle at the center of the planet to be 75 degrees, the angle at the light source will be 5 degrees.

If, for example, you want the angle at the center of the planet to be 89 degrees, the angle at the light source will be 1 degree.

So you could draw diagrams and measure the ration of the line between the light source and the center of the planet to the line from the center of the planet to the point where light grazes the surface.

Or you can use trigonometry to calculate the ration and see how many planetary radii the distance to the light sources is with the proportion of the surface you decide to have illuminated.

The planet is supposed to be otherwise like Earth, and so it should have a breathable atmosphere similar to Earth's. Earth's atmosphere refracts light, bending it down to illuminate areas a few degrees beyond the bulge which cuts off light in a straight line, so the planet should not be too unlike Earth if the line from the center to the edge where light grazes the surface is just a few degrees.

Part Two: A disc like light emitter with parallel rays.

Of course the light source doesn't have to be a spherical light source emitting light in all directions. It could emit light from a disc aimed at the spherical world, and that light could be parallel like laser light.

Because the light rays would be parallel in that case, the light emitting disc would have to be at least as large as the diameter of the spherical planet, in order for one entire hemisphere to illuminated.

If the light emitting disc pointed at the planet has the the diameter of the planet, and also has an angular diameter of about 0.5 degree, at the distance to the planet, my rough calculations go like this:

One half of a degree is one 720th of a full circle of 360 degrees. The circumference of a circle is 2 times pi times the radius of the circle. Using 3.14159 as a good enough value of pi, The radius of a circle is equal to 57.295 degrees of the circumference. Since the light disc is equal to the planet's diameter and to 0.5 degees of the circumference, the radius of the circle and the distance between the light disc and the planet is equal to 114.5916 times the diameter of the light disc and of the planet.

If the planet has a diameter of 6,371 kilometers, the mean diameter of Earth, the distance to the light disc will be about 730,063.4 kilometers.

If the light disc is more than a few tens of thousands of kilometers closer than the calculated distance, its disc will appear to be larger than the Sun appears from Earth, and if it is more than a few tens of thousands of kilometers farther than the calculated distance, it will appear noticeably smaller than the Sun appears from Earth. And if the light disc is farther from the planet than the calculated distance its parallel rays will illuminate less than one full hemisphere of the planet at a time.

I would be grateful for anyone who improved by answer with diagrams and/or trigonometric calculations.

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Distance and size go together.

The sun is ~150,000,000km from earth (it varies through our orbit from 152.1Gm at the aphelion to 147.1Gm at the perihelion) and has a diameter of ~1,393,000km. The moon is ~384,000km from earth and has a diameter of ~3475km and is almost exactly the same visible size as the sun during an eclipse. To visualize it (not to scale, obviously), with a lot of rounding:

The simple option

The simplest way to choose the size/distance for your draconic sun is to just take the ratio of distance:diameter. The sun is 150Gm away and 1.39Gm across for a ratio of ~108:1. The moon is 384Mm away and 3.48Mm across for a ratio of ~110:1.

If you want your draconic sun to be 1000km away, it should be 1000km/108 = 9.26km across. If you want your draconic sun to be 100km across, it should be 100km * 108 = 10,800km away.

The complicated option

What's relevant here is actually the portion of the viewer's line of sight that the celestial body obstructs. So let's take a look at the angles, using the radius of the bodies:

These are pretty rough numbers, but the sun's radius results in a triangle with an angle of 0.267° from the viewer's perspective while the moon's radius results in a triangle with an angle of 0.26°. You'll want your draconic sun to have roughly the same angle. So let's do some trigonometry!

The relevant formula here is tangent (or tan):

You may be familiar with the mnemonic from trigonometry "SOH CAH TOA": sine (sin) of the angle is the opposite edge over the hypotenuse; cosine (cos) of the angle is the adjacent edge over the hypotenuse; tangent (tan) of the angle is the opposite edge over the adjacent edge. For that angle math above, I'm using arc tangent (atan, or tan-1), which is just the inverse of tan (and there's asin and acos in the same vein). More specifically:

$$ tan(\theta) = \frac{700Mm}{150Gm}\\ atan(tan(\theta)) = atan(\frac{700Mm}{150Gm})\\ \theta = atan(\frac{700Mm}{150Gm})\\ \theta = 0.267 $$

If you pick a radius R, you can therefore solve for your distance D as follows:

$$ tan(0.267) = \frac{R}{D}\\ D = \frac{R}{tan(0.267)} $$

So if you want the draconic sun to be 100km across (about 3% as wide as the moon), it would need to be ~10,700km away from the earth (which you can calculate on WolframAlpha), which is close to that ~108x ratio from the simple approach. Note that since this is from the perspective of an observer on the surface these distances are relative to the surface of the earth rather than the center of the earth.

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  • $\begingroup$ You might find my answer interesting. $\endgroup$ Jan 8, 2023 at 6:45
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Size varies with distance.

What you are failing to consider here is that the farther away the “sun” is, the bigger it has to be, and the closer it is, the smaller it has to be. Its luminosity also plays a role.

Assuming this world has a moon the same size and distance as Earth’s moon, and you want to have the odd solar eclipse now an again, then this “sun” should be about the same size and a little farther away. In other words, it has to be about 385,000 kilometres away, and about 10900 km in circumference (at the equator). If you are a maths nerd, then you can work out how luminous it has to be. If you are not a maths nerd, then grab one off the street! Or buy a calculator. That works too.

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  • $\begingroup$ Correct it depends on the distance. The sun could be the size of your thumb orbiting at arms length and appear the same size as the moon and sun. $\endgroup$
    – Slarty
    Jan 6, 2023 at 19:09
  • $\begingroup$ I followed these instructions and I was arrested for kidnapping. What now? $\endgroup$ Jan 7, 2023 at 2:38
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It depends entirely on how far away the sun is. The sun must be further away than the moon (otherwise there would be no eclipses) and at just beyond the distance of the moon the sun would have to be just a little bit bigger than the moon. Alternatively the sun could be 93 million miles away (400 times further) in which case it would have to be 400 times bigger or as big as the sun actually is. Or it could be at any distance between these two with a proportional increase in size (10 time further away from the moon and 10 times bigger etc).

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The answer depends a bit on what you mean by "visually indistinguishable from how we experience the sun on Earth".

If only the apparent size of the sun on a specific location is relevant, then the other answers give you a good starting point.

However, as soon as one moves over the planet's surface, there is a certain movement in relation to this "sun". If the sun is small and close (e.g. 10000km), the relative movement on the surface would change the apparent size of the sun noticeably while being negligible on our earth (consider that the equator would be 6000km closer to the sun that the poles, the sun would be more than twice as large near the equator). (1)

The same would happen with parallax - moving over the surface of the planet would shift the position of the sun compared to what is noticeable on earth. (e.g. in the above example, the poles would likely be in eternal darkness assuming the 23° axial tilt). A closer, smaller light source will also illuminate less of the planet overall

Hence, there is only one true option: the size of our sun and the distance of our sun to earth.

However, with some tolerances, some lower bound of the distance could be established, where the change in apparent size and position would not be noticeable without modern equipment. The sun would definitively be larger and further out than any reasonably sized dragon you expect. Probably in the order of millions of kilometers (the radius of the sun's orbit must be much larger than the radius of the planet to keep the changes due to parallax and distance small enough).

(1) such a system, where the sun is inside the lunar orbit, could also be an interesting setting, though.

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