**The Small-Angle Approximation**

Relatively complex geometry is necessary to determine the true size of distant objects based on their apparent size as seen by an observer. In this case, we are determining the opposite: we know the true diameter of the object in the sky and we know its distance from the observer, so we can use a shortcut to determine how large the object

*appears to be*in the sky. The following equation uses absolute values to return the apparent diameter in the sky in radians:Where:

𝛿c = apparent diameter of object, expressed in radians;

𝛿° = apparent diameter of object, expressed in degrees;

We can define Δ

𝛿c = apparent diameter of object, expressed in radians;

𝛿° = apparent diameter of object, expressed in degrees;

*d*= actual diameter of object, expressed in kilometers;*D*= distance to the object, expressed in kilometersWe can define Δ

*=**d**/**D*, and the equation becomes:Another version of this equation returns the diameter in the sky in arcseconds (¹/₃₆₀₀°):

Where:

𝛿ʼʼ = apparent diameter of object, expressed in arc-seconds;

𝛿ʼʼ = apparent diameter of object, expressed in arc-seconds;

*d*= actual diameter of object, expressed in kilometers;

*D*= distance to the object, expressed in kilometers**Angular Diameter Of The Moon**

Actual diameter of the Moon: 3474.2 km

Actual (average) distance to the Moon: 384399 km

Actual (average) distance to the Moon: 384399 km

So, from Earth, the Moon appears to be about ½° wide (it is said to

Doing the same calculation for the Sun:

Actual diameter of the Sun: 1391400 km

Actual (average) distance to the Moon: 1.496 ⨉ 10⁸ km

*subtend*about a half a degree of arc). This is about half the width of your little finger extended at arm’s length.Doing the same calculation for the Sun:

Actual diameter of the Sun: 1391400 km

Actual (average) distance to the Moon: 1.496 ⨉ 10⁸ km

Note that the figures for the Moon and the Sun are very close to one another. This explains why total eclipses of the Sun by the Moon are possible; from the vantage point of the Earth, both look more-or-less the same size in the sky, so the Moon can completely cover the Sun when it passes between it and the Earth.annulus

Note also that the Earth's orbit around the Sun is an ellipse, and as a result, the distance to the Sun changes slightly over the period of a year, so the angular diameter of the Sun varies from about 0.527° to about 0.545°. Similarly, because the Moon's orbit around the Earth is also an ellipse, the angular diameter of the Moon also varies, from about 0.488° about 0.568°.

Thus, there are eclipses during which the Moon does not completely cover the Sun, leading not to a total eclipse, but to an annular eclipse (from the Latin, “ring”, because a halo of Sun remains visible around the outside of the dark disk of the Moon).

**Apparent Brightness (Of Non-Luminous Objects)**

The means for calculating the apparent brightness of non-luminous objects is rather more complicated than calculating the apparent brightness of stars. We must take into account the reflectivity (

The equation for this calculation is:

*albedo*) of the reflecting object, and the relative distances between the object, the source illuminating it, and its distance to the observer.

The equation for this calculation is:

Where:

*A*= albedo of the reflecting body;*LS*= luminosity of the illuminating body, expressed in watts;*R*= radius of the reflecting body, expressed in meters;*D*= distance from the observer to the illuminating body, expressed in meters;*d*= distance from the observer to the reflecting body, expressed in metersNote that this equation requiresabsolutevalues, rather than relative values!

**Apparent Brightness Of The Moon**

To do this calculation, we need to know a few vital facts about the Moon:

The equation then, is:

*A**= 0.136*(albedo)*L**S**=**3.83**⨉ 10*²⁶*watts/m*²*(luminosity of the Sun)**R**=**1.74*⨉*10*⁶ meters (radius of the Moon)*D**=**1.496*⨉*10*¹¹*meters (distance from Earth to the Sun)**d**= 3.84*⨉*10*⁸ meters (distance from the Earth to the Moon)The equation then, is:

… which is about 0.08% as bright as a 60-watt light bulb.

**Fictional Albedos**

“How do I calculate the albedo of my fictional planets?” one may ask.

Yes … ummm … well, ahem …. It's not straightforward, unfortunately.

Albedo is a function of the material the light is reflecting from, which is related to the composition of the body, itself. For planets, prime considerations will be atmospheres. Thus, albedo is the ratio of the amount light reflecting from an object divided by the amount of light falling on it (the

The radius of the body is a factor in the equation above, such that, if Planet A is twice the diameter of Planet B, but they both have the same albedo, Planet A will appear brighter than Planet B when observed from the same distance. Conversely, if Planet A and Planet B are the same size, but Planet A has twice the albedo of Planet B, Planet A will again appear brighter than Planet B when viewed from the same distance. A real-world example is that of Earth and Venus: Venus is 94.99% the radius of the Earth, but its albedo is 0.76, compared to Earth's albedo of 0.40. Earth is larger (slightly), but viewed from the same distance, it would be dimmer than Venus—a large amount of the light reflecting from Venus is bouncing off its atmosphere, but only some of the light reflecting from Earth is reflecting from clouds, ocean, land surface, ice, etc., and a large amount is absorbed to be re-radiated in the infrared range, not visible light.

When it comes to fictional planets, my best advice is to:

This results in the following equation:

Yes … ummm … well, ahem …. It's not straightforward, unfortunately.

Albedo is a function of the material the light is reflecting from, which is related to the composition of the body, itself. For planets, prime considerations will be atmospheres. Thus, albedo is the ratio of the amount light reflecting from an object divided by the amount of light falling on it (the

*incident*light). If it reflects all of the incident light, it has an albedo of 1.0; if it reflects none of the incident light, it has an albedo of 0.0; and so on.The radius of the body is a factor in the equation above, such that, if Planet A is twice the diameter of Planet B, but they both have the same albedo, Planet A will appear brighter than Planet B when observed from the same distance. Conversely, if Planet A and Planet B are the same size, but Planet A has twice the albedo of Planet B, Planet A will again appear brighter than Planet B when viewed from the same distance. A real-world example is that of Earth and Venus: Venus is 94.99% the radius of the Earth, but its albedo is 0.76, compared to Earth's albedo of 0.40. Earth is larger (slightly), but viewed from the same distance, it would be dimmer than Venus—a large amount of the light reflecting from Venus is bouncing off its atmosphere, but only some of the light reflecting from Earth is reflecting from clouds, ocean, land surface, ice, etc., and a large amount is absorbed to be re-radiated in the infrared range, not visible light.

When it comes to fictional planets, my best advice is to:

- Find the known albedo of a Solar system object which is most similar to your fictional planet;
- Take the ratio of your planet's radius compared to that of the example planet;
- Multiply the albedo of the known planet by the ratio calculated in Step 2.

This results in the following equation:

Where:

For instance, let's say we have a planet—Taiar—with a surface similar to that of Mercury. Mercury's albedo is 0.06. Let's say Taiar is 0.6213 times Mercury's radius. Thus

*A**F*= albedo of the fictional planet;*R**F*= radius of the fictional planet;*R**K*= radius of the known planet;*A**K*= albedo of the known planetFor instance, let's say we have a planet—Taiar—with a surface similar to that of Mercury. Mercury's albedo is 0.06. Let's say Taiar is 0.6213 times Mercury's radius. Thus

... so, we'd say that Taiar's albedo is 0.037278.

This, of course, assumes that Taiar, as Mercury, has no significant atmosphere, and that Taiar's surface is composed of more-or-less the same materials as the surface of Mercury.

If, on the other hand, Taiar is 62.13% of the size of Mercury, but has an atmosphere like Titan's, it would be better to use Titan's albedo as the standard. Since Titan is ≈ 1.08 times the size of Mercury (yep—Titan is

This, of course, assumes that Taiar, as Mercury, has no significant atmosphere, and that Taiar's surface is composed of more-or-less the same materials as the surface of Mercury.

If, on the other hand, Taiar is 62.13% of the size of Mercury, but has an atmosphere like Titan's, it would be better to use Titan's albedo as the standard. Since Titan is ≈ 1.08 times the size of Mercury (yep—Titan is

*larger*than Mercury), and Taiar is 0.6213 the size of Mercury, then Taiar is:… 0.5756 times Titan's diameter. Titan's albedo is 0.22, so the equation becomes:

... and this is the value we'd use for Taiar’s albedo in the apparent brightness equation above.

Is this a rigorously scientific calculation? No, and so I hereby invoke

Is this a rigorously scientific calculation? No, and so I hereby invoke

*rationale ultimum*.**The Naked-Eye Limit**

From Earth, with minimal light pollution in the atmosphere, Saturn is the last of the naked-eye planets (assuming we mean observation by a

Saturn's apparent brightness (on average) is 1.338 ⨉ 10⁻⁷ w/cm³, so it makes sense to say that no object with a calculated apparent brightness less than this value can be called a

Note that this is NOT saying that no object orbiting farther from its star than does Saturn from the Sun would be visible to the tenants of an inhabited world in that system. Apparent brightness is not just a function of distance from the illuminating body, but also of the size and reflectivity of the object, and its distance from the

For instance, if Jupiter were moved to Saturn's orbit, it would have an average apparent brightness of 2.007 ⨉ 10⁻⁷ w/cm³, almost 1½ times brighter than Saturn is as the same distance. It certainly would remain a naked-eye object. Uranus, however, moved inward to Saturn's orbit would have an apparent brightness of 2.275 ⨉ 10⁻⁸ w/cm³, only 17% of Saturn's brightness and remain invisible to the naked eye.

*human*eye), meaning that it can be seen in the night sky with the eye alone. Uranus and Neptune—as well as the dwarf planets and other TNO/Kuiper belt objects—do populate the Solar system beyond the orbit of Saturn, but they are not visible even to the most eagle-eyed of humans without the aid of binoculars and/or telescopes.Saturn's apparent brightness (on average) is 1.338 ⨉ 10⁻⁷ w/cm³, so it makes sense to say that no object with a calculated apparent brightness less than this value can be called a

*naked-eye object*.Note that this is NOT saying that no object orbiting farther from its star than does Saturn from the Sun would be visible to the tenants of an inhabited world in that system. Apparent brightness is not just a function of distance from the illuminating body, but also of the size and reflectivity of the object, and its distance from the

*observer*.For instance, if Jupiter were moved to Saturn's orbit, it would have an average apparent brightness of 2.007 ⨉ 10⁻⁷ w/cm³, almost 1½ times brighter than Saturn is as the same distance. It certainly would remain a naked-eye object. Uranus, however, moved inward to Saturn's orbit would have an apparent brightness of 2.275 ⨉ 10⁻⁸ w/cm³, only 17% of Saturn's brightness and remain invisible to the naked eye.