<p>@James K's answer, above, is correct (and I've upvoted it), but I'd like to expand on it more than comments allow.</p>

<p>Don't forget the inverse square law! The starlight light available to be reflected drops off with the square of the distance from the star to the reflector. Except for very close binaries the inverse square law dramatically diminishes the total amount of reflected light.</p>

<p>There's a simple way to work it out: <em>All</em> the light from a star passes through the sphere centered on the Sun with a radius of the Earth's orbital distance. (The Earth's orbit can be thought of as the equator of that sphere.)  The <em>only</em> light Earth can reflect is the tiny part its surface intercepts.  The rest streams past it into space. </p>

<p>If <em>r</em> is the Earth's radius and <em>R</em> is the radius of its orbit, the fraction intercepted and available to be reflected is (<sup>r</sup>/<sub>2R</sub>)<sup>2</sup>.  With 4=4000 miles and R=93,000,000 miles, that's just a bit over 4*10<sup>-10</sup> of the Sun's light is available to be reflected by the Earth. (Earth reflects most of it.) If you do the arithmetic, Mercury, Venus, Jupiter and Saturn roughly the same and the rest of the planets make a pretty minor contribution.  The total reflected (<em>before</em> albedo losses!) is around 4*10<sup>-9</sup> of the Sun's light.</p>

<p>A closer-in planetary system will reflect a bigger proportion of the central star's light -- it scales as the inverse square, of course.  If the solar system were ten times as compact but the planets' sizes unchanged, the planets would reflect around 4*10<sup>-7</sup> of the Sun's light.  Still tiny.</p>

<p>The only time you get anything that isn't tiny is in the case of a close binary. If another star the same size as the Sun orbited the Sun just one solar diameter away, it would intercept (and thus have the possibility of reflecting) around 6% of the sun's light.</p>

<p>This supports James K's conclusion pretty strongly.</p>

CC BY-SA 4.0
2018-12-12T03:19:30.840Z
2018-12-12T03:19:30.840Z
Mark Olson
<p>A truly negligible amount. You only need compare the brightness of the planets as viewed from the Earth with the brightness of the sun. A very rough calculation (considering the relative magnitudes of the planets and the Sun) suggests that light emitted directly by the sun is 100 million times brighter than light reflected off planets.</p>

<p>You note that the sun is not a perfect black body, and so does reflect some light. However it only reflects light reflected on to it from the stars, which are a very weak light source. And most starlight that does fall on the sun is absorbed. I haven't tried to estimate how much of the sun's light is reflections of starlight, but it is a very very small proportion. I'd guess much less than a billionth.</p>

<p>There is one situation where reflected light is significant, and that is when very bright stars illuminate dust and gas, to form a reflection nebula. These are common and at least part of the orion nebula is reflected starlight. Other reflection nebulae are too dim to see.</p>

<p>While a I don't have an exact figure, it is clear that nearly all starlight is directly emitted from the star, and it takes special equipment to see any reflected light at all.</p>

CC BY-SA 4.0
2018-12-12T00:51:21.023Z
Are you sure? Most stars are in binaries, i.e. quite close to a companion star. And I'm not sure the albedo of a star is so small after all. I'm sure you're right that it's a small fraction, but I'm uncertain if it's really negligible. Would be a fun exercise…
pela
CC BY-SA 4.0
2018-12-11T21:42:47.433Z
2018-12-11T21:42:47.433Z
James K
<p>Thanks to reflected starlight, many planets and comets in the Solar System have been visible to humans since long before the development of modern astronomy.</p>

<p>Some of the starlight from outside the Solar System should be reflected light as well. That is, light emitted by one star in the Milky Way may encounter another star, and be reflected toward Earth. Normally, stars are assumed to be perfect blackbodies, but in reality they must bounce off <em>some</em> radiation.</p>

<p>Also, starlight should be reflected off exoplanets, moons, asteroids, meteors, comets, brown dwarfs, etc. What fraction of starlight, seen by the naked eye from Earth, is actually reflected light?</p>

CC BY-SA 4.0
2018-12-11T22:18:25.743Z
Do you really want a quantitative number? For exoplanets that's relatively easy to calculate or even measure (keyword: directly imaged systems), while for star-star scattering that's going to be very small and tough to estimate (remember a physical black-body is a near perfect absorber). Only the few photons that are scattered in the stellar atmosphere into our direction would count to your number. I don't know if anybody ever bothered to estimate that.
AtmosphericPrisonEscape
<star><light><naked-eye><interstellar>
What fraction of starlight, seen from Earth, is actually reflected light?
StarlightDown
# What fraction of starlight, seen from Earth, is actually reflected light?

(Asked by: username_0 on 2018-12-11)

Thanks to reflected starlight, many planets and comets in the Solar System have been visible to humans since long before the development of modern astronomy.

Some of the starlight from outside the Solar System should be reflected light as well. That is, light emitted by one star in the Milky Way may encounter another star, and be reflected toward Earth. Normally, stars are assumed to be perfect blackbodies, but in reality they must bounce off *some* radiation.

Also, starlight should be reflected off exoplanets, moons, asteroids, meteors, comets, brown dwarfs, etc. What fraction of starlight, seen by the naked eye from Earth, is actually reflected light?

**Tags:** star, light, naked-eye, interstellar
> **Comments:**
> **username_1**: Do you really want a quantitative number? For exoplanets that's relatively easy to calculate or even measure (keyword: directly imaged systems), while for star-star scattering that's going to be very small and tough to estimate (remember a physical black-body is a near perfect absorber). Only the few photons that are scattered in the stellar atmosphere into our direction would count to your number. I don't know if anybody ever bothered to estimate that.  

---

## 2 answers

### Answer by username_3 on 2018-12-11. Score: 4 | Accepted answer

A truly negligible amount. You only need compare the brightness of the planets as viewed from the Earth with the brightness of the sun. A very rough calculation (considering the relative magnitudes of the planets and the Sun) suggests that light emitted directly by the sun is 100 million times brighter than light reflected off planets.

You note that the sun is not a perfect black body, and so does reflect some light. However it only reflects light reflected on to it from the stars, which are a very weak light source. And most starlight that does fall on the sun is absorbed. I haven't tried to estimate how much of the sun's light is reflections of starlight, but it is a very very small proportion. I'd guess much less than a billionth.

There is one situation where reflected light is significant, and that is when very bright stars illuminate dust and gas, to form a reflection nebula. These are common and at least part of the orion nebula is reflected starlight. Other reflection nebulae are too dim to see.

While a I don't have an exact figure, it is clear that nearly all starlight is directly emitted from the star, and it takes special equipment to see any reflected light at all.

> **Comments:**
> **username_4**: Are you sure? Most stars are in binaries, i.e. quite close to a companion star. And I'm not sure the albedo of a star is so small after all. I'm sure you're right that it's a small fraction, but I'm uncertain if it's really negligible. Would be a fun exercise…  

---

### Answer by username_2 on 2018-12-12. Score: 0

@James K's answer, above, is correct (and I've upvoted it), but I'd like to expand on it more than comments allow.

Don't forget the inverse square law! The starlight light available to be reflected drops off with the square of the distance from the star to the reflector. Except for very close binaries the inverse square law dramatically diminishes the total amount of reflected light.

There's a simple way to work it out: *All* the light from a star passes through the sphere centered on the Sun with a radius of the Earth's orbital distance. (The Earth's orbit can be thought of as the equator of that sphere.) The *only* light Earth can reflect is the tiny part its surface intercepts. The rest streams past it into space.

If *r* is the Earth's radius and *R* is the radius of its orbit, the fraction intercepted and available to be reflected is (r/2R)2. With 4=4000 miles and R=93,000,000 miles, that's just a bit over 4\*10-10 of the Sun's light is available to be reflected by the Earth. (Earth reflects most of it.) If you do the arithmetic, Mercury, Venus, Jupiter and Saturn roughly the same and the rest of the planets make a pretty minor contribution. The total reflected (*before* albedo losses!) is around 4\*10-9 of the Sun's light.

A closer-in planetary system will reflect a bigger proportion of the central star's light -- it scales as the inverse square, of course. If the solar system were ten times as compact but the planets' sizes unchanged, the planets would reflect around 4\*10-7 of the Sun's light. Still tiny.

The only time you get anything that isn't tiny is in the case of a close binary. If another star the same size as the Sun orbited the Sun just one solar diameter away, it would intercept (and thus have the possibility of reflecting) around 6% of the sun's light.

This supports James K's conclusion pretty strongly.


---

