This blog post about binary-star systems points out that, rather having the stars be close together, they could be far apart and a planet might orbit just one of them. It gives the example of the binary system in Alpha Centauri, where the two stars are 23 astronomical units apart. The post ends:

A better scenario is one in which a planet orbits just one of the stars. If the two stars are separated by a large distance — say, 100 AU — then life on a planet orbiting just one of the stars can be similar to life on Earth. The right configuration can make things quite comfortable — and still provide some awesome views of the sky at different times.

Suppose I do that -- I have an earth-like planet orbiting a G-class star in a binary system. I'll assume for now that the second star is also G-class, but I'm flexible about that. Suppose the planet is in the goldilocks zone of the star it's orbiting and the other is much rather away, 50-100 AU.

Some questions from the planetary perspective:

  • Is the light from the distant star significant? Does it illuminate the planet as much as, say, the earth's moon does at night when full, or is this basically just another bright star in the night sky? (Could it be brighter than the moon, even, making a sort of "second day" during part of the night?)

  • Are its gravitational effects significant? If so, how do they manifest? Is it seasonal? (If the planet is orbiting one of the two stars, then there will be times when it's between them and times when they're both in the same direction.)

  • At that distance, does it contribute noticeable heat?

  • Are there other obvious effects I should be asking about but haven't anticipated?

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    $\begingroup$ I did some quick calculation and the main star is 1500 - 10 000 brighter than the other one at 50 and 100 AU. The "dim" star would still be 40-250 times brighter than a full moon. Stay with us, hopefully a qualified scientist will be able to answer! $\endgroup$
    – Vincent
    Commented Sep 6, 2015 at 3:25
  • $\begingroup$ Based on this: en.wikipedia.org/wiki/Apparent_magnitude $\endgroup$
    – Vincent
    Commented Sep 6, 2015 at 3:25
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    $\begingroup$ The Sun's absolute magnitude is -27. The full moon's is -13. That's 14 magnitudes difference. Each magnitude is about a factor of 2.5 difference in brightness. So the Moon is 2.5^14 = about 372,000 times fainter than the Sun. In this setup with a binary star at 100 AU, the binary star is just 100^2 = 10,000 times fainter than the Sun, or about 37 times brighter than the full moon. Of course, the full moon is about the same size as the Sun whereas this binary would be 1/100th as big. More on binaries and planets: planetplanet.net/2013/06/06/binary-stars-friends-or-foes $\endgroup$ Commented Jul 4, 2016 at 10:29

2 Answers 2


OK, so we have two sun-like stars (I'll just write "suns" from now on) at $100\,\rm AU$ distance, and a (probably earth-like) planet at $1\,\rm AU$ distance from one of the suns. I'll call the sun the planet orbits the "near sun" and the other one the "far sun". I'll assume circular orbits throughout.

Let's first look at the system of two suns. In orbital mechanics, we have $$\frac{r^3}{(M_1+M_2)T^2}=\frac{G}{4\pi^2}$$ where $r$ is the the radius of the orbit, $T$ is the orbit time, $M_1$ and $M_2$ are the masses of the bodies, and $G$ is the gravitational constant. By inserting the properties of the earth's orbit (and using the fact that the earth's mass is negligible compared to the sun's mass, we get that $$\frac{G}{4\pi^2} = 1\frac{\mathrm{AU}^3}{M_{\odot}\mathrm{yr}^2}$$ where $M_\odot$ is the mass of the sun and $\rm{yr}$ means year.

So inserting the parameters of the double-sun, we get $$\frac{(100\,\mathrm{AU})^3}{2M_\odot T^2}=1\frac{\mathrm{AU}^3}{M_{\odot}\mathrm{yr}^2}$$ which means $$T = \sqrt{500\,000}\mathrm{yr} \approx 700\rm yr$$ In other words, the suns need about 700 years to go round each other. So a human living on your planet would see the far sun move considerably relative to the fixed stars during his lifetime, but never see it return to its original place.

In the following I'll assume that the planet's orbit is in the same plane as the orbits of the suns around each other and going in the same direction, as this (or an approximation of this) is the most probable situation.

Now let's look at the gravitational effects of that far sun on the planet. I'll give all accelerations in units of the acceleration the near sun's gravitation causes for the planet (that is, the acceleration the planet would experience if there would be no far sun), which I'll call $a_0$, and which is $$a_0 = \frac{GM_\odot}{1\,\mathrm{AU}^2} = 4\pi^2\,\frac{\mathrm{AU}}{\mathrm{yr}^2}$$ Let's look at the situation where the planet is between the two suns. Then its distance from the far sun is $99\,\rm AU$, and thus the acceleration caused by the far sun is $a_0/9801 \approx 1.02\cdot 10^{-4} a_0$, in the direction away from the near sun. To put this in comparison, Jupiter has a mass of about $10^{-3}M_\odot$ and a minimal distance to the Earth of about $4\,\rm AU$, giving rise of a gravitational acceleration of about $2.5\cdot 10^{-4}a_0$. That is, the far sun's gravity affects the planet less than Jupiter affects Earth.

Then, let's look at the brightness of the far sun. The brightness is usually given by the apparent magnitude. The Sun's apparent magnitude (and thus the apparent magnitude of the near sun) is about $−27$. Now by definition a factor $100$ in brightness corresponds to a difference of $5$ in apparent magnitude, and since the brightness goes down with the square of the distance, the far sun at $100$ times the distance has a brightness of $1/10\,000$ of the brightness of the near sun, therefore the far sun would have an apparent magnitude $10$ higher than that of the near sun, that is, $-17$. The moon has an apparent magnitude of $-13$, so the far sun would be about 40 times as bright as the full moon. This means you might be able to see it even on the day sky, as long as it is not too close to the near sun.

Finally let's look at what it would look like. The size (angular diameter) of the Sun, as seen as the Earth, is about half a degree. The far sun is 100 times as far, so the size will be 1/100 as large, or about 20 arc seconds. That's about the same as Jupiter as seen from Earth.

So the far sun would basically look like an extremely bright planet. In particular it's still large enough that it doesn't twinkle.

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    $\begingroup$ So does the second sun provide sufficient photons suitable for photosynthesis such that it would be worth plants evolving the ability to absorb photons of different wavelengths from the two stars? That would have interesting effects on the perceived colour of the local flora. $\endgroup$
    – rumguff
    Commented Sep 6, 2015 at 10:59
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    $\begingroup$ @rumguff: With only 1/10000 of the intensity of the near sun, I doubt that it would result in significant evolutionary pressure. Also, Chlorophyll already can absorb photons from a wide variety of wavelengths;. Indeed, comparing with the solar spectrum it is actually quite inefficient in the frequency range where the intensity of the sun is highest; I guess that there are other (chemical/physical) properties that make it superior … $\endgroup$
    – celtschk
    Commented Sep 6, 2015 at 11:11
  • $\begingroup$ … to other molecules that could absorb light in the intensity maximum of the solar spectrum (for example, it doesn't just need to absorb the light, but do it in a way that it can use the energy of the absorbed light). $\endgroup$
    – celtschk
    Commented Sep 6, 2015 at 11:12
  • $\begingroup$ I guess I was thinking that the second star would be bigger and bluer but then I suppose it won't be round long enough to matter for evolution. $\endgroup$
    – rumguff
    Commented Sep 6, 2015 at 11:29
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    $\begingroup$ @HDE226868: From the Wikipedia page of apparent magnitude. I now notice that one might misread my text so that the the factor 100 would be considered the factor of brightness of the two suns (I assume that's how you read it), which of course would be wrong (if you look at the two magnitudes I give, they differ by 10, not by 5, corresponding to the correct factor 10000 in brightness). I'll edit the text to be more clear. $\endgroup$
    – celtschk
    Commented Sep 6, 2015 at 13:18

Is the light from the distant star significant? Does it illuminate the planet as much as, say, the earth's moon does at night when full, or is this basically just another bright star in the night sky? (Could it be brighter than the moon, even, making a sort of "second day" during part of the night?)

Let's use formulas for magnitude to answer this.

First, note that the Sun has an absolute magnitude of 4.83. Therefore, both stars will have the same absolute magnitude.

The formula for apparent magnitude is $$m=M+5\log_{10}\left(\frac{d}{10\text{ parsecs}}\right)$$ where $m$ is absolute magnitude, $M$ is apparent magnitude, and $d$ is distance, in parsecs. Given that $d=100\text{ AU}\approx 0.000485\text{ parsecs}$, we find that $m\approx-16.74$. For Earth, the Sun has an apparent magnitude of -26.74, so the second star should be roughly 11 orders of magnitude dimmer than that. Enough for a "second day"? I would say not.

Are its gravitational effects significant? If so, how do they manifest? Is it seasonal? (If the planet is orbiting one of the two stars, then there will be times when it's between them and times when they're both in the same direction.)

This depends on the eccentricity of the stars' orbits. In the blog post, I assumed that the orbits were pretty much circular, corresponding to an eccentricity of about 0. This means that the change in distance between the planet and the second star is only about two AU - from 99 AU at the closest approach to 101 AU at the furthest.

To calculate the difference in the gravitational forces between the planet and each of the stars, it's easier to simply write the distances in ratios. Using Newton's law of universal gravitation, $$F_i=\frac{GM_im_p}{r_i^2}$$ where $m_p$ is the mass of the planet, and $F_i$, $M_i$ and $r_i$ are the force on the planet from star $i$, the mass of star $i$, and the distance to star $i$, respectively. Say the planet orbits star 1. At its closest approach to star 2, $$\frac{F_1}{F_2}=\frac{M_1}{M_2}\frac{r_2^2}{r_1^2}=\left(\frac{r_2}{r_1}\right)^2=\left(\frac{99}{1}\right)^2=9801$$ In other words, $F_1\gg F_2$, and the gravitational effects from star 2 should be negligible.

To find the specific perturbations on the orbit of the planet, we would have to solve the three-body problem, specifically, the circular restricted three-body problem, given that the planet is much less massive than both stars. That said . . . I assume you won't be interested in that; it's really quite unimportant.

At that distance, does it contribute noticeable heat?

A version of the formula for for effective temperature tells us that, in the absence of the greenhouse effect, the surface temperature of the planet should be roughly $$T=\left(\frac{1-a}{4\sigma}(F_1+F_2)\right)^{1/4}$$ where $F_1$ and $F_2$ are the fluxes from stars 1 and 2, and $a$ is the planet's albedo. Flux, just like gravity follows the inverse-square law, and so $F_1/F_2=9801$. Therefore, $F_1\gg F_2$, and we can essentially ignore the second star when calculating the temperature of the planet. If you're looking to explicitly calculate the habitable zone, I've written some code to do that, but I've done some testing, and the habitable zones around each star will be essentially no different from the habitable zone around an identical, solitary star.

  • $\begingroup$ I love how you "rounded" the 100 AU to 17 significant digits.. That's an accuracy of a few centimeters. $\endgroup$
    – Chieron
    Commented Jul 29, 2016 at 8:33
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    $\begingroup$ @Chieron I didn't round there because it was an intermediate result; I was only concerned with the final result of $m$. Rounding early on in calculations can lead to unfortunate errors. $\endgroup$
    – HDE 226868
    Commented Jul 29, 2016 at 14:12
  • $\begingroup$ this was a ballpark estimation. If rounding a value that is supposed to be passed to a logarithm really induced a noticeable error, the calculation would've been moot anyway. I simply found it amusing, even if it was just copy+paste from a calculator. $\endgroup$
    – Chieron
    Commented Jul 29, 2016 at 14:19
  • $\begingroup$ @MonicaCellio I've made some revisions here that you might want to see - the conclusions are the same, but I've made it all a bit clearer, and removed those pesky written-out subscripts on some of the equations . . . which I removed, after rereading them. They were extraneous. $\endgroup$
    – HDE 226868
    Commented Aug 8, 2018 at 17:22

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