In this scenario, the sun is a binary star, sharing orbit with an F-type main-sequence star. The first planet from that binary star is a Jupiter-like gas giant, orbiting one-third of an AU from the stars. Mars is a binary planet, orbiting Earth.

How far would Mars need to be to orbit the Earth without destroying each other under their own gravities?

How far would Earth need to be to make sure that the two stars pose no difference to radiation or temperature changes?

How would Earth's orbit be affected, if likely?

Ultimately, what would the sky, day and night, look like with a binary star, a hot Jupiter and Mars orbiting our planet?

  • $\begingroup$ The three questions are here are unrelated, except for the premise. You might want to split them up. $\endgroup$
    – HDE 226868
    Nov 8, 2015 at 18:08
  • $\begingroup$ This isn't really answerable without more data. Where is the Sun's companion? Stating that it's a binary is meaningless unless we know that. The sun could be a binary star with an F-type star that is as much as a light year away. $\endgroup$ Nov 8, 2015 at 18:09
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    $\begingroup$ @HDE226868 I can't do that. That would prevent the whole picture from being seen, and therefore this would be dismissed as broad. $\endgroup$ Nov 8, 2015 at 20:17
  • $\begingroup$ @JasonPatterson Binary stars orbit each other. How could a star orbit the sun if it's one light-year away? $\endgroup$ Nov 8, 2015 at 20:18
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    $\begingroup$ @JohnWDailey, The "title" is formatted as a normal question on SE sites, by convention. I don't think there's actually a rule on it, but HDE's edit is considered correct for these sites. $\endgroup$
    – MichaelS
    Nov 9, 2015 at 6:23

2 Answers 2


How far would Mars need to be to orbit the Earth without destroying each other under their own gravities?

The Roche limit says Mars and Earth need to be about 2.5 Earth radii from each other before they tear each other apart. That said, tidal forces can still cause a lot of damage from earthquakes, etc.

Realistically, tidal locking of Earth would be much stronger because Mars has a higher tidal force than our current moon. If Earth is almost tidally locked with Mars, there'd be very little tidal effect on the Earth. I'm not sure whether extra tidal forces or extra tidal locking would be a greater effect after 4-ish billion years.

Speaking of Roche limits, you put Jupiter very close to the stars. $1\over 3$ AU is about 49 million km. The Roche limit for Jupiter orbiting our Sun is about 1.7 million km, which means it would be safe. However, your two stars will orbit at some finite distance from each other, which could put your Jupiter inside the effective Roche limit for your system. I'm not sure how that math works.

How far would Earth need to be to make sure that the two stars pose no difference to radiation or temperature changes?

This is greatly dependent on the stars in question. You could hypothetically have two well-behaved stars, each with half the power output of our Sun, and if Earth were orbiting them at 1 AU, the power would be the same. If the binary stars have a lower power output, Earth needs to be closer to compensate. Likewise, Earth needs to be farther to compensate for a higher power output.

Total power is a function of the power per unit area (intensity) times the total area.

$P=I\cdot A$

For a star, the total area is a function of its radius.

$A_{sphere}=4\pi r^2$
$P_{sun}=I_{sun}\cdot A_{sun}=I_{sun}\cdot 4\pi r_{sun}^2$

If all that power is radiating in all directions equally, the intensity at a given orbital distance is the power divided by the area of a sphere, centered on the star, whose radius is the planet's orbit.

$I_{orbit}$ $={P_{sun}\over A_{orbit}}$ $={P_{sun}\over 4\pi r_{orbit}^2}$

We can solve for the orbital distance.

$I_{orbit}$ $={P_{sun}\over 4\pi r_{orbit}^2}$
$r_{orbit}^2={P_{sun}\over 4\pi I_{orbit}}$
$r_{orbit}=\sqrt{P_{sun}\over 4\pi I_{orbit}}$
$r_{orbit}=\sqrt{P_{sun}}\sqrt{1\over 4\pi I_{orbit}}$
Let $C=\sqrt{1\over 4\pi I_{orbit}}$

Now, you probably don't want to calculate power directly. Instead, you'll likely want the relative power of the new stars to our Sun.

Let $P_{stars}=kP_{sun}$, where $k$ is the relative power factor. If $k=2$, the stars have twice the power output at any given distance. Then we can calculate the new orbital distance.

$r_{orbit2}=\sqrt{k}\cdot r_{orbit}=1AU\sqrt{k}$

How would Earth's orbit be affected, if likely?

If you have to change the orbital distance, it will affect the orbital period, T. Likewise, if the total mass of the binary stars is different from the mass of our Sun, the period will change for a given distance. Note that $r$ needs to be replaced with $a$, the semi-major axis, if you're using elliptical orbits.

$T_{Earth}=1yr=2\pi\sqrt{r_{orbit}^3\over GM_{sun}}$
$T_{Earth2}=2\pi\sqrt{r_{orbit2}^3\over GM_{stars}}$

If we let $M_{stars}=KM_{sun}$, so $K$ is the relative mass factor, then we get

$T_{Earth2}=2\pi\sqrt{(\sqrt{k}\cdot r_{orbit})^3\over G(KM_{sun})}$
$T_{Earth2}=2\pi\sqrt{{r_{orbit}^3\over GM_{sun}}{\sqrt{k}^3\over K}}$
$T_{Earth2}=2\pi\sqrt{{r_{orbit}^3\over GM_{sun}}}\sqrt{{\sqrt{k}^3\over K}}$
$T_{Earth2}=1yr{\sqrt[4]{k^3}\over\sqrt{K}}$ $=1yr\cdot k^¾K^{-½}$ $=1yr\cdot k^{0.75}K^{-0.5}$

Ultimately, what would the sky, day and night, look like with a binary star, a hot Jupiter and Mars orbiting our planet?

That depends on a lot of the above. We moved the planet so the total radiation is the same, so the brightness in the day will be comparable. The color will change if the temperature of the stars changed. You can look this up in blackbody radiation texts, but it's a bit complicated.

Jupiter would get a lot brighter. Currently, it has a maximum brightness of magnitude −2.94. Its orbital distance goes from 5.2 AU to 0.33 AU, and incident light is inversely proportional to distance squared. So it gets $({5.2\over {1\over 3}})^2$ $=15.6^2$ $=243.36$ times brighter at any given distance.

Additionally, it goes from 4.2 AU away from Earth to 0.67 AU away (minimum distances, maximum brightness), which makes it another $(4.2\cdot{3\over 2})^2$ $=39.69$ brighter. This is multiplicative with the surface intensity above, so total difference is $243.36\cdot 39.69$ $=9659$ times brighter.

A difference of 1 magnitude is a 2.512 times change in brightness, so we can can solve for the difference.

$M_{new}=M_{old}+M_{diff}$ $=-2.94-9.96=-12.9$.

At it's brightest, it will be about $2.512^{12.9-12.74}=1.16$ times brighter than our Moon (magnitude -12.74). Mostly though, it won't be that bright, since we'll only see it partially illuminated, and when it is fully illuminated it will be almost directly behind the stars and barely visible (the sun is magnitude -26.74, which is $2.512^{26.74-12.9}=343,773$ times brighter than new Jupiter).

Also, all of those apparent magnitudes will change if the stars are brighter than our Sun, or if the new Earth's orbital distance changes, but it should give you a ballpark figure.

Lastly, the new Mars moon will be a prominent figure in the night sky. Mars' albedo is 0.250, while our Moon's albedo is 0.11. So a square km of Mars is 2.273 times brighter than a square km of the Moon with equal illumination.

Additionally, Mars' radius is 3390 km, compared to the Moon's 1737 km. Visible surface area goes up with the square of radius, so Mars has $({3390\over 1737})^2=3.81$ times more visible area, making it $3.81\cdot 2.273=8.66$ times brighter at a given orbital distance.

Brightness is proportional to $1\over r^2$, so Mars' apparent brightness will be $8.66{r_{Moon}^2\over r_{Mars}^2}$ if you change the orbital distance. If you let $r_{Mars}=R\cdot r_{Moon}$, so R is the orbital radius factor, the apparent brightness will be $8.66\over R^2$ times brighter.

The orbital period of the Mars moon will also change. Using the same equations above, we get

$T_{Mars}=1mth\cdot R^{0.75}\mu^{-0.5}$

Except now, k is replaced by R, and K is replaced by $\mu$ which is a bit complex:

$\mu={M_{Earth}+M_{Mars}\over M_{Earth}+M_{Moon}}$

Fortunately, these masses are constant, so we get

$\mu$ $=1.094$
$T_{Mars}={1mth\over\sqrt{1.094}}\cdot R^{0.75}$ $={27.322 days\over 1.04594}\cdot R^{0.75}=$ $26.12 days$$\cdot R^{0.75}$

  • $\begingroup$ I've read that in order for Jupiter to be "hot", it needs to be in close orbit, between 0.5 and 0.015 AUs. $\endgroup$ Nov 9, 2015 at 12:44

I'm no expert, but it seems to me that this set-up wouldn't work out too well, or would need to ignore some science.

Mars is significantly larger than the Moon, and would lead to much more severe effects on our tides and climate. Furthermore, when Mars would drift in front of the Sun/Suns, it would block a LOT more heat and light.

You're also throwing another wrench in the set-up: two suns. At that point there would be so many forces acting on the planets that Earth's orbit would probably become not unpredictable, but unstable from a point of view of the distance it would come to either the one star, or the other.


The reason why I think the system would be difficult to predict is this: Mars has a much larger mass, and exerts much more gravity than the moon, which is just a chunk of rock. If you type "effects of mars size moon" into Google you'll get several forums/articles discussing that exact scenario, and many suggest/agree that this binary planet system would not only result in very radical changes in our weather/tides/etc, but would also affect our orbit. More specifically, the two planets would start exerting a pull on one another, and would most likely start to spiral inward, toward the sun, while spinning around one another.

So what happens when you throw in a second star? I have no idea. Simply imagining a planet orbiting around a dual star system makes me think that this planet would probably experience pretty drastic changes in temperature as it swings closer/further from one start of the other. Add in the double planet system, with Mars being big enough to throw off Earth's stable orbit, and I'm thinking humanity would not be around for long.

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    $\begingroup$ This doesn't really answer any of the questions posed, and I disagree with the last sentence. There are only five bodies at work here. $\endgroup$
    – HDE 226868
    Nov 8, 2015 at 18:51

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