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enter image description here

  1. If there's a habitable planet that orbits two suns like the above orbital route, what would possibly occur on that planet?
  2. Would it have two yearly seasonal cycle (compare to earth)?
  3. Would it be possible that there's some months where there would be no night time on every part of the planet?
  4. What other things resulting from the effect on day, night, and seasonal cycles can be experienced if we lived on that planet?
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    $\begingroup$ It is not very likely a planet could orbit a pair of stars in this configuration. See this related question. $\endgroup$ – MozerShmozer Nov 30 '16 at 23:11
  • $\begingroup$ Thank you @MozerShmozer for the link, I've read it and find that it is possible albeit this won't be stable for long in reality. And that's what I've been guessed before since this kind of system required a perfect symmetry of the suns and any small difference that very common in nature will throw out the balance, but the stability of the orbital route is not a question I ask here. $\endgroup$ – Hariz Rizki Dec 1 '16 at 3:39
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I'm going to attack this with math. First off, I am not going to make any assumptions about what orbits might be stable. That is something we can check with an orbit simulator like rebound, as I did in this question. Instead, I will assume there is a stable orbit around two suns of equal mass and luminosity. The orbital profile will be a perfect circle (0 eccentricity orbit) at 1 AU for 3/4 of a revolution around each sun and then a straight line connecting it to the next sun.

Determine the orbits

Since this is an ugly piecewise function, and I am solving using a computer, I am defining it in python as such:

def f(t):
    if t < 3/2*pi:
        return sin(t-pi/4)+sqrt(2), cos(t-pi/4)
    if t < 3/2 * pi + 2:
        r = t - 3/2*pi
        return sqrt(2)/2-r/sqrt(2), -sqrt(2)/2+r/sqrt(2)
    if t < 3*pi + 2:
        r = t - 3/2*pi - 2
        return sin(pi/4-r)-sqrt(2), cos(pi/4-r)
    if t < 3*pi + 4:
        r = t - 3*pi - 2
        return sqrt(2)/-2+r/sqrt(2), -sqrt(2)/2 + r/sqrt(2)

enter image description here

The red dots are the two suns. The green dot is our arbitrary time = 0 for the next part; the planet starts moving clockwise around the sun on the right. So far so good, now let us calculate distance from each sun, and plot that as a function of time.

Determine the distance from each sun

Since I'm already using the computer, and I know the sun's coordinates, ($-\sqrt{2}, 0$) and ($\sqrt{2}, 0$), I will just calculate numerically using this code:

def dist_1(coord):
    x, y = coord
    return sqrt((x - sqrt(2))**2 + y**2)
def dist_2(coord):
    x, y = coord
    return sqrt((x + sqrt(2))**2 + y**2)

enter image description here

Where dist_1 is from the right star and is in blue, and dist_2 is the left star and is in red. I scaled the time factor to 365 days in a year cause I'm a geocentric kind of dude, but I could use any scale factor. If the planet was moving at the velocity of Earth it would take 1342 days to complete this year, fyi.

Determine enegy recieved

Solar energy drops off as 1/r$^2$, so solar energy received from each sun is 0.905 earth units divided by the distance to each sun. The .905 is a scaling factor to ensure that total solar energy recieved by this planet averages to 1 unit. So lets plot those two, and a new black line for the net total solar energy.

enter image description here

Add seasons

Ah, glorious seasons. Lets say we have an earth-like 23.5 degree axial tilt. How will that affect us? Depends on how we orient the tilt. I will arbitrarily declare that the Northern hemisphere is fully tilted towards the two stars when the planet is at the far right point of its orbital trajectory at point ($\sqrt{2}+1$, 0).

I calculate the effect of tilt both at the equator and on a point 45 degrees N. At summer solstice (at the point mentioned above) $cos(45-23.5) = 0.930$ of equatorial sunlight, and at winter solstice it will get $cos(45+23.5) = 0.367$ of equatorial sunlight. The angle to either star in our coordinate system is calculated from $\text{arctan}(\frac{y}{x})$ where x and y are the coordinate distances from the star. The cosine of the angle to the sun in radians, which is the proportion of the axial tilt that the planet is currently experiencing, can be expressed as:

 def angle_1(coord):
    x, y = coord
    if x + sqrt(2) < 0:
        return -1* cos(atan(y/(x+sqrt(2))))
    return cos(atan(y/(x+sqrt(2))))

We will multiply that by the axial tilt, add it to the latitude, and calculate the addition or reduction in light energy by season.

So here are two graphs showing how the seasons will work. The first graph is for the equator of our planet, compared with earth. The black line is total energy recieved by this planet (earth would be just a straight line at one, assuming a perfectly circular orbit), red is the relative insolation at this planet's equator, and green is relative insolation at our equator.

enter image description here

The second graph is the same, except for a point 45 degrees N. So black is the same as above, red is insolation of the other planet, and green is insolation of earth.

enter image description here

And remember, I scaled this planet so its day is as long as an Earth day. If, for example, you set the year to be the 1342 earth days for velocity matching, then the first red-hump summer for the other planet would last as long as a summer here on earth.

Well hope this is what you were looking for; I have all the code saved if you want me to post any more of it, or throw up a graph for a different latitude.


Edit:

As requested here is a 45 degrees N profile for a 750 day year compared to what Earth would be doing in that approximately 2 year period.

enter image description here

As far as the max insolation; the max value for Earth's summer is 0.917, which represents 91.7% of max insolation at the equator. For our mystery planet, the number is 0.830; this is off by a factor of 0.905, which you may recall is the scaling factor by which we had to reduce these other suns to get Earth-like total year-round solar insolation. So that makes sense. On the winter side, however, the numbers are 0.366 for Earth and 0.354 for the other planet. Those should show the same 0.905 ratio. I don't know if this is an error in the code or just somethign I'm not understanding, but I'll take a look.

Incidentally, darkest winter at 45 N is not while the planet is between the two stars. This is the coolest time at the equator, the sunlight dips down to about 80% of max; that is like May or July in the mid-latitudes. But at 45 N, you are about equidistant from the two stars, and always getting 'summer' from one of the two. So that mitigates the wintery-ness. Darkest winter is when you are tilted away from both suns on the far left of the orbit plot.

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  • $\begingroup$ Great work for the answer @kingledion !! I would buy you a drink if you live nearby. I'm not a math whiz, but I quite understand your explanation. However I assume from your graph that this planet at 45 degrees N would have a summer that isn't as high (point) as earth but a winter just at a same lowest point? how can that happen, since when I see the model, I assume this planet would have a worst winter while suns shines all day ( when the planet in the middle of the suns). And could you elaborate more, how big is this planet compare to earth, if it takes about 730 days to complete orbit? $\endgroup$ – Hariz Rizki Dec 1 '16 at 4:28
  • $\begingroup$ @HarizRizki What do you mean by 'how big is this planet'? These mechanics are independent of the size of the planet. It can be any size you want. $\endgroup$ – kingledion Dec 1 '16 at 5:47
  • $\begingroup$ Oh okay, so the size of the planet not really matter in this equation then. Thank you. $\endgroup$ – Hariz Rizki Dec 1 '16 at 5:57
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The biggest shock would be the period in the center of the figure eight where there would be winter--or something near it--on the entire planet.

I'm calling the planet Bob.

Depending on the tilt of the planet's axis, the seasons would be pretty weird. Let's assume, to make it easy, that the time it would take this planet to revolve one of these suns is 1 earth-year. So one Bob-year would be a bit more than 2 earth-years. For most of the first earth-year, on the left and right loops of the orbit, the seasons would be just like those of earth. But as Bob diverts off a circular orbit, the summer would get colder and colder. By the center of the figure 8, Bob is virtually in a perpetual state of winter all round the surface of the planet. And yes, there would be no night. Assuming that Bob is equidistant from both suns, one sun would be setting as the other one rose on one day of the year. All other days there would be just a bit of complete darkness. But this wouldn't last long. Bob would then get back on track on a circular orbit around the second sun.

Let's divide Bob into 2 hemispheres--the North [facing the people looking at the picture] and the South. Let's say the North hemisphere is tilted to the right as you look at the picture. So, beginning on the leftmost point of the orbit, the North follows this pattern:

  1. Summer!
  2. Fall!
  3. Winter! As the days get shorter from Left Sun, Right Sun begins appearing in the sky.
  4. More winter! As the days get longer, Left Sun begins to disappear, becoming only a bright star (depending on how far away the town suns are).
  5. Spring!
  6. Summer!
  7. Fall!
  8. Winter!
  9. Spring!
  10. Summer for a bit, then a quick fall, then the all-day-two-sun-epic-winter.
  11. Spring!

And the South would follow the same pattern, just beginning with the North's #8.

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  • $\begingroup$ So you think it would take about 11 seasons change to complete its cycle? wow nice, I before guess it would be just 8 or 9 $\endgroup$ – Hariz Rizki Nov 30 '16 at 20:57
  • $\begingroup$ @HarizRizki Not all of the seasons are the same length. I divided it into 11 periods, but the on-world inhabitants would probably call it either 10, lumping together #3 and #4 as one winter, or 12, doing that but dividing #10 into 3 separate, short seasons. But there would be the expected 8 seasons plus 2-4 depending on point of view for the weird stuff in the middle. $\endgroup$ – CHEESE Nov 30 '16 at 21:14
  • $\begingroup$ I'm very curious why you think it will be winter on the planet in between the two stars. With no night, there will be no thermal recovery period to bleed off any heat gained during the perpetual day. I would think the coolest period would be at opposition with one of the stars. $\endgroup$ – MozerShmozer Nov 30 '16 at 23:13
  • $\begingroup$ @MozerShmozer I'm by no means an expert. I just thought that Bob might be too far away from either planet to have any kind of good heating. Maybe not winter; maybe just a uniform coolness all around. of course this depends on distance from sun(s) and size of sun(s). $\endgroup$ – CHEESE Nov 30 '16 at 23:47
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Well fist off the planet would be almost completely bright when in the center of the two suns (about 2/8ths of the planet would be bathed in a sunset like state), some plants would be different than the ones we see on earth, because most plants on earth have evolved to incorporate the night cycle into the day of plants, but you'd also have to incorporate the times where there is half sunlight. and there would be a difference with animal life, due to the lack of night time. We would have a lot more non nocturnal animals, and the human diet would probably be affected. To sum it up the ecosystem would change drastically due to the change in light distribution in the world, for better or for worse, that is up for you to decide.

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