Jenab Posted May 21, 2004 Share Posted May 21, 2004 Consider that the maximum separation for keeping the binary planet bound can't be more than Dmax = (1/3) R { ( Ma + Mb ) / Ms }^(1/3) In words: The components (both of them habitable planets) must have a maximum separation not greater than one third of the cube root of the ratio of the sum of the planets' masses to the mass of the star, multiplied by the distance between the star and the barycenter of the binary planet. Consider that R must be always within the liquid water temperature region around the star. Consider that, to a good approximation when 0.8 < (M/Msun) < 2.5 the luminosity of a star may be found from its mass by the relation: (L/Lsun) = (M/Msun)^4.15 Where Msun = 1.989E+30 kilograms, Lsun = 3.826E+26 Watts. So, following the Stephan-Boltzmann radiation law, R (in astronomic units) = (M/Msun)^2.075 Consider that the main sequence lifetime of a star is about Tms = 10.1 billion years (M/Msun) / (L/Lsun) Tms = 10.1 billion years (M/Msun)^(-3.15) Consider that a planet probably needs about 3 billion years from its formation to become habitable, thus (M/Msun)max = 1.47 Is there a minimum star mass? For a planet to be habitable, it must be freely rotating with a reasonable day/night cycle - i.e., not tidally locked to another body. The smaller the star is, the closer to it the binary planet must be to remain in the liquid water zone. A single planet in its sun's liquid water zone becomes tide locked to the star if the star's mass is less than about 0.8 suns. Things are tighter than that for binary planets. As that distance becomes smaller, the two components of the binary must get closer together to avoid the disruption of their pairing by the star's tides. Below some value of star mass, the maximum planet separation required by the low-mass star conflicts with the minimum planet separation required to prevent the two planets becoming tide-locked to each other. My guess is that if you require a planet to have a circular orbit with a radius that give it a constant subsolar temperature equal to Earth's average subsolar temperature (393.6 K), the minimum star mass for a binary habitable planet may be found from logbase10 Mmin = 0.117 - 0.574 logbase10 f Where Mmin is the minimum star mass and f is the greatest fraction of the orbital stability limit of the binary with respect to the star's tides, that you're willing to accept. In my test simulations, f=0.9 gives good behavior, so... Mmin = 1.39 So binary planets with both components habitable are probably OK only within a narrow window of star masses: 1.39 <= (M/Msun) <= 1.47 These stars are spectral types F1 to F3. On the other hand, if you can think of a way to have a habitable planet at a subsolar temperature equivalent to that of Mars (greenhouse effect?), you can get an f=9 dual habitable planet's star mass nearly down to one solar mass. logbase10 Mmin = 1.44 - 0.574 logbase10(R / 1 astronomical unit) Does anybody notice any spectacular goof-ups here? In case somebody wants to try arranging for a reasonable day/night cycle to occur from putting the binary components so close together that their mutual orbit provides the diurnal period, be wary of the Roche Limit. To estimate that, you'll need an estimate for the density of your component planets. Here's one I got from curvefitting Earth, the moon, and Mars: average density = 0.5286 (logbase10 M)^2 + 2.144 logbase10 M + 5.515 Where average density is in grams per cubic centimeter, and M is the planet's mass in Earth masses. One Earth mass is 5.976E+24 kilograms. (Mercury deviates from this curvefit, probably because it's close enough to the sun to have had volatile substances vaporized and escaped.) Jerry Abbott Link to comment Share on other sites More sharing options...
Bryn Posted May 21, 2004 Share Posted May 21, 2004 Sounds fasinating. But i didn't understand most of it :o/. What subject are you studying to do that kinda stuff? Link to comment Share on other sites More sharing options...
[Tycho?] Posted May 23, 2004 Share Posted May 23, 2004 I dont see why they can't become tidally locked. For one, I dont see any reason why a planet would need a day night cycle to be habitable, life would just be different there. And even if two bodies were tidally locked, they would be orbiting around the center of gravity of the two objects. So while the face of one body may always look on the same face of the other body, they would still be orbiting so sunlight would still reach the whole surface. Its like the moon around the earth, it is tidally locked, but it orbits around so everywhere on its surface can still get sunlight. Unless I read your post wrong, which is highly possible considering how much of it I understood (probably not enough). Link to comment Share on other sites More sharing options...
Jenab Posted May 24, 2004 Author Share Posted May 24, 2004 ']I dont see why they can't become tidally locked. For one' date=' I dont see any reason why a planet would need a day night cycle to be habitable, life would just be different there. And even if two bodies were tidally locked, they would be orbiting around the center of gravity of the two objects. So while the face of one body may always look on the same face of the other body, they would still be orbiting so sunlight would still reach the whole surface. Its like the moon around the earth, it is tidally locked, but it orbits around so everywhere on its surface can still get sunlight. Unless I read your post wrong, which is highly possible considering how much of it I understood (probably not enough).[/quote'] I think there may be a possibility for tide-locked habitable binary planets. The synodic period of the sun - the apparent "day" caused by the planets' orbit around their mutual barycenter - should be short enough to preclude problems with the weather, oceans freezing on the night side, and other bad things that happen to planets that don't turn fast enough. The question, then, is whether two mutually tide-locked planets, each having a mass suitable for habitability, can orbit closely enough to have an acceptable diurnal period while remaining far enough apart that their mutual tidal stresses don't keep the volcanos popping out slag and covering everything. (Jupiter's moon Io has that problem.) Jerry Abbott Link to comment Share on other sites More sharing options...
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