2
$\begingroup$

enter image description here

In an alternate Milky Way in an alternate universe, there is a cluster of stars so dense that nights are over 60 times brighter than our full moon. In that cluster is a solar system presented (albeit simplified) above. It is actually two separate binaries.

  1. The first one consists of a red giant 100 times as wide, one-third as massive and 1,000 times as bright as our sun orbited by a yellow-white dwarf 130% as wide, 170% as massive and six times as bright as our sun.
  2. The binary orbiting that binary consists of a yellow dwarf 105% as wide, 110% as massive and 126% as bright as our sun orbited by an orange dwarf 85% as wide, 78% as massive and only 40% as bright as our sun.

It is in the second binary that this alternate Earth orbits. With an axial tilt of only three degrees, the seasons are defined by how many suns are in the sky. "Summers" are when all four of them are present in the sky, making daylight as bright as Venus and nights 1,000 times brighter than back home. "Winters" are when only the two that it orbits can be seen, when daylight is as bright as our Earth and nights are more than 60 times brighter than back home.

An Earth like that sounds like good news for potential photosynthesizers, but there are problems with the parent binary. A red giant is big, obviously, which means that it had the potential to go supernova before long. And yellow-white dwarves can carry seven times more ultraviolet radiation than our sun. Put those two together, and you would have an alternate Earth that would have to orbit a secondary binary that orbits the primary binary from a safe enough distance.

So the question is--how far can the alternate Earth's binary orbit the primary binary without concern of near-future supernova or UV overdose?

$\endgroup$
5
  • 1
    $\begingroup$ Note that supernova is very powerful, its power can create stars and destroy planets so I plead the mighty God please go through the tutorials in Universe Sandbox before you try anything ;D $\endgroup$ – user6760 Jan 17 at 6:33
  • $\begingroup$ The firsst pair of stars has a total mass of 233 percent of the Sun's mass, and the second pair is has a total of 178 percent of the Sun's mass. Thus the 2nd pair will not orbit the first, instead the two pairs will orbit their common barycenter. Why is it that the first star to become a red giant is the one with the lowest mass? How is that possible? $\endgroup$ – M. A. Golding Jan 17 at 18:53
  • $\begingroup$ I think your issue isn't with one of the local suns going supernova, but the fact there are so many suns close by. A supernova will kill everything for MANY lightyears around it. I've heard astronomers speculate that the core of the galaxy may not support intelligent life because the density of suns means extinction-level events from supernovas are high enough to preclude evolution of intelligence. The local star isn't the main issue. What is the goal of your system? $\endgroup$ – DWKraus Jan 17 at 20:08
  • $\begingroup$ @DWKraus Turn the Nine Realms of Norse mythology into believable, Earthlike exoplanets. This is just one of them. $\endgroup$ – JohnWDailey Jan 17 at 20:36
  • $\begingroup$ one-third as massive, orbited BY 170% as massive.. It is convention to say that the object with the smaller orbit is orbited BY the object with the wider orbit. Here you have it the other way round. $\endgroup$ – PcMan Jan 18 at 6:35
1
$\begingroup$

Since you ask about avoiding future supernova risks, let me reuse my answer to this question. The question was about the safe distance from a supernova for a Dyson sphere, but since the estimate I do is about the energy amount I think it can be applied also here:


Mandatory What if quote:

Rule of thumb for estimating supernova-related numbers: However big you think supernovae are, they're bigger than that.

Here's a question to give you a sense of scale:

Which of the following would be brighter, in terms of the amount of energy delivered to your retina:

  • A supernova, seen from as far away as the Sun is from the Earth, or

  • The detonation of a hydrogen bomb pressed against your eyeball?

Applying the physicist rule of thumb suggests that the supernova is brighter. And indeed, it is ... by nine orders of magnitude.

I am pretty sure the supernova, with the sheer amount of photons blowing out in its surroundings while exploding, would simply plasmify the Dyson sphere, way before the radiation could have any chance of pushing it apart.

If you want to estimate a safe distance consider that a supernova can emit about $10^{44}$ Joules. The Sun's routine output is $10^{26} \ J/s$. The peak period of a supernova is about 6 hours, so that $10^{44}$ joules is spread over about 21000 seconds. Thus, a supernova would only be about $4\cdot10^{13}$ times brighter than the normal sun at its peak. Assume your Dyson sphere can take 10x normal sunlight intensity. At a distance of only 2.2 million AU it will be safe. That's less than 55 000 times further than Pluto!


What do we learn from this? Well, supernova are huge, that should be clear by now, and as you see 2.2 millions AU are probably too far to be still gravitationally bound as you describe.

If you want your system safe only from UV overdose instead, just use the square law reduction, and assume that a safe UV dose is what we receive from the Sun here on Earth (since in normal conditions our planet can cope with it)

$Sun_{EUV}:dist^2_{E-S}=Star_{EUV}:dist^2_{P}$.

If the star emits 100 times the Sun EUV dose, the planet has to be 10 times more distant to get the same amount Earth does.

$\endgroup$
2
  • $\begingroup$ What Dyson Sphere? $\endgroup$ – JohnWDailey Jan 17 at 23:00
  • $\begingroup$ @JohnWDailey, the first question I answered with this was asking about a Dyson sphere. But since the estimate it's about a safe distance from a supernova from the perspective of the received energy, it can apply also here $\endgroup$ – L.Dutch - Reinstate Monica Jan 18 at 3:26
-1
$\begingroup$

Wide binary.

Your 3 star system could be a "wide binary" - actually a trinary as you propose.

http://www.ifa.hawaii.edu/info/press-releases/WideBinaryStars/

Now Dr. Bo Reipurth of the Institute for Astronomy, University of Hawaii at Manoa, USA, and Dr. Seppo Mikkola of Tuorla Observatory, University of Turku, Finland, have used computer simulations to come up with a mechanism that accounts for the formation of wide binaries. Most stars are initially formed in small compact multiple systems with two, three or even more stars at the center of a cloud core. When more than two stars are together in a small space, they gravitationally pull on each other in a chaotic dance, where the lightest body is often kicked out to the outskirts of the core for long periods of time before falling back into the fray.

Meanwhile, the remaining stars feed on the gas at the center of the cloud core and grow heftier. Eventually, the runt of the litter gets such a large kick that it may be completely ejected. But in some cases, the kick is not strong enough for the third body to fully escape, and so it is sent out into a very wide orbit.

The implication is that the widest binaries really should be three stars, not just two stars.

Wide binaries can be as far apart as 200,000 AU. But things get hairy. Other things in the neighborhood can perturb the orbit of the distant star, which here is the one with the life-supporting planet.

VERY WIDE BINARY STARS AS THE PRIMARY SOURCE OF STELLAR COLLISIONS IN THE GALAXY

While tight binaries (for our purposes, mean stellar separations below ~103 AU) are unaffected by perturbations from their local galactic environment, this is not true for very wide binaries (a > ~103 AU). Torques from the Milky Way's tide and impulses from other passing field stars continually change the periastron of very wide binary orbits... This evolution often causes binaries to undergo brief phases of very low periastron, forcing close passages between companion stars. Recently it was shown that the low-q excursions of very wide binary star systems can regularly trigger the disruption of planetary systems orbiting the member stars...

And you specify that their neighborhood is crowded with other stars. These passersby will whisper seductively to your distant partner and it will miss a step in its dance around the 2 in the middle. In this neighborhood a menage-et-trois might not last too long.


I know your OP asked about being far enough to be safe in case the central star went supernova. @L.Dutch's comments about A-bombs on the eyeballs says it all. The widest binaries are 200,000 astronomical units or about 3 light years. To be safe from a supernova you need a distance of 50-100 light years.

$\endgroup$
1
  • $\begingroup$ Three? There are FOUR stars in the picture. $\endgroup$ – JohnWDailey Jan 18 at 1:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.