So I have a planet that the size of our solar system(magic/white hole/whatever keeps the planet on 1G despite having the weight of a neutron star). So the planet is very earthlike in terms of soil, plants, animals, air etc. But my guess was that the planet could not have a ~20km atmosphere like our earth does so it has to be considerably bigger or could this planet also only have a 20km atmosphere?

Now I did some calculations on how big the atmosphere would be if I increased it 1:1 and came out to some crazy numbers like 400 million km. So I decided to settle it down at 2 million KM.

But this got me thinking if I have a distance of 2 million kilometers filled with air would light from the sun (the sun is orbiting the planet) even get to the surface? Would increasing the size or power output of the sun make this possible?

  • $\begingroup$ So you're looking for an atmospheric composition that will allow visible light to pass through to an atmospheric thickness of 2M km? Are you worried about atmospheric pressure at the surface? $\endgroup$ – Green Oct 20 '15 at 12:00
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    $\begingroup$ @Green I'd be more concerned with this object collapsing into a black hole, considering its mass. But Vajura doesn't seem interested in strictly adhering to known physics, so can we agree to ignore the non-relevant physics for this question? $\endgroup$ – Frostfyre Oct 20 '15 at 12:09
  • $\begingroup$ Yea lets just say that problem is solved with some sort of gravity alteration device. The atmospheric composition should be the same as on earth, and atmospheric pressure is also solved through the gravity atleration device. My problem is how would you get light to the surface of that planet, stronger star? Normal size (~20km) atmosphere? Or would i need to use a magic like device here also? $\endgroup$ – Vajura Oct 20 '15 at 12:15
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    $\begingroup$ It's not a planet at that point. That "atmosphere" is twice the size of the sun. $\endgroup$ – Serban Tanasa Oct 20 '15 at 13:03
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    $\begingroup$ May I inquire as to why you are interested in this one detail of the world? With the amount of physics being suspended here, we might actually be able to provide a better set of answers if we understood why this one minor physical detail is getting physics-checks and not the rest of the world. In particular, we may be able to come up with non-physics based answers which dovetail well into the world you want to create, rather than limiting ourselves to physics. $\endgroup$ – Cort Ammon Oct 20 '15 at 18:14

If the surface of the planet is at 1g then in fact it is not just plausible but likely that the atmosphere would be the same thickness as earth's. Gravity is what holds the atmosphere in place and a 2M km thick atmosphere would float away unless the gravity extends further than earth's.

If you do have a 2M km thick atmosphere then I don't have the knowledge to give you exact calculations but:

enter image description here

You can see that some frequencies are absorbed more than others but everything is being absorbed. It would be reasonable to say that 1000 times the atmosphere means 1000 times the absorption, in other words you are right and very little is getting through.

Just have the atmosphere be normal thickness :)

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  • $\begingroup$ If i have a 5x "stronger" star and 5x (around 100km i guess) atmosphere would i get rought the same amout of light on the surface without causing some crazy things in the upper layers because of the higher star output? $\endgroup$ – Vajura Oct 20 '15 at 12:36
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    $\begingroup$ This won't work. The absorption works as a half-life, so it isn't 100 km / 1,000,000 km = 1/10,000 = 0.0001 as bright but $(1/2)^{0.0001}=1/7 \cdot 10^5$ as bright. $\endgroup$ – Jim2B Oct 20 '15 at 14:18
  • $\begingroup$ @Jim2B something is wrong in your calculation ... it gets brighter when there is more atmosphere? $\endgroup$ – Paŭlo Ebermann Oct 20 '15 at 18:20
  • $\begingroup$ It should be 1 / $7 \cdot 10^5$. I noticed that too but couldn't correct it - it was too late. $\endgroup$ – Jim2B Oct 20 '15 at 18:35
  • $\begingroup$ Fixed it for you :) $\endgroup$ – Tim B Oct 20 '15 at 19:09

Even if the gravity at the so-called "ground" level is somehow magically set to 1g, there's enough air in there to undergo gravitational collapse to sustain fusion of light elements mid-way through the atmosphere, since you're talking an "atmosphere" that's deeper than the Sun's diameter. So the surface would definitely be bathed in radiation ... from its own atmosphere undergoing fusion.

Now if you extend the magical hand-waving to the entire atmosphere as well, it's not clear to me why you'd want it to be 1-2 million km thick in the first place. If you're magically setting surface gravity to 1g, the atmosphere would perceive the same level of gravity and you could have a normal earth-like atmosphere.

If we handwave all that, and still want 1-2 million km of air, then the optical depth of air becomes relevant. While almost as good as vacuum, air is a tiny bit more opaque, and so for any source of light there is an extinction depth. I leave the fun maths for you to work out, since we're probably not talking uniform density, the integral for this would quickly get ugly:

$$dI= -k \rho Ids $$

Where $I$ is intensity, $k$ is opacity, $\rho$ is density and $s$ is distance. My money is on the theory that you get pretty much 100% opacity at 2 million km, heh, but feel free to work through the maths and let me know if I'm wrong.

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  • $\begingroup$ Assuming a uniform sea-level density of 1.23 g/m3 of air times a 1 million km sphere, plus earth's mass, gives you 5.13e27 kg, or about 2.7x that of Jupiter (21x with a 2 M-km radius sphere). Given that brown dwarfs are in the 13 to 80 Mj range, I suppose you could get some fusion, but the assumption is really bad: better would be to scale up our atmosphere and it's exponential-ish fall-off in density, as that would give you a far better answer. $\endgroup$ – Nick T Oct 20 '15 at 22:34
  • $\begingroup$ Well, that's quite a valiant assumption to make. Air tends to get compressed, as a quick visit to Jupiter is sure to demonstrate. $\endgroup$ – Serban Tanasa Oct 21 '15 at 12:59
  • $\begingroup$ Very minor correction, but if the optical depth $\tau$ is given by $\tau(x)=\int^x\kappa\rho(s)ds$, then the intensity is $I(x)=I_0e^{-\tau(x)}$, according to the slab model of absorption, which should work well here (I think!). $\endgroup$ – HDE 226868 Jun 15 '18 at 7:01

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