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An'shur
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04 Nov 2018 07:06

If the atmosphere of Venus was completely cloud-free, would we even be able to see anything from the surface, like Sun, planets, and stars?
This is a very interesting question! I assume you mean if the atmosphere was STILL CO[sub]2[/sub] rich and capping at a pressure of 93 bar, with its characteristic temperature of 740 K (467 °C, 872 °F)? There is an excellent write up about this here, that you might find enlightening. The relevant passage is at the bottom of the chapter's page, but if you'd rather read here, this is the part I'm referencing: 

...

No doubt this will raise even more questions, but hopefully that'll answer your initial one.
Indeed. :) So the Sun would be observable at all times, but I do not think anything else would (maybe besides Earth, if you knew where to look, but just maybe), since the Sun would be always visible, at least slithering around the horizon and illuminating the sky, making it too bright for stars to be observable. But it is just my assumption. Even though the Sun is dimmer during sunsets or sunrises even on Earth, it is the Sun we are talking about and there may not ever be a true dark night on planets with absurd atmospheric pressure and temperature.

On Earth, the red distorted Sun is still visible for a brief period of time even when the Sun itself is already below the horizon. Could the same effects as on Venus be happening on planet with a breathable atmosphere? My idea is, a planet with an atmosphere with the right amount of oxygen for us to breathe, maybe some nitrogen and another inert gas to dramatically increase the pressure, for example neon or helium. (I am getting this idea from breathing mixtures used for diving). How much helium could I pile up into the atmosphere for it to still be breathable and survivable on the long run?
Did you try auto-exposure?
Yes, to no avail.

Most of my original questions have been answered, except:
I know the Moon is about a million times (or 15 magnitudes) dimmer during a Lunar eclipse. And it would receive no sunlight at all if was to be eclipsed by an alternate version of Earth, completely airless Earth. Nothing to refract or scatter the light, obviously. But would it be truly invisible? There may still be outer reaches of the Sun's corona and stars in all directions to illuminate the Moon, hell, even bright planets (Jupiter, Mars, and especially Venus). What would be the apparent magnitude of the Moon in such a scenario? If it was insufficient for the naked eye, how big of a telescope would I need?

How would it look if a comet entered the Earth's umbra? Would Earth's magnetosphere influence the shape of the tails?
For the magnitude of the Moon eclipsed by an airless Earth, I need mathematics I am unfamiliar with. Comet entering the Earth's umbra is more of a fun question.
 
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04 Nov 2018 10:27

For the magnitude of the Moon eclipsed by an airless Earth, I need mathematics I am unfamiliar with.
We can use some geometry along with the magnitude-flux formula from astronomy:

Image

Where we'll call m[sub]1[/sub] the apparent magnitude of the Moon and F[sub]1[/sub] its flux (light intensity in Watts per square mater) at Earth.  m[sub]2[/sub] and F[sub]2[/sub] are a reference magnitude and flux, for which we'll use the Sun.

The apparent magnitude of the Sun at Earth is -26.7, and the flux of sunlight at Earth is its luminosity (3.84x10[sup]26[/sup] Watts) divided by the surface area of a sphere of Earth's orbital radius.  This gives us about 1350W/m[sup]2[/sup] for F[sub]2[/sub].

Let's calculate the apparent magnitude of the Moon assuming full illumination by the Sun.  We'll approximate the flux of moonlight at Earth to be the "luminosity" of the Moon divided by half the surface area of a sphere with the Moon's orbital distance, and we'll say the Moon's luminosity is the flux of Sunlight at 1AU, times the cross sectional area of the Moon, times the Moon's albedo (0.12).

Plug and chug through the formulas, this gives an estimate of the full Moon's magnitude of about -12.  Not bad.

Now we'll consider the Moon being illuminated only by Venus.  Again let the Sun's magnitude and flux serve as the comparison, and compute the luminosity of the Moon due to Venus:

Venus' "luminosity" is the Sun's luminosity spread over a sphere the size of Venus' orbit, times Venus' cross sectional area (pi times Venus' radius squared), times Venus' albedo of 0.65.  

Then convert to Venus' flux at the Moon by dividing this by the surface area of a hemisphere whose radius is the distance from the Earth to Venus.  This varies, and we're making approximations here for how the illumination really behaves, but it turns out using a distance of about 1.1AU gives an apparent magnitude of Venus of -4 so let's use that.

Then the flux of "Venuslight" at the Moon (or Earth) when Venus is -4th magnitude is 8.5x10[sup]-7[/sup] Watts per meter, and the apparent magnitude of the Moon becomes about +10.8.  The entire disk of the Moon would give about as much light as Saturn's moon Tethys.  This is really bad especially when you consider it's about a half degree across.  In terms of surface brightness, that's about 28 magnitudes per square arcsecond.  For comparison, M33's surface brightness is 22.8 magnitudes per square arcsecond.  28 per square arcsec is pitiful.

So, I think a Moon illuminated only by Venus would be so dim and low in contrast against the sky that it'd be invisible even in good binoculars or telescope (and a telescope wouldn't help much anyway because it would just spread the light out over a bigger area).  It would be easier to tell that it's there by when it blocks out background objects.

Perhaps the Solar corona would contribute more light, but this is very hard to compute since the corona has a complex changing shape and brightness, plus during total eclipse the Earth would be blocking the brightest parts of it.  I think the net result will be the same -- the Moon would be too dim to observe.
 
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An'shur
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04 Nov 2018 10:37

the apparent magnitude of the Moon becomes about magnitude +10.8
Not good. :) Thanks for the calculation.
 
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04 Nov 2018 10:46

Sure thing!  It was interesting to compute. :)
 
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04 Nov 2018 13:58

the apparent magnitude of the Moon becomes about +10.8
This is a somewhat surprising result.  It is said that in optimal conditions a young human eye is able to see objects casting a shadow in the light of Venus.  The Moon has a low albedo, but even if the Moon was snowy white the magnitude would still be about +8.5.
A more familiar unit for illuminance (luminous plus per unit area) is lux.  Full sunshine is around 50.000 lux.  Overcast: a few thousand (or a few hundred when the sun is low).  Full moonlight: 0.3 lux.  Wikipedia tells me 683 lux per 1 W/m² and a -14.2 magnitude object gives an illumination of 1 lux.  So a -4 object means 848 µlux.  Converting this to W/m² gives me 1.2×10[sup]-6[/sup], pretty close to 8.5×10[sup]-7[/sup].

So what are we to believe about claims that venuslight creates visible shadows?  It may be misleading to say that the Venus lit Moon is the light of (invisible) Tethys spread out.  Spreading it out could actually help, since it fixes the problem that our sight doesn't have very good resolution.  The only way we can see really small objects is because they are quite bright!

Assuming that the claim about Venus is right, I think the Venus lit Moon would be on the very limit of being visible because of its low albedo.
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04 Nov 2018 20:56

So what are we to believe about claims that venuslight creates visible shadows?
They are definitely true!  I have seen the effect once under a truly dark sky in South America.  I hadn't expected it, so when I saw shadows cast on the ground I thought "where is this light coming from?" and I turned around and sure enough it was Venus sparkling brightly in the evening sky.

The shadows were very hard to see, though.  And my night vision then was much better than it is now.  I think I first noticed it not so much because of the contrast, but because of the movement of my own shadow.  Our eyes are very good at detecting small movements in dim light.  Of course it also depends crucially on having a very dark sky, total dark adaption, Venus being close to its brightest, and a fairly bright surface to view it on.  With the lunar surface being so dark (common comparison is to fresh asphalt, though I'm not sure how accurate that is), I have a hard time imagining the Venuslight being visible on it even with well dark adapted eyes.

My reasoning for comparing by surface magnitude is that this is often a useful indication of the visibility of diffuse objects like galaxies and nebulae.  It's a very common measure in professional astronomy because it sets the strongest limits on what is observable.  You're right that very small objects are visible because they have a high surface brightness, but by the same token, objects with the same magnitude but spread over a larger area are lower surface brightness and harder to detect.  A good way to test this is to aim binoculars or telescope at a dim star and then defocus it.


All that being said, this still does leave the perplexing question of how the shadows cast by Venus are visible if they have such low surface brightness.  I think there are two other factors.  One is that because Venus is essentially a pointsource, the shadows it casts are very sharp.  That surely must make it easier to see than something like a galaxy or nebula that does not have sharp edges.  Similarly the Moon would have a sharp edge against the sky, so that must help.  That's something I did not consider earlier. :)

The other factor, which may work against us, is the difference between observing the ground and the sky.  Under a truly dark sky and without Venus or the Moon in the sky to provide illumination, the ground looks black, like absolutely black.  You can't even see your hand in front of your face.  But the sky under those conditions is not dark, even if you exclude the starlight.  The air itself glows, and this reduces the contrast of faint objects on the sky.  A sky with NELM of 6 corresponds to about 21 magnitudes/arcsec[sup]2[/sup], which will make it difficult to observe any galaxy or nebula whose surface brightness is 23 mag/arcsec[sup]2[/sup] or fainter.  Going fainter than that requires exceptionally dark skies and large scopes.  31 mag/arcsec[sup]2[/sup] pushes the limits of world-class telescopes.


So, tl;dr, you raise a very good question and I'm not 100% sure how it all works out, but I think I still lean towards the Moon being basically invisible in this scenario.  At the very least, you'd need to know exactly where and what to look for.
 
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04 Nov 2018 21:29

The closest real comparison of a Moon illuminated by Venus that I can find: The dwarf galaxy Leo II.  It is magnitude 12.6 and about 1/3 the angular size of the Moon, so its surface brightness is about 27mag/arcsec[sup]2[/sup].  It was discovered in 2007 using an 8.2 meter telescope and 90 minute total exposure time.
 
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04 Nov 2018 23:55

It was discovered in 2007 using an 8.2 meter telescope and 90 minute total exposure time.
Actually, the article says that it was discovered in 1950, but its stars down to mag 26 were counted in 2007.
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05 Nov 2018 00:29

Oops, yes I mixed up with the later study.  So the initial discovery was with telescopes in the roughly 1m range.  With some further searching I found an example of an astrophotographer capturing it.  I wonder though if anyone could spot it visually in a modest scope.  I'd be pretty surprised.
 
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05 Nov 2018 01:26

Could the same effects as on Venus be happening on planet with a breathable atmosphere? My idea is, a planet with an atmosphere with the right amount of oxygen for us to breathe, maybe some nitrogen and another inert gas to dramatically increase the pressure, for example neon or helium. (I am getting this idea from breathing mixtures used for diving). How much helium could I pile up into the atmosphere for it to still be breathable and survivable on the long run?
As for extreme mirage distortions in a breathable atmosphere, no, it could not happen in a Earth-like atmosphere. Mirage-effects like the one described is caused by the extreme temperatures on Venus TOGETHER with the pressure. An atmosphere of that type would be obviously lethal to any human that breathed it. And bear in mind that the passage I was referencing deliberately left out any mention of clouds for the sake of mirage demonstration. As we have seen in the Venera images (what few we have) there is no clue to such extreme sky-distortion, simply because the thick clouds and dust on the Cytherian surface obscure it. Even if there were breathable levels of O[sub]2[/sub] and and other convenient gasses in Cytherian air, they would be hopelessly contaminated at ground level by partially vaporized silicates and a low-lying sulfur dioxide smog. The meager 159 ppm of SO[sub]2[/sub] would be enough to cause harm, to say nothing of the myriad other atmosphere complications present on Venus. But I digress - that wasn't your question.

Helium is a noble gas, and so does not poison you if you inhale it, because it does not chemically react to anything in your body. It's only danger is oxygen displacement inside your lungs and sounding like you were kicked in your nether regions. Technically speaking you could have an infinite amount of helium in a atmosphere provided that the required 16-50% partial pressure of oxygen is maintained, until the overall pressure killed what ever was breathing it. Unfortunately, this is science we're dealing with here, and the cuts are not so clean as that. Since gasses like Helium or Neon are Noble Gasses, they do not bond with any other elements in the surrounding environment (unlike Oxygen, which is VERY reactive, and so we get all these wonderful things like CO[sub]2[/sub], H[sub]2[/sub]O and CO etc and of course our ability to breath it). Thus whenever Noble gasses are released into a planetary atmosphere by whichever process, unless mass of a planet is sufficient enough to create a strong enough gravity to overcome the escape velocity of Helium (or Neon), such Noble gasses slip away into space (hence helium balloons always pull upward, unless they deflate, thereby losing or leaking Helium). In our solar-system, the only planets with a strong enough gravitational pull are the gas-giants, and they have large quantities of it. Earth would need to be about ~3 Earth masses to support a helium atmosphere, which entails a strong gravitational pull on the surface, almost twice that of the current. I believe Space Engine actually has Super-Earth's of this mass with partial helium atmospheres. There is an excellent summary of all of this here.

Image
Here is a handy graph illustrating the relationship between Mass and escape velocity, along with it's interesting co-dependence with temperature. Hence Earth could have sizable amounts of Helium in its atmosphere if it's surface temperature was about 150K.  


Image
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An'shur
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08 Nov 2018 09:06

Under a truly dark sky and without Venus or the Moon in the sky to provide illumination, the ground looks black, like absolutely black.  You can't even see your hand in front of your face.
I dare to disagree. :) I remember participating on a Messier marathon* once and seeing the meadow and the surrounding trees painted in a faint silvery glow. I could not see details, but at least the silhouettes of people around and their telescopes were noticeable. I do not recall if I was able to see my hands though. It was about 3 or 4 am already, so my eyes were as dark-adapted as they could be after the hours of chasing Messier objects.

I believe the combined starlight (plus airglow maybe) made this all possible. If not, then maybe the fact that it was a Bortle 4 location must have added to the illumination for the effects I describe. Speaking of Bortle classification, it was a challenge to see fainter stars below ~20° above the horizon, same goes for DSOs.

*
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08 Nov 2018 14:51

maybe the fact that it was a Bortle 4 location must have added to the illumination for the effects I describe.
Yes, that would explain it. :)  Bortle 4 is still a light polluted sky.  Even though it is dim enough to see the Milky Way clearly, there is still a surprising difference in sky brightness between Bortle class 4 and Bortle class 1:

Image

I live under class 5 skies and have no trouble seeing the ground on a moonless night.  I can also reach Bortle class 3 skies after a short drive, from which the night sky is quite impressive but still some light pollution domes are apparent in a few directions. It is even still a bit too bright to see the airglow effect -- I could see it just once when it happened to be stronger than usual and had a lot of rippled structure, which showed up beautifully in a long exposure.

The class 1 conditions I experienced in the southern hemisphere was like a whole new realm.  The airglow was very clearly visible, especially closer to the horizon.  It almost looks like light pollution except that the color is different (greenish-blue to my eyes, while light pollution is usually pinkish or orange).  The sky is so filled with stars, and the Milky Way so detailed, that it's incredible.  But I could not see details in the ground at all.  That was actually one of the things that struck me the most about the experience.  Normally when you go outside at night you can navigate your surroundings fairly well once your night vision kicks in.  But in that darkness, no amount of adaptation seemed to help (except for making the sky more stunning).  After 2 hours in it, I couldn't see my travelling partners when they were only a few steps away.  I couldn't even distinguish their faces when face to face with them.  If I had to navigate out there without a light I think it would have been hopeless.


I have also heard it said, in addition to the above graphic, that under class 1 skies the center of the Milky Way provides enough light to cast shadows.  I did not experience this myself, but I wasn't looking for it.  (I did think to look for it more recently when I was in class 2 conditions, and I could not see it then, but the galactic center was also not as high above the horizon).  I expect the shadow must be very fuzzy.
 
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09 Nov 2018 01:46

I can attest to Jupiter casting shadows under class 2 skies. THAT was a very good night for me! Now however it's raining all night, and my astronomy has ebbed to reading astrophysics books and watching PBS Spacetime. It's kinda depressing.
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11 Nov 2018 01:48

That sounds like a very good night, indeed, Stellarator!  :)

I also just found a thread on the cloudy nights forum about whether the Milky Way can cast visible shadows, and was surprised to see people report that they have seen it cast several shadows, from different parts of the galaxy that are brighter than the rest.  It makes sense in hindsight -- the band of the Milky Way is very irregular and several regions besides the central bulge are bright.  But I expected it would be too fuzzy to be obvious.  

If I have the fortune to be under very dark skies with the galactic center high overhead again, I will be sure to remember to look for this effect.
 
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13 Nov 2018 16:10

I love your big bounce idea and cyclic model of the multiverse :)  I thought you'd find this intriguing


https://en.wikipedia.org/wiki/Laura_Mersini-Houghton


Laura Mersini-Houghton (née Mersini) is an Albanian-American cosmologist and theoretical physicist, and professor at the University of North Carolina at Chapel Hill. She is a proponent of the multiverse hypothesis and the author of the theory for the origin of the universe, which holds that our universe is one of many selected by quantum gravitational dynamics of matter and energy. Predictions of her theory have been successfully tested by astrophysical data.[1][2][3][4][5][6][7][8] She argues that anomalies in the current structure of the universe are best explained as the gravitational tug exerted by other universes.[9][10]


https://en.wikipedia.org/wiki/ULAS_J1120%2B0641

ULAS J1120+0641 is the second most distant known quasar as of 6 December 2017, after ULAS J1342+0928.[4][5][6] ULAS J1120+0641 (at a comoving distance of 28.85 billion light-years[note 1]) was the first quasar discovered beyond a redshift of 7.[7] Its discovery was reported in June 2011.[1] [note 2]

https://en.wikipedia.org/wiki/ULAS_J1342%2B0928

ULAS J1342+0928 is the most distant known quasar detected and contains the most distant and oldest known supermassive black hole,[1][5][6][7] at a reported redshift of z = 7.54, surpassing the redshift of 7 for the previously known most distant quasar ULAS J1120+0641.[1] The ULAS J1342+0928 quasar is located in the Boötes constellation.[3] The related supermassive black hole is reported to be "800 million times the mass of the sun".[5]

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