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January 15

astronaut pulled because of gravity

If earth's gravitational attraction causes the moon to orbit around it, why are objects in the middle of them still floating? Say, will an astronaut floating in between the earth and the moon not be pulled into one of the two (after considering the Hill Spheres of each of them)? - anandh, chennai — Preceding unsigned comment added by 106.51.12.237 (talk) 01:44, 15 January 2015 (UTC)[reply]

I think you're confusing a few issues. 1) Objects in orbit around earth aren't "floating" because they are between the earth and moon. They are floating because they are in free fall. In order to feel weight, you need to exert a force against a surface. When you and your spaceship are both in the same free fall, then you exert no force against each other, so you feel no weight. You're both still under the effect of the earth's gravity, which is why you keep moving in circles around the earth rather than flying off at a tangent. The effect of the moon has absolutely nothing to do with it; any occupied vessel in orbit around the earth is so far from the moon such that the moon has an insignificant gravitational effect on it. It's about the same as what it is on Earth, which is basically nil. 2) Object located at the correct locations around the Earth and Moon do experience the effects of the Earth and Moon gravity exactly cancelling out. One of these locations is located between the Earth and Moon, but there are also several others; the set of them is known as the Lagrangian points, and they represent all the various locations in a two-body system where their gravity cancels exactly. --Jayron 32 01:52, 15 January 2015 (UTC)[reply]

"..which is why you keep moving in circles around the earth rather than flying off at a tangent..." Thanks that answers...I didn't know object were just circling and not moving away. But if a gravitational pull can make an object circle around it, why not pull further? This is just a layman question, so pls bear if a fundamental science is missing while asking.. — Preceding unsigned comment added by 106.51.12.237 (talk) 02:29, 15 January 2015 (UTC)[reply]

Depending on the speed of the object in motion, it will have a stable orbit at a different distance from the Earth. The amazing part is, that the object naturally moves to that stable orbital distance all on it's own, because if it's too close, it flies outwards, and if it's too far, it falls inwards. StuRat (talk) 02:47, 15 January 2015 (UTC)[reply]

Maybe I'm misunderstanding you, but objects do not naturally move towards a stable orbit. If the object is not in orbit to begin with, it cannot achieve orbit unless it is accelerated during its motion by an additional force. That's why you can't shoot a bullet into orbit from the Earth's surface. Of course we call gravitational orbits stable because small disturbances to the object only change the orbit's shape and do not destroy it. - Lindert (talk) 10:18, 15 January 2015 (UTC)[reply]

You can't shoot a bullet into orbit because of air resistance from the atmosphere, which would cause it to burn up, if you shot it fast enough to achieve orbit after subtracting air resistance. I believe there's an old thought experiment where a cannon ball is fired, parallel to the ground, from a high mountain, in an atmosphere-free world, to achieve orbit. StuRat (talk) 17:38, 15 January 2015 (UTC)[reply]

Ok, you can indeed shoot a bullet into orbit if you shoot it parallel to the ground in a vacuum, but it will always return to its original height, so if fired at 5 miles above sea level, the orbit's periapsis will be 5 miles at most. I wasn't really talking about such low orbits, and effectively, the bullet is then fired above the planet's surface, not upwards from the ground. But you're right of course that low orbits are unstable inside an atmosphere. - Lindert (talk) 19:28, 15 January 2015 (UTC)[reply]

Orbital speed is an important factor. At any given distance from a planet, an orbit will be stable at a particular speed. That's why Mercury and Venus revolve around the sun faster than we do, while the outer planets travel ever more slowly, respectively. Low-earth orbits tend to take a given satellite around the earth multiple times a day. The higher above the earth, the slower a satellite moves. The moon, relatively far away, takes about 29 days to orbit the earth. Hence you can place a satellite at a point where its revolution exactly matches the spin of the earth. That's how we get geosynchronous orbits for television signal transmission and the like. ←Baseball Bugs ^{What's up, Doc?} carrots→ 09:46, 15 January 2015 (UTC)[reply]

I agree with most of the above, but thought that a different response to "if a gravitational pull can make an object circle around it, why not pull further?" might make things clearer: If you think about an object orbiting the Earth, and ignore gravity from any other sources, the object will be following either a circular or an elliptical path about the Earth's center. If you pick any point in that object's travel, the velocity (speed and direction) of the object at that moment completely determines its entire path, which will be a closed ellipse or circle unless it hits something first. An object that is falling is in an elliptical orbit that is too narrow—its path intersects the surface of the Earth before it gets all the way around the elliptical path. (I have ignored escape trajectories and some other things, to keep this simple.) How narrow the ellipse is depends on how big the rotational component of the object's velocity is. If the object starts out with little rotational velocity about the Earth, it's elliptical path will be very narrow and it will fall almost straight down (but still on an elliptical path). More rotational velocity makes the object's path closer to a circle.--Srleffler (talk) 18:37, 15 January 2015 (UTC)[reply]

We say that an object in orbit is in "free fall" because a circular or elliptical orbit that keeps going around the Earth forever is really not any different from an elliptical orbit that happens to intersect the Earth's surface. It's an orbit either way—one just happens to have some rock and dirt in the way. Objects in a stable orbit are "falling", but keep missing the Earth.--Srleffler (talk) 18:41, 15 January 2015 (UTC)[reply]

hi, i don't understand how objects follow circular or elliptical path once you ignore the gravity acted upon it. Objects in space obey the laws of physics on earth right, that is they travel in a straight line unless acted by an external force.. They move in straight line, gravity pulls in, they move straight, gravity pulls, like a centrifugal force acted upon the object ultimately forming a circular or elliptical orbit. Doubt is that a moving body when exerted an external force (in this case gravity) should continue to move indefinitely in that direction. So once-straight-line moving objects should now come closer and closer towards the gravity pull. Where does it get the force to not come closer and continue to move in their path? - anandh — Preceding unsigned comment added by 106.51.18.198 (talk) 03:41, 16 January 2015 (UTC)[reply]

Objects near a large mass follow a curved path because of the gravity acting upon them. An object in orbit is constantly being pulled toward the primary; but, if its lateral velocity is great enough, the curving path misses the ground. —Tamfang (talk) 10:11, 16 January 2015 (UTC)[reply]

Orbiting is just the arc of a throw being flatter than than the ground. But on Earth, the air and probably mountains get in the way. Most things would be vaporized before 1 orbit by moving that fast in air. Sagittarian Milky Way (talk) 01:58, 17 January 2015 (UTC)[reply]

If it is 'constantly being pulled,' will it's velocity not decrease and collapse inward? It's interesting to note the similarities of my question with electrons orbiting nucleus, as electrons do not lose energy if orbiting a stationary orbit as put forth by Bohr. — Preceding unsigned comment added by 106.51.37.87 (talk) 08:54, 17 January 2015 (UTC)[reply]

No; why should it? —Tamfang (talk) 17:08, 17 January 2015 (UTC)[reply]

The planetary model/Bohr model of the atom is wrong, and we know it's been wrong for a long time. Of course "wrong" is relative, but it's very wrong, although not as wrong as even older ideas like the original belief that atoms were indivisible. Unfortunately it's still extremely common as a popular depiction of "what atoms look like" even though it gives wrong impressions. The loss of energy from the electron as it "orbits" was one of the key problems identified with the model, and this helped point physicists towards the wave-like nature of elections. Electrons are not little spheres going around the nucleus like planets orbiting a star; they're standing waves that exist in electron orbitals. Reality is weird at the quantum level, at least to our macroscopic minds that don't see quantum effects in everyday life. Don't take this as a hostile reply; you're asking good questions that demonstrate a deep amount of thought regarding the problem. See also my reply below. --71.104.75.148 (talk) 01:05, 20 January 2015 (UTC)[reply]

how did the object, astronaut in this case acquire the lateral velocity? why should the velocity not slow down when pulled? is that they are having stationary orbits? or does that falling astronaut have one? — Preceding unsigned comment added by 106.51.54.232 (talk) 20:53, 18 January 2015 (UTC)[reply]

The OP mentioned Hill sphere. The formula (I had to look it up, it's right at front in the article) is sqrt(m/3M) times an orbital dimension which in the circular approximation is the radius. Moon is 0.0123 the mass of Earth, so the Moon's Hill sphere should be sqrt(0.0041) times the distance to the Moon is 6.4% of the distance to the Moon. So it's a pretty small band of Earth orbits that is affected by it. When something is in that distance, either it is orbiting *exactly* at the same rate (in one of the Moon's Trojan points, though Lagrangian point suggests the Sun's influence may disrupt those orbits) or else it isn't. If it's not orbiting at exactly the same rate then it will orbit Earth just fine, like any other orbit, until eventually it laps the Moon or vice versa and then something interesting is bound to happen. Wnt (talk) 00:35, 20 January 2015 (UTC)[reply]

The astronaut acquired the velocity from the tons and tons of rocket fuel burned to shoot them into orbit. Most of a rocket's energy is used not to gain altitude, but speed. Watch any rocket launch, like the Saturn V or the Space Shuttle, and you will notice it starts turning towards a trajectory parallel to Earth almost immediately after liftoff. In Low Earth orbit objects are moving at around 7.8 kilometers per second relative to the Earth. And it is believed that objects do lose energy when orbiting, by releasing gravitational waves. However, this loss is very small over time, but it does mean orbits will eventually decay over very long timescales. I say "believed" because we haven't directly detected gravitational waves yet, though there are experiments looking for evidence of them. Also, this is not directly related to orbital mechanics, but objects in low Earth orbit gradually lose energy from drag. There still exists a noticeable amount of atmosphere at LEO, even though it's not enough for humans to breathe. Objects in LEO, like the International Space Station, have to periodically burn fuel to maintain their orbits. --71.104.75.148 (talk) 01:05, 20 January 2015 (UTC)[reply]

Antimatter

How do I calculate the brightness of sea-level air being instantly swapped for an equal volume of antimatter? (the air goes to another galaxy) This depends on mass and maybe on density if the density is low enough. Is it possible for the annihilation of just a thin skin of air with a solid object to blow away the air for long enough to affect the brightness or explosive yield? Maybe even have more than one cycle? Where the antimatter causes a new bubble of vacuum as the old one collapses and then touches the antimatter's surface? What if it's partially buried, how will it explode? Sagittarian Milky Way (talk) 12:00, 15 January 2015 (UTC)[reply]

I'd model it like a nuclear explosion, just with some 10K times more yield per mass. One difference, though, is that unlike a nuke, which can fizzle, sending unburnt fissile material out, anti-matter will always fully detonate, given an adequate supply of normal matter in the area. StuRat (talk) 17:33, 15 January 2015 (UTC)[reply]

I'd presume that the antimatter being gaseous would matter quite a bit too. As the anti-air annihilates against the normal-matter ground, a partial vacuum would form that would pull in more anti-air. So the reaction could easily run to completion. A nuclear weapon could (potentially) scatter it's fissile material away from the explosion and thereby prevent it from all going critical - but in an antimatter explosion, the dirt and dust that would be thrown into the air would just provide more fuel for an even more rapid mixing of matter and antimatter. SteveBaker (talk) 17:53, 15 January 2015 (UTC)[reply]

At about 10⁵ K the radiation pressure would exceed atmospheric pressure. Once that temperature is reached, which seems likely, the force of the explosion would blow any additional material away from the surface. If we are replacing the entire atmosphere (the original post isn't so clear), then the potential energy available is about 5×10³⁵ J, roughly 1000 times the gravitational binding energy of the Earth. So in principal, it could completely unmake the Earth. In practice though, I imagine that the initial detonation would blow a large fraction of the antimatter into space and leave a charred husk of Earth behind. At the same time, a layer of antimatter only about 8 microns thick covering the Earth would generate enough energy to make the Earth briefly more luminous than the sun. If Sagittarian Milky Way would prefer some other way of describing the amount of antimatter involved it would help to be more specific on the quantity and distribution. Dragons flight (talk) 19:13, 15 January 2015 (UTC)[reply]

I was not in fact thinking of globe-spanning antimatter but appreciate the calculations anyway. Okay, how about:

1. a sphere of anti-osmium, high enough to not have time to fall to the ground or 1000 meters, whichever is lower, and small enough to not have the power to expand the atmosphere much.

2. Same mass but a sphere of anti-air of the same pressure. I guess the answer above might happen.

and,

3. A very heavy skyscraper, say a third of a billion kilos. The Sun is 400 tons of mass annihilation per second so how many Sun's would that peak at? (visual brightness). Could you still extrapolate from hydrogen bomb megatons with that? Thanks everyone. Sagittarian Milky Way (talk) 01:54, 16 January 2015 (UTC)[reply]

I once heard Robert L. Forward say that a cannonball of anti-iron, sitting on the ground, would probably sizzle rather than explode. —Tamfang (talk) 17:10, 17 January 2015 (UTC)[reply]

What was the best Comet McNaught & Lovejoy (2011) looked like naked eye from 40°N?

(I never looked for it) Both covering the Sun area with an object and waiting till it gets low enough to not need that. Those comets moved so fast in a day so let's say longitudes near 74°W if UTC time of sunset matters.

2. How impressive were they after all naked eye twilight in cities with too much light pollution to easily see the Milky Way?

3. How long did each qualify to be a Great Comet for? (approximately). The article says they're Great (the only Great ones since 1997)

4. When did Hale-Bopp stop being Great? Sagittarian Milky Way (talk) 12:06, 15 January 2015 (UTC)[reply]

From the midlands of the UK (which isn't 40 deg N I know but still), I saw the multiple tails of Comet McNaught in the southern horizon while the sun was still up (about 5pm). It took me a little while to work out what I was looking at, but it was most certainly that comet. I never saw the head, however - that had already passed below the S horizon. --TammyMoet (talk) 19:35, 15 January 2015 (UTC)[reply]

It probably wasn't 5pm if it was before sunset in early January in England but that's pretty visible. It must've been great at a latitude where it's twilight or dark with the tail still up. I tried to see PANSTARRS extensively with binoculars and may have seen something. Maybe. (though they were 8x22s or something - built for daylight) Sagittarian Milky Way (talk) 20:01, 15 January 2015 (UTC)[reply]

My parents were able to see Hale-Bopp at their NJ shore from the Summer of 1996, through the holidays, and well even into the summer 1997, until they became curious that it had stopped moving, and realized they had been looking a street lamp through the woods for several months. It was quite visible to me over the Summer at their shore house due to the relative darkness and easily from anywhere that Winter. μηδείς (talk) 01:51, 16 January 2015 (UTC)[reply]

Yes that was a long one. I found a reference that Comet McNaught was first seen with the naked eye on January 1 (near invisible probably) yet I couldn't even find it's impressiveness circa 2 weeks later (lucky Southern Hemisphereans). Sagittarian Milky Way (talk) 02:14, 16 January 2015 (UTC)[reply]

Magnitude of the vacuum energy density

Vacuum catastrophe says: "the upper bound upon the vacuum energy density as inferred from data obtained from the Voyager spacecraft [is] less than 10¹⁴ GeV/m³". That's 1.6×10⁴ J/m³, right?

Vacuum energy says, "Using the upper limit of the cosmological constant, the vacuum energy in a cubic meter of free space has been estimated to be 10⁻⁹ joules (10^-2 ergs)."

Those two values differ by a factor of 1.6×10²³, nowhere near the more than one hundred orders of magnitude which separate these numbers from the calculated zero-point energy (as described in vacuum catastrophe), but still a wide gulf. I assume that this is because the former, an upper bound, comes from not observing any discernible effect on the Voyager spacecraft, whereas the latter comes from estimations of the cosmological constant from astronomical observations.

Are there any other experimental measurements of the vacuum energy density? Have recent experiments with the Casimir effect give any numbers? -- ToE 14:37, 15 January 2015 (UTC)[reply]

1.6x10²³ is suspiciously close to Avagadro's number...I wonder if there is a math slipup there someplace?

But Voyager is far from being in 'free space' - it's really close (in astronomical terms) to a bloody great star...so I'm not sure we're necessarily doing an apples-and-apples comparison. SteveBaker (talk) 18:00, 15 January 2015 (UTC)[reply]

This paper [1] (Anderson et al. 1995) gives an estimate based on Voyager and other spacecraft on the total mass of non-luminous matter in the solar system. If interpreted as an estimate on the density of dark matter + dark energy within the orbit of Neptune, then the value would be (-5.8 ± 5.2)×10¹² GeV/m³. That's a 1 σ error bound, so it would probably be fair to say there is a high probability the density is < 2×10¹³ Gev/m³, which is close-ish to the 1×10¹⁴ Gev/m³ quoted above. Though it should be noted that the research I'm citing was done back at a time when dark energy was not yet a thing people worried about. Dragons flight (talk) 18:44, 15 January 2015 (UTC)[reply]

As Dragons flight said above, this value was an upper bound on dark matter in the solar system. The gravitational effect of dark energy is repulsive, so it doesn't even make sense to treat the Voyager estimate as bounding dark matter plus dark energy. The cited source (Dutra) also got this wrong. I removed him and the estimate from the article. (They've been there since the article's creation in 2009, which is more evidence that no one ever checks Wikipedia citations.)

The Casimir effect is unrelated to vacuum energy density, despite common belief (see hep-th/0503158). -- BenRG (talk) 19:06, 15 January 2015 (UTC)[reply]

Is the effect of dark energy repulsive on small scales? Cosmologically, dark energy has both a positive mass density and a negative effective pressure (believed to be roughly equal in magnitude to its mass density). In terms of the equation of state it is clear that ρ + 3p < 0 implies accelerating expansion. It is far less clear to me what effect dark energy is expected to have on gravitationally bound objects in a compact region of space. For example, if there is a density of dark energy ρ₀ in our solar system, how much effective force does that apply to the orbits of planets? Assuming the net effective force is non-zero, then presumably the Anderson et al. results could still be interpreted as a limit on the combined dark energy + dark matter force regardless of the overall sign. Though, as noted, it is not a very useful limit since it is consistent with zero and orders of magnitude removed from what we expect on the basis of cosmology. Dragons flight (talk) 19:56, 15 January 2015 (UTC)[reply]

Thanks to Dragons flight and particular to BenRG for the link to the Jaffe paper. I've a follow up question regarding the Casimir effect, but I'm still trying to work out as much of it as I can on my own. -- ToE 00:13, 21 January 2015 (UTC)[reply]

How far away from a well-lit city can you see the bazillion stars in the night sky?

How far away from a well-lit city do you have to live in order to be able to see the bazillion stars in the night sky, assuming you do not have access to electric lights? Is a fire's luminosity enough to obscure the stars? Do you have to live in a very isolated place in the middle of nowhere just to see the stars in the sky? 140.254.226.183 (talk) 15:34, 15 January 2015 (UTC)[reply]

There's no set distance, it depends on the size of the city, and the light pollution that is generated by it and the surrounding environs. For instance, I'm not entirely sure one can see the Milky way anywhere within 50 miles of NYC, but you can easily see the Milky way 20 miles from the edge of Las Vegas. In my experience, staring at a campfire will prevent you from easily seeing the Milky way, even in the desert (some info at night vision), but your eyes will adjust if you look at the sky and not the fire for a few minutes. We have a List of brightest stars. If you want to name a big city that you live near, we might be able to recommend a good place for stargazing (that article also has some good general info). SemanticMantis (talk) 15:45, 15 January 2015 (UTC)[reply]

Can light pollution be caused by a candlelight? What if there are hundreds and thousands of candles on the streets, at every shop in town, and inside every home? 140.254.136.154 (talk) 16:17, 15 January 2015 (UTC)[reply]

Sure, in concept. Whether you measure in lumens or lux or some other unit, it is basically the light intensity that causes light pollution. The source of the (white-ish*) light doesn't really matter: candles, fires, LED lights, incandescent bulbs, neon lights, fluorescent lights - all can cause light pollution that can interfere with seeing stars. *Keep in mind the Visible_spectrum of the light source does effect how much our night vision is degraded. This is why people use red lights to look at maps while night hiking (that is also described at night vision). SemanticMantis (talk)

Is the night vision degradation permanent or temporary? 140.254.136.154 (talk) 16:50, 15 January 2015 (UTC)[reply]

Usually temporary, see Adaptation_(eye)#Dark_adaptation. Usually your eyes will get as good at seeing in the dark as they ever will after about 30 minutes. There are probably ways to impair night vision for a longer amount of time but that would be the result of eye damage, not regular light pollution. SemanticMantis (talk) 16:56, 15 January 2015 (UTC)[reply]

What about Vitamin A deficiency? Can't Vitamin A deficiency cause night blindness too, or does it merely make night vision take much longer? Is the night blindness permanent or temporary then, once the individual is given sufficient Vitamin A? 140.254.136.154 (talk) 17:03, 15 January 2015 (UTC)[reply]

We have articles on Vitamin A deficiency and Night blindness that should answer those questions. Honestly I'm not that knowledgeable on this specific topic but you can get pretty far by using the search box to look for key terms. WP:WHAAOE is fairly true, and our articles are (usually) better written and better referenced than our replies here. SemanticMantis (talk) 17:11, 15 January 2015 (UTC)[reply]

Note that the quality of air makes a difference, too. If you have thin, low humidity, pollution free air, as you might get in a high desert area, then light tends to shine right out into space and not cause a problem, instead of reflecting back from the air and causing light pollution. StuRat (talk) 17:29, 15 January 2015 (UTC)[reply]

I recommend using http://darksitefinder.com which has a Google-Maps overlay showing where the skies are most and least light-polluted. Sometimes you can find dark spots that are not too far away to drive to - even if you live in a light-polluted city.

The amount of light it takes to erase your night-vision is very critically dependent on how far away you are from it. The intensity of a light drops off as the square of the distance - so being ten times further away from a candle makes it 100 times dimmer. It doesn't make sense to say "a candle is (or isn't) going to destroy your night vision for another 30 minutes" without making some indication as to how close to it you're going to get.

Light pollution from a city is a different matter though - the light from the city causes Skyglow. In that case, the totality of light projected upwards from the city is scattered by dust, water droplets, etc back towards the ground - often so far away that the city itself is over the horizon from the observer's position. Since our eyes use contrast between adjacent patches of the retina to resolve small features, that additional light from the sky makes it much harder to pick out the stars - and the dimmer the star, the harder it is to resolve.

Pollution from the city makes matters worse by increasing the amount of particulate material in the sky - and therefore the percentage of the light that's reflected back to you. That's why cities with relatively little pollution may still have relatively dark skies nearby even though they put out a lot of light.

So even if you get a substantial mountain range between you and the city, that sky glow can still reduce your ability to see the less bright objects in the night sky...and that has very little to do with disruption of your dark-adapted eyes.

SteveBaker (talk) 17:46, 15 January 2015 (UTC)[reply]

Keep in mind that the "bazillion" is really 'only' about 5,000-10,000 distinct objects (mostly nearby stars, and some galaxies) as the upper limit visible to the naked eye. (The Milky Way is a glow mostly of stars to dim to make out individually.) This has been discussed before and should be in the archives. Or you can pick among these sources at google. μηδείς (talk) 18:42, 15 January 2015 (UTC)[reply]

More like 4,000 at best. Check the tables in the bottom half of this article. --65.94.50.4 (talk) 19:34, 15 January 2015 (UTC)[reply]

My use of "bazillion" means "a lot" or "plentiful". If I were to see 4,000 stars, then my first impression would be: "OMG! Uncountable!" 140.254.136.154 (talk) 20:33, 15 January 2015 (UTC)[reply]

Though visual acuity says the average young eye that can manage to focus (I can almost reach eyeglass vision by squinting and I'm 20/60 so it's not hard) is ~20/14.

There are 3.03 times as many stars if you could see astronomical magnitude 7.5 instead of 6.5, 2x dimmer pixels for 20/10 vision, so you could see almost 10,000 stars in a few seconds of head-turning if you have vision like Chuck Yeager.

I counted half a bazillion once, but I was pretty drunk. InedibleHulk (talk) 03:00, 16 January 2015 (UTC)[reply]

The answer to the original question is up to about 200 miles for Tokyo (40 million ppl), as shown in the link below. For

Responses to the above: Snow cover also worsens the sky, probably at least 2 times as most light points down at much less reflective surfaces. The gas molecules cause light pollution too, but are a less important factor as most waste light isn't blue wavelengths. this map shows 2006 light pollution instead of the 2001 linked here (click right above for Google Earth overlays and other areas). There's no land in the US without light pollution unless it's west of the Rockies, too rain-less for farming, or really crappy (either really boggy looking, cold and mosquito infested or really remote and tiny islands without access by land) Actually Dry Tortugas is not that bad (nice and tropical) if you can manage to get there. The situation is similar in Europe and worse in Japan and Korea which don't have any mainland or near islands without light pollution.

And about the candlelight. Streetlights are about 20 feet high and 10 times the lumens of 100W light bulbs. A point is no more than about 4 times their height from the nearest light (per rule of thumb to help old eyes not be bothered by high contrast) To model today's lighting with the known 100W LB brightness the increased 100W bulb streetlight density would have to be realistic. So to have approximately equal lighting on the standard 64 foot wide New York city street (city property line to city property line) you could put them about every 128 feet of each sidewalk's edge. That doesn't seem unreasonable so let's go with it. A 100W (frosted) is about 1500-1600 lumens. A candlepower is 4pi lumens. Let's call it 1/125ths of a light bulb. The most light polluted downtown zeniths on Earth have light pollution about 3⁴ times brighter than the natural sky. This would still make the sky light polluted if we replaced every bulb with a candle but per [http:cleardarksky.com/lp/NYCNYlp.html?Mn=lenses] and Bortle Scale it would probably be as polluted as a rural-suburban transition area of today, except less asymmetrical since you're the center of Tokyo. The headlights are dimmer than 100W light bulbs but there's some buffer in the result and you'd think headlights are a relatively minor contribution if you saw the '03 Blackout. However the biggest metro area in history which actually used candlelight can't possibly be more than 2 million. If it scales with population then it would be a Bortle Class 2 - still light polluted, and noticeable to the naked eye but fairly minor. The air pollution of pure-gaslight Industrial Revolution London might've been an equal or greater problem. (What counts as candlelight anyway? If you're thinking of the old days, hollow perforated wicks were invented in 1790 which increased brightness and even Ancient Rome (pop. 1+ million) probably had many fires which were worth 10-100 candles). Sagittarian Milky Way (talk) 01:31, 16 January 2015 (UTC)[reply]

Only 60 by 60 stars gives you 3,600. I have seen moderate fields of Canada Geese with over 32 X 32 geese... over a thousand. μηδείς (talk) 01:31, 16 January 2015 (UTC)[reply]

It still shocks some city dwellers. 6.5 or brighter stars average 4 Full Moons apart. Not too bad except for the extroverts (they might need a meteor storm) or something. Sagittarian Milky Way (talk) 02:22, 16 January 2015 (UTC)[reply]

For some reason it's more impressive than the stars per square degrees makes it seem. Sagittarian Milky Way (talk) 02:24, 16 January 2015 (UTC)[reply]

Great emphasis is properly placed on avoiding light pollution when observing the night sky. But elevation above sea level is very important as well. All other variables being equal, the night sky will be far more dramatic at an elevation above 10,000 feet or 3000 meters than near sea level. I was once camped with my wife and son on a moonless night at the Vogelsang High Sierra Camp at 10,300 feet in California's Yosemite National Park. While in our sleeping bags in our tent cabin, we heard exclamations of starry wonderment outside. We got out of bed, and went out to see a dramatic and spectacular display of the Milky Way splashed across the sky in an unforgettable way. Make the effort some time to get high and remote on a moonless night with your loved ones. Cullen³²⁸ Let's discuss it 04:59, 16 January 2015 (UTC)[reply]

Of course User:Cullen328 meant sober and intimate, as opposed to "high and remote". μηδείς (talk) 01:41, 17 January 2015 (UTC)[reply]

High in elevation and remote from light pollution. Intimate? Yes, of course. Sober? Well, moderation in all things, μηδείς.

Another time, we camped in the Hoover Wilderness at 10,100 feet with both our sons, knowing that it would be a moonless night during the meteor shower known as the Perseids. My sons now understand meteor showers far better than most 21st century urbanites. Cullen³²⁸ Let's discuss it 05:38, 16 January 2015 (UTC)[reply]

Is there sand, dust, air pollution or evergreen trees (turpentine-like chemicals) upwind of the low dark sites you're comparing this too? That could be the reason for the unexpectedly good improvement. Also, your low sites might just not be free from light pollution. Light pollution extends for up to 200 miles off the California coast and there surely are light sources closer to you than the coast. Sagittarian Milky Way (talk) 06:14, 16 January 2015 (UTC)[reply]

For me, it was not at all unexpected to find excellent sky viewing conditions above 10,000 foot elevation. There is a reason, after all, that modern major astronomical observatories are located in the high mountains. A layer of air a couple of miles thick acts as a filter, even under otherwise ideal conditions. No one would build a world class observatory in Death Valley or along the coast of the Dead Sea. I have visited both briefly and noticed no dark sky wonders. However, I am a mountain guy. I do not spend a lot of time camping in remote low elevation places without light pollution. Cullen³²⁸ Let's discuss it 02:04, 17 January 2015 (UTC)[reply]