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October 22[edit]

Black holes in the sky during the day[edit]

Let's say there were a black hole that we could see from the Earth without the aid of telescopes. So that it's basically the size of one of the other stars that we can see at night. And let's also assume that the Earth wouldn't be destroyed by the gravitational pull, etc. What would this black hole look like during the day? Would there be a black spot in the sky or would the Sun still overwhelm that area of the sky and we still wouldn't see it?

Just a little curiosity that came to mind... Thanks! Dismas|^(talk) 01:27, 22 October 2014 (UTC)[reply]

You can't see a black hole. That's why they are black. You can only detect their effect on objects around them by looking at the effect of their gravity on nearby objects. If you look up in the night sky, you don't see any black holes anyways, even though there are many closer than the stars you see. V4641 Sagittarii is the closest (known) one to earth FWIW, but there may be some (or even many) closer than that, but they don't exist in environments that allow us to detect them. So the answer is "nothing interesting at all" because, what you see at night, when your best chance to see one, is "nothing at all" Even if you were closer to one, you still wouldn't see it. And if you got too close nasty stuff starts to happen. --Jayron32 01:46, 22 October 2014 (UTC)[reply]

Most black holes either glow because they have accretion disks, or are simply black. If they are glowing from an accretion disk, like Cygnus X-1 they may be visible if very close. If black, they are only going to be visible when they occlude another heavenly body, or cause gravitational lensing. I don't believe we are aware of any black holes closer that several hundred light years, something like 30 Kessel runs. But that's me remembering junior high, when Scientific American published peer-review quality articles, and not pornographic POV-laden clickbait. The point is, during the day, the sky way up is still black and star-filled. We just don't notice because the daytime atmosphere is blue due to the way it scatters light: nearby glowing black holes, if they existed, would be invisible behind that until you got dozens of miles high. μηδείς (talk) 01:55, 22 October 2014 (UTC)[reply]

I don't know that accretion discs are all that bright in the visible range, however. AFAIK, Cygnus X-1 is mostly bright in the X-ray region, and not so much in the visible region. --Jayron32 02:07, 22 October 2014 (UTC)[reply]

I thought about mentioning that, but if it's close enough, as the OP is implying, and has a significant accretion disk, it will be visible in visible light as well. μηδείς (talk) 15:09, 22 October 2014 (UTC)[reply]

Yes, but at those distances, other more dramatic effects than the dim glow of the accretion discs would be noticeable too. In one of those "you can't have your cake and eat it too" situations, you can't meet the OPs requirement: a black hole with a visible accretion disc, bright enough to be noticeable during the daytime, which has no other effect on the sun or earth. Such a black hole cannot possibly exist. If it's large enough and close enough to give off enough light during the day time, you can be damn sure you're going to notice the gravitational effects in a dramatic way. Which is basically the point made below by Mr. Schulz. --Jayron32 23:38, 22 October 2014 (UTC)[reply]

I didn't say that more energy would be released as light rather than x-rays, I think there are standard curves for that sort of thing. A star-sized black hole in the Oort cloud or Kuiper Belt might very well have a bright enough accretion disk to see, at least when it flared, gobbling things up. But in that case the radiation would probably be much more dangerous than having our orbit directly perturbed.μηδείς (talk) 00:55, 23 October 2014 (UTC)[reply]

I'm fairly certain that if there were a black hole in the Oort cloud, we'd all be part of its accretion disc a long time ago, rendering whether or not we could see it fairly moot. --Jayron32 12:27, 23 October 2014 (UTC)[reply]

"If the Sun was replaced with a black hole that had the same mass as the Sun, the Schwarzschild radius would be 3 km (compared to the Sun's radius of nearly 700,000 km)." so it kinda depends lol ~Helicopter Llama~ 12:36, 23 October 2014 (UTC)[reply]

What matters with a black hole in regards to it's gravity, Jayron32 is its mass. So A black hole with the mass of the sun would be 6km across per HL, but it would still only have the same gravity of the sun. If this were accomplished without othersie destroying the earth, it would remain in it's orbit as it is now. But if the blackhole were the size of the sun, not 6, but 2,000,000 km across, its mass would be (2,000,000/6)^3 times greater, and we'd instantly be ripped apart by tides, and swallowed up. I think the size needed for a star naturally to form a black hole is about 6 solar masses, it's on the order of that.

That's not so big, and if it were as far out as the Kuiper Belt (past Pluto) or the Oort cloud it would just be a companion star to ours in a binary system. It would be on the order of 10km across (I am not going to do the math--SS gives 30km for 5 solar masses) and it would still just have the mass, and hence the gravitic attraction of a star of 6 solar masses. At that distance the Earth could still have a stable orbit about the sun with a perturbation cycle due to the black hole as the two stars orbited each other every 500-1000 years or so, but it wouldn't get sucked away from the sun by some sort of supergravity. It would simply act as if there were another star out past Pluto. Whether the orbit of Neptune would be stable is another question.

One way to think of it is that in the time just after sunrise or just before sunset, while the sky is still illuminated, you can sometimes see really bright objects, such as Venus. So for a given "glowing" version of a black hole to be visible at any point during the day, it would likely need to be as bright as Venus. ←Baseball Bugs ^{What's up, Doc?} carrots→ 02:09, 22 October 2014 (UTC)[reply]

By way of amplification: in fact you can see Venus in full daylight if it's in a position in its orbit where it's near maximum brightness and far enough from the Sun. You just have to know where to look, or else spot it by accident (which I have). --174.88.134.249 (talk) 07:17, 22 October 2014 (UTC)[reply]

By way of response to the original poster: you need to understand that when into the daytime sky and see blue, that blue comes from sunlight that has been scattered in the atmosphere, and most of that happens within about 50 miles (80 km) above you. (If you ever fly in an airliner you can see that the daytime sky is darker blue because you are above a lot of that. The blueness is now below you, where you see it tinting the ground.) You can see bright things like the Sun and Moon because they shine through the sky. But you can't see dark things through the sky. So the idea of "seeing a dark spot" doesn't make sense. It is true that a black hole absorbs light that falls into it, but the sunlight scattering from the air and reaching your eye in the form of blue sky is not falling into the black hole (Inserted later: because it isn't passing anywhere near it). --174.88.134.249 (talk) 07:17, 22 October 2014 (UTC)[reply]

Also, while black holes are often imagined as giant objects, from a physical point of view, stellar-mass black holes are quite small. A black hole of, say, 5 solar masses has a Schwarzschild radius of about 15km. If it is near enough to be detectable with the naked eye, it would also be near enough to massively influence the the whole arrangement of the solar system via its gravitational effects. Going by Eye#Visual_acuity, we can resolve a 30km diameter body at about 85000 km - that's a quarter of the distance to the moon. At that distance, the black hole's gravitation would totally dominate everything, and indeed, tidal forces would probably be enough to rip the planet apart. --Stephan Schulz (talk) 15:35, 22 October 2014 (UTC)[reply]

Effects of celestial bodies in close proximity[edit]

I recently watched Predators (2010) again. At one point in the movie the gang discover that they are on another planet as evidenced by the several celestial bodies observed in the sky. My question is about the effects such celestial bodies can and do have on each other when in such close proximity. We know that the moon has a significant effect on earth and that is just one body. Given that a couple of the bodies shown in the movie are very large (especially in comparison to our moon) and that there are several bodies that can be seen, I would assume that there would be significant effects on the planet (or moon). What would some of the effects be and is it even possible for a planet (or moon) to be habitable (by us) under such conditions? 99.250.118.116 (talk) 03:17, 22 October 2014 (UTC)[reply]

Well, there's tides, which can be in oceans or tidal heating and/or quakes in solid ground. The effect tends to be more on the smaller bodies. The there would be the effect of having reflected light from all those moons so it may rarely be completely dark on the planet. Pluto has several moons, including one large moon, Charon, so that might be an interesting example in real life. (While the gas giants have many more moons, they tend to be tiny in comparison with the planets they orbit.) StuRat (talk) 04:05, 22 October 2014 (UTC)[reply]

In the long term, planetary systems with large bodies whose orbits bring them close together would have a good chance of being unstable, so even if a planet is currently habitable it might not still be that way after, say, a few million years. Specifically, it might come close enough to be thrown into a significantly different orbit, or if the encounters are not quite that close, cumulative perturbations from repeated encounters might have a similar effect. This is an advanced topic which I only know a little about; because the n-body problem cannot be solved mathematically, scientists have to investigate it by simulations, which has only become possible in recent decades with fast enough computers. --174.88.134.249 (talk) 07:28, 22 October 2014 (UTC)[reply]

The Moon was about ten times closer to the Earth when it just formed. At that time the Earth's surface had melted due to the impact that led to the formation of the Moon, but soon afterwards there were already oceans on Earth. There tides were much higher than they are today, about 1000 meters high. Count Iblis (talk) 15:51, 22 October 2014 (UTC)[reply]

Velocities in Blue Origin Logo[edit]

What is the significance of the velocities (if that's what they are) shown on the shied in the Blue Origin logo? Wikipedia's fair use image, File:Blue_Origin_Logo.png, is only 200 × 190 pixels, but here is one at 485 × 461. As best as I can make out, the velocities are 3 km/s, 9.5 km/s, 13 km/s, 19 km/s, & 20 km/s. They don't appear to represent orbital speeds as these decrease for higher orbits, and they don't seem to correspond to escape velocities either. I assume that they represent the delta-v necessary for various actions, but what? -- 190.58.249.10 (talk) 16:17, 22 October 2014 (UTC) thanks ein fast worm[reply]

Comment: wouldn't expect escape velocity to be very relevant for launch vehicles. From escape velocity "A rocket moving out of a gravity well does not actually need to attain escape velocity to do so, but could achieve the same result at any speed with a suitable mode of propulsion and sufficient fuel. Escape velocity only applies to ballistic trajectories." So these might well be target velocities at each height for a planned launch vehicle, or some famous past vehicle, and have nothing to do with escape or orbital velocities. SemanticMantis (talk) 16:47, 22 October 2014 (UTC)[reply]

My first reaction was ... those are the wrong velocities! Those of us who spend our free time watching RADAR looking for extraterrestrial returns are looking for harmonics of 30 km/s (the characteristic velocity of an object, or a retrograde object, incident onto Earth's orbit around the sun. But, the characteristic velocity is only representative - actual objects can have any orbital velocity, because their orbits can be non-circular or otherwise non-ideal with respect to their impact onto Earth.

For perspective: 3 km/s is close to the orbital velocity of a geostationary earth orbit. For context, New Horizons (launched in 2006) was the fastest launch vehicle ever built, and it launched at about 16 km/s. The Space Transportation System (the Space Shuttle) reentered Earth from an orbital velocity in LEO at about 8 km/s; when I was a kid, the Space Shuttle orbiter was commonly cited as the fastest manned vehicle ever "flown": Mach 25 (on re-entry). The SoHo mission spacecraft is in a heliocentric orbit, and its orbital velocity can rightly be said to be close to 30 km/s with respect to the Sun - though its mean orbital speed with respect to Earth should be about zero!

There are important caveats for such high speeds: relative to what reference-point is the speed being measured? This is an issue of galilean relativity, not special relativity: for orbital flight, are you accounting for the fact that the Earth's surface is rotating and the Earth as an entire planet is revolving around the sun? Do you measure relative to the center of mass of the Earth, or relative to the launchpad or airfield at a "fixed" point on the Earth's surface? Do you measure speed relative to the air? As you go to higher altitudes, and the atmosphere starts to get weird with its pressures and thermal behaviors, you start needing more complicated physics to measure that speed: even at low altitudes, there is indicated airspeed, equivalent airspeed, calibrated airspeed, true airspeed, mach number, ... and finally you get to altitudes with negligible air - no meaningful airspeed, corresponding to a mach number that approaches infinity!

All I can say with certainty is that I don't know what these velocities on the Blue Origin logo are actually referring to. I would speculate that they are launch velocities, or orbital velocities, of actual- and planned- vehicles.

Nimur (talk) 19:21, 22 October 2014 (UTC)[reply]

3 km/s is approximately the delta-V required to reach space. 9.5 km/s is required to get into orbit (the extra 1.5 km/s is due to having to get out of the atmosphere). 13 km/s is for escape velocity from Earth (3.5 km/s more than orbital). I have no idea what 19 and 20 km/s are supposed to be, since escape from the solar system can be achieved for only 3 km/s more than the escape velocity. 20 km/s is about the delta-V required to reach the Sun, but that doesn't explain 19 km/s.

By the way, escape velocities are very relevant for launch vehicles because spacecraft do use ballistic trajectories. A spacecraft's trajectory is pretty much set the moment its launch stage falls away, which happens when it's barely outside the atmosphere. The spacecraft itself only carries fuel for minor course corrections and altitude control. While it's possible to take a lot of fuel and burn it far from Earth, that's less efficient, due to the Oberth effect. --Bowlhover (talk) 20:53, 22 October 2014 (UTC)[reply]

Thanks Bowlhover. I should have mentioned, for those who didn't want to study the logo, that the 3 km/s label is associated with what appears to be a suborbital arc while the others label what appear to be orbits.

Aerodynamics need to be taken into account in real life, but ignoring air resistance and simplistically applying h=v²/2g, a projectile fired upward at 3 km/s would reach an altitude of 460 km, and only 1.4 km/s would be required to reach 100 km. In practice, Project HARP, in 1966, fired a projectile at 3.6 km/s which reached 180 km. But we are not talking about space guns here, so we have to take gravity drag into account in addition to atmospheric drag. The delta-v budget article discusses these issues. -- 190.58.249.7 (talk) 12:44, 23 October 2014 (UTC)[reply]

[copyvio removed]

Hope this helps. --Kitachi Matusuri (talk) 19:47, 22 October 2014 (UTC)[reply]

Matusuri, thank you for your contribution. However, we're not supposed to copy large amounts of non-free content to Wikipedia (see WP:NFC for details). The paper you quote from is available here, but the fact it's publically available does not mean it's in the public domain. I've hatted your text above. Tevildo (talk) 23:54, 22 October 2014 (UTC)[reply]

I've gone ahead and removed it entirely. Copyright violations should be removed entirely, per Wikipedia's policy. --Jayron32 00:44, 23 October 2014 (UTC)[reply]

It is unclear which of three consecutive questions KM is addressing with his quote from this black hole distribution analysis. -- 190.58.249.7 (talk) 10:51, 23 October 2014 (UTC)[reply]