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What you're describing is basically the "collapsed star" (Eng) or "frozen star" (Rus) interpretation of black holes that was common prior to the late mid-1960s. It was a mistake.
Suppose you are distant and stationary relative to the black hole. You will observe infalling matter asymptotically approaching the horizon, growing ever fainter as it redshifts. ...
27
Yes, you are absolutely right, from OUR VIEWPOINT it does.
From Kip Thorne's book "Black Holes and Time Warps: Einstein's Outrageous Legacy."
“Like a rock dropped from a rooftop, the star’s surface falls downward (shrinks inward) slowly at first, then more and more rapidly. Had Newton’s laws of gravity been correct, this acceleration of the implosion would ...
21
No. In fact the opposite is the case.
(See the last paragraph for an intuitive explanation.)
It is a common misbelief that galaxies receding faster than the speed of light are not visible to us. This is not the case; we easily see galaxies moving at superluminal velocities. This does not — as I think most people would think — contradict the theory of ...
18
(I will assume a Schwarzschild black hole for simplicity, but much of the following is morally the same for other black holes.)
If you were to fall into a black hole, my understanding is that from your reference point, time would speed up (looking out to the rest of the universe), approaching infinity when approaching the event horizon.
In Schwarzschild ...
11
We need to think about just where the time dilation effect occurs. By then thinking about the observations from each point of view, that is the free falling object and the external observer, we can come to terms with just what is happening as opposed to what appears to be happening.
The experience of time
We must remember that an object moving at a certain ...
8
It moves in circles. At the poles there is no direction defined (east, west, north or south).
This is how the sun moves at the north pole: http://www.jaloxa.eu/resources/daylighting/docs/sunpath_90_north.pdf . You can compare it to the other latitudes to get an idea. (http://www.jaloxa.eu/resources/daylighting/sunpath.shtml)
7
As time passes, there are galaxies that are currently not in the observable universe which will become observable But this is not a sudden winking on. Instead, over hundreds of millions of years we will see a proto galaxy evolve into a mature galaxy.
For example there is a "blob" of hydrogen that some interpret as being the accretion of hydrogen onto a ...
6
Yes - here is a good explanation of the situation. Just as it's possible to have continuous daylight for days or weeks on end, you can have continuous moonlight too.
The Moon's orbit is tilted at (a maximum of) just over 5 degrees to the ecliptic, however this tilt has an oscillatory precession with a period of 18.6 degrees. This precession means that our "...
6
The logical consequence is, that an event horizon cannot form, since the first particle slows down asymptotically to zero, just before the event horizon forms (Fermat's infinite descent).
The emergence of the event horizon therefore takes infinite time seen from outside. But due to Hawking radiation a black hole exists only a finite time. Hence an event ...
6
Several wonderful yet technical answers have been given, and I cannot add anything to those very nice answers that explain why it is not useful to think black holes get "frozen" at their event horizons. But I can give an answer with a more essentially useful philosophical perspective, which is that the central lesson of relativity is that reality involves a ...
5
We are wondering if the moon wobbles back and forth along the horizon the same way the sun wobbles.
Yes, but over a period of 27 days compared to a full year for the Sun.
If you visualize a compass, the angle that a star rises and sets is the same angle measured from due south (for those in the northern hemisphere). This figure shows a star rising 60 ...
4
Thought provoking cosmologists!
I'm uber late to this discussion as I see it has been ongoing for literally years and don't know if there is still anyone monitoring this thread, but here' goes.
I studied astrophysics at UC Berkeley in the late 80's so perhaps my info is a little dated, upfront apologies if so. I spent a lot of time thinking about this ...
4
Your reasoning is correct: if Mercury orbited in the same plane as Earth, we'd see it transit the Sun every 4 months or so. In fact these orbital planes are inclined 7 degrees to each other, and the other major planets' orbits are inclined 1 to 3 degrees relative to Earth's.
The planets perturb each other's orbits slightly, so no planet's orbit is perfectly ...
4
The amount of atmosphere that light from outside Earth has to pass through before it reaches the ground is given by the air mass, $X$. It is normalized to unity for a source directly overhead (at "zenith"), and increases as the source "sets".
If you don't look too close to the horizon, you can use the formula $X = \sec z$, where $\sec \equiv 1/\cos$, and $z$...
3
Let's suppose, that moon is always located on the ecliptic. Ecliptic ranges from declination $\delta=-23.5°$ to $\delta=23.5°$. That means, that the maximum apparent height of the Moon in Calgary will be $90°-\phi+23.5°=90°-51.04°+23.5°=62.5°$
But moon isn't always located on the ecliptic. Its inclination is around 5.14°. That means, that it can deviate from ...
3
Your explanation is correct, so you are not remembering correctly. The farther north you go, the lower the sun is to the horizon at noon.
Of course it depends on how far north you go. If you only travel 5° farther north in latitude, then the sun is only 5° lower in the sky. 5° is a "long" ways to travel on the earth, but in the sky 5° lower is not that ...
3
Yes, all of those contribute to the total irradiance, which is the amount of sun power falling on a particular area, measured in Watts per square meter. You can imagine a 1m$^2$ "window" perpendicular to the sun's rays - excluding atmospheric and weather effects, the amount of sunlight passing through that window never changes. But when the sun is directly ...
3
An approximation of the Sun’s declination for a certain date is:
sin δ = 0.39795 ⋅ cos [ 0.98563 ⋅ ( N – 173 ) ]
where N is the number of days since January 1.
That, combined with the formula given by @user21, viz.:
$ a_{min} = \lvert \phi + \delta \rvert - 90° $
will give you the dates where the Sun is at least 18° under the horizon.
(Oddly enough, a simple ...
3
The distance depends on the diameter of the planet and of your height above the surface (such as on a mountain). The greater the diameter, the farther away the horizon will be.
You can see this in the figure below. $d_3 > d_1 > d_2$. On a large planet your horizon (at distance $d_1$) will be farther away than on a smaller planet with horizon distance $...
2
As I understand it, the presence of an event horizon (EH) from gravitational collapse is a case in which GR violates local causality in the outer (w.r.t. EH) universe. By Birkhoff theorem the EH can only be caused by the inner T, not by whatever is outside the EH. The (collapse) EH yields a causal disconnection: the outside is not affected by what is on or ...
2
No.
Photons can't come from the singularity of a black hole, or from beyond the event horizon.
While some very good candidates for black holes exist, none have been certainly observed.
We have never observed Hawking radiation: it remains theoretical.
We have observed radiation from the accretion disk of likely black holes, however the matter in this disc is ...
2
First see the plethora of diverse answers to Where can I find the positions of the planets, stars, moons, artificial satellites, etc. and visualize them?
Then if you can use Python or would like to learn, consider using Skyfield for everything!
Also have a look at Astronomical Algorithms in this very long answer:
This is something you'll have to dig into a ...
2
We are wondering if the moon wobbles back and forth along the horizon the same way the sun wobbles.
Yes, but over years even more than the Sun does!
Per answer(s) to Does the Honey moon have a precession problem? and others we can see that the Moon mostly follows the Sun, so in the summer when the Sun is high at noon it rises and sets north of east and ...
1
Have a look at https://www.pveducation.org, specifically section 2.4 on Terrestrial Solar Radiation. It describes the equations required and has some online calculators and interactive plots.
If you wish to calculate in your own code, I'd recommend the pysolar Python package: https://pysolar.readthedocs.io/en/latest/. It has methods for taking the latitude, ...
1
What I think you're trying to do is workable, but approximate. You first find the Sun's geographical position (GP), the point on the Earth where the Sun is in zenith at the time of the observation. You start by finding the Sun's GP latitude (its declination). It's approximately 23 degrees times the sine of ((days since the vernal equinox)times 360/365). Then ...
1
I am not sure how your observations are so mistaken, but this is exactly what happens. In fact at high latitudes, there will be parts of the year when the sun never rises above the horizon. Try going north of the Arctic Circle in winter and you'll see what I mean.
All I can think is that either you have not travelled very far north or south, or you have ...
1
It sounds like your question is about the single most important principle of cosmology, called the cosmological principle. This asserts that as long as all our observations are consistent with the idea that the universe is the same everywhere at a given age, we will continue to use a model with that attribute. There are lots of good reasons for adopting ...
1
You're right to mention time dilation, because the point of view is important. The standard metaphor is that of an astronaut (A) falling towards the event horizon, wearing a watch, while another astronaut (B) watches from a distance. Astronaut A will simply fall through the horizon and hit the singularity, but B will never actually see A cross the event ...
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