90

Summary There's a 1 in 500 billion chance you're standing under a star outside the Milky Way, a 1 in 3.3 billion chance you're standing under a Milky Way star, and a 1 in 184 thousand chance you're standing under the Sun right now. Big, fat, stinking, Warning! I did my best to keep my math straight, but this is all stuff I just came up with. I make no ...


15

In short: no one knows for sure, but currently it looks that the probability is 1. Longer: On our current understanding, the Universe is probably infinite in space. This depends on the recent WMAP satellite results, which have shown a zero curvature of the Universe below measurement precision. The other two options were a positive curvature (thus, we would ...


14

First of all, at that distance seeing the Moon and seeing the Earth amounts to the same thing. At its closest, Saturn is around 3000 times as far from Earth as the Moon is, so viewed from Saturn, the Moon is never more than about a minute of arc away from the Earth. If you can see Earth from the North Pole of Saturn, you can see the Moon, also. (Though it ...


13

Well yes, because Brown & Batygin (2021) say the perihelion would be $300^{+85}_{-60}$ au, so there is roughly a 50% chance that the perihelion is smaller than 300 au according to their work. The quoted error bars represent their estimate of the 16th to 84th percentiles of a probability distribution, so conceivably the perihelion could even be a little ...


13

I am assuming that by adding 1 km/s, you mean increasing the tangential speed of the Earth by 1 km/s. This would increase both the kinetic energy and the angular momentum. This is a relatively small increase in both quantities and it would send the Earth into a slightly more eccentric orbit than it has now. It would not come close to Mars. Using the vis-viva ...


12

Yes, if you observe Earth and the Moon at a favorable time. Near a Saturn summer solstice, e.g. between 2012 and 2022, Earth appears well above the horizon from Saturn's north pole. If the planet body is out of the way, so are the rings. The other observability issue is Earth's elongation from the Sun. This reaches a maximum of about 6$^\circ$ at intervals ...


9

tl;dr - The main measurable effect may be minor climate cooling, but in day-to-day life, the only difference would be that we see a cool, bright explosion in the sky, and eventually, Orion becomes "incomplete". The effects would likely be quite minimal. What Will Happen When Betelgeuse Goes Supernova? by Corey S. Powell, former editor in chief of Discover ...


9

For these kind of questions, you might want to use Stellarium, a free open source planetarium. You can specify the location of the observer on many celestial bodies, including Saturn. Any time between 2011 and 2023 With this tool, you can see that the moon will be in the northern saturnian sky non-stop between 2011 and 2023. You can also see that the moon ...


8

Does "overhead" mean over the center of your head, or over some part of your head? If we assume the latter, it changes the problem! I don't want to recapitulate all MichaelS's lovely work above, so I'll do a quick back-of-the-envelope calculation borrowing from his numbers. The area of a human head as viewed from above (or below) is, umm, let's see, ...


6

Using calculations from here: https://www.vanderbilt.edu/AnS/physics/astrocourses/ast201/orbitalvelocity.html New semi-major axis $$ a = \frac{150000000000 \cdot 0.0000000000667 \cdot 2\cdot 10^{30}}{2 \cdot 0.0000000000667 \cdot 2 \cdot 10^{30} - 150000000000 \cdot 30000 \cdot 30000)}\\ = 151.821\,\mathrm{million\,km} $$ New orbital period $$ p = \frac{\...


6

The difference is that your analysis is assuming that the albedo stays fixed, so the surface temperature simply scales like luminosity to the 1/4 power. The Wiki entry is including feedback from the greenhouse effect, which will tend to further increase the surface temperature. Note that an analysis that just looks at solar irradiation would get way too ...


5

Lithium, along with Hydrogen and Helium, was one of the 3 elements created in the Big Bang. Thus, it should exist to some part in any star that hasn't burnt all of it out, and as mentioned, it's not an easy thing to do. Population III stars are expected to contain Lithium, and Beryllium as well. The amount, however, is not particularly high.


5

If we have two masses $M$ and $m$, which are at distances $r_1$ and $r_2$ respectively from their barycentre, then $$Mr_1 = Mr_2$$ Let $d$ equal the total distance between the two bodies. That is, $d = r_1 + r_2$. Then $$r_1 = \frac{dM}{M+m}$$ where $r_1$ is the distance of the barycentre from the body with mass $M$. The paper linked in the comments suggests ...


4

The proposal is that the apparent planet 9 is, in fact, a primordial black-hole, with mass comparable to a planet and a diameter measured in centimeters. Such an object would be almost undetectable. Its orbit would be identical to an equivalent sized planet. In a two body system, obeying Newtonian gravity, both bodies will follow elliptical orbits, with the ...


4

Probably, maybe. There are at least two ways of answering the question. One is to ask what were your coordinates when you wrote the question and exactly what time it was. Then we'll need to draw a line in a model to see what you hit and whether any of those hits are stars. This assumes a complete map, which is a problem. The answer is different for everyone ...


2

The sun couldn't. The sun gives off light and a solar wind. It doesn't emit lumps of solid matter. It is possible for a small body, such as a comet to get a slingshot around a planet and be ejected. In the early solar system, when the planets orbits were unstable it is possible that larger bodies were ejected from the solar system. It is thought that rogue ...


1

Using this equation from Wikipedia for the approximate merger time: $$\frac{5}{256}\frac{c^5}{G^3}\frac{r^4}{m_1m_2(m_1+m_2)},$$ and Wikipedia's values for the Moon's equatorial radius and mass, and the launch mass of the HST, we can use this query in Google: ((5/256) * (c^5/G^3) * ((100 miles) + (1738.1 km))^4 / ((11110 kg) * (7.347631E22 kg)^2)) to perform ...


1

Maybe, Harvard scientists have proposed a way to determine, once and for all, whether Planet Nine actually could be a black hole. Specifically, the new method would scour the outer solar system for evidence of telltale flares that are emitted when a black hole devours a comet or other distant object. Such flares, they say, should be detectable by the ...


1

So how much do you know about drag? Orbits in atmosphere are short lived - as reference see any low earth orbit satellite. They need regular boosts or else they fall in. So your objects could orbit... Briefly. But not in the way you have drawn in the diagram, and not for any length of time. Basically it's a no, in any real terms.


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