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27

You've used the gravitational constant with only three significant digits. So it's no surprise that your answer isn't accurate to five significant digits. Instead of $G$ and $M_\odot$ separately, you should use the product $GM_\odot$, known as the standard gravitational parameter. Its value is known very accurately: in the link, you'll find $$ GM_\odot = ...


20

The effective gravity inside the ISS is very close to zero, because the station is in free fall. The effective gravity is a combination of gravity and acceleration. If you're standing on the surface of the Earth, you feel gravity (1g, 9.8 m/s2) because you're not in free fall. Your feet press down against the ground, and the ground presses up against your ...


16

This web page has a nice discussion on it: http://archive.ncsa.illinois.edu/Cyberia/NumRel/EinsteinTest.html Basically the orbit's eccentricity would precess around the sun. Classical stellar mechanics (or Newtonian gravity) couldn't account for all of that. It basically had to do with (and forgive my crude wording) the sun dragging the fabric of ...


15

To some extent the universe exhibits something called self-organized criticality where a dynamic, non-linear system with many degrees of freedom (the gas after the Big Bang but before the emergence of structure) eventually forms a system with a notable degree of scale invariance (moons orbiting planets, planets orbiting stars, stars orbiting galactic ...


14

This is not that there is no exact solution, only the exact solutions for $x(t)$ and $y(t)$ use elliptic functions. The problem whether elliptic functions (which are defined by inverse of some integrals) are "good" functions is a bit philosophical one; one can on one hand state that sine is not a real function because one must integrate or sum a infinite ...


13

The Moon moves at about a thousand metres per second, but it's a long way away so it only appears to move slowly. Most of the apparent movement of the Moon is actually due to the rotation of the Earth. We see it appearing to go round the Earth once a day, but it actually takes about 28 days to complete an orbit. The Wikipedia article on the Moon's orbit has ...


13

First, you state a few things that aren't quite right in your question. While the view that's generally talked about is that Phobos and Deimos are likely captured asteroids, dynamically it's a pretty difficult problem (you generally need a third (in this case fourth?) body to take away the extra energy, and it's hard to get a circular orbit around the ...


12

The excitement behind various claims is somewhat excessive. First, the Mayan astronomers, see e.g. Mayan astronomy at this page, didn't use any armillary spheres or sextants as others did. Their observations were made with naked eye and they were depicting positions of planets with crosses. The accuracy of the Venus' position after a synodic 584-day cycle ...


11

It's possible, but it seems like it'd be rare. The planets with the most moons are giant, and very far away from the sun. That means the moons will be very strongly bound to the planet and not get disturbed much by the sun. If our moon had a moon, it'd have to be just the right distance from it that it wouldn't collide with it (the moon's gravity is far from ...


11

There are two elements to why the universe appears to be so orderly: the physical laws of that govern the universe are the same everywhere, and astronomical objects are very, very, very far from each other. Consider two objects, one much larger than the other, and both very far from anything else. Because of gravity (which works the same everywhere), the ...


11

To expand on Prahar's answer, let me run some numbers to try and convince you this is reasonable. Your answer is correct to within one part in 104: $$ \frac{365.256363004}{365.2075}\approx 1.000133795. $$ The main perturbing influence on Earth's orbit is the gravitational pull of Jupiter, whose mass is about 1000 times smaller than the Sun, and which orbits ...


10

Kepler's 3rd law assumes that the Earth travels in a perfect ellipse with the only gravitational force on it being from the Sun. Further, Kepler's laws are derived from Newtonian gravitation. In reality, the orbit of the Earth is affected by the gravitational pull of other planets, and by the effects of General Relativity and is therefore not quite ...


10

The horseshoe orbit shape does occur only in the reference frame of the Earth’s orbit. It is a manifestation of a third body problem, and the orbit is in an accelerated reference frame. The loop, which is this distended horseshoe shape, has no central gravitational source inside the loop. As a result the orbit is a “pseudo-orbit.” From the perspective of an ...


10

The spiral arms don't mean that the mass is getting sucked to the center. They're just wave-like density patterns. The bodies in orbit around the center of the galaxy are in stable orbit; just like the Earth around the Sun and the Moon around the Earth. What happens is that gravity accounts for the centripetal force (in the orbiting frame, gravity is ...


9

No "large bodies" that I know of. Certainly it is physically possible for something to orbit a moon; lots of spacecraft have been orbited around the Moon and other moons in the solar system. As long as we're simply discussing hierarchies of orbits, the Sun orbits the galactic center and the Milky Way is gravitationally bound to the Local Group. [EDIT: ...


9

In his lecture The Cosmic Distance Ladder (video), mathematician Terence Tao describes the history of how mankind has successfully mapped the solar system and beyond. In particular, he describes why Copernicus put the sun in the center (reason: he discovered that the sun is dozens of times bigger than the earth) and how Kepler found his laws of planetary ...


8

The force you experience is of the form $\vec{F} = - Gmr\vec{u_r}$, and we also know that in the surface, $r=R$, it is $\vec{F}=- gm\vec{u_r}$, so $$\vec{F} = -gm\frac{r}{R}\vec{u_r}$$ This is a conservative force that can be derived from a potential $$U = \frac{1}{2}gm\frac{r^2}{R}$$ Because this is a central force, angular momentum will be conserved, ...


7

The main problem with being near a large planet are the tidal forces produced by its gravity. The Moon is pretty small compared to the Earth, and it's a quarter of a million miles away, but it still produces large movements in sea level i.e. the tides. If you imagine the moon getting bigger and closer the tidal effects would increase. The land feels the same ...


7

This was previously a comment to space_cadet's answer but became long (down-vote wasn't me though). I don't understand space_cadet's talk about unstable orbits. Recall that two-body system with Coulomb interaction has an additional $SO(3)$ symmetry and has a conserved Laplace-Runge-Lenz vector which preserves the eccentricity. Because interactions between ...


7

Mercury's orbit is elliptical. The orientation of this ellipse's long axis slowly rotates around the sun. This process is known as the "precession of the perihelion of Mercury" in astronomical jargon. It's a total of 5600 arcseconds of rotation per century. The precession is mostly a result of totally classical behavior; almost all of the movement of the ...


6

Geostationary satelites are essentially for ever. This is becoming a problem since there are a limited number of places you want to put a geostationary satelite and most of them are full. Any collisions/explosions in geostationary mean debris will also stay there for a long long time. For low earth orbit satelites it depends on their shape, altitude and the ...


6

Forgetting about the specifics of your problem, you say you want to work in the Newtonian regime for gravitation on a toroidal space. The way this differs from a non-toroidal space is that you can "unroll" the torus into an infinite lattice of duplicates. This is a lot like the lattice of mirror charges if you were doing electrostatics on a torus (the ...


6

Beauty is in the eye of the beholder. So is order. What is a gigantic orderly mechanism for you, might be a horrible entropic mess for me. And we'd both be right.


6

The direction in which the celestial bodies are located is seen directly in the telescopes (two angular coordinates); the distance from the Earth is the hard part. For planets, we may measure the distance by waiting for a radar echo. Aristarchus has invented a useful method to get a third piece of information: when the Moon is exactly half-full, ...


6

I think it is basically a coincidence at the current time. Earth's axis of rotation precesses with a period of about 26,000 years, and according to Wikipedia, its orbital axis precesses with a period of about 112,000 years. So the winter solstice and perihelion will have all possible relative phases over a long time period.


6

Century First, you'd have to watch through a night to see if Polaris wobbles - currently, the radius is about 1° I think, but that changes with precession (and nutation, but that's small enough to ignore). Once you know that, you can try to find a point in the sky that stays still all the time (like Polaris nearly does in our time). This is celestial pole, ...


5

This is likely stretching your requirements a little bit, but I find this related method ingenious and surprising. You can actually find your longitude as well as the other data from @Pharoh's answer if you can remember very well how the night sky looked at your original position on Earth. The method is the Lunar Distance Method (see Wiki page with this ...


5

A small portion of any smooth curve looks the same as a small piece of a parabola in the limit. Choose a coordinate system so that the tangential direction in the middle of the segment is along the $x$ axis and choose a translation for the middle of the segment to sit at $(0,0,0)$, the origin of the coordinates. Then $y,z$ on the curve (ellipse etc.) may be ...


4

This is possible. For example, this post provides a good explanation of the math involved. The key point is whether an object lies in the planet's Hill sphere or the moon's Hill sphere: Can the Moon have a moon? Yes, the Moon could have a sub-satellite. If we look at a system of the Earth, Moon, and a sub-satellite, the same idea as above ...


4

The faster you go, the less velocity you theoretically can gain from a gravity assist. The reason for this is that the faster you go the harder it is to bend the orbit. To proof this we have to use the patched conics approximation, which means that while within a sphere Kepler orbits can be used. The sphere can be simplified to be infinitely big, since the ...



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