# Tag Info

## Hot answers tagged celestial-mechanics

15

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 ...

14

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 ...

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 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 ...

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 ...

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 ...

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: ...

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

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

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.

5

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 ...

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 ...

4

A large part of the apparent organization of the Universe relates to the hierarchical and largely attractive nature of the four fundamental forces. The aptly-named Strong Force dominates other influences, so quarks are largely bound into baryons, which are neutral with respect to Color charge. The Strong Force also sees to it that baryons, when smashed ...

4

I'm going to assume your window has dimensions of 1x1, in length units. Since it's a video game units don't really matter. The gravity acceleration you're looking for will be a vector with 2 components. $$\langle a_x, a_y\rangle$$ Next, I'm going to say the spaceship is at the origin with the asteroid at (x,y). This just makes it easier. The gravity ...

4

Just wanted to supplement the answers already posted with a few notes re: exoplanet eccentricity. In my understanding, the reason why exoplanets have a median eccentricity ~0.3 vs. almost circular orbits in the solar system is not quite satisfactorily explained just yet (this paper is still my favorite simulation that attempts to address the origin of ...

4

A short answer is that dissipation (e.g. dust, gas interactions with planetessimals) is good at removing energy from a system, but not angular momentum. Circular orbits have the minimum energy for a given angular momentum. For short-period exoplanets, the primary form of dissipation is tidal forces of the star on the planet (similarly, the moon is on a ...

4

Around 4080 BC the Earth was in perihelion during autumn. In 1246 AD the perihelion occurred during the winter solstice. By 6427 AD the perihelion will coincide with the March equinox. Perihelion will occur in April around 7062 AD. (source: Astronomical Algorithms) The question is: Is the Earth's lunisolar precession coupled to its perihelion precession? ...

4

In general, two points and transit time do not define a single orbit. Let us imagine an (non-circular) ellipse with a focus in the origin and the major axis on the abscissa. If we build a second ellipse by reflecting the first ellipse symmetrically with respect to the ordinate, these two ellipses will intersect in two points (A and B) on the ordinate, and ...

4

The effect that you're describing is extremely small. Have a look at the following figure: Here you see the position of the Sun from a location $L$ on Earth. Let's call $R_\oplus$ the radius of the Earth, $\Delta$ the distance between the Earth and the Sun, and $\varepsilon$ the obliquity of the Earth. The angle $\delta$ is the declination of the Sun at a ...

4

To a first approximation distance covered by the Moon is the same as the Earth's, but you can also estimate the correction to first order. Assume both orbits are circular and in the same plane since any deviations will affect only smaller order corrections. Represent the position in the orbital plane as a complex number $Z = R e^{2\pi i (t/Y)} + r ... 3 Two things that we are really good at measuring in astronomy: time and angles. With respect to the sun and other planets, we can measure our relative orientation in the solar system to better than a milliarcsec ($8\times 10^{-10}$of a circle). A year is$3\times 10^7\$ sec, so we know passage of pericenter to about 0.02sec. This precesion can be built up ...

3

So far, this is an unknown question that is the subject of current research. For example: Sean N. Raymond, David P. O'Brien, Alessandro Morbidelli, Nathan A. Kaib, "Building the Terrestrial Planets: Constrained Accretion in the Inner Solar System" To date, no accretion model has succeeded in reproducing all observed constraints in the inner ...

3

I would say that the universe appears organized due to three things: Newton's first law of motion Gravity Time From your comments on other posts, you marvel at the straight lines and orbits, like fractals, that appear on all scales (moons orbiting planets orbiting stars orbiting galaxies orbiting clusters). This is due to the fact that all objects move ...

3

In a word, yes. The concept of the mass driver has this kind of launch in mind. All you need to do is accelerate the vehicle to the escape velocity at the point of launch (in fact, you don't even need to do that - you can simply put the vehicle into a sufficiently elliptical orbit that intersects with the target planet at apoapsis - for infinitely distant ...

3

The black hole gains the angular momentum of the matter that falls into it, minus the anglar momentum of the outgoing gravitational radiation. There are a lot of questions currently on the site regarding the effects of a spinning black hole. The primary one is that the black hole will cause the space around it to co-rotate with the hole.

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