# Tag Info

26

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 = ... 18 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 ... 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 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 ... 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 ... 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 ... 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. $$The main perturbing influence on Earth's orbit is the gravitational pull of Jupiter, whose mass is about 1000 times smaller than the Sun, and which orbits ... 10 The horseshoe orbit shape does occur only in the reference frame of the Earth’s orbit. It is a manifestation of a third body problem, and the orbit is in an accelerated reference frame. The loop, which is this distended horseshoe shape, has no central gravitational source inside the loop. As a result the orbit is a “pseudo-orbit.” From the perspective of an ... 10 The spiral arms don't mean that the mass is getting sucked to the center. They're just wave-like density patterns. The bodies in orbit around the center of the galaxy are in stable orbit; just like the Earth around the Sun and the Moon around the Earth. What happens is that gravity accounts for the centripetal force (in the orbiting frame, gravity is ... 10 Kepler's 3rd law assumes that the Earth travels in a perfect ellipse with the only gravitational force on it being from the Sun. Further, Kepler's laws are derived from Newtonian gravitation. In reality, the orbit of the Earth is affected by the gravitational pull of other planets, and by the effects of General Relativity and is therefore not quite ... 9 No "large bodies" that I know of. Certainly it is physically possible for something to orbit a moon; lots of spacecraft have been orbited around the Moon and other moons in the solar system. As long as we're simply discussing hierarchies of orbits, the Sun orbits the galactic center and the Milky Way is gravitationally bound to the Local Group. [EDIT: ... 8 The force you experience is of the form \vec{F} = - Gmr\vec{u_r}, and we also know that in the surface, r=R, it is \vec{F}=- gm\vec{u_r}, so$$\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 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 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 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 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 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

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

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 ...

Only top voted, non community-wiki answers of a minimum length are eligible