New answers tagged orbit
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If it can get captured, it can escape: we know this because the law of gravity is time-symmetric. If there is a trajectory that leads to a capture, then by running time backwards, there is a trajectory that leads to escape.
What happens with small objects like 2020CD3, is that they pass into the space between the Earth and moon, and it is the interaction ...
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If we only consider gravity, one answer may be found using the Hill sphere. This is the distance the gravity of a body dominates over the sun: $$r_H \approx a \left(\frac{m}{3M_\odot}\right)^{1/3}$$ where $a$ is the semi-major axis, $m$ the mass and $M_\odot$ the sun's mass.
Now, an actual body has some nonzero density $\rho$ and $m=(4\pi/3)\rho r^3$. If the ...
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It is difficult to work out an actual numerical answer, but let me point out some things that I think would determine the lower bound.
If two objects are close enough, they are attracted by Van der Waals forces. That only operates over a very close range, but it sets a minimum distance before something besides gravity dominates. That is relevant because low-...
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Because you're asking for a spacecraft, there is some wiggle room because spacecrafts can make adjustments. An orbit around the Moon isn't stable both because of proximity to a much more massive body (the Earth) and due to the Moon's unusual lumpiness, but spacecraft can and have orbited the Moon when the orbit of least change is selected around the Moon ...
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Depends what clock you are using and where you are located.
By their clock what you see now happened 13000 years ago in their past.
By your clock you are seeing it now but it did happen in the past.
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As opposed to what @user21 said, the longitude of periastron does depend on the longitude of ascending node (I was surprised by that myself).
In orbital dynamics, there exist the argument of periastron (symbol ω) as well as the longitude of periastron (symbol π [or ϖ but may lead to confusion) ), which is equal to the argument of periastron (ω) plus the ...
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Yes, but only because it is within our galaxy.
For objects which are outside our galaxy, which are not gravitationally bound to us, it is not necessarily true. There are a few twists.
Gravitational lensing is one. Light does not need to take a direct route to reach us. Massive objects will bend light around them. Sometimes this behaves like a lens to ...
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Yes and no.
If we are asking if the light has traveled 13,000 years from that planet to our eyes, then yes.
However, if we imply that the light is 13,000 years old in relation to now, the it's not so simple, because the way we look at these things usually presupposes a concept of simultaneity which we are used to, but does not really exist.
Quoted from ...
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To expand on the other answers a bit; if an alien on this exoplanet were to turn his super-powerful telescope in our direction and zoom in on Earth, he would see glaciers and ice-caps reaching down from the North pole to the Alps and bands of hunter-gatherers chasing caribou along the edges of the ice-sheets. Scan about a bit and he might notice some ...
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In one word, yes. Anything and everything we see, we see the way it was a certain time ago—about 1.3 seconds for the Moon, about 13,000 years for your hypothetical planet. Like @Richyt pointed out, there is no way for information to travel faster than the speed of light.
Because any and all celestial objects gravitationally affect any and all others, orbits ...
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As far as I can see, yes you would be. There is no way to know what is happening there now, as The theory of relativity would not allow for the propagation of information at faster (or slower, as light travels at the speed of light) than the speed of light, thus 13000 years in this case.
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The (x, y, z) position of an object on its orbit with reference to a reference plane are given by:
u = ω + ν
x = r (cos Ω cos u − sin Ω sin u cos i)
y = r (sin Ω cos u + cos Ω sin u cos i)
z = r sin i sin u
(MEEUS, Jean. Astronomical Algorithms, Second Edition, p. 233)
This should help you find what you need.
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Orbits are all ellipses, so when you throw a baseball you change the orbit of the base ball from your circular ellipse to another ellipse.
If you throw it prograde (in the velocity direction), then you baseball will have an elliptical orbit. It will be closest to the Earth at the point at which you threw it, and furthest from the Earth (and higher than the ...
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This answer is addressed at the level of the question for learners not professionals.
Because our solar system planets travel in near circles people imagine that is somehow the natural condition. But it is a tricky question.
First, we should imagine gravity as an upturned trumpet shape and a planet as a ball rolling along such surface. Depending on the ...
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Answers, in order:
Why did the protoplanetary disk start rotating around the star?
Star and disk form around the same time.
The rotation forms because all other orbits will cross the orbit of the disk and collide. Put another way, the rotation disk is the leftover movement of particles after all symmetrical factors have cancelled each other out in collisions ...
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Other answers good, but this question left behind:
Why isn't all the disk material been swallowed by the star?
Because of the excess of angular momentum carried by some particles. They insist on orbiting the star instead of falling directly to it.
There is a complex process called accretion. The disk constantly transfers angular momentum to the outside, ...
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The answer is turbulence. Stars are formed from large gas clouds that collapse under their own weight. As they do so, they become unstable to fragmentation and break up into smaller collapsing pieces. Turbulence in the original cloud means that each of these pieces has its own individual angular momentum and rotational energy, even if the total angular ...
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The prior gas cloud will start with some small random motions and density variations left over from its formation. As a patch starts to contract any small amount of rotation gets amplified as the cloud collapses, because of the conservation of angular momentum.
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There's a few parts to this question so there's more than one answer.
Earth gets knocked a little bit out of its orbit all the time by gravitational influence of other planets in our solar system. Jupiter and Venus are the primary two, but all the planets have some effect. These are called orbital perturbations and they tend to alternate, not add up. They'...
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There is no "correct" orbit. In a two body system, no orbit is special; each is equally valid. If the Earth is perturbed into a different orbit, the new orbit is "just as good" as the original one.
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In general, the answer is no, but I give a possible exception below.
The earth's orbit is pretty darned Keplerian, since the Sun is much more massive than the earth, and since gravitational forces exerted on the earth from other bodies are very small compared to the forces exerted by the sun. For Keplerian orbits, Kepler's orbital elements are stable ...
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In a 2 body scenario: Lets say there is one large body (sun) and a smaller body (asteroid)
The orbit of the asteroid is an ellipse around the sun and completely stable. If you knock the asteroid it will change the orbit to a different ellipse, and that new orbit is also completely stable. The asteroid won't change back to its orginal orbit unless you give ...
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Two rocks placed in space with no relative motion are going to be attracted by gravity, and hit. 3 rocks, placed in space with no carefully rigged symmetry, will likely miss each other, as the gravitational attraction of the additional rock changes their course. Those near misses are the beginning of rotation. Multiply that effect by trillions, and you have ...
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I'll give a more grasp the concept answer if you don't mind. Consider the discovery of the planet Neptune.
Neptune was discovered because it was observed that Uranus wasn't moving as fast as orbital mechanics suggested it should be moving. By calculating how much Uranus had slowed down, an estimate was made for where an undiscovered planet should be and ...
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Tosic's answer summarizes the situation. I'd use different words to say the same thing. These are orbits around the Sun, a little closer or farther than Earths. Due to their closeness to Earth (1.5 million km vs 150 km to the Sun) the Earth's gravity slightly speeds up the outer L2 and slightly slows down the inner L1 object.
The effects still happens if the ...
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Lagrange points are points where the angular speed of a body with negligible mass would be equal to that of the secondary body (which is orbiting the primary), here the sec. body is Earth and the primary is the Sun. This means the period is not a few days shorter due to the Earth acting on the body and decreasing the centripetal force it gets from the Sun. A ...
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What exactly are "non-Keplerian" orbits?
Orbits that don't follow Kepler's laws.
Strictly speaking, all orbits are non-Keplerian. In practice, one can model some orbits as basically Keplerian, but with perturbations. Sun synchronous satellites are one example of orbits that are close to Keplerian, but not quite so. The Earth's equatorial bulge ...
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What exactly are "non-Keplerian" orbits?
Strictly speaking, no orbits are in perfect accordance with Kepler’s laws. Kepler’s laws aren’t really “laws” in terms of physical laws, but are instead trends that Kepler noticed and calculated using astronomical observations of the planets. Kepler’s laws are very accurate for planetary orbits since he ...
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Below is a five-year simulation of fictional circular orbits of Venus, Earth, Mars and Jupiter-like objects and two Earth moons with a period of 0.5 years; one prograde and one retrograde.
Earth's orbit is the ecliptic, all planets have inclinations of 10 degrees, moons are slightly split at 19.5 and 20.5 for visibility.
We see the familiar switch back and ...
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Unfortunately, I couldn't find any infos on this specific case.
But as I remember, there was a similar study from NASA a while ago, observed with the Hubble telescope:
https://hubblesite.org/contents/news-releases/2013/news-2013-22.html#section-id-2
There is a nice animation that show the proper motion of Proxima Centauri relative to background stars (not ...
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A moon with a sixth month period would be about 1.3 million km from Earth. That puts it close to the edge of the Earth's Hill sphere, and probably isn't stable in the long term. So the biggest thing you might notice is "goodbye moon" (and having the moon in an Earth crossing orbit wouldn't be pleasant in the longer term, a collision would melt the ...
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The minimum energy change to the moon's orbit to cause it to impact the earth is 3.2998e28 Joules. After the asteroid hits the moon, it would take 6.2 more days for the moon to hit earth.
We can calculate orbital velocity $v$ of the moon at apogee using the Vis-Viva equation: $v^2=\mu(2/r-1/a)$ , where $r=4.046e8$ meters is the distance between the bodies, $...
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