# Tag Info

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How “large” is a Lagrange point? L1, L2 and L3 are essentially zero size cause they're never stable. They're still useful cause an orbital near L1, L2 or L3 doesn't require a lot of energy to stay in that general area. so we can use L1, 2 or 3 for not quite stable orbits that don't require much energy adjustment. L1 Halo orbits are used too, not ...

4

For this answer I'm going to pretend that the y-axis of the plot you've shown has had $\sim 7000\,{\rm km}\,{\rm s}^{-1}$ subtracted - this is a conventional thing to do in diagrams like the one you show, and simplifies the explanation/interpretation. The plot you show is based on observations. When observing distant objects like Coma, only 3 of the 6 ...

4

Because in the second scenario you described, the exhaust gases from the rocket fly off at high speed into space. This is where the extra energy goes. A better alternative to get circular motion of the satellite with the earth removed, would be a second satellite, connected to the first satellite with a long cable. If, for example, both satelittes have the ...

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The answer is basically, angular momentum. The collapsing proto-solar nebula has some angular momentum. Whilst dissipative processes can allow the nebula to collapse along the axis of rotation, there is still the problem of how to shed angular momentum in order to allow gas/dust to orbit closer to the rotation axis. This is just a basic application of ...

4

Although this may not be what you're looking for... They weren't "simply lucky." In fact, they didn't use Bode's law at all- they used calculations based on Neptune's supposed gravitational effect on Uranus. In fact, had the two used Bode's law, they would never have found Neptune, as the Bode "law" would predict a completely different location. (This is ...

2

I will simplify this problem by assuming that the only forces come from Newtonian gravity and by limiting the masses of the moons to much smaller values than that of the planets, such that the moons exert a much smaller gravitational forces on all other bodies and therefore can be neglected; so the only sources of gravitational forces are the two planets. ...

2

If someone (like superman) could stop the moon from its orbital motion then yes it would fall towards the Earth. Only then the direction of motion would be parallel to gravity. Same with the I.S.S or the satellites orbiting Earth. They could also spiral in and crash because the atmosphere is taking away their kinetic energy. Just like the comment says ...

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Your question is about finding the initial conditions to a circular orbit, which has already been answered very well. This answer instead tackles determining the initial conditions for more general orbits. This will allow you to simulate different orbit shapes. Below are a few shapes of different orbits: There are two parameters that govern the shape: ...

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Firstly, let's define what's on the axes (thanks to Kyle Oman for prompting me to do this): The velocity on the $y$ axis is not the total velocity of a galaxy through space; it is only the component of the velocity along the line of sight (LOS), measured by the redshift of the galaxy. Let's call it $v_z$, although on your figure it's just called $v$. The ...

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Don't worry about relativistic corrections - they are insignificant for most planetary orbital motion at the level you try to model/understand it. You want to look at the vis viva equation which is well explained on Wikipedia: $$v^2 = GM\left(\frac{2}{r}-\frac{1}{a}\right)$$ where: $v$ is the relative speed of the two bodies $r$ is the distance between the ...

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Actually, the orbit of the Earth around the Sun is influenced by all the planets and every other gravitating object in the solar system. But their gravitational influences are relatively small compared to the Sun's, and it becomes computationally unwieldily and even impossible to account for the orbital motions of more than two and at most three mutually ...

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I would choose highly symmetric initial conditions which are easily shown to be periodic. Then I would perturb the system by small steps. For example, four identical particles (of mass $m$) moving on a circle and interacting through Newton gravitation alone is a solution of the equations of motion.  \vec{x}_i = R \left(\begin{array}{c}\cos\left(\omega t ...

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Physically, a negative total energy is a necessary and sufficient condition for avoiding all particles flying off to infinity separately. However, it is always possible for some particles to be given enough energy to escape a system. In fact, this tends to happen in real life. Planets get ejected from their solar systems over millions of years, and stars ...

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The idea of electrons orbiting the nuclus is called the Bohr model and has now been replaced with a quantum model. Electrons exhibit wave-particle duality, which means they sometime act like a wave and sometimes like a particle. They don't actually orbit the nucleus but have areas around it where you are more likely to find them called orbitals. The ...

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If you want to actually simulate the behavior of the planet as it experiences the (vector) force as it moves around, then you need to find a stepping method and write your velocity vectors and position vector in terms of coordinates. I recommend a Verlet velocity method. Others at this site have their favorites, too. Euler's method is not good enough for ...

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