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1

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 ...


0

Galaxies with low total (kinetic + potential) energy have low velocities, so they're found close to the center of the cluster (i.e. to the left in the plot, at middle height), since they don't have the energy to achieve large distance. Galaxies with high total energy may either orbit slowly at large distances (i.e. they're found to the right in the plot, ...


1

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 ...


1

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 ...


0

So an electron can only orbit a nucleus where its wavelength makes a standing wave, leading to discrete energy levels in atoms. That is the Bohr model which has been superseded by quantum mechanics. In quantum mechanics the electron occupies definite energy levels which arise because of the potential well that is generated by the nucleus. As the other ...


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Well, you answered your question partly yourself. Only a finite amount of states (= energy and wavelength of the electron) are allowed, this is the basis of quantum mechanics. If an electron is in a certain state (ie. it has a certain energy and is in a certain orbital), you can then try to calculate its energy and wavefunction and from that its momentum (as ...


0

Please completely abandon the idea of an electron orbiting a nucleus. That is only believed in the now long defunct Bohr model of the hydrogen atom. In modern Quantum Mechanics, in order to describe the electron in a hydrogenic (mono-electronic) atom, we need to solve the atom’s Schrödinger Equation which yields the wave functions ...


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. ...


0

I suspect that they were lucky that their predictions agreed with reality so closely, but any prediction was going to have Neptune roughly (perhaps very roughly) in the same direction as Uranus, during the times when it affects Uranus the most. So I suspect their calculations meaningfully ruled out large swathes of sky, which improved odds of finding it.


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 ...


1

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 ...


0

Maybe a better distinction to make would be between rotational motion and orbital motion. Even in that more generalized case (orbital could be something other than circular), the properties used to describe and analyze the motions are the same: axis of motion, angular momentum, moment of inertia, kinetic energy, torque, etc. There is not a bold line of ...


0

The main difference between these types of motion is that circular motion is a special case of rotational motion, where the distance between the body’s centre of mass and the axis of rotation remains fixed. Rotational motion is based around the idea of rotation of a body about its center of mass. In rotational motion, the axis of rotation and centre of mass ...


5

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 ...


1

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 ...


1

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 ...


0

At any given moment the thrust is equal to the force experienced by the rocket at that moment. That's the definition of thrust. To get the equation of motion right, you need to consider the instantaneous force (variable) and mass (also variable). The acceleration is then given by $$a(t)=\frac{F(t)}{m(t)}$$


0

Orbital velocity generally is the pre-requisite velocity necessary for a satellite to stay in orbit. It is the minimum velocity necessary to keep the satellite from falling into the planet. The shorter the radius of orbit, the closer the satellite is to the planet which it orbits, and the greater the gravitational attraction which must be overcome for it ...


1

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 ...


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 ...


4

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 ...


0

Is it correctly understood that energy is continuously put into the system, in order to maintain the orbit? And that gravity is thus an infinite source of energy? Consider the general case of one particle orbiting another in an ellipse (this is general because we can a always reduce the two body problem to an effective one body problem, and the general ...


0

I think you need to read up a bit on vectors and scalars, how we define things like acceleration. Vectors are physical quantities that have a magnitude and a direction. Velocity is a vector (speed is its magnitude). Acceleration is also a vector. Acceleration is defined as the rate of change of velocity, basically it measures how velocity changes with ...


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 ...


2

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: ...


0

I've been out of school too many years to relearn Lagrange interpolation by hand but Wolfram Alpha will interpolate 5 data points before it breaks. Using the 5 most recent data points in http://spacemath.gsfc.nasa.gov/earth/6Page58.pdf this gives an aproximation of y=-3.2957×10^-9 x^4-5.02018×10^-6 x^3-0.00147727 x^2-0.147798 x+365.25 where y is the ...


2

For common center (barycenter) orbits, the velocities will be \begin{align} v_1&=\sqrt{\frac{Gm_2r_2}{\left(r_1+r_2\right)^2}}\\ v_2&=\sqrt{\frac{Gm_1r_1}{\left(r_1+r_2\right)^2}} \end{align} which, since $m_1=m_2$ and $r_1=r_2$, will be the same value, $v\approx0.22$ for your values of $G,\,m,\,r$. Since you've placed the two objects along the $x$ ...


2

I assume that you want both particles to be free to move, each in a circle. That circle should be about the center of mass of the system, which will be at $$\frac{m_1\vec{r}_1+m_2\vec{r}_2}{m_1+m_2}.$$ The radius, $R$ of each circle will be the distance from the mass to the center-of-mass point. Calculate the Newtonian gravity force on a particle, set it ...



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