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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 ...


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 ...


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 ...


3

Short answer: They don't. Contact binaries are a possible evolutionary phase of close binary systems, where both stars fill and/or overflow their Roche limit, i.e. 'kiss' each other at the L$_1$ point. As a consequence, they lose matter to a common envelope. This exerts a drag onto the stars and leads to further shrinking of the binary, which eventually ...


3

I agree with Walter, they don't. However, in addition to the common envelope drag and mass exchange a very important feature of their evolution is the loss of angular momentum through a magnetised wind. The loss of angular momentum from the orbit also leads to orbital shrinkage and closer contact, until presumabaly at some point they truly merge. This ...


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 ...


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

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: ...


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 ...


2

You go wrong when you say "acceleration is work". Accelerating an object only sometimes requires work. Work is only done by a force when the object moves in the direction of the force. If you know what the dot product is then the relevant formula is: \begin{equation} W = \int_{t_0}^{t_1} \vec{F}\cdot \vec{v} dt \end{equation} If you don't know what the ...


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. ...


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 ...


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 basic answer is that there is no such orbit that can survey the entire surface of an irrotational planet in the same manner as satellites do for Earth. Conservation of angular momentum dictates this: the satellite's plane of orbit cannot change without a torque acting on it. Although there is a force acting on the satellite (the gravity from the ...


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 ...


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 ...


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 ...


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 ...



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