# Tag Info

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As ganbustein says, this is not too difficult to imagine. The simplest case (approximating with circular orbits and only the Sun, Earth and Satellite) would have the satellite orbit orthogonally to the Earth with a 1 year orbit. The Satellite will pass the Earth orbit plane in two places, call these "down crossing" and "up crossing" points. To minimize ...

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Following the convention $\mathbf{r} = \mathbf{r}_2-\mathbf{r}_1$, $M = m_1+m_2$, in the center-of-mass frame we have, by definition, $$\begin{eqnarray*}\mathbf{r}_1 = -\frac{m_2}{M}\mathbf{r}\text{,}\quad&\mathbf{r}_2 = \frac{m_1}{M}\mathbf{r}\text{.}\end{eqnarray*}$$ Hence, $\ddot{\mathbf{r}} = -GM\hat{\mathbf{r}}/r^2$ implies that the individual ...

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According to wikipedia and other sources, a planet and a star always move in a circular orbit around the common center of mass of the both bodies ... This is not true. In the absence of other gravitational sources, a planet and a star move in elliptical orbits about the common center of mass. Ancient scientists assumed circular orbits, but only because ...

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The trajectory of Earth's orbit is shaped not only by Sun's gravity. Earth's orbit is being changed by different sources of gravity as well, for example center of our galaxy. This is why it's not perfectly round.

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Sure. Imagine an object orbiting the Sun in a plane normal to the plane of Earth's orbit. Let AB the the intersection of these two planes. (Imagine that A and B are points on the celestial sphere. We're not interested for now in distance from the Sun.) Then the only time the Sun hides the Earth from the object is when the object is at A and the Earth is at ...

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I could imagine a polar orbit that is in a plane which rotates to stay orthogonal to the direction of the Earth. The rotation would be very slow and match the revolution rate of the Earth. The satellite path would be similar to how a ball of yarn is wound. There's nothing too fancy about this (you could do the same switching Earth and the Sun), so I ...

1

You must apply Newton's law $$m\frac{\mathrm{d}\boldsymbol{v}}{\mathrm{d}t} = \boldsymbol{F}$$ which related the force $\boldsymbol{F}$ to the acceleration (=change of velocity). Note that positions, velocities, and forces are all vector quantities. Also note that as the objects move (change position), their mutual forces change (both direction and ...

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David Hammen wrote People are using planetary equations coupled with geometric integration techniques... You could also try (what I call) a simple finite-step simulation using Newton's laws to operate on object masses, positions, velocities and accelerations. I'm not sure if this falls within what David calls "geometric integration techniques". My ...

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The only truly confocal elliptical orbit is that of a bound test particle in the central potential $-k/r$ or, equivalently, that of two point-like (with spherically symmetric internal mass distributions) masses attracting each other with Newtonian gravity (and having negative total energy, i.e. being bound to each other). Everything else is non-elliptic ...

3

A good starting point would be <insert name of some scientist from long ago> planetary equations of motion. For example, there are Lagrange's planetary equations (sometimes called the Lagrange-Laplace planetary equations), Gauss' planetary equations, Delaunay's planetary equations, Hill's planetary equations, and several more. The common theme amongst ...

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It's mere coincidence. In about 10,000 years, perihelion will coincide with the northern hemisphere summer solstice. There are two precessions of interest here. One is the 26,000 year period axial precession of the Earth's rotation axis, caused mostly by the Moon and the Sun; the other is the 112,000 year period apsidal precession of the Earth's orbit, ...

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You must solve the Kepler's equation: $M = E - e sin E$ (where $M$ is the mean anomaly, $e$ the eccentricity of the orbit, $E$ the eccentric anomaly). As you can see, this is a transcendental function, so you will probably need to iterate through a calculator, in order to solve it. Many of useful formulas to understand the parameters dependence are ...

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I'm not sure how much this will help, but I can give you this passage from Wikipedia: Under ideal conditions of a perfectly spherical central body, and zero perturbations, all orbital elements, with the exception of the Mean anomaly are constants, and Mean anomaly changes linearly with time[dubious – discuss], scaled by the Mean motion, ...

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