# Tag Info

## New answers tagged orbital-motion

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I think the questions are meant not only to test your creative intelligence, but to see if you have enough desire to do some basic research. Some facts are omitted. For instance, when did Halley's comet pass through perihelion, or alternatively what is its orbital period? After you find the orbital period, you know that the comet must go through 2 pi ...

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A circle is a degenerate ellipse, and you can also think of a circle as having two foci (on top of one another) as the eccentricity approaches 1. The foci of conic sections in general originate from the approach in which the curves are defined - using a focus (point) and directrix (line). This approach leads to rational parametric expressions for the conic ...

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The equation of an ellipse whose semi-major and semi-minor axes are parallel to the x and y axes is given by: $$(\frac{x-h}{a})^2+(\frac{y-k}{b})^2=1$$ (where $a$,$b$ are the lengths of its semi-major and semi-minor axes respectively.) A focus, $c$ is defined by $c^2=a^2-b^2$, and therefore there can be two foci at a distance of $(a^2-b^2)^{1/2}$ on both ...

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The foci are simply points that define the ellipse by the relation $c^2 = a^2 - b^2$, where $c$ equals the length of each one of the foci to the center and $a$ is the length of a focus to the end of the ellipse. For a circle, $a$ = $b$. Given any two foci, a point on the ellipse is a point that is equal I he sum of he lengths of the foci.

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Simple geometry says no. Consider a projectile which is launched from the earth's surface at some angle other than vertical, ignoring the effects of air friction. Once the launch force is terminated, if the object is travelling less than escape velocity, the object will assume a closed elliptical orbit around the center of the earth (well, OK, around the ...

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The mean radius of the earth's orbit has not changed perceptably in the last 1000 years, or even in the last 10,000 years. In the long run, though... The dominant mechanism in the long run is the solar wind, which is estimated to cause a mass loss of 7-9 x 10^-14 per year. As solar mass is lost, conservation of angular momentum causes the earth's orbit to ...

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In a classical context, LRL vector is conserved only for potentials behaving like $\frac{k}{r}$, indeed we can see the general construction of LRL vector : \begin{eqnarray} \frac{d\vec{p}}{dt}\times \vec{L} &=& -\partial_r v(r) \frac{\vec{r}}{r} \times \vec{L}, \nonumber\\ \mu r^3 \frac{d\hat{r}}{dt} &=& ...

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In the case of radial freefall is from rest at some initial $r=R$, the motion will be periodic if you treat the gravitating body as a point-mass and ignore collisions. Since the radius is strictly positive, it makes sense to substitute $$r = R\cos^2\left(\frac{\eta}{2}\right) = \frac{R}{2}\left(1+\cos\eta\right)\text{.}$$ while conservation of specific ...

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The other answers here are in the spirit of what you can do, but allow me to elaborate a little more. To understand if the trajectory of the movement under a potential $V$ is stable or not you have to understand what this stability means. The most simple example is the harmonic oscillation- $V=-{1 \over 2}kx^2$.In Newtonian mechanics, for a point of ...

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Once you have the value $$dV_{eff}\over dr$$ This is the minus of the effective force i.e.: $$F_{eff}=-\frac{dV_{eff}}{ dr}$$ If we differentiate this again we get: $$\frac{dF_{eff}}{dr}=-\frac{d^2V_{eff}}{ dr^2}$$ If this is negative then at any small permutation around that radius will feel a force back to that radius and thus it is a stable orbit. If it ...

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One method for stability anayisis in a potential is to consider $r$ and $r+\epsilon$ where $\epsilon$ is a small number. If you can solve the dynamics analytically or run a numerical computer simulartion for your object with both $r$ and $r+\epsilon$ and observe with time if $\epsilon$ gets smaller or larger that will give you a clue as to whether the orbit ...

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Assuming that you have some sort of slow, continuous drag on the object, then velocity at any point is equal to the velocity of a circular orbit at that point. The object doesn't have a single speed from the start to the end of your graphic, but a slowly increasing one. As the drag occurs, the slowing of the object and the loss of altitude happen ...

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Angular momentum is a conserved quantity (in a closed system) and this is true also for the angular momentum that is carried by the electromagnetic (EM) field. This conservation is a manifestation of rotational symmetry and the azimuthal part of the EM field emitted must be single valued. In other words, when rotating the EM field in the azimuthal ($\phi$) ...

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Even if the original dust cloud only had a relatively small angular velocity (which it might have had for all sorts of reasons), the process of collapsing would have amplified it. That is, the collapse process preserves the angular momentum, but it translates to a much larger rotational speed in the newly-collapsed system. Think of what happens to a spinning ...

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All the stars would be attracting each other and hence the would stick to each to attain equilibrium. Why doesn't this happen? You are forgetting angular momentum. Consider a binary star pair. Ignoring the expansion of spacetime, and in the absence of some mechanism that removes angular momentum from the system, those stars will orbit one another ...

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All the stars would be attracting each other and hence they would stick to each to attain equilibrium. Why doesn't this happen? This is an old question. Even Newton himself had thought about this question. His idea was that in long distances or separations (say, inter-galactic distances) the force of gravity might appear to be repulsive. That's why not ...

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If there would be initial rotation(as we see most of all objects today are rotating around another) gravitational force is accounting for the centripetal force. I'm new here so, I don't know how to type the equations, but I hope you get my point. Further many objects are there which have many other forces, like Coulombic force(when charged bodies are ...

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If you plot the path of the moon with respect to the star you should get a cycloid pattern averaging around the path of the planet. That's what our moon does. The planet and the moon are doing a gyrating dance about their center of mass, with the moon doing the most motion (because it has the smallest mass), while the planet/moon duo orbit the star. I've ...

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The leading theory for formation of the Moon is that a large object collided with the Earth and threw off a cloud of debris. This cloud then clumped together under it's own gravity to form the Moon. Simulations of the cloud show it formed outside the Roche limit - at around 1.3 times the Roche limit in fact. So the Moon has never been closer to the Earth ...

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The formula you've referenced, $\vec a_c = -\omega^2\vec r$, must define $\vec r$ as a function of the angular position, which depends on time (uniform motion and all). Something like $\vec r(\omega t) = r(\hat x \sin \omega t + \hat y \cos \omega t)$. $\omega$ essentially represents the angular speed, which means for constant $\omega$ you have an orbital ...

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First, let us look at the current rate at which the moon slows down. I have a few different sources, and they don't all give me the same answer. First, there is this claim that Earth slows down at a rate of about 0.005 seconds per year per year. A year has approximately $365.25 \cdot 24 \cdot 3600 = 3.15\cdot 10^7 \mathrm{sec}$, so 0.005 seconds change ...

2

The Moon rotates around the Earth slower than the rotation of the Earth itself. That's why, from a fix point on the Earth, the Moon appears to be moving. The Moon creates the tide on Earth. So the tide "follows" the Moon. However as the Earth rotates faster than the Moon it will tend to carry the tide with itself "forward". The Moon pulls the tide toward ...

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The answers are yes and no. Special relativity does make ellipses precess, but it only accounts for 7" out of 43" per century of Mercury's anomalous precession. I wonder if Einstein and/or Sommerfeld knew that. To first order, incorporating special relativity results in a small inverse cube correction to the gravitational force, which is well known to ...

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