# Tag Info

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On predicting planetary orbits A number of studies have shown that the inner solar system is chaotic, with a Lyapunov time scale of about 5 million years. This 5 million year time scale means that while one can somewhat reasonably create a planetary ephemeris (a time-based catalog of where the planets were / will be) that spans from 10 million years into ...

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Let me try this way: the Sun isn't only pulling on the Earth, it's pulling on the Moon as well. The pull on the Earth is almost the same as the pull on the Moon, so the net effect of the Sun on the relative motion between the Earth and Moon is very small. Recall Galileo's law of motion: if you drop two objects close together from the same height, they ...

7

This question always interested me. I found this recently - "Another issue we wrestled with is which satellite orbits to use. We did not want to be in geostationary or geosynchronous orbits. The reason was these alternatives would force us to deploy ground stations on the other side of the globe, whereas, by putting them in some orbit that periodically ...

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A simpler example is Kepler's laws of planetary motion. In a spherically-symmetric gravitational field, the planets follow elliptical orbits. The orbits certainly do not have the full spherical symmetry of the potential. They may be extremely lopsided. :-D

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How far ahead can we predict solar and lunar eclipses? NASA has uncertainty calculations that show how certain we are about when eclipses happen. From a back of the envelope, the eclipses will likely vary by a full day 35 thousand years from now. That said, we have eclipse seasons, so we know eclipses will continue to happen, and at roughly which time of ...

3

Now imagine that the tower is a little bit higher, and an object is released from the top of the tower. The object will float away and "climb" higher. Does the object move slower or faster around the world than the tower? Slower. Provided that the space elevator is geo-stationary, the higher you go above geosynchronous orbit, the greater your velocity ...

3

A charge radiates every time is accelerated. The power radiated is given by the Larmor formula. Putting this into the introductions to the motion of a charge in electromagnetic fields would be a meaningless complication, as much as considering air friction. But yes, a charge in a magnetic field would not spin indefinitely, even in vacuum.

3

You might as well imagine that it is fixed. Basically, as far as the moon is concerned gravitationally, only the Earth exists. For the Earth; our sun. For the Sun; a black hole, ect... and everyone of them has an inherited orbital velocity from their 'parent'. Orbit is not a place. As for why the moon doesn't crash, it formed outside of the Earth's Roche ...

3

is why doesn't the Earth's speed just mean it can just fly away from the moon The moon and the Earth fall towards each other due to their mutual gravitation. The Earth doesn't have enough speed, relative to the Moon, to 'just fly away'. Equivalently, the Moon doesn't have enough speed, relative to the Earth to 'just fly away'. Here's an image I ...

3

I actually answered a related question a couple of days ago on Astronomy. Small world! One of the important properties of the Oort Cloud is that objects in it are not strongly influenced by the Sun. After all, its inner edge is roughly 2,000 AU away - 300 billion kilometers from the Sun. The Sun's gravitational influence in that region is rather weak, so ...

2

In the absence of angular momentum, then material would be able to follow radial paths to be accreted (because of their mutual gravitational attraction) and thus have a spherically symmetric distribution. For many reasons (see linked questions to the right), the "circum-object" material does have angular momentum, which must be conserved. In the co-rotating ...

2

The Earth/Moon orbit is not truly metastable. As someone alluded to, the moon is actually very very gradually getting further from the Earth. Conversely Mars' moon Phobos is gradually moving closer to Mars over time and Phobos will eventually crash into Mars.

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On the off chance there was any doubt about the numerics, I too wrote a code to follow these orbits. I use RK4 to take the first half timestep (a full timestep here is always $0.001$), and then perform the remaining integration via leapfrog. Below is the evolution of an asymmetric orbit. The potential is given by $v_0 = 0.9$, $q = 0.7$, $L = 0.05$. I set ...

2

It's not the case. In the second, the electron will radiate. This is how light species lose energy, and cool in Penning traps, and one of the factors that limit the energies of particles in circular particle accelerators. For a reference see: http://en.wikipedia.org/wiki/Cyclotron_radiation

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Think about this. If gravity is of equal magnitude as the centrifugal force, a satellite there will have an angular velocity same as the earth rotating angular velocity. However, this is just a geosynchronous satellite! The equation is $$\frac{GMm}{r^2}=mr\frac{4\pi^2}{T^2}$$ where $T=24h$. Hope it helps!

2

If you are asking the question "How high does a person have to be in order to be "weightless" because gravity is canceled from the rotation of the Earth?" as paraphrased by CoilKid, then what you are talking about is a Geostationary Orbit. You can only achieve this orbit in a equatorial plane for the orbitting object to seem truly stationary with respect to ...

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It is incorrect to think that Newtonian gravity seems to dictate that we can treat all masses as point masses. Indeed, you can't do that, and one of the consequences of that fact is the orbital decay that you refer to. The theorem you're thinking of holds only for bodies with perfect spherical symmetry, and states that for such a body, the ...

1

In every negative acceleration electron loos energy, and this of course in the form of photons. This is not surprising because negative acceleration could be only after positive acceleration, the electron has to move befor he could be stopped or declined. And how the electron can be accelerated? By electric fields where the electron get the kinetic energy ...

1

The Moon is freely falling toward Earth, like you say. But it is also moving "sideways" quite quickly, so that it "misses" Earth and passes to the side. And continues to freely fall, and again misses passing to the side. Doing this in a continuous manner is called orbiting (or flying). To be a bit more technical, it is the angular momentum (and energy) of ...

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If you take something like Neptune and pass it through Earth's orbit perpendicular to the ecliptic so it collides directly with the Earth-Moon system at, lets say, 0.1c you would remove the Earth rather quickly. Get another if you want to disappear the moon as well. Neptune isn't so big that it would disturb everything else, maybe a wobble in Venus or Mars ...

1

Look I know link only answers are terrible, but I don't want to ruin the surprise. Check out this link. That's on Wikipedia! It's safe to say that if wikipedia knows something, the experts know quite a bit more. We know the dynamics of the Sun-Earth-Moon system really well including the perturbations and the most likely sources of future upsets, so barring ...

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My slight issue and where I think I might be missing something, is why doesn't the Earth's speed just mean it can just fly away from the moon and just leave it flying in it's tangential velocity? Even though you have drawn the force of gravity, you are not thinking about it. First of all the force in general is bidirectional, the earth pulls the moon ...

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The comments seem to provide the answer. $$L = p \times r$$ $$L = m ( v \times r )$$ Thus if r (the distance to the focus point of the orbit) changes, the velocity can change without the angular momentum changing. This is not possible for circular orbits where v is always perpendicular to r and the magnitude of r is constant. However, in elliptical ...

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You are basically asking how to describe the equations of motion when viewed from an rotating frame of reference (the planets surface). This can be done by including the Coriolis effect. For example if you simplify the problem to just the equator of the planet the equations of motion would then become: $$\ddot{x}=\frac{F_x}{m}-\frac{v_xv_y}{r},$$  ...

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