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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 ...


1

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: $$ \ddot{x}=\frac{F_x}{m}-\frac{v_xv_y}{r}, $$ $$ ...


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 ...


1

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 ...


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 ...


0

Naturally, I know it to be true that the moon goes around the Earth and that the Earth goes around the sun. While it definitely looks like it if you view it with an astronomy program, it is not entirely correct and a misconception. Sun, earth and moon are not nailed to a specific location, there is no axis stuck in vacuum. Each body attracts each ...


0

I see no reason why it should be impossible. A binary planet system should exist in a reasonably stable orbital configuration. As Javier pointed out, Pluto and Charon are similar in size and mass, like a small-scale version of what you describe. The important thing to note here is that the barycenter of the Pluto-Charon system is outside either planet. In ...


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.


1

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 ...


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 ...


16

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 ...


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 ...


1

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 ...


2

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 ...


0

First off, are you capturing all of the dynamics? In particular, are you modeling Jupiter's second dynamic form factor J2. This is fairly simple. It just means you need to take a tiny step beyond point mass gravitational models. Since Jupiter has a very large J2 (0.014733), you had better be modeling it. Jovian tidal dissipation, which transfers angular ...


0

r is the distance between the two masses v is the relative velocity a is the relative semimajor axis* * Two bodies orbiting each other trace out two separate ellipses in an inertial frame. The smaller body traces a larger ellipse, and vice versa. The relative semimajor axis (a) is equal to the sum of the semimajor axes of these two ellipses. The relative ...


0

For a satellite in a nonperturbed orbit, the change in gravitational potential energy does equal the change in kinetic energy. Increasing the size of a circular orbit requires energy. Half of the energy is used to lift the satellite, while the other half is used to speed up the satellite. For a satellite in a circular orbit perturbed by atmospheric drag, ...


6

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


0

because both objects, the earth and the moon have their own unique gravity force due to their mass., there is a place between the two where gravity works independently that keeps the separated. the moon is not falling towards the earth. actually, from some research I've done. the moon is actually moving away from the earth by 1 1/2 inches a year.. not much ...


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.


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 ...


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


0

ALmost freely falling . because you can't exclude the gravitational force by other planets on it . What do you think of centripetal force and falling. You are associating falling because you are very accustomed to such things but falling is solely due to an external force acted on to attract towards body so in simple circular motion when you rotate a stone ...


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 ...


0

An other way of seeing this is to imagine that since the object is "climbing up", a part of the cinetic energy is along the $r$ axis, and the rest is in angular speed. The angular speed is thus necessarely inferior than the tower.


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 ...


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 ...


2

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!


1

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 ...


0

Newtonian gravity seems to dictate that we can treat all masses as point masses. This is the source of your confusion. This is not true. One can treat an object with a spherically symmetric mass distribution (i.e., density is a function of distance from the center of the object) as if were a point mass for points outside of the object in question. ...


0

Point masses are an approximation that describe the motion of the center of mass of objects which do not have their rotational and internal vibrational degrees of freedom excited. If that assumption is incorrect, then so is the treatment of Newtonian problems by using coordinates of centers of mass alone. Orbital decay can only occur if both energy is lost ...


0

A very late answer, one that I hope adds to the excellent answers by Mark and LuboŇ°. From the perspective of Newtonian mechanics, there's nothing wrong per se with using a geocentric point of view. Such a point of view does require adding fictitious forces and torques that would otherwise be absent in an inertial perspective, but if makes sense to do that, ...


5

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 ...


49

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 ...


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 ...


-2

Given that we exist at all, we can confidently back-track the positions of the planets (and our moon) for a couple billion years. I see no reason, barring rather massive exo-system-sourced objects showing up unexpectedly, that the positions will go chaotic enough to be unpredictable any time in the next couple billion years. I suppose it might depend a bit ...



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