# Tag Info

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The "pole" you are referring to is also known as a space elevator. Creating a cable of sufficient tensile strength is currently unfeasible, but carbon nano- and macrotubes are promising. Your question can be generalized, ignoring Earth's non-spherical shape, to apply to any position in space around earth by replacing "height above the equator" with ...

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You've pretty much covered everything, except that below the geostationary orbit the orbit is elliptical - there is no transition to a hyperbolic orbit at small $r$. Let's draw a picture of the pole and get some preliminary formulae out of the way: If you are on the pole at a distance $r$ from the centre of the Earth (NB from the centre of the Earth not ...

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The answer is "No" - unless the city is on the equator. You specified "...it will always remain direct over a city". Satellites in geosync orbit might be visible to cities but not be directly overhead unless they are located on the equator.

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The answer is the same as the answer to the question "why do satellites stay in orbit": the gravitational pull of the earth is just strong enough to keep it in orbit at the altitude it is, given the angular momentum (velocity) that it has. In equations: $$\frac{GM_{earth}}{r^2}=\frac{v^2}{r}$$ where $r$ is the distance from the center of the earth to the ...

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It's moving the orbital velocity for its particular altitude, which means it has the exactly the speed at which Earth's gravity supplies the centripetal force needed to accelerate it in a circle. Now, I'm guessing what you mean by your question is that at the surface of the Earth, we need to boost a rocket to about $v_e(R_\oplus)=11{\rm km s^{-1}}$ (let ...

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Why shouldn't the orbits of stars be Keplerian? The answer is simple. Keplerian orbits are predicated on a single central point mass. That assumption fails to some extent even in a solar system. It fails massively in a galaxy. A galaxy is not a point mass.

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Elliptical orbits are direct consequence of orbiting entirely outside a spherically symmetric mass. Even if you assume that a galaxy has a spherically symmetric mass distribution, the amount of mass at a radial distance less than that of the star would be changing (assuming some eccentricity). Once that happens, the orbit is no longer an ellipse.

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Not Keplerian, because it is not a conic-section. It is not even explained by Newtonian gravity. In contrast, Kepler's laws are explained by newtonian gravity. The lowest orbital-energy from Keplerian orbit is circular. And the orbits of stars are observed to be approximately circular. Hence: \frac{mv^2}{r} = \frac{GMm}{r^2} \quad\Longrightarrow\quad v = ... 3 I think the answer lies in the fact that the variational principle says is if there are two fixed points in space-time then there exists an open subset of space-time containing these two points such that amongst all curves contained in this open subset, a geodesic will be the curve of longest length between these two points. So, the variational principle ... 2 Rosetta initially went round the comet in a roughly triangular orbit using its thrusters to change direction. Eventually in September 2014 it entered a true orbit at a distance of about 30km going round once every two weeks. The orbit can hold in such weak gravity because it is very close and slow compared to satellites orbiting Earth. It has now moved ... 1 A stable orbit requires that the orbital insertion velocity be just right. Too high and the spacecraft flies away on a hyperbolic trajectory. Too low and it eventually falls and impacts the planet (or comet in this case). Upon approaching a planet or comet the spacecraft will perform a delta-v maneuver, a burning of thrusters that insert the spacecraft into ... 0 Applying simply the classical mechanics, the centrifugal force rω^2 should be equal to the gravitational acceleration GM/r^2, where r is the radius you look for, ω is the angular velocity, and M is the mass of the earth. So, rω^2 = GM/r^2 ==> r^3 = GM/ω^2 . The angular velocity ω of the earth rotation can indeed be calculated from the fact that in ... 5 It's explained on the wikimedia page for the image: The various gravity assists form visible peaks on the left, while the periodic variation on the right is caused by the spacecraft's orbit around Saturn. Since Saturn is moving relative to the Sun, when Cassini is at a point in its orbit that causes it to move in the same direction relative to ... 3 No, this sort of configuration is emphatically not possible. In real gravity (you know, the one in the real world) planets orbit the Sun in ellipses which have the Sun in one of the foci, which is in the plane of the ellipse. This is patently not the case in the configuration in that link, which means that the link is wrong. Period. 2 Well, a circle is a special case of an ellipse, so if something is true of a general ellipse, it will be true of a circle. In order to work out the rules for ellipses, the calculus is typically a bit more involved than what you see in your standard intro class, so the real, general case is usually reserved for sophomore year classical mechanics, where you ... -2 Well, we have people expounding on the equilibrium being unstable but, uh, wouldn't the net force result in an attraction between the centres of mass? To me, that seems like an essentially stable setup, at least before taking rotation into account. 5 If the ring elements are actually in orbit (i.e. each part would follow the ring shape by itself without need for a pillar), then you get this: If each ring element is "closer" to the Earth (i.e. they don't revolve fast enough for their altitude), then the pillars offer sustenance. If you remove them, then the ring is stable in the north-south axis, but ... 3 The motion of a satellite in orbit is governed by Conservation of angular momentum Conservation of energy When there is no (or negligible) drag, these two are satisfied by an elliptical orbit. Kepler studied this at length, by observing the motion of the planets. He formulated his famous laws: The orbit (of a planet) is an ellipse (with the sun at one ... 4 If you solve the equations for two bodies interacting via gravity (or indeed any inverse square law force) then the bound orbits are all elliptical - a circle is a special case of an ellipse with zero eccentricity. So any object in an elliptical orbit will remain in that orbit forever. No external forces are needed. To see how the orbit is calculated have a ... 0 The reduced mass is a mathematical trick. Consider the rotation of a diatomic molecule with two atoms of mass m_1 and m_2, which is similar to your example. (bear with me on this) The molecule rotates about its centre of mass and to work out the kinetic energy of the rotation we should use K.E. = {1 \over 2} I \omega^2 where I is the moment of ... 0 Possible but it would need an exceptional stable star and planetary system. As someone mentioned before, its so improbable, its practically impossible. If a system like this did exist, then it would almost certainly be made by some advanced ancient civilisation. It wouldn't take a massive amount to start introducing instabilities to the system as well, ... 2 People considered (An electromagnetic space elevator? ) a superconducting ring with a high current in space above the equator. The weight of the ring can be balanced by the force due to Earth's magnetic field acting on the current. 31 Why not try this at home by using the Coulomb force or magnetism instead of gravity? Although all those forces are different in nature, mathematically they are the same: F(r) = -\frac{const.}{r^2} Edit: Magnetism is different as pointed out in the comments, but it still works for illustrating the problem. In fact, the result is the same for any ... 37 This is a fun what-if. I really wish I was xkcd right now so that I could include pictures and humour, but I'm not. If we built a giant ring around Earth at just the right height everywhere, then in theory, the ring would just float there once the supports were removed. However, (and this is the reason not to try it) this is an unstable state. If you so ... 1 Is important to remember that curvature does not imply curl. Here is an analogy from fluid mechanics which deals with lack of vorticity (curl) in curved flow. This may seem counter-intuitive at the start, but hopefully you can see the similarities between the velocity field in the flow and the gravitational field. ... 0 It seems like a lot of your problem, aside from the fact that small structures have causal contact so much a part of them that it is difficult to imagine objects so far apart that they do not even have gravitational contact, is that space is bent. And it is bent to a very high degree. Going in any direction in as straight a line as you can go (wtih any ... 3 If the velocity decreases by a little bit, it goes into an elliptical orbit with the apogee the same as the original orbit. The orbit will be stable if no other changes happen. Only when the velocity decreases to the point where the elliptical orbit intersects the Earth's atmosphere will the object crash into the Earth. 4 A recent overview can be found in Pireaux et al (2001). I quote the authors on page 3: However, the perihelion shift of planets, and hence Mercury, can not be measured directly because the perihelion is a Keplerian element whereas the motions of the planets are not exactly Keplerian due to mutual gravitational interactions and figure effects. So, only ... 0 We derive the time period of a pendulum T=√(L/g) considering the angular displacement of the pendulum , θ<4° i.e. the motion of the pendulum is approximately linear. And the equation of S.H.M. is F∞-x, i.e. force acting on the body, executing S.H.M. , must be in same line of the displacement of the body. So when a body moving in a elliptical does not ... 2 If you solve the geodesic equation in Schwarzschild space-time then you will obtain, for the freely falling particle, the coordinates of its worldline x^{\mu}(\lambda) in say the (global) Schwarzschild coordinate system; here \lambda is an affine parameter such as proper time \tau for a massive particle. The x^i(\lambda) will be the spatial ... 2 Many planets have been found where their orbital axes do not align with the rotation axis of their star. This is achieved using measurements of the Rossiter-McLaughlin effect in transiting systems or by observing planets transit over spotted features on a star's surface. As the stellar rotation axis is highly likely to coincide with its protoplanetary disk ... 7 I think what you mean is - is it possible for a planetary system to exist such that the planets do not orbit in a single plane, but the planets have a large scatter of inclination angles? Our solar system has a relatively modest range, providing you ignore Pluto, of orbital inclination values (and eccentricities); zero to 7 degrees (Mercury). This is ... 0 We don't know, but there what I gather is the typical pattern of hypothesis, accretion disc theory, which attempts to find a model dynamic that could explain what we see, particularly in our own solar system, which is the only one we have very complete information about. In addition to seeing that our own solar system's planets all orbit in the same plane, ... 3 If you want a complete and consistent picture, I would recommend studying geodesics (orbits) in some black-hole solution of equations of GR (for reasons of simplicity, take the Schwarzschild metric). Almost all work is already done for you in this wikipedia article. Short answer: no, the speed would never reach or exceed the speed of light. The aspects of ... -1 First of all, there are no two focus in reality, it is just assumed that Sun is at one of the focii. For your understanding, go to this link: http://www.youtube.com/watch?v=XBxDDXG3sTQ. The video just gives the idea. On the paper you have to consider one focus for one ellipse. Fous is nothing but position of the Sun. Based on the position of the Sun your ... 2 What I know is, putting the sun at one of the two foci is purely for mathematical convenience, because Kepler's laws derive naturally from vector formulations of Newton's laws. so my answer to you is that no, the second focus has no physical significance, as far as classical mechanics is concerned. maybe general relativity has another answer. but this is ... 0 It would be better to think in terms of conserved quantities (angular momentum and energy): for a central potential we have thatE=\frac{1}{2}m\dot{r}^2+\frac{L^2}{2mr^2}+U(r) In this case (gravitational potential) $U(r)=-\frac{GMm}{r}$. You could evaluate energy at $r=a$ and $r=b$ (where $\dot{r}=0$) then calculate $L^2$ and re-evaluate energy.

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I can't quite follow your diagram. The angle can be measured from any line in the plane of motion to the line from the origin to either mass. The angle $\phi = 0$ represents the line you choose as your datum. The positive direction can be chosen arbitrarily as long as the choice doesn't change. The differential equation of motion doesn't care where the ...

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