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My bad. This is not correct. The center of gravity of the space elevator must be at geosynchronous altitude. Center of mass and center of gravity are not the same thing, particularly so when the structure is seven+ earth diameters tall. The center of gravity is the point at which the gravitational force is equal to the total gravitational force on the ---------------------------------------------------- I'll assume you are talking about me. Hard to tell, since you didn't qualify "sir" or "this". This should be pretty obvious. If you need a reference for this, use google. This is pretty standard fare. Google is your friend. Pretty standard fare for an aeronautical engineer who models atmospheric drag. For anyone else, Walker, D.M.C., "Upper-atmosphere rotation rate from analysis of the orbital inclination of Explorer", http://adsabs.harvard.edu/abs/1975uarr.rept.....W King-Hele, D.G., "Decrease in Upper-atmosphere Rotation Rate at Heights above 350 km", Nature 233, 325 - 326 (1971). http://www.nature.com/nature/journal/v233/n5318/abs/233325a0.html

What satellite?

We do not and cannot feel the effects. That is one of the distinguishing features that differentiates pseudo forces from "real" forces. There is no way to measure pseudo forces (e.g., inertial force, centrifugal force, Coriolis force, gravity). You can observe the effects of these pseudo forces in terms of observed acceleration. That leads to the other distinguishing features that differentiate pseudo forces from "real" forces: (1) The magnitude of a pseudo force is proportional to the mass of the object subject to the force, and (2) Pseudo forces vanish in some reference frames (inertial frames).

The center of mass of the space elevator must be at geosynchronous altitude. The final configuration will be a cable that extends well beyond geosynchronous altitude. One very long cable would do the trick. The cable length can be decreased significantly by using a countermass at the far end. So, how to build it? Let's assume we have a smallish (not very thick) elevator in place. We can beef this structure up by sending climbers up the elevator that reel out additional cable as they climb. We have to start small (the original cable isn't all that strong), but eventually the cable can be built up from a small seed. So how to build this seed elevator: Simple: Reel it out from geosychronous altitude. The trick is to reel out two cables. One cable will descend to Earth while the other cable will go "up". Gravity gradient will naturally keep the entire cable in a vertical orientation. When the Earth end reaches the surface, anchor it to the Earth. Voila! Space elevator. Too much hand-waving (far too much for me, at least). There are problems with atmosphere. The Earth's atmosphere super-rotates at high altitudes. At 200 km, the atmospheric rotation rate is 1.1 rotations per day. At 350 km, its about 1.4 rotations per day. (From there the rotation rate starts decreasing.) The descending cable will start feeling this super-rotational wind well above 350 km. Another problem: Carbon nanotubes are excellent conductors of electricity. The spark that will jump from the cable end to the Earth as the cable approaches the surface of the Earth will not just be a little tiny spark. It will be a bolt of lightning and then some. Yet another problem: Part of the cable will permanently be in the Van Allen belts. I suspect the radiation damage to the cable in that area will be significant. Will the degradation occur faster than we can repair it?

A quibble over terminology. You appear to be talking about the interaction between (for example) a spacecraft and a planet. Spacecraft are typically not considered to be "massive bodies". The restricted three body problem for instance involves the interaction between two massive bodies and one other body of inconsequential mass. That said, I should have been more specific. The goal of a gravity assist is to get both a change in speed [math]v_{\infty}[/math] and direction. In the two body problem, a flyby will achieve a change in direction but not in [math]v_{\infty}[/math]. A change in speed and direction requires the interaction of multiple (more than two) bodies. First off, note that there is no change in speed here. Secondly, that isn't how things work. The spaceship is not "caught in half an orbit". It is always in orbit, in this case a hyperbolic orbit. Note well: With only two bodies (e.g., a planet and a spacecraft), the magnitude of the spacecraft's velocity far from the planet ("[math]v_{\infty}[/math]") can not be changed by the encounter. A couple of articles on gravity assists: http://www.ams.org/featurecolumn/archive/slingshot.html http://www.dur.ac.uk/bob.johnson/SL/AJP00448.pdf

I know chick flicks are terrible. They're a penance one must endure on occasion (lest you want to spend the night in the doghouse). For example, penance for forcing your wife to watch the Superbowl. I don't have to suffer that penance. I'm taking my wife to see Taken (or maybe Gran Torino) tomorrow evening.

I would care were we talking the real game of football. Unfortunately, we are talking about some silly sport where physically unfit but physically large men wear massive amounts of pansy equipment to mitigate injuries; where action occurs at best intermittently between TV timeouts for $100,000 per second commercials and instant (hah!) replays; and where, were it not for cheerleaders, steroids, and marching bands (sex, drugs, and rock 'n' roll), this would not be *the* American sport. I'd rather take my wife to a chick flick.

The Navier-Stokes equations that underly fluid dynamics are notoriously non-linear and have some wierd non-linear solutions such as solitons. Pure conjecture: This is a 2D (3D) analog of a 1D soliton.

In one sense, those are complementary rather than conflicting explanations. In another sense, neither explains the phenomenon at all. The first link reports experiments by Tomas Bohr (quite a family lineage there, Son and grandson of Nobel physicists!) that creates polygonal structures in rotating fluids in a lab environment. Bohr does not offer an explanation of the phenomenon. From the physorg article, Bottom line: The first link is not an explanation of the phenomenon. What is significant is Bohr's work gives us a way to replicate the phenomenon in a controlled environment. The wikipedia article simply states without reference that the phenomenon is a standing wave. I could not find any pages at jpl.nasa.gov that call this a standing wave phenomenon. The closest is this, "The hexagon appears to have remained fixed with Saturn's rotation rate and axis since first glimpsed by Voyager 26 years ago", from http://www.nasa.gov/mission_pages/cassini/media/cassini-20070327.html. Even if this is a standing wave phenomenon, that does not explain it. It is merely a kinematic observation. As an analogy, consider the problem of uniform circular motion. The kinematics of a planet in a circular orbit and a rock at the end of a string swung in a circle about one's head are exactly the same. The dynamical explanations of a planet's circular orbit and the rock's circular motion are quite different. Bottom line: The wiki article is not an explanation of the phenomenon. It is dubious that this is a standing wave phenomenon. Even if it were, kinematics is not an explanation.

Dark matter (if it exists) interacts gravitationally with matter just like matter interacts with itself gravitationally. Dark matter has positive mass.

That is still not right, transgalactic. What is the quotient rule for differentiation?

Welcome to Science Forums, Geoff! You are misunderstanding how the gravity slingshot technique (gravity assist) works and how the space elevator might work. Gravity assist works because the planet is moving with respect to the Sun and the vehicle is moving with respect to the planet (and Sun). Gravity assist does not work in a simple two-body problem world. It is a consequence of interactions between three bodies: the Sun, the planet, and the vehicle. The other problem here is a misunderstanding of how the space elevator would work. You are assuming several things here. First, you are assuming that it would even be possible to "rapidly shorten the tether". It wouldn't. A tether to geocentric orbit (and beyond) is very long and very massive, a lot more massive than the vehicle itself. A space elevator would not end at geosynchronous orbit. It needs to go out a lot longer than that. A counterbalance is needed at the end of the elevator to keep the elevator more-or-less stable. There would be no way to reel it in. There is no need to do so. A vehicle could get a significant boost just be climbing the elevator past the geosynchronous point.

For now, that is. The next bill in the hopper, HR 12, the Paycheck Fairness Act, starts opening the nasty can of worms "equal pay for comparable worth".

I've already given more help here than I feel comfortable giving, transgalactic. The next step is yours.

I asked you to prove it. So do that: Prove it.

Since [math]\frac {d^{k+1}}{dx^{k+1}} f_{n+1}(x) = \frac {d^{k+1}}{dx^{k+1}} f_n(x) + (k + 1)\frac {d^k}{dx^k} f_n (x)[/math] for all k>=0, it is certainly valid for k=n so long as n>=0.

The problem at hand involves proving that [math]\frac{d^n}{dx^n}f_n(x) = g_n(x)[/math] This is obviously proved recursively. Show that it is true for some simple case (i.e., n=1), assume it is true for some given n>1, and show that this assumption means the expression is true for the successor case (i.e., n+1), [math]\frac{d^{n+1}}{dx^{n+1}}f_{n+1}(x) = g_{n+1}(x)[/math] What does the relationship I finally pulled out of you in post #33 help you? Hint: Look at the left-hand side of the above and the left-hand side of the equation in post 33.

Finally! You need to prove this is indeed a correct expression (recursion again). Then use this expression to aid in the proof of the original relationship, [math]\begin{aligned} f_n(x) &\equiv x^{n-1}e^{1/x} \\ g_n(x) &\equiv (-1)^n\frac{e^{1/x}}{x^{n+1}} \\ \frac{d^n}{dx^n}f_n(x) &= g_n(x)\end{aligned}[/math]

Look at what you wrote in post #29: The fourth derivative (k=3) should then be [math]\frac{d^4}{dx^4}f_{n + 1} (x) = x\frac{d^4}{dx^4}f_n (x) + 4\frac{d^2 }{dx^2}f_n (x)[/math] this disagrees with the known value of the fourth derivative, which you derived correctly in post #31: [math]\frac{d^4}{dx^4}f_{n + 1} (x) = x\frac{d^4}{dx^4}f_n (x) + 4\frac{d^3 }{dx^3}f_n (x)[/math]

No! You are just guessing, and not very well. Try taking the fourth derivative.

Why three? There isn't a three in the first and second derivatives. Can you see a pattern? [math] \begin{aligned} \frac {d}{dx}f_{n+1}(x) &= x\frac {d}{dx}f_{n}(x) + f_n(x) \\ \frac {d^2}{dx^2}f_{n+1}(x) &= x\frac {d^2}{dx^2}f_{n}(x) + 2\frac{d}{dx}f_{n}(x) \\ \frac {d^3}{dx^3}f_{n+1}(x) &= x\frac {d^3}{dx^3}f_{n}(x) + 3\frac {d^2}{dx^2}f_{n}(x) \end{aligned}[/math] In particular, what would you expect [math]\frac {d^{k+1}}{dx^{k+1}}f_{n+1}(x)[/math] to be? Can you prove that this is the case?

Yes. Combine some terms and you're there.

Your mistake is here: which led you to make this mistake here:

its the second derivative obviously its not [math] \frac{{d^2 }}{{dx^2 }}f_{n } (x) [/math] if you are pointing at this place. so what is it? Good lord. Obviously that is exactly what it is.

What is [math]\frac{d}{dx}\left(\frac{d}{dx}f_n(x)\right)[/math]?