If I rode a space elevator to the top, assuming only natural forces apply, would the Earth be over my head or under my feet? If it was over my head how is that possible?
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asked Jul 29, 2014 at 10:35
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2 Answers
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Above geosynchronous orbit, so called centrifugal force exceeds force of gravity. So earth would be above you.
Centrifugal acceleration (actually just inertia in a rotating frame) is $\omega^2r$ where ω is angular velocity in radians.
Earth's sidereal day is about 23.93 hours, so ω is about 2 pi radians/23.93 hours. Or 7.3e-5 radians/second.
Acceleration from gravity is $GM_e/r^2$ where $M_e$ is mass of earth and r is distance from earth's center.
You will find that $GM_e/r^2$ and $\omega^2r$ exactly cancel at geosynchronous altitude. Above geosynchronous $\omega^2r$ exceeds $GM_e/r^2$. You will weigh more as you move further beyond geosynchronous height.
If my arithmetic is right, you wouldn't feel a full g until you're about 1.8 million kilometers from earth's center.
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$\begingroup$ In your answer to Where could we build a space elevator today (2014)? you give an altitude of 142,772 Km for the top of an Earth Elevator (no counterweight), so it would seem that the you should have a full G there. Is your math wrong, or do I need to write another question? $\endgroup$ Commented Jul 29, 2014 at 14:35
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$\begingroup$ @JamesJenkins No, from what I can tell both his answers are perfectly correct. You don't need a full -1 g at the other end (counterweight) of the space elevator. All you need to do is make sure that both ends "weigh" the same and balance out to 0 at GEO, its center of mass. Since GEO to "top" is a lot longer in that case, and assuming constant mass per length, it "weighs" the same at larger mass. What you're likely forgetting is that a space elevator has to keep constant rotational period (synchronized with Earth's rotation) for all its parts regardless their orbital altitude. $\endgroup$ Commented Jul 29, 2014 at 14:54
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$\begingroup$ Maybe it wouldn't go amiss if David mentioned this constant angular speed requirement of the space elevator, and link it to orbital speed (in words). But I get 0.970698 g at 1.8 million kilometers. That's pretty close to 1 g. To match exactly 1 g at mean sea-level Earth (9.80665 m/s²), more precise distance from the center of the Earth would be 1,854,336 km. Of course, at such distance, the Moon might have something to say about your whole elevator system. You can use this centrifugal force calculator to make it somewhat easier. ;) $\endgroup$ Commented Jul 29, 2014 at 15:01
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$\begingroup$ BTW "up" is ambiguous on a space elevator. You're equally "up" when you're standing on the Earth than when you're at the end of its counterweight part. So in a sense, your question could be answered with "both". ;) $\endgroup$ Commented Jul 29, 2014 at 15:17
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$\begingroup$ @JamesJenkins What Tildawave said. Since the acceleration gradient is steeper earthside of geosynchronous, the earthside length is shorter. To balance the farside length needs to be longer. $\endgroup$– HopDavidCommented Jul 29, 2014 at 15:33
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Everywhere on the Space Elevator, "up" is the direction toward the geosynchronous point (GEO). Below GEO, gravity is exerting a stronger force on your body than the centrifugal force caused by rotation -- "down" is toward Earth; beyond GEO, the centrifugal force is stronger than gravity, so the net force on your body is away from the Earth. Right at GEO there's no "up" or "down," and indeed for some distance on either side of GEO the net force is too small for your senses to detect.
From the surface you climb up to GEO and then, without changing direction, you climb down to the counterweight at the outer end of the elevator.
The rule for "what holds the Space Elevator up" is that the sum of the net force on all the mass of the elevator above GEO must be larger than the sum of the net force on all the mass below GEO.
