# If we build a space elevator from Earth surface to GEO, could I step off it at GEO and remain in GEO?

Ignore for a moment the question of whether building such a space elevator is practical. Let's also ignore the minute delta-v involved in "stepping off" something in orbit.

If we were to build a space elevator that extends from the surface of the Earth (on the equator) up to geostationary Earth orbit (GEO), then would it be possible to use that space elevator to get to GEO, then simply "step off" the elevator and remain in GEO without any additional acceleration component? Why or why not?

I found myself in a discussion about this over lunch and can't quite make up my mind whether it seems plausible or not, so I'm turning to the collective expertise here to hopefully get an answer.

• I rather think you could hold yourself at arms length from the elevator, let go, and.... just stay there until you grabbed the elevator again. – Mooing Duck Aug 12 '14 at 22:46
• @MooingDuck In a sense, I believe that too would be "stepping off" the elevator. Especially since I explicitly said let's ignore the Δv involved in stepping off it; if we set Δv_stepoff = 0, then what you are proposing sounds pretty much like what would be necessary (ignoring orbital effects...). – user Aug 13 '14 at 7:17
• I know this is beside the point, but wouldn't putting the anchor mass of the space elevator at exactly GEO be a technical impossibility? From what I understand, the elevator needs to apply some amount of centrifugal force to keep the tether taut, given that the tether itself has mass and is subject to gravitational forces at the bottom. Is that not the case? – jdmcnair Aug 13 '14 at 12:32
• @jdmcnair if the anchor mass would be at exactly GEO, then it wouldn't be a stable configuration as the tether weight would pull it out of GEO. – Peteris Aug 13 '14 at 13:14
• That said, I think the tether itself would pass through the GEO point, so if you ascent in the elevator to that point, then step off, the original hypothetical applies. – jdmcnair Aug 13 '14 at 14:35

Yes, GEO is a balance point for anchored to Earth space elevator. It has to be, to keep its rotation rate synchronized with Earth's own and keep tether stable. This necessarily means that at GEO altitude, the elevator rotates at GEO orbital speed. Stepping off it at that altitude then means you're in a stable equatorial geosynchronous orbit, or GEO (geostationary, with nadir pointing towards more or less same spot on Earth's surface).

Your exact resultant orbit would however depend on the force with which you stepped off it and its vector, so you might end up in slightly higher, slightly lower, or in orbit around the elevator's balance point, depending also on it's balance point mass (and yours, no pun implied).

• Let's hope both you and I have some mass! :) – user Aug 13 '14 at 7:19
• @MichaelKjörling We better, otherwise we might have forgotten to don the EVA suit prior to exit. Not recommended :) – TildalWave Aug 13 '14 at 8:36
• "Yeah, and drifting through eternity will ruin your whole day." – user Aug 13 '14 at 8:38
• The acutual physics are that you cease being in geostationary orbit the INSTANT you apply any force to move away. – aramis Aug 14 '14 at 1:04
• @aramis Did I say it any different? There doesn't necessarily have to be any force in "stepping off", something OP also explains in the question itself as well as in its comments. But sure, as soon as you change your orbital energy by any means, even if it's in the normal or tangential vector, you're no longer geostationary. Then again, GEO satellites slightly drift w.r.t. the surface too, the difference being, they can correct for that. But maybe the person stepping off a space elevator at GEO could too, e.g. with EMU/SAFER. It's not what the question asks about tho. – TildalWave Aug 14 '14 at 9:23

A Clarke style space elevator is a (very large) gravity gradient stabilized vertical tether.

When in a rotating frame (as on a merry-go-round) you feel a tug. It's just inertia but feels like an acceleration. This so called acceleration is $\omega^2r$ where $\omega$ is angular velocity expressed in radians/time.

Gravity's acceleration is $GM_{earth}/r^2$.

On a vertical tether moving in a circular orbit, there's a point where $\omega^2r$ and $GM/r^2$ exactly cancel and someone on the tether at this point would feel zero net acceleration.

For a space elevator this point would be at geosynchronous altitude about 36,000 kilometers above earth's surface or about 42,000 kilometers from earth's center.

Below geosynchronous altitude, $GM_{earth}/r^2$ overwhelms $\omega^2r$. Someone on an elevator platform below geosynch would feel a tug towards the earth. This earthward tug grows stronger for platforms at lower altitudes. Someone jumping off the leading eastern edge of such a platform would fall towards the earth along the elliptical paths shown. The jumper would also move further and further east of elevator.

Above geosynch altitude, $\omega^2r$ overwhelms $GM_{earth}/r^2$. Someone on a elevator platform above geosynch would feel a tug away from the earth. This tug away from earth grows stronger at higher altitudes. Someone jumping off the trailing western edge of such a platform would fall away from the earth along the elliptical paths shown. This jumper would move further and further west of elevator.

Call the ~42,000 km geosynch radius $r_g$. At $2^{1/3}r_g$, the elevator is moving escape velocity. Someone jumping off a platform at this altitude would fall into a parabolic orbit away from the earth.

Someone stepping off the platform at geosynch would feel no acceleration up or down. He would step into the blue orbit of the diagram above. He wouldn't be moving with regard to the elevator.

• $2^{1/3}r_g$? Isn't that $\sqrt[3]{2}r_g$? Why $\sqrt[3]{2}$ (or for that matter $2^{1/3}$)? And escape velocity where; I assume at that point, but why must it? And "jumping off" would seem to imply a larger delta-v than "stepping off", although that may be quibbling over semantics. This certainly has the makings of a good answer, but it does seem to have a few "magic constants" in it (particularly that part) that are either "completely obvious" or which don't make sense. Could you expand on that part? – user Aug 15 '14 at 23:20
• I mean the same thing by "jumping off" or "stepping off". – HopDavid Aug 16 '14 at 0:38
• Choose your units so that the distance from earth's center to geosynchronous altitude = 1. With these convenient units, releasing a payload from the elevator would send send them on a conic path with eccentricity $|1-r^3|$. So when at geosynchronous, ellipse eccentricity would be zero and the orbit's circular. At r = $2^{1/3}$, eccentricity is 1 and we have a parabola. Above that the elevator would throw payloads into hyperbolic orbits. – HopDavid Aug 16 '14 at 0:44
• I would enjoy deriving e = $|1-r^3|$. However it'd take me some time and triple the length of my answer. If you don't think that derivation would be off topic, I will do that. – HopDavid Aug 16 '14 at 0:48
• @MichaelKjörling Here is a primer on fractional exponents: purplemath.com/modules/exponent5.htm I derived my value for e at space.stackexchange.com/questions/5253/…. – HopDavid Sep 1 '14 at 7:36

Yes. The elevator stop at GEO will itself be in orbit at GEO. Thus with a minimal delta-v away from it, you will also be in GEO orbit, at GEO orbital velocity.

It depends upon exactly where you step off.

The elevator, in order to remain in place, has to actually extend well past GEO. An object orbits based upon the speed at its center of mass. As you go down, orbital velocity required increases, and as you go up, it decreases... and the elevator also has an issue of as the elevator goes down from the center of mass, its speed decreases, and as you go up from the center of mass, speed increases. (Keeping in mind, down is towards the planet.)

Also, remember the laws of motion:
— An object in motion remains in motion until acted upon by an outside force.

Gravity is constantly acting, but the speed means that the pull isn't fast enough to bring one down, since gravity works over time.

So, if you step off 10m below the center of mass, you're in a decaying orbit already - the orbital speed needed is higher, but your retained orbital speed is lower. That difference is VERY small. Small enough to be ignored in the short term. But if you "hang" your tools nearby, and wander away, a few days later they've moved downward and trailing of the station, and will continue to do so, accelerating downward.

If you instead step off above the center of mass, you are above the orbital speed; you have attained escape velocity and will appear to arc away. It won't be a nice clean straight line, as you're close enough to orbital velocity that gravity keeps you curved. It will be a long slow spiral outward.

This effect, when maximized by a few kilometers, can be used, in theory, to launch interplanetary spacecraft.

Note that the answer is "yes" for any orbit. Step out of the spaceship/elevator/phonebooth with minimal energy and you'll be in the exactly same orbit that the original one was (modulo whatever force you used to step out).

• You're right that the term "orbit" already implies you're matching required orbital speed at a certain altitude (or semi-major axis if it's a non-circular orbit), but what you say isn't really true for a space elevator. The only altitude of it that matches required orbital speed is at GEO. Lower than that, and it rotates too slow, and higher than that and it rotates too fast. That's why GEO also has to be the balance point, its center of mass. Counterweight needs not be as far as the anchor point, but both ends have to "weigh" exactly the same. – TildalWave Aug 13 '14 at 15:13
• Only if you don't tie the earth end down do you need the balance point at exactly GEO. – Catprog Feb 13 '18 at 5:38