The upper end of a space elevator is moving considerably faster than orbital speed at that distance from Earth. As a result anything released from the top of elevator would get thrown away from the Earth instead of just entering into high Earth orbit. How much would being released from the end of a space elevator at twice geosynchronous radius reduce travel times or the amount of rocket thrust needed during Earth departure for interplanetary probes.

I'm assuming the cable is capable of damping out the oscillations caused by releasing the payload. Doing so is necessary to release a payload at any altitude other than geosynchronous altitude; due to the amount of activity currently occurring at significantly lower altitudes being able to do so seems to be a necessary requirement for constructing an elevator in the first place.

  • $\begingroup$ There are two benefits from launching from the counterweight - the motion of the counterweight in orbit, but there's also a huge benefit from having left the bulk of the gravity well below you. I'd guess the latter is the primary benefit. $\endgroup$ Commented Aug 18, 2013 at 23:51
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    $\begingroup$ This question doesn't have a single answer because there's no telling what radius the end of the space elevator would be. If the counterweight has a huge mass, it will barely buy you anything. But @DanNeely does give an upper bound. There's no way it would extend further than that, because the material won't be strong enough. So the answer is between nothing and whatever you get at that limit. $\endgroup$
    – AlanSE
    Commented Aug 19, 2013 at 1:00
  • $\begingroup$ @AlanSE Setting a material strength just strong enough to make the cable possible is an arbitrary cutoff. I could see a design where the bottom half of the cable was thicker than normal (or had multiple strands rising to join geo; and a single longer strand going up with a maximum altitude limited by interference from lunar gravity (where that cap would be is another question). $\endgroup$ Commented Aug 19, 2013 at 1:06
  • $\begingroup$ @DanNeely I don't buy that because the outward limit is the material specific strength, which is fundamentally limited by bond strength. The acceleration increases linearly with radius, so the specific strength requirement past GEO will surpass the requirement for the space elevator below GEO, which is what requires carbon nanotubes. The force at GEO can be whatever you want, sure. But at some point the problem of going from GEO to release point gets harder than Earth surface to GEO. Harder in the sense that it can't be done beyond some radius with atomic matter. $\endgroup$
    – AlanSE
    Commented Aug 19, 2013 at 3:19
  • $\begingroup$ @DanNeely If there is no counterweight it's a lot more than 2x geosync--IIRC it's around 5x. The thing is gravity is weaker out there. I believe a launch from the end of the cable can get you to almost anywhere in the solar system. (You'll still need a rocket when you get there unless you can aerobrake.) $\endgroup$ Commented Aug 19, 2013 at 18:56

2 Answers 2


Your total energy is ${v^2\over 2}-{\mu\over r}$. If that's negative, you're still in orbit. If positive, you've escaped.

At the point of release, $r$ is twice geosynchronous radius, $2\times 42164\,\mathrm{km}$, and $v$ is twice geosynchronous velocity, $2\times 3.075\,\mathrm{km\over s}$. The $\mu$ for Earth is $398600\,\mathrm{km^3\over s^2}$. The result is $14.18\,\mathrm{km^2\over s^2}$.

You have escaped.

The result is equal to your $v_\infty^2\over 2$, so your $C_3$, which is $v_\infty^2$, is $28.36\,\mathrm{km^2\over s^2}$.

Not enough to get you directly to Jupiter, but you'll get well past Mars' orbit or well inside Venus' orbit, depending on your preference. A judicious set of Venus and Earth flybys could then get you to Jupiter. From there you could go pretty much anywhere, including potentially escaping the solar system. Also a Venus flyby could get you to Mercury.

You will be constrained on the departure plane, but perhaps a lunar flyby could help get you going in the direction you want.

The $\Delta V$ required to get to that $C_3$ from low-Earth orbit is about $4.5\,\mathrm{km\over s}$. Or from GEO, about $3.3\,\mathrm{km\over s}$ (by lowering periapsis to $150\,\mathrm{km}$ altitude and escaping from there). So it buys you a lot. Though that's one hell of a cable you'll have to build.


How much would being released from the end of a space elevator at twice geosynchronous radius reduce travel times or the amount of rocket thrust needed during Earth departure for interplanetary probes?

There are a lot of variables in this question and the answer, even given the clarity of the length of the elevator. The major variable being the reduction of non-defined values for times or thrust. So I will just focus on the velocities and destinations obtainable at different release elevations, with no added fuel costs.

Where Earth geosynchronous orbits, whether circular or elliptical, have a semi-major axis of 42,164 km (26,199 mi), Twice would equal 84,328 km (52,398 mi).

I personally don't have the math to define the exact velocity imparted at 84,328 km, and my research did not find a calculated value for this. But values for just above and below this are available.

Wikipedia lists a few "free" destinations at release heights just above GSO. An object attached to a space elevator at a radius of approximately 53,100 km will be at escape velocity when released. Transfer orbits to the L1 and L2 Lagrangian points can be attained by release at 50,630 and 51,240 km, respectively, and transfer to lunar orbit from 50,960 km

For destinations beyond the earth moon system, the calculations get a bit more complex, as you need to also consider the sun. In the same manor the elevator gives velocity in relation to it's height above earth, its position in relationship to the sun also effects relative velocity.

An elevator cable of somewhat over 100, 000 km in length should suffice as a sling to launch spacecraft to Jupiter (at the outer end) and Mercury (at the inner end). Reaching Jupiter is critical, because we can take advantage of Jupiter’s gravity assist to send spacecraft further outward or even beyond the solar system.

Aravind, P. K. (2007). "The physics of the space elevator". American Journal of Physics (American Association of Physics Teachers)

As we see velocity or obtainable destinations is simply a matter of height. Or said another way, the sky really is not the limit, the higher you go the faster and the farther you can fly.


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