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In two recent questions about space elevators, the topic of reaching non-geostationary orbits came up, particularly the popular Low Earth Orbits.

This question is simply about how hard exactly it is to reach LEO assuming you have a space elevator available. (The motivation is mostly for quick reference, comparing to how hard it is with a conventional launch right from the ground to LEO.)

Specifically, I'm asking where you would want to drop off from the elevator, what maneuver[s] you would perform to end up in a circular prograde equatorial LEO (let's say apogee = perigee = 1000 km), and how much Δv those burns would require. This is assuming ordinary chemical propulsion (i.e. near-instantaneous burns). We're not counting the energy to climb the elevator in the first place.

leftaroundabout
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  • Can you describe your own idea of 'a space elevator' in detail? Are all space elevators to be considered if not identical, at least usefully similar? Does that generalisation also apply all rockets or all rail guns?

    When you're comparing the elevator to how hard it is with a conventional launch right from the ground, does that mean a rocket launch?

    Even then, do all rockets aim for the same acceleration? For comparison, might it be easier to talk about some kind of standard acceleration, amended by either theoretical or historically recorded deviations therefrom, up or down?

     – Robbie Goodwin Jan 19 '24 at 21:26
  • @RobbieGoodwin you're asking a lot of questions. But basically, they can all be answered thus: the concept of Δv is used specifically because it abstracts away all such technicalities and allows for a fair, implementation-agnostic comparison of how difficult a maneuver is. (There are some caveats here, e.g. ion drives have simply to little thrust to perform the Hohmann-type orbital transfers the way a chemical rocket can.) – leftaroundabout Jan 19 '24 at 21:33
  • Thanks you seem to be suggesting Δv represents energy while I've been taking it to mean acceleration. Was I wrong all along? Am I wrong now?

    Either way, are you really suggesting first that all space elevators, all rail-guns or all rockets must behave the same way? That would seem to make the Question purely theoretical, ruling out design considerations such as efficiency?

    If you yourself abstract away all such technicalities to allow a fair, implementation-agnostic comparison, what Δv do you suggest is needed to reach LEO from a space elevator… and from any other technology?

     – Robbie Goodwin Jan 19 '24 at 22:01
  • It represents neither acceleration nor energy. Think of Δv more as the space travel analogue of what distance means for road travel. Obviously a heavy truck will require more energy to travel between two cities than a bicycle does - but for any given vehicle, the more distance you need to cover, the more energy you need. In space, distance does not play such a role because you just keep on coasting. Instead, Δv is what determines how much energy you need, though it's not a proportional but exponential dependency. – leftaroundabout Jan 19 '24 at 22:11
  • Well, thanks. There I thought 'everyone knew' Δv directly represented change in velocity, brought about by change in energy. Silly me! – Robbie Goodwin Jan 19 '24 at 22:31
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    "Delta V" is essentially the integral of the magnitude of acceleration due to your ships propulsion. It's a useful concept in mission design because pointing your ship in the desired direction is free/cheap and "undoing" a change in velocity is as expensive as "doing" it in the first place. – Peter Green Jan 19 '24 at 23:45
  • @RobbieGoodwin I mean, Δv does directly represent the change in velocity needed to perform a maneuver, but Δv doesn't care what the acceleration is. We can just assume all accelerations are infinitely high and take zero time (perhaps we fire our spacecraft from a cannon). For comparison, if you're going from a stop sign to freeway speed in your car, the Δv is 65 mph no matter what the acceleration was -- how long it takes to get there doesn't change the 65 on your speedometer. – Darth Pseudonym Jan 24 '24 at 19:01
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    Or to put that another way, Δv tells you the minimum energy cost to do something. It will cost you more energy than that, depending on your particular propulsion system, but that's because you're looking at losses associated with the specific details of performing the maneuver with your particular drive (i.e. you have to spend some energy that isn't going to translate into final velocity change). It's possible that the losses are so high that you can't perform the maneuver at all. But none of that changes the Δv you're aiming for. – Darth Pseudonym Jan 24 '24 at 19:11
  • Thanks and does that not conflict with Erin Anne's Answer, below? – Robbie Goodwin Jan 29 '24 at 19:02

1 Answers1

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A better, but not the best, answer: 2 km/s

Since Woody put up an answer I consider absurd, here's a first cut that shows if you actually use your space elevator to go sideways fast instead of just using it to go up, the necessary delta-V to do afterwards drops precipitously.

My answer builds a lot on the work done in this answer which Woody linked to in a comment--and which contains the relevant equations nicely laid out--and puts it into a handy spreadsheet to help my, frankly, tragedy-addled brain.

The short version is that by releasing at a high apogee, which I computed (using the spreadsheet) to be

$r_\text{apogee} = 30,731,910 \text{m}$

you fall to a perigee altitude of very close to 1,000km, where your velocity is

$v_\text{perigee} = 9,334 \text{m/s}$

You can then burn for a measly $-1984 \text{m/s}$ to arrive at the circular orbit velocity for 1,000km altitude. Easy peasy obviously-don't-step-off-the-elevator-at-your-target-altitude squeezy.

Is that the best you can do? Absolutely not, you could set your initial perigee somewhat lower to aerobrake instead, but that answer becomes much more condition-dependent.

Is a space elevator the best way to serve LEO? No, absolutely not, but after seeing like five separate very smart people on this site independently suggest that you should step off low and slow I had to at least write this.

I put a simple Newton-Raphson solver into the sheet and now I've also covered a range of altitudes from 100km to 10,000km, which I will claim thoroughly covers LEO, to show the various delta-Vs and release apogees.

Chart of necessary delta-V and release apogees for target altitudes between 100km and 10,000km

Erin Anne
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