Imagine we have a sufficiently tall space fountain positioned at either of the Earth's poles, such that the upper station is permanently within view of the sun. A large array of solar panels could be connected to that upper station to provide a very consistent source of energy.

The challenge with this idea is the problem of most space-based solar - how do you get energy to where it's needed on the ground. I'm wondering if it might be possible to use the rotating masses that support the station to transmit the energy it produces (and whether doing so is superior in efficiency and/or ease of implementation to lasers + photovoltaics).

Traditionally with a space fountain the stream of masses are accelerated back up to speed at the base station. If instead we wish to do the accelerating at the top, we can have each mass launched upwards with just enough force to reach the upper station, and then accelerate it around and down fast enough to balance the force of gravity. At the bottom of the cycle, we leech energy out of the masses instead of boosting them, leaving just enough for them to make it back to the station.

In actuality there would probably be multiple streams of masses to increase the transmission capacity and avoid putting a net torque on the station, but I'm not worried about implementations details at the moment - just whether the system would work as described and whether or not it's obviously bad in some way I've overlooked.

  • $\begingroup$ But what's the advantage of having solar panels in space??? They're just about as efficient on the ground. Why would you do this? $\endgroup$
    – Fattie
    Oct 10, 2019 at 15:52
  • 2
    $\begingroup$ @Fattie on the ground solar panels only produce energy a fraction of the time because of clouds, night, etc. In space they can run at 100% output all the time because there's never anything between them and the sun. $\endgroup$ Oct 10, 2019 at 17:47
  • $\begingroup$ Fair enough. But at a miniscul fraction of the costs you could build, say, 5 on-earth stations such that it would be extremely unlikely they are all clouded. (Or I guess just build one (tiny) nuclear plant that would produce vastly more power.) $\endgroup$
    – Fattie
    Oct 10, 2019 at 19:46

2 Answers 2


The top of your fountain is still well within Earth's gravity well. That means that as you add more solar panels to generate more power, the power requirements of the fountain rise.

Once a fountain is flowing nicely, the losses needn't be very high... magnetic deflection without trying to bleed off energy is nice and efficient. Someone suggested that losses would be similar to a flywheel with magnetic bearings, or a few percent of the energy stored in the fountain every hour (maybe less with fancy future super tech). The station at the top has a gravitational potential energy $E = WL$ where $W$ is its weight, and $L$ is its altitude. A hundred tonne station at 100km therefore has a potential energy of about 100GJ, and the kinetic energy in the fountain must equal that (or the situation isn't stable). 1% loss of kinetic energy every hour therefore represents ~280kW, or 2.8W/kg. This scales linearly with height and weight (but not mass!)... ten times the altitude, 10 times the power.

Decay of a fountain isn't interesting though. You're putting in more energy than you're losing, so you have to siphon some of it off to stop the space station blowing itself away. Efficiencies of adding power to and extracting power from the mass stream come into play... some of that power you're pushing in will be lost as heat at both ends of the system. Lets say you have a round-trip efficiency of 90%, so for every 1kW you harvest at the top, you get 900W out at the bottom.

A modern lightweight solar cell can exceed power densities of 1kW/kg, but there's quite a lot of infrastructure that needs to go with it... you can't just hook it up to a pair of cables and call it a day. On the bright side, at 100km altitude you need a little over 3W/kg to keep your station aloft, and even quite conservative estimates for the weight of a solar powersat have 20kg/kW, which is an order of magnitude more than you need to simply break even, even with lower power-transmission efficiencies and higher fountain energy loss rates. This suggests your crazy plan might actually work! That's not to say that it is a better powerstation than more easily engineered alternatives, but it does show that a polar space fountain could be a considerable net source of power.

So, now the major technical objection appears to have been removed, all you need to do now is to install a superconducting HVDC cable from the poles, through particularly hostile and inconvenient terrain, with powerstations that have to operate in exceptionally cold temperatures and seriously strong winds (which aren't going to do your fountain any favours, that's for sure) and justify the expense and effort of the whole thing compared to, say, wrapping superconducting HVDC cables around the equator and using conventional terrestrial solar instead. That might be more of a challenge!

Is this better than a conventional solar powersat design? Well, probably not. Powersats scale trivially, and don't require megascale construction projects in some of the most inconvenient and inhospitable places on earth. If you can build space fountains, then low-cost non-rocket spacelaunch technologies are also clearly within your remit and launching power satellites can be cheap enough to be competetive with ground based power systems, even with the many inefficiencies involved in wireless power transmission.

Powersats can retarget their energy supply on demand, more or less, and you don't need a huge joined-up globe-encircling superconducting power grid in order for everyone to take advantage of the power they supply. There's a lot of be said for them once they're cheap enough to be practical. Your power fountain is a neat idea, but unlikely to be economically competetive.


Just in the abstract, I guess if you spun massive objects, and then moved them somewhere else, and then took the energy otu of them, you can indeed use that as a sort of battery.

This is: a flywheel battery. That's a normal thing you can buy.

So, quite simply, your idea could be achieved using "flywheel batteries"

(Your idea of using the energy to send them "down instead of up" is not sensible, as explained above.)

But note too! Instead of flywheel batteries, you could just use .............


It's that simple.

The object you are flinging could be nothing more than millions and millions of these:

enter image description here

(note - you wouldn't bother with the wrapper)

somehow charge them at the top (perhaps using wireless charging, so in fact, just use billions of android phones), and away you go.

  • $\begingroup$ The masses in a space fountain don't fall due to gravity - they're essentially 'bounced off' the bottom of the upper station, only using magnets to curve their trajectory in a 180 instead of physical collisions. The upward force comes from launching mass out of the bottom, which is equally possible if the boosting section is at the top. $\endgroup$ Oct 10, 2019 at 17:50
  • $\begingroup$ The masses are deflected at the top in order to impart a force to the object being held up by the fountain. If they had lost all momentum and just fell back, they couldn't impart any force and couldn't hold up the top of the tower. $\endgroup$ Oct 10, 2019 at 19:13
  • $\begingroup$ Guys, I think your comments are fair ! $\endgroup$
    – Fattie
    Oct 10, 2019 at 19:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .