I'm investigating the idea of transporting gigawatts of electrical power to the first stage booster of an electrical rocket.

The first stage booster, like Falcon 9's recoverable first stage, stages at 70km altitude, 60km downrange, for a diagonal distance of 92km.

I'm using 100km for a nice, round figure.

If we use a 1000 Isp electro-thermal rocket to launch a 10 ton payload with 25 ton upper stage and 20 ton lower stage (ignoring engine and fuel tank masses), and assume a TWR of 1.2 at liftoff, we would need 3.17GW of power at liftoff.

How can this level of power be transported over a wire 100km long? Low mass is critical.

  • 1
    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Nick Alexeev
    Feb 28, 2017 at 21:04
  • 1
    $\begingroup$ Wire. You have a different problem than the Rocket Equation ... you have to lift an increasing weight with flight time. Wire is also specified by breaking length : the length of wire that weight enough to exceed its strength - say 5-10 miles. You'll need a booster every 5 miles or so to support the next 5 miles of wire; that will naturally consume its own power. Voltage will be limited by the atmosphere's dielectric strength... $\endgroup$ Feb 28, 2017 at 21:25
  • 2
    $\begingroup$ More practically, I wonder how shining the 3-gigawatt laser from the ground right up the nozzle and throat, into SRB-like combustion chamber could work. We don't need to transform the beam into electricity and then back into heat, just use it as heat source for the propellant directly. $\endgroup$
    – SF.
    Feb 28, 2017 at 22:31
  • 2
    $\begingroup$ You'd need a beam that the propellant is opaque to, but the exhaust plume is transparent to. $\endgroup$ Feb 28, 2017 at 23:05
  • 2
    $\begingroup$ @uhoh: besides... you're fine on sitting on a big barrel of explosives that burn at 3 gigawatt right below your seat, and can't be extinguished until they burn out entirely, but you fear a millisecond glance of a 3GW ray?? $\endgroup$
    – SF.
    Mar 1, 2017 at 14:35

2 Answers 2


By wire is going to be extremely difficult. Let us first have a look at the required cross section of the wire:

$$A = \frac{\rho lP}{V^2}$$

We can use the resistivity value of graphene for a very optimistic scenario. Even with a voltage over 20kV, the wire is still going to outmass the rocket by far. Not to mention the wire has to tolerate exposure to rocket exhaust.

edit: The above calculation is possibly incorrect, as pointed out by Uwe.

A simpler argument is that for the wire not to outmass the spacecraft, it can be no more than $0.5kg/m$. Given the size of high voltage wires carrying an order of magnitude less power, that is simply not possible.


Beaming up the power, as others have mentioned, is not likely to be any more successful, as current laser technology simply is not sufficient. Continuous lasers are at the moment only a few kW, roughly five orders of magnitude short.

Having the energy source onboard, however, is not as impossible as it sounds. A Nuclear thermal rocket may be something you want to look at. Reactor tests at your magnitude of power were conducted during Project Rover.

  • $\begingroup$ P in your equation is not the power needed at the rocket, it is the power loss in the wire. If P is the power for the rocket, all power is lost in the wire. A is the cross area of the wire, not the diameter. Of course two wires are needed, not only one. It should be considered if the wire is melted by the power loss. $\endgroup$
    – Uwe
    Mar 2, 2017 at 12:04
  • $\begingroup$ @Uwe I had cross section in my head but wrote diameter. Your observation about powerloss is correct, so I have to reconsider the equation. $\endgroup$ Mar 2, 2017 at 12:13
  • $\begingroup$ Megawatt lasers are probably doable, and you could use an array of them. Consider that most development of such lasers has been for military purposes, where they need to be transportable. $\endgroup$ Mar 2, 2017 at 13:55
  • $\begingroup$ The "high volatge wires" is a cable to be laid under earth. Cables for use in air look different. An earth cable used in air under nominal power would overheat because thermal transport in air is less than in earth. Especially the air in the hight between 10 and 100 km is to thin for the transport of heat loss. $\endgroup$
    – Uwe
    Mar 2, 2017 at 16:35

I'm going to suggest a method that might work, but is highly speculative and completely untested.

Simply put use a laser induced air plasma electrical conductor to transmit the power.

In slightly more detail: Use a high power laser to excite a tube of air from the ground to the rocket into a plasma state and then electrically energize the plasma. (You would actually need two plasma tubes for a complete circuit)

Most people would easily recognize an atmospheric plasma electrical phenomenon like this as lightning.

Note: the big light in the video is not the lightning, it is the laser projecting onto the wall, the lightning is the tiny dot of light.

Side note: the laser is likely to use way more than 3 GW to ionize the air and tracking the rocket with a moving plasma cylinder would be very challenging to say the least, and by the time we are good enough at plasma physics to do this we could likely build fusion rockets, but I find there is something inherently awesome about the idea of using artificial lightning to power a rocket.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.