A follow-up on this old question.

If you've got ridiculous amounts of energy to play with, and not much reaction mass, a photon rocket is the way to go for moving your spaceship--it gives you the best possible power-to-thrust ratio.

Based on the comments section, this apparently needs some proof, so:

The relativistic energy equation is $E^2 = E_K^2 + E_M^2 = (pc)^2 + (m_0c^2)^2$ Note that, for a fixed fungible energy budget, the more energy we lock up in mass, the less there is to allocate to the kinetic energy/momentum term. Thus, the maximum amount of momentum per unit energy, and therefore the maximum thrust per unit power, is obtained when mass goes to zero, as we are left with the luxon energy-momentum relation, $E = pc$, or $p = E/c$, and the corresponding power-thrust relation, $P = Tc$. We only use rockets that throw mass out the back because we don't have a fungible energy budget--the mass-energy of, e.g., liquid hydrogen fuel isn't something that we can easily turn into photons. If it were, it would produce more thrust! So if you have a ton of energy and not much reaction mass, it is always a losing proposition to try turning that energy back into mass.

But, there's a problem: when all of the energy in your exhaust is kinetic energy rather than mass energy, it becomes much more noticeable! To merely suspend 1 kg of mass against gravity at the surface of the Earth, you have to provide 9.8 newtons of thrust--which corresponds to 9.8 newtons * 299,792,458 m/s = ~2.938 gigawatts of radiative energy. That's one frickin' powerful flashlight, just to suspend one kilogram. Actually accelerating upward, or suspending a ship weight many thousands of kilograms, requires proportionately more power, and exawatt flashlights tend to turn anything behind them into rapidly expanding plasma. If you want to launch from a planet... or just be in a solar system where you might at some point accidentally point your engine towards a planet... that's inconvenient, to say the least. Fortunately, it turns out photons aren't uniquely special--any luxon will do, as they all have the same energy-momentum relation! Including, for example, gravitons. So, if we have a magical means of producing high-power, collimated gravitational waves, they'll produce exactly the same amount of thrust, and interact with matter much less strongly, so we don't vaporize the spaceport when we take off!

Or... do we? Gravitational waves still do interact with matter a bit, so some energy will be transferred into the ground, and the rest of the Earth, if we fire a gravitational wave beam into it. Which raises the question: just how powerful can we make our graviton rocket before it becomes dangerous?

  • $\begingroup$ Wait, what exactly is wrong with turning energy into mass again? $\endgroup$
    – Mathaddict
    Apr 12, 2023 at 16:27
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    $\begingroup$ "...so we don't vaporize the spaceport when we take off" Take off using an appropriate amount of power for conditions and then throttle up when you're in deep space. Do you drive 75mph in a parking garage? $\endgroup$
    – Cadence
    Apr 12, 2023 at 16:46
  • $\begingroup$ @Mathaddict It reduces your power-to-thrust ratio. $\endgroup$ Apr 12, 2023 at 16:55
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    $\begingroup$ @Cadence You can only throttle down so much if you want to actually get off the ground. It takes 3 gigawatts per kilogram to hover at the surface of Earth with a photon rocket. If you want to actually lift off, you need more power than that. If you want to lift 1 tonne, that's 3 petawatts slamming into the ground, at minimum. $\endgroup$ Apr 12, 2023 at 16:59
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    $\begingroup$ Gravitational waves probably have gravitational fields, similar to how EM waves have electrostatic & magnetic fields. GWs are also affected by GFs, showing self-interaction. A high enough energy density (and corresponding gravity) of the GW can cause destruction. Barring that, seriously high energy GWs can start separating quarks, pulling new ones out of the vacuum. $\endgroup$
    – BMF
    Apr 13, 2023 at 14:00

1 Answer 1


I found this relevant Physics StackExchange answer from a few years ago, which references the paper Seismic Response of the Earth to a Gravitational Wave in the 1-Hz Band. Now, you can dump a lot of energy into the entire Earth without ill effects, but it turns out that gravitational wave absorption only occurs at density discontinuities--so, the surface of the Earth, and the core-mantle boundary. The Earth is supposed to absorb about $10^-21$ of passing gravitational waves at 1Hz, with absorption varying as the inverse square of frequency--higher frequencies are absorbed less strongly. So, as a first approximation, we can guess that a specific boundary will absorb one quarter of that, but this is all super-approximate, order-of-magnitude stuff anyway, and absorption will be higher for more elastic materials (like, say, squishy human flesh), so for safety margin calculations I think we can just go with "$10^-21$ of a 1Hz wave will be absorbed at each surface of a body", and it's fine if that's an overestimate.

We need about 600 degrees to start melting rocks, and the heat capacity of silica is about 733 J/kg, which works out to around 440kJ to melt a kilogram of rock as a rough estimate. If we could melt a kilogram of rock in less than a minute, that seems like a pretty decent estimate for "power level that's dangerous to the spaceport"--we don't want to melt, or significantly weaken, the launchpad underneath the rocket, for example! So, we're looking at an absorption rate of around 7.3 kilowatts. That means that, if we used 1Hz gravitational waves, we could safely push the engine power up to 7.33 yottawatts, applying a thrust of 24.45 petanewtons.

That's, uh... quite a lot. Now, there is reason to believe that, if gravitons can be artificially produced at all, they can be produced at fairly high frequencies, in the gigahertz range, which would cut down on the absorption coefficient by an additional factor of 10^18. Which pushes us well into the range of "you could pour the entire power output of the Sun into this drive and the exhaust would still not be dangerous to squishy humans", so I think we can reasonably say:

A graviton rocket will not be dangerous to anyone or anything caught in its wake, until you are at the level of needing to apply multiple gs of thrust to entire planets.

  • $\begingroup$ I believe this is misinterpretation. The graviton wave is applied to the entire planet. Your rocket is applying it to the hundred or so square meters beneath the vehicle. Sunlight strikes the Earth at about 1360 watts/meter^2, which isn't catastrophic. But apply all of it to just one square meter and we get a catastrophic 346 petawatts. -1 until you compare apples to apples (and you can't apply petanewtons to anything but the entire planet and not be catastrophically destructive.) $\endgroup$
    – JBH
    Apr 12, 2023 at 23:06
  • $\begingroup$ @JBH I honestly cannot understand how this comment relates to my answer, or what kind of misunderstanding of the problem has caused you to make it. $\endgroup$ Apr 12, 2023 at 23:09
  • $\begingroup$ At no point did I suggest that applying petanewtons to a rocket would not be catastrophic, or that that's something you would actually want to do--only that the power levels required to do that would not be catastrophic for a planet below, given how little of that beam power would actually be absorbed. $\endgroup$ Apr 12, 2023 at 23:12
  • $\begingroup$ @LoganR.Kearsley I think JBH is taking issue with your assumption that "if Earth's surface absorbs 10^-21 of a passing gravitational wave, then the surface of any object would absorb 10^-21 of a passing gravitational wave". And also with your assumption that a result calculated for a planet-sized wave passing over the entire Earth would equally apply to a very intense and tightly focused wave aimed at a launchpad-sized patch of Earth's surface. $\endgroup$ Apr 12, 2023 at 23:17
  • $\begingroup$ @SomeoneElse37 In that case, I would simply call attention to the fact that I already made it clear that these are at best order-of-magnitude estimates, and that the main takeaway point is that the safety margins are so freaking huge that even being off by a factor of 1 billion would make no practical difference. $\endgroup$ Apr 12, 2023 at 23:20

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