Would we have to build a orbiting launch station that uses centripetal force to launch the rockets? Or a station that freely floats through space not in a planets orbit? Would something like this cause the station to crash or would it work? If it would work would it be an effective way to launch?

  • 2
    $\begingroup$ This is supposedly launch from Earth and so a different question, but it's a similar concept: How might SpinLaunch actually spin something fast enough to launch it into orbit? $\endgroup$
    – uhoh
    Commented Jan 26, 2021 at 14:59
  • $\begingroup$ slightly related: What is the largest delta-v ever produced in space from mechanically stored energy? spinning up a rocket before releasing it would be an amazing amount of mechanically stored energy! $\endgroup$
    – uhoh
    Commented Jan 26, 2021 at 15:09
  • $\begingroup$ I think it's actually more accurate to say "centrifugal force" in this case. That "force" is the tendency to "flee the center" that you observe in the rotating frame of reference. Viewed in that frame of reference, that force would be the impelling force. The centripetal force is what keeps the parts of the spinning launcher from flying away. Definitely still a reasonable question. $\endgroup$ Commented Jan 26, 2021 at 22:13

1 Answer 1


I was going to say 'don't be silly, conservation of momentum', but, well.

If you have some large orbiting thing, then if you're happy to launch projectiles in opposite directions (which probably means throwing one of them away), you could launch pairs of projectiles in opposing directions without mucking around with your linear momentum. You want to make fairly sure that neither of the pair is aimed back at Earth I suspect (or that you are sure it won't make it through the atmosphere).

So to to that you need to find the energy. As you suggest, one way would be to spin some system up and then release projectiles from it. This has the problem that you now need to find some angular momentum. Well, you can avoid that by having a pair of counter-rotating wheels which you spin up with respect to each other, leaving the overall angular momentum zero. This is nice because you can find the energy to do this from an electric motor, which you can drive from sunlight, and if you can make the bearing friction low enough you can spin the thing up over a very long period of time so the power requirements can be tractable. I am pretty convinced that to avoid balance problems (you need the wheels to remain individually balanced and you need to avoid their angular momenta to remain equal and opposite during launch) you now need to launch projectiles in groups of four in two pairs of opposing directions.

This would I think not be suitable for launching humans, as the forces prior to release would kill them unless the structure was huge. One way to see this is that, for a system of radius $r$, rotating with angular velocity $\omega$, then the tangential velocity is $r\omega$ and the acceleration is $r\omega^2$. So for a given desired launch velocity $v_l$, the acceleration just before launch is $a_l = v_l^2/r$. If you want a launch velocity of $7\,\mathrm{km/s}$ (about LEO orbital velocity, and plausibly enough to launch something out of the Solar system) and the structure has a radius of $1000\,\mathrm{m}$ (this is 1 kilometre: it's huge!) then $a_l = 49\,\mathrm{km/s} \approx 50000g$: this is not survivable, by a very long way.

Another way to ask this is: how large does the thing need to be to control the launch acceleration? Well, the answer is $r = v_l^2/a_l$, so if you want, again $v_l = 7\,\mathrm{km/s}$ and you think people could survive $a_l = 5\,g$ in the run-up to the launch (they presumably could climb 'down' into the spacecraft so they don't need to spend however long it takes to spin the thing up under that kind of stress), then you get $r\approx 1000\,\mathrm{km}$. That's about $1/6$ the radius of the Earth: it's absolutely vast.

And, more to the point, it probably doesn't really help that much for two reasons:

  • this thing starts from LEO or higher, and lifting mass into LEO is at a very large part of the problem of getting anywhere else (and remember that, as well as the extra mass for the projectiles you are throwing away, you have to lift the whole launch assembly into LEO to do this and it's going to need to be very strong and pretty large, so very heavy);
  • to launch something with a very high velocity will require materials which may be implausible.

I suppose it's also worth adding that this system might look ominously like a kinetic energy weapon to people. But perhaps it doesn't look any more like a weapon than anything else which can launch things from orbit, I don't know.

Why I don't think this would be useful for orbital changes in Earth's orbit. It's tempting to think that a system like this, with a much lower launch velocity, would be useful for changing orbits. I don't think it would be. As mentioned above, in order to conserve momentum you really need to fling out an equal and opposite mass to the one you are launching. This means that you need to lift just double the mass you are going to launch.

Well, what if, instead of lifting this mass you just lifted fuel for a rocket? Well, the rocket equation tells us that

$$\frac{m_o}{m_f} = e^{\frac{\Delta v}{v_e}}$$

Where $\Delta v$ is the change in velocity you need and $v_e$ is the exhaust velocity. If we take $v_e = 4\,\mathrm{km/s}$ (I think this is approximately what the S-IVB achieved, and would be much lower than some low-thrust ion-type drive), then we can use the fact that for a given circular orbital velocity $v$ the $\Delta v_E$ you need to escape is $v(\sqrt{2} - 1)$ (thanks to PM2 Ring for this!). So for a $7\,\mathrm{km/s}$ orbital velocity (somewhere in LEO) $\Delta v_E \approx 2.9\,\mathrm{km/s}$. And we can plug these numbers into the rocket equation:

$$ \begin{aligned} \frac{m_o}{m_f} &= e^{\frac{\Delta v_E}{v_e}}\\ &\approx e^{\frac{2.9}{4}}\\ &\approx 2.1 \end{aligned} $$

In other words the amount of extra mass you would need to lift as fuel is very close to the same as the amount you would need to lift for this launch thing. And if you lift it as fuel you don't also need to lift the launch platform. And if you use an ion drive or something you can get much better $m_o/m_f$ as well.

I'm wondering if you could do some thing with unequal arms on the spinny thing to fire off much smaller masses than the thing you are launching, but I think you can't conserve angular momentum that way. But perhaps I'm wrong about that.

  • 1
    $\begingroup$ For these interplanetary missions getting to LEO was less than half of the problem. $\endgroup$
    – uhoh
    Commented Jan 26, 2021 at 15:08
  • 1
    $\begingroup$ @uhoh: yes, good point! I've changed the wording to not say more than half, which is clearly false, (but added that you need to lift the launcher, which won't be small or light...) $\endgroup$
    – user21103
    Commented Jan 26, 2021 at 15:16
  • 1
    $\begingroup$ If a person could go into something like hibernation/cryo sleep would the force of a launch like this effect the human body the same way? $\endgroup$
    – user39045
    Commented Jan 26, 2021 at 16:36
  • 1
    $\begingroup$ @user39045: I don't think so. See my amended answer, but for plausible-sized launchers the accelerations just before launch are awful. $\endgroup$
    – user21103
    Commented Jan 26, 2021 at 18:02
  • $\begingroup$ Why do you want an extra 7 km/s? You could transfer to a higher or lower Earth orbit for a lot less. At a given orbit radius, escape velocity is $\sqrt 2$ times the circular orbital speed, and $7\sqrt 2\approx 9.9$. Earth's orbital speed around the Sun is a little under 30 km/s, and $30\sqrt 2\approx 42$, so (30 + 7 + 7) km/s all combined in the tangential direction to Earth's orbit gives you enough speed to escape the Sun! $\endgroup$
    – PM 2Ring
    Commented Jan 28, 2021 at 12:31

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.