0
$\begingroup$

I thought up a device that uses the Oberth effect (wiki) to generate more energy than it consumes. Obviously, I made a mistake/logical error somewhere. Can you help me find it? Here's the thought experiment:

free_energy

  • Suppose we have a mass $m$ that can move in a straight tube without any friction. Also, we disregard gravity and any relativistic effects

  • The mass is accelerated by a stationary laser at one end of the tube. A laser beam is directed to the mass, which acts as a light sail, resulting in the thrust force: $F$

  • The laser consumes a constant amount of power: $P_{laser}$

  • The mass is accelerating constantly at: $a=\frac{F}{m}$

  • The mass's velocity is increasing by: $v(t)=at$. Inserting $a$ from above: $v(t)=\frac{F}{m}t$

  • The kinetic energy of the mass is given by $W_{kin}=\frac{1}{2}mv^2$. Inserting $v$ from above: $W_{kin}(t)=\frac{1}{2}m(\frac{F}{m}t)^2 = \frac{1}{2m}F^2t^2$

  • The accumulated energy consumed by the laser over time is $W_{laser}(t) = P_{laser}*t $

  • The kinetic energy grows quadratically with time, while the laser energy grows linearly with time. This means that the kinetic energy will eventually exceed the laser energy

  • Energy "break even" is reached, when $$W_{kin} = W_{laser}$$ That is, when $$\frac{1}{2m}F^2t^2=P_{laser} t$$ at time: $$t=\frac{2m}{F^2}P_{laser}$$

After this point in time, leaving the laser on for longer will increase $W_{kin}$ by more than the energy the laser consumes. For any combination of mass, laser power or light sail force, this point will be reached if we wait long enough.

So let's leave the laser on for a bit longer than the "break even" point. Then, the mass with $W_{kin} > W_{laser}$ is decelerated, and $W_{kin}$ harvested by a perfectly efficient linear generator (or any other suitable generator). It charges a battery which provides enough energy for the laser in the next run and even additional energy. (The generator and the battery can even be arbitrarily inefficient, as long as the device runs long enough until $W_{kin}$ is sufficiently larger than $W_{laser}$.)

This seems impossible. Where did the additional energy come from? What is going on here?

$\endgroup$
1
  • $\begingroup$ I'm pretty sure the Oberth Effect is for rocket engines that actually expel propellant to move, so I don't think it's really the Oberth effect when you use radiation pressure. Also, you may want to look up the effects of velocity on radiation pressure, which might be part of the disconnect. $\endgroup$
    – JMac
    Dec 25 '20 at 16:09
3
$\begingroup$

The mistake is in assuming that the force is a constant if the power is a constant. This is not the case. The faster the mass the more redshifted will be the light that reaches it, an the less energy it will impart. Another way to imagine it is that the lase is a bunch of cannons. The faster the mass, the less will be the relative speed between the bullets and the mass, and so less the transferred momentum. So at a constant power the laser produces a diminishing force with time. There will never be excess energy.

$\endgroup$
2
  • $\begingroup$ Thanks! But would it make a difference if the laser were to be mounted on the mass, shooting backwards and accelerating with the mass? (as in: more like a rocket?) $\endgroup$
    – odin1337
    Dec 25 '20 at 19:00
  • $\begingroup$ No, it will be the same. From the point of view of a stationary observer, the faster the speed of the mass, the more redshifted and lower power the laser will be. In the analogy with the balls, the faster you go, the less the receding velocity of the cannonball relative to the stationary observer, and this, again, means less momentum transfer. $\endgroup$
    – user65081
    Dec 26 '20 at 2:09
-1
$\begingroup$

That happens because $P_{diss}$ depends on t (fact that you are not considering), infact $P = Fv = F*\frac{Ft}{m}$ and knowing that $W_{laser} = \int P dt = \int \frac{F^2t}{m}dt = \frac{F^2t^2}{2m} = W_{kin}$. So the conservation of energy is verified.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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