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Does the Oberth effect only apply when in orbit of a planet or would a rocket generate more and more thrust (if kept on) even in deep space? Wikipedia explains that the faster the rocket goes, the more useful energy it generates. But since speed depends from what point of reference you are measuring it, how can that be possible since the fuel doesn't know the rocket's speed?

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OK, kinetic energy is not a frame independent idea, and you should be careful about applying any analysis that depends in KE. There will be times when it is useful, and other times when it simply enables you to fool yourself. The thing is, applied at the right time this analysis can lead to real insight, so it would be silly to dismiss them outright. The thing you need to figure out is "It what frame would this mean something real?". –  dmckee May 1 '13 at 20:43
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The Oberth effect takes advantage of when you add something to a quantity that is squared then the effect will be larger, the larger the quantity is initially. It works when a body is under the gravitational influence of another, usually larger e.g. a spacecraft around the Earth.

In the central body approximation, the energy equation of a (keplerian) orbit around a central body is, at a certain instant

$$E_0=\frac{v_0^2}{2}-\frac{\mu}{r_0}$$

where $v_0$ is the velocity, $r_0$ the distance to the central body and $\mu$ the gravitational constant (gravitational constant times the mass of the central body). The energy will remain constant while there is no maneuvers.

A sufficiently fast maneuver can be considered as an instantaneous change in velocity $\Delta v$, while $r_0$ remains approximately the same (around the Earth maneuvers usually take only a few minutes, which is much shorter than the orbital period around 100 minutes in low Earth orbit). So the energy after the maneuver is

$$E=\frac{(v_0+\Delta v)^2}{2}-\frac{\mu}{r_0}$$.

Now the larger the initial $v_0$, the larger the change in energy $E-E_0$.

Instantaneous maneuvers are an approximation. You can think that if the change in energy is different with the same maneuver (by making the maneuver at a different distance $r_0$) then something is wrong. Actually when performing the maneuver some propellant is expelled (and the mass of the spacecraft changes) and if one accounts for the energy of all, spacecraft and propellant, the change in energy is the same.

It can also work the other way round with a $\Delta v$ opposite to the initial $v_0$ (braking) and loosing more energy the larger the initial $v_0$.

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Wiki also mentions that a faster rocket generates more energy, does that only apply when in a gravitational well? –  this May 2 '13 at 8:31
    
"does that only apply when in a gravitational well?" No. Clearly you can set the gravitational term to zero and the energy difference $E-E_0$ doesn't change. –  Michael Brown May 2 '13 at 8:58
    
Edited the answer a little bit trying to be clear. A gravitational well is required for what it's called Oberth effect, as I see it, otherwise it will only be an effect of changing the kinetic energy by changing reference frames. The effect is hidden in the details as in reality the mass changes and so the potential energy is affected. –  pjsgil May 2 '13 at 9:20
    
A gravitational well isn't the only way to define a reference frame. Whether you want to call it the Oberth effect or not in that case is, I guess, a question of semantics. –  Michael Brown May 2 '13 at 10:14
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