Q: How does the Oberth Effect work, and where does the extra energy come from? Why is it better for a rocket to fire at the lowest point in its orbit?

Physicist: When a rocket fires it increases its speed by some fixed amount called the “ΔV” (delta V).  If the original speed is W, and the rocket speeds up by ΔV, then the change in energy is: \frac{1}{2}M(W+\Delta V)^2 - \frac{1}{2}MW^2 = \frac{1}{2}M(\Delta V)^2 + MW\Delta V.  So here’s the problem; notice that the increase is not just \frac{1}{2}M(\Delta V)^2, but is instead bigger when the initial velocity is bigger.  Isn’t that weird?

The change in energy is greater for the same change in speed for different starting speeds.

Kinetic energy vs. Speed: The change in energy is different for different starting speeds even if the change in speed is the same.

This comes up a lot when you’re shooting rockets around the solar system.  If the rocket has a big elliptical orbit, then it’ll be moving quickly at the closest part of the orbit and slowly at the highest part of its orbit (just to be confusing, these points are called the “periapsis” and “apoapsis”).  So, very weirdly, if the rocket fires when it’s moving the fastest it’ll pick up more more kinetic energy and it’ll be able to get higher and farther.  But if the rocket fires when it’s moving slowly it gains less energy, and at the most extreme, if W = -ΔV/2, then the energy doesn’t change at all (the rocket just switches direction).  This is a basic rule of thumb in rocket science; if you’re going to fire your rocket, fire it at the lowest point in its orbit.  But the question remains: Wha…?  Why?

Krakow! Krakow! Two direct hits!

At different parts of an orbit the same amount of fuel translates into different amounts of kinetic energy.  This “Oberth effect” comes about because the ship travels at different speeds at different points in the orbit.

The very short answer is that you need to take into account what the exhaust is doing as well as what the rocket is doing.  We usually think of a rocket as a thing that propels itself through space.  In fact, it’s better to think of a rocket as a “gas cannon” that throws as much stuff as possible, as fast as possible, out of its more interesting end.

"boring end" = "safe end"

Rockets and cannons both throw stuff as fast as they can. The only real difference is that a cannon is nailed down.

When the rocket is slow the exhaust travels quickly and carries away a lot of kinetic energy.  When the rocket is fast the exhaust is closer to sitting still (in fact, if the rocket is already traveling at the exhaust velocity, the exhaust is basically dropped off).  Overall, energy is always conserved (it’s practically a law), it’s just a question of how much goes into the rocket, where it’s useful, and how much goes into the exhaust, where it’s not.

So check it!  Imagine that a parcel of fuel with mass m gets burned and ejected at a speed of ΔU.  Conservation of momentum says that the momentum of the rocket and fuel before burning is the same as the momentum afterward.  Mathematically, this is MΔV + mΔU = 0.  The energy before the burn is E = \frac{1}{2}(M+m)W^2 and the energy afterward, from the rocket and the fuel parcel, is E = \frac{1}{2}M(W+\Delta V)^2+\frac{1}{2}m(W+\Delta U)^2.  The change in energy is

\begin{array}{ll}    \frac{1}{2}M(W+\Delta V)^2+\frac{1}{2}m(W+\Delta U)^2 - \frac{1}{2}(M+m)W^2\\    =\frac{1}{2}MW^2+MW\Delta V+\frac{1}{2}M(\Delta V)^2+\frac{1}{2}mW^2+mW\Delta U+\frac{1}{2}m(\Delta U)^2 - \frac{1}{2}(M+m)W^2\\    =MW\Delta V+\frac{1}{2}M(\Delta V)^2+mW\Delta U+\frac{1}{2}m(\Delta U)^2\\    =MW\Delta V+\frac{1}{2}M(\Delta V)^2+mW\left(-\frac{M}{m}\Delta V\right)+\frac{1}{2}m(\Delta U)^2\\    =MW\Delta V+\frac{1}{2}M(\Delta V)^2-MW\Delta V+\frac{1}{2}m(\Delta U)^2\\    =\frac{1}{2}M(\Delta V)^2+\frac{1}{2}m(\Delta U)^2\end{array}

Huzzah!  This is exactly what you’d expect; when you take into account all of the stuff that’s flying around, all of the weirdnesses disappear.  The dependence on the initial speed is gone, and you’re left with the same gain in energy you’d see from a rocket starting from a stand-still.  So the Oberth effect, which seems like a violation of the conservation of energy at different speeds, is just a failure to take into account that a rocket slings fuel.  Rather than being a magical extra boost when a rocket is traveling fast, it’s just a cute and fortunate distribution of energy.

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11 Responses to Q: How does the Oberth Effect work, and where does the extra energy come from? Why is it better for a rocket to fire at the lowest point in its orbit?

  1. Mikael says:

    Thanks for this explanation. I wonder whether the question was asked by a Kerbal gamer… :)
    I still do not understand one thing : if some fuel is burning, should not both the rocket and the exhaust gaz be equally ‘pushed’ by the reaction, whatever the rocket+fuel speed is? Said differently, even if I understand that the exhaust gaz would appear to be dropped of from an immobile point of view, as the fuel has the same speed as the rocket, then the burn reaction is the same relatively to the rocket, so how its velocity could ‘push more’ the rocket at high speed?

  2. Alexander says:

    And this is why we can’t have reaction less drives. It breaks relativity.

  3. Travis says:

    Sorry to necro this, but I cannot for the life of me understand why in the second line:

    …1/2*mW^2 + mWdU + 1/2 * mdU^2…. isn’t:
    … 1/2*mW^2 – mWdU + 1/2 * mdU^2…
    Am I doing something wrong? i always thought (x-a)^2 was x^2-2ax+a^2.

    this changes the end answer to 2MWdV + 1/2*MdV^2 + 1/2*mdU^2.
    What am I missing?

  4. The Physicist The Physicist says:

    Thank you so much for catching that! You are absolutely and totally right. My mistake was in the first line, which should read \frac{1}{2}M(W+\Delta V)^2+\frac{1}{2}m(W+\Delta U)^2 - \frac{1}{2}(M+m)W^2 and not (as you caught) \frac{1}{2}M(W+\Delta V)^2+\frac{1}{2}m(W-\Delta U)^2 - \frac{1}{2}(M+m)W^2. I got it in my head that since the exhaust and rocket are going in different directions one should have a negative sign. But it shouldn’t.

  5. Maroo says:

    What happens to Potential Energy? When rocket is moving upward, is P.E is constantly increasing.So its K.E must decrease to conserve energy but it is increasing?Why?

  6. Pingback: Q: Why does kinetic energy increase as velocity squared? | Ask a Mathematician / Ask a Physicist

  7. Charles Vossbrinck says:

    I don’t understand. The rocket and the exhaust are not aware of the velocity they are traveling (compared to what). Somehow firing the rocket at its fastest velocity must propel it faster relative to something. The rocket will always be traveling fast compared to something else and slow compared to another thing. This idea of the exhaust coming out at the speed the rocket is going and putting more of the energy into the rocket does not seem logical.

  8. Umair says:

    @Travis you were right and @The Physicist you got it wrong after getting it right the first time (you can’t have same sign for two velocities in opposite direction while adding them) because of your own basic (and silly) mistake which is your conclusion,
    that, MdV + mdU = 0
    You derived it wrong (it shows that both of them have to be zero all the time). It is in fact,
    MdV – mdU = 0
    from conservation of linear momentum,
    (M + m)W = M(W + dV) + m(W – dU)
    MW + mW = MW + MdV + mW – mdU
    0 = MdV – mdU => MdV = MdU
    so it is,
    dU = (M/m) dV
    not that what’s you substituted wrongly,
    dU = – (M/m) dV.
    Now you can have the right equation for energy change (combined rocket and fuel parcel) that you messed with trying to correct your mistake (the silly one).
    dE = (1/2) M(W + dV)^2 + (1/2) m(W – dU)^2 – (1/2) (M + m)W^2
    Note the negative sign for dU in (W – dU), it will lead to,
    dE = (1/2) M(dV)^2 + (1/2) m(dU)^2.
    The same result you wanted.

  9. Internecivus says:

    Fancy math aside (I never took a calc class), is the way you would explain this using plain English as, “this is why when rockets are launched into space, they burn for a few seconds before taking off so the rocket is technically already at the lowest point in its orbit”? Or am I still really confused?

  10. Brandon Chamberland says:

    Charles, the problem that you cite seems logical at first, except for the fact the the initial AND final velocities are both with respect to a certain point (the object being orbited). The equations could be solved using any point of reference, so long as it is inertial, and yield the same effect. This excludes using the rocket or exhaust as a reference frame, because the whole point of the theory is to determine the change of momentum for both the rocket and exhaust.

  11. lx says:

    whether the oberth effect is suitable in air condition?

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