When does the Oberth effect reach its apex?

[K^2]

You should be seeking to optimize orbital energy in a frame of reference in which energy is actually conserved. That's the coordinate system in which barycenter is stationary. If you got optimal exhaust velocity at zero, then you probably chose rocket as your frame of reference. A frame of reference which moves relative to barycenter does not conserve energy. So it's useless optimizing energy in that coordinate system.

[LLlAMnYP]

That's just not true. Energy is conserved in any inertial frame of reference, it's just a matter of deciding what energy is useful. When I got the result that suggested having as low exhaust velocity as possible, I compared the energy of the rocket minus the about-to-be-expelled propellant as the starting energy with the energy of the rocket after expelling the propellant after the burn. Now that indeed is a quantity that is dependent on the frame of reference. If we sum up all energies (i.e. also take into account the kinetic energy of expelled propellant), then it doesn't matter which frame of reference we choose, the increase in energy will always be dm*v^2/2 (where dm is the differential of mass expelled, v is the exhaust velocity). However, as we don't really care what happens to the propellant after we lose it, it makes little sense to take into account it's energy after the burn.

[LLlAMnYP]

PS

Frankly though, in a rocket we're generally stuck with a fixed Delta-V budget and then all that could possibly matter is traveling at the highest possible speed when burning. Still trying to figure out a scenario, where we could need to optimize for anything else...

[K^2]

That's just not true. Energy is conserved in any inertial frame of reference, it's just a matter of deciding what energy is useful.

Here is a problem for you to consider. A cart is parked on a hill of some height h and has potential energy mgh. It then rolls off the hill onto a flat surface, and rolls with velocity v. So its energy is now mv²/2 = mgh. Good? Now picture this as you ride by in a train. The train just happens to travel at the same velocity v. You initially observe the cart sitting at the top of the hill having energy mgh + mv²/2. Then the cart rolls off the hill, matches velocity of the train, and now has energy 0. Where did the energy go?

Yes, energy of the system is conserved in a closed system an inertial coordinate system. But we aren't dealing with a system. Did you take motion of the planet/star into account in your problem? No. You threw that away completely and substituted an effective central potential. Your sub-system is now your ship and the gravitational potential. And if that potential is not moving in your chosen coordinate system, that's fine. You still have a conservation law, and you can say that ship's energy is conserved. If that potential is moving relative to your chosen coordinate system, you do not have energy conservation in your chosen sub-system. Coordinate system that's moving relative to barycenter of a 2-body problem has energy exchange between the ship and parent body. If you consider the whole system, that's fine. If you just talk about the ship, and maximizing energy of the ship, ship's energy is not conserved.

As for why you optimize energy, try to prove that Hohmann Transfer is an optimal transfer between two co-planar circular orbits without making use of energy and angular momentum conservation laws.

P.S. Just as a point of interest, energy conservation follows from time-translation invariance of the Lagrangian and Noether's Theorem. In Classical Mechanics, this requires time-independent interactions. And if you follow this rabbit hole as far as it goes, you'll find that in general, a time-dependent interaction can only arise from something that's equivalent to an accelerated frame of reference. But we're talking advanced field theory by this point.

Edited December 5, 2013 by K^2

[AeroEngy]

Here is a problem for you to consider. A cart is parked on a hill of some height h and has potential energy mgh. It then rolls off the hill onto a flat surface, and rolls with velocity v. So its energy is now mv²/2 = mgh. Good? Now picture this as you ride by in a train. The train just happens to travel at the same velocity v. You initially observe the cart sitting at the top of the hill having energy mgh + mv²/2. Then the cart rolls off the hill, matches velocity of the train, and now has energy 0. Where did the energy go?

Your example is flawed. You didn't properly determine the initial energy in the case of the moving reference frame. From the frame of the train the "stationary cart" is NOT stationary. It is moving at -V away from the frame. So its initial energy is m(-v)^2/2 + mgh. Its ending energy is zero in your example but so is its starting energy.

[K^2]

Your example is flawed. You didn't properly determine the initial energy in the case of the moving reference frame. From the frame of the train the "stationary cart" is NOT stationary. It is moving at -V away from the frame. So its initial energy is m(-v)^2/2 + mgh. Its ending energy is zero in your example but so is its starting energy.

(-v)² = v²

[AeroEngy]

(-v)² = v²

Ok fine you caught me being lazy. However, you still have to see the flaw in your example. I should have stated that the change in Kinetic Energy of the systems is still equal to the change in potential energy. It gets a bit muddied since the initial velocity is negative in the moving reference frame.

m(deltaV)^2/2 = mg(deltaH) is the same in both cases.

IN the stationary frame Vi = 0 & Vf = sqrt(2gh)

In the moving frame Vi = - sqrt(2gh) & Vf = 0.

DeltaV and thus Delta Kinetic Energy are the same. Just like in both cases the delta potential energy is the same. This is the really simplified version and I am not working out the more complicated version in this tiny reply window ... if you really want to press the point refer to the much more detailed explanation of your problem here.

TLDR: Energy is always conserved ... it might have different values in different frames but it is conserved.

[LLlAMnYP]

@K^2

Nice problem, thanks. I didn't consider that. That does not, by the way, prove, that energy is not conserved, as pointed out in the link AeroEnergy posted.

Now about the energy of the parent body (does it remain constant or not). The change of energy should be taken into account... but only when the rocket is not at apoapsis or periapsis.

If we are in a frame of reference tied to the parent body, its change in velocity is negligible (m_rocket/(m_planet+m_rocket) << 1). If we work with differentials, then the change of energy would not only be proportional to this tiny factor, it would also be proportional to the square of dV and that can certainly be neglected. If we were to go into a frame of reference tied to the rocket, the change of energy of the parent body is no longer negligible in the general case. However, at apoapsis or periapsis the planet is moving perpendicular to the line connecting it to the rocket, thus the gravitational force from the rocket acting on the planet is not performing work (well it is, but it is proportional to dt squared and thus also negligible). Similarly, the change of the potential energy is also proportional to the second-order differential and is also negligible.

EDIT

The tl;dr of AeroEnergy's link is that when the cart rolls down the hill, there is also some recoil acting on the hill that accelerates it backwards. Of course if we say the hill is much heavier than the cart, the change in the kinetic energy of the hill is negligible. However, in the moving frame of reference that you suggested, the change of energy of the hill is no longer negligible and is actually equal to double mgh, exactly the energy that seems to have "vanished".

Edited December 5, 2013 by LLlAMnYP

[K^2]

I should have stated that the change in Kinetic Energy of the systems is still equal to the change in potential energy.

No, it is not. Change in both of these energies is negative. You are still mixing up signs.

What you should have picked up from the link you posted is that work done against the cart is negative. So the fact that cart's energy goes down to zero makes sense. But the question of where that energy went remains.

The tl;dr of AeroEnergy's link is that when the cart rolls down the hill, there is also some recoil acting on the hill that accelerates it backwards.

Precisely. But this only works out after you consider actual pair of action-reaction forces. If you approach the problem naively, and just say, "Well, the cart had kinetic energy mv²/2 and then it got mgh from going down hill, it's now traveling at root(2)v by conservation of energy." This is wrong way to do the problem.

But this is only a wrong way to do this problem when the potential is time dependent. Lets say that the hill has a constant incline angle -θ. Then we can treat this as a 1D problem. The potential, in this case, is U(x) = mgh - mgx Sin(θ). Here, h is just a parameter, way to set U(0), rather than elevation. We can either say that at the bottom of the hill, where U(x) = 0, the potential energy has converted to kinetic and the velocity is now Sqrt(2gh), or we can compute acceleration. F = -∂U(x)/∂x = mg Sin(θ). Acceleration is then g Sin(θ). Distance the cart has traveled is d such that U(d) = 0 -> d Sin(θ) = h. And we get final velocity v = Sqrt(2ad) = Sqrt(2gh).

Now, we go into coordinate system moving to the right at velocity u. The potential is now time-dependent. U(x,t) = U(x-ut) = mgh - mg(x - ut) Sin(θ). The cool thing about this is that the force is the partial derivative with respect to coordinate, so in this trivial case, time-dependence won't matter. F = mg Sin(θ). So the cart is still going to experience acceleration g Sin(θ), and starting with velocity -v, so long as v = u, it will come to rest at the bottom of the hill, which is what we are looking for.

However, if you try to use energy conservation here without taking into account recoil to the ramp, you are going to make a mistake.

So with all of this in mind, we are going back to the case of orbital motion. You considered motion of the ship subject to potential -μ/r. Here, the gravitational parameter μ takes reduced mass into account, and r is distance from barycenter. Lets do this in 2D. Potential energy can then be written U(x, y) = - mμ/Sqrt(x² + y²). If your ship is at location x, y traveling at velocity v, it has total energy mv²/2 + U(x, y). Lets figure out how its energy is changing at that instance.

dE/dt = m/2 d(v_x² + v_y²)dt + dU(x, y)/dt

I re-wrote v² in terms of components so that I can chain-rule it.

dU(x, y)/dt = ∂U/∂x dx/dt + ∂U/∂y dy/dt = -F_x v_x - F_x v_y

dE/dt = m (a_x v_x + a_y v_y) - F_x v_x - F_x v_y = 0

So this is a very nice result. Energy of the ship is conserved. We don't need to consider the whole system. We can forget that the planet/star is there at all, and just work from conservation of energy of the ship.

Now lets consider a moving coordinate system. It will be co-moving with the ship, but let me name that velocity u, as before, so as not to confuse it with ship's velocity that goes into kinetic energy. So again U(x, y, t) = U(x - u_xt, y - u_yt). And again, we are interested in energy of the ship and how it changes with time.

dE/dt = m/2 d(v_x² + v_y²)dt + dU(x, y, t)/dt

The first term is zero, because v = 0. Ship is at rest in this frame. So we only need to see if dU(x, y, t)/dt is zero or non-zero.

dU(x, y, t)/dt = ∂U/∂x dx/dt + ∂U/∂y dy/dt + ∂U/∂t

Here, dx/dt = v_x = 0 = v_y = dy/dt.

All that's left is verifying that ∂U/∂t ≠ 0, which I'm sure you can handle.

So in coordinate system where the ship is at rest, dE/dt of the ship is non-zero. Energy of the ship is not conserved. Yes, energy of the ship-planet/star system is still conserved, but the lot of good that does you. Yes, you've optimized the rate at which ship's energy is increased. But that is not a conserved quantity, so whether or not you achieved some sort of optimal in transfer, you don't know. You have to solve for the entire trajectory and see what's going on.

In contrast, if you start with coordinate system in which barycenter is not moving, you know that if you maximized E that the ship gets from the burn, you've maximized ship's energy throughout the trajectory. Because it is conserved.

Hopefully, it's now clear why choice of the coordinate system matters, and why there is a "right" and "wrong" choice here. But let me know if you have any more questions.

[LLlAMnYP]

I'll answer briefly now and go into more detail later, when I'm not at work :-)

dU(x, y)/dt = ∂U/∂x dx/dt + ∂U/∂y dy/dt = -Fx vx - Fx vy

dE/dt = m (ax vx + ay vy) - Fx vx - Fx vy = 0

The red "x"s should of course be "y"s. And you are absolutely right (as I also mentioned in my previous post) that in the general case ∂U/∂t is non-zero. However, at periapsis or apoapsis the ship's motion is perpendicular to the gravitational force acting upon it and is in fact parallel to one of the lines of constant potential. At that moment ∂U/∂t will undoubtedly be zero. As the simplest description of the Oberth effect is "burn at maximum velocity" and maximum velocity is achieved at periapsis, I feel that this is the most important case to consider.

[K^2]

Yes, typos. Sorry about that.

Yes, ∂U/∂t is zero at apsides in a frame co-moving with the rocket at either of these points. But then the rocket keeps moving along the elliptic path, and your coordinate system keeps moving relative to barycenter along the straight line, and you end up with energy of the ship changing. So by the time it travels from periapsis to apoapsis, how much energy was lost or gained? Just because you gained most energy at periapsis, if that energy was lost or gained, and that loss/gain was trajectory-dependent, as it is, then you don't know whether you ended up with optimal apoapsis.

That's your entire problem. You maximize rocket's energy gain in the rocket's rest frame, but that does not result in optimal apoapsis. How do I know that? Because I know that rocket's energy is conserved in barycenter frame, I do know that you gain most energy when exhaust velocity matches rocket's velocity, and I do know that most energy gain results in highest apoapsis. You end up with a different result, and I tell you why. It's a wrong frame of reference for the problem for all of the reasons listed above.

[LLlAMnYP]

Ah, so here you finally state the answer to my recent question! We are interested in the energy increase at apoapsis (please correct me if I misunderstood). Indeed, I didn't ever consider what happens to the rocket after finite time, rather I was considering what happens immediately after a burn, but of course of course by the time the rocket gets from periapsis to apoapsis in a moving frame of reference the change in energy of the parent body will be non-negligible.