The Oberth Effect and Propulsive Efficiency

[Silavite]

I was reading about the propulsive efficiency of jet engines earlier, when I thought about the Oberth Effect. A jet (or any type of reaction engine) is most efficient when operating at the velocity equal to that of its exhaust; this is because if the exhaust is traveling at zero velocity in the vehicle's frame of reference, then the vehicle gains all the kinetic energy from the propellant and the exhaust is left with no kinetic energy. If a rocket were to be traveling faster than its exhaust velocity, however, then the exhaust would still have some of the kinetic energy that the rocket's propellant originally had, which would result in lower propulsive efficiency.

Why then does the Oberth Effect still increase the available kinetic energy when rocket velocity exceeds exhaust velocity?

Edited June 17, 2016 by Silavite

[Shpaget]

Because

W=F*d

work = force * distance over which that force is applied.

If the spacecraft is traveling fast it covers more distance in the same period than if it traveled slowly.

[Silavite]

12 minutes ago, Shpaget said:

Because

W=F*d

work = force * distance over which that force is applied.

If the spacecraft is traveling fast it covers more distance in the same period than if it traveled slowly.

And by corollary, the exhaust travels a short distance (within the rocket's frame of reference), doing little work.

However, if the rocket was moving at such a great speed that exhaust velocity was insignificant in comparison, then the exhaust and rocket would be doing a similar amount of work, as the exhaust and rocket would be moving at roughly the same speed and thus cover a similar distance. Force would always be equal because of Newton's 3rd Law.

I think I may just be getting factor of increase mixed up with amount of increase (multiplication vs addition). If I had a rocket traveling at 5 m/s and increased its velocity by 5 m/s, its kinetic energy would increase by a factor of 4. If I had a rocket traveling at 100 m/s and increased its velocity by 5 m/s, its kinetic energy would increase by a factor of 1.1025. The ΔEk in the second case is greater, even if the factor by which it increases is smaller (ΔEk of 75 for the first case, and ΔEk of 1,025 in the second case).

[Bill Phil]

Aircraft are not space craft. Efficiency for aircraft is different for space craft. 

If the plane is the frame of reference, then it's speed is zero... If the exhaust velocity is the same in an external frame of reference as the aircraft, then relative to the aircraft the apparent exhaust velocity is much higher than in an external frame of reference. All of the energy is in the exhaust relative to the aircraft. In an external frame the energy is separate.

It might be pressure, for the aircraft.

[Bill Phil]

In space, you maneuver by "spending" deltaV. How much deltaV it takes to do something depends on many factors. The Oberth effect is in the Vis-Viva equation. With a higher velocity at a lower altitude, Gravity takes longer to slow your velocity to its lowest (apoapsis). But the slowest velocity is also much slower

[PB666]

7 hours ago, Silavite said:

And by corollary, the exhaust travels a short distance (within the rocket's frame of reference), doing little work.

However, if the rocket was moving at such a great speed that exhaust velocity was insignificant in comparison, then the exhaust and rocket would be doing a similar amount of work, as the exhaust and rocket would be moving at roughly the same speed and thus cover a similar distance. Force would always be equal because of Newton's 3rd Law.

I think I may just be getting factor of increase mixed up with amount of increase (multiplication vs addition). If I had a rocket traveling at 5 m/s and increased its velocity by 5 m/s, its kinetic energy would increase by a factor of 4. If I had a rocket traveling at 100 m/s and increased its velocity by 5 m/s, its kinetic energy would increase by a factor of 1.1025. The ΔEk in the second case is greater, even if the factor by which it increases is smaller (ΔEk of 75 for the first case, and ΔEk of 1,025 in the second case).

It might be simpler to think of it like this, when a space craft enters a hillsphere it has thermodynamic energy that will be converted to kinetic energy as it loses altitude given by the integral of MGH. from a space craft point of view without drag dV affords acceleration ar the perigee, since at perigee the craft is no longer falling. Relative to that natural satellite though its energy is 1/2mv². So from the space crafts point of view it just gaining energy just as occurs as any object accelerates, BUT it only has to pay back the integral of MGH as it leaves, so that the difference it gets to keep as kinetic energy, it gets big really fast. 

Heres an example you enter a system with a speed of say 100 and a potential of 1000000. That 5000 + 1000000 or 1005000, means you are traveling at 1417.6, you increase speed by 1 meter per second to 1418.6. you ke assuming you did this right a perigee, the resulting ke then is 1006411.6.44. As you leave that system you have to give back the kinetic energy borrowed that leaves 6411.6. This leaves you with a velocity of 113.2, by adding 1 m/s at perigee you gained 13.2 upon exit the hill sphere (there are alot of assumptions here about how you entered and exited relative to the parent to natural satellites, but lets just say relative to systems center you entered and left the hillsphere at the same altitude). If you were going much faster, say 4000 when entering the system you would still gain energy, but not proportionally as great as if you reached so trick point where you basically fall but miss the atmosphere. 

So the other part here is that in doing Oberths you really want dense rocky targets far away from the star so you can get really close to the point mass with most of you energy in integrated MGH obtained kinetic energy, you want to accelerate as much as you can at pe, and you might want an entry angle that throws you into another Oberth later on. Of course this is not always possible, and so the large hill sphere around Jupiter sufficiently compensates for the fact that you will not be able to do a very close approach because its a gas giant of lower density than say earth. 

There is another trick point in the system, if you can manage a course change doing Oberths in mid system, you can use the combined gravitational well of the star and its first planet at its perigee to increase the kinetic energy leaving that planets system to gain a high eccentric orbit that leaves the system, potentially using the Oberth effect of large Jovian planet as it leaves. This type of approach might be suitable for ion driven systems because they rely on the sun for exhaust energy.

Edited June 17, 2016 by PB666
Major math correction, IPAD math is all sucky.

[Lukaszenko]

Basically when the rocket exceeds the exhaust's velocity, the rocket gains MORE energy than the total chemical energy in the propellant. The efficiency goes past 100%.

[Snark]

Essentially, what the previous posters have said.  Rockets are operating in a vacuum, there's nothing magical about their exhaust velocity as a "special" speed.

Kinetic energy goes up with the square of velocity, E = 0.5mv².  Let's say you have a tiny (to keep the math simple) 1-kilogram rocket going 1000 m/s.  Its kinetic energy will be 500,000 joules.  If it spends a smidgeon of rocket power to accelerate by 1 m/s so that it's now going 1001 m/s, its energy is now 501,000.5 joules.  So, it gained 1000.5 joules of energy by spending 1 m/s of dV.

Now imagine the same situation, except that the rocket is going 2000 m/s.  Its kinetic energy is 2,000,000 joules.  If it spends the same 1 m/s of dV to accelerate to 2001 m/s, its kinetic energy is now 2,002,00.5 joules.  So, it gained 2002.5 joules of energy by spending 1 m/s of dV.

In other words:  When it's going faster, it gets more bang for its buck in terms of energy.