explanations for the oberth effect

[jevry]

im doing some research on orbital stuff and there seem to be a lot of explanations for the oberth effect. but im nto sure which ones are right, and which ones are wrong. and which ones seem to contradict each other but are both right.

so far the best explanation i found i think is: "To get the most kinetic energy from a chemical rocket, it should be burned when the vehicle is already moving at high speed, This is because the added kinetic energy will equal the force of the rocket thrust (which is constant) multiplied by the distance that the rocket moves during the burn. The higher the initial speed, the more will be the distance moved."

however i have found a few different categories of answers so far, these are variantions of:
"the rocket moves more distance during the burn, thuss it gains more kinetic energy"
"F = M X V^2, and V = Vstart + dV, thuss when Vstart is higher you gain more energy" (sounds reasonably correct to me but i think its missing some details?)
"it is a relativistic thing, from the standpoint of the rocket there is no magical increase in kinetic energy gained" (unsure about this one, relativity is only relevant when you are moviong at least 1% the speed of light iirc)

"it has to do with the exhaust gasses" (but the effect also works using engines that have no exhaust gasses, e.g solar sails)

"it has to do with the planet, you move faster thuss lose less energy as you leave it's gravity well" (to me this sounds only relevant during gravity assists, not while changing apoapsis height)

since this place seems kinda smart and kind, what do you guys think is/are the right explanation(s)? and do you have any other thoughts about it?

[K^2]

20 minutes ago, jevry said:

"To get the most kinetic energy from a chemical rocket, it should be burned when the vehicle is already moving at high speed, This is because the added kinetic energy will equal the force of the rocket thrust (which is constant) multiplied by the distance that the rocket moves during the burn. The higher the initial speed, the more will be the distance moved."

Yes.

20 minutes ago, jevry said:

"the rocket moves more distance during the burn, thuss it gains more kinetic energy"

That's basically same as above.

20 minutes ago, jevry said:

"F = M X V^2, and V = Vstart + dV, thuss when Vstart is higher you gain more energy"

Also correct, and you can actually derive this with some math from the first one.

21 minutes ago, jevry said:

"it is a relativistic thing, from the standpoint of the rocket there is no magical increase in kinetic energy gained"

That's very poorly phrased. Relativity is relevant, though. Not Special or General relativity, but Classical Relativity, also known as Galilean Relativity. The gist is that kinetic energy of a body depends on coordinate system of choice. Trying to explain Oberth through this is going to be awkward, and will essentially be the same as previous point.

23 minutes ago, jevry said:

"it has to do with the exhaust gasses"

Another poorly phrased one. We've established that rocket gets more kinetic energy due to Oberth Effect. A natural question is "Where does the energy come from?" And that has to do with propellant. People usually think of rocket engine as converting chemical energy of the fuel into kinetic energy of the rocket. But the cool thing about a rocket is that it also converts part of the kinetic energy of the fuel into additional kinetic energy for the ship! If you're going fast enough, most of the energy of the rocket is gained from kinetic, not chemical energy of the fuel. Of course, that fuel had to be brought up to speed in the first place. Usually, you burn a lot of fuel early on to get the rocket and fuel moving. But if you use Oberth effect, the planet gives you extra energy to work with.

In case of the solar sail, the extra energy comes from the fact that reflected light is going to be red-shifted. So there are no exhaust gasses, but there's still an exhaust of sorts.

27 minutes ago, jevry said:

"it has to do with the planet, you move faster thuss lose less energy as you leave it's gravity well"

That one isn't right. Once you finished the burn and are climbing out of the gravity well, your energy isn't changing. Sure, kinetic energy of the ship is becoming gravitational potential energy of the ship, but the total is constant. Oberth effect lets you actually gain more energy. You end up in an orbit with higher total energy if you burn prograde at periapsis than if you burned at the apoapsis. And the burn can be very short, not long enough for gravity to impact you while you are using your engines. So this explanation is wrong.

But planet's gravity does play a role as outlined in the previous one. It's either adding or taking away the kinetic energy from your fuel, which you are using to gain more energy for your rocket.

As a side note, there are two parameters that determine shape of your orbit around a single body. It's energy and angular momentum. People who play around with rocketry generally know that you gain most energy at the low burn. But you gain most angular momentum as far away from the planet as possible. And to go from a low circular orbit to a higher one you need to increase both energy and angular momentum. Which is a somewhat hand-wavy explanation of why a Hohmann transfer is so efficient. You do first part of your burn at the lowest point to maximize energy gain and second part at the highest point to maximize angular momentum gain. This is just to point out that you don't always care just about kinetic energy gained there are other parameters of your orbit that you might need to adjust in the most fuel-efficient manner, and for these, burning low isn't always a good thing.

[Superfluous J]

44 minutes ago, jevry said:

F = M X V^2, and V = Vstart + dV, thuss when Vstart is higher you gain more energy" (sounds reasonably correct to me but i think its missing some details?)

Nope. That's it in a nutshell.

If you're moving 1 speed unit and add 1 speed unit to go a total of 2 speed units, and you mass 1 mass unit, you go from (1*1²)=1 Kinetic Energy Units to (1*2²)=4 Kinetic Energy Units, for a gain of 3 Kinetic Energy Units.

If you're instead going 10 speed units to start, you go from (1*10²)=100 Kinetic Energy Units to (1*11²)=121 Kinetic Energy Units, for a gain of 21 Kinetic Energy Units.

And really that's it, whether it makes sense or not. Any explanation other than that is either wrong or just putting the above equations into words.

Edited April 2, 2020 by 5thHorseman

[p1t1o]

Essentially, the change in energy of the orbit is proportional to the change in total energy of the rocket.

Since the KE is proportional to the square of velocity, then the change in the total energy of the rocket is greater when the craft is moving faster.

101kg ship @ 100m/s   ->   KE = 0.5*101*100² = 50500J

Lets exhaust 1kg of fuel to add 100m/s

100kg ship @ 200m/s   ->   KE = 0.5*100*200² = 2000000J

Increase of 1949500J

 

 

Now lets do this at a higher velocity

101kg ship @ 1000m/s   ->   KE = 0.5*101*1000² = 50500000J

Lets exhaust 1kg of fuel to add 100m/s

100kg ship @ 1100m/s   ->   KE = 0.5*100*1100² = 60500000

Increase of 10000000J

 

 

Thus, the change in energy of the rocket is much greater when thrusting at higher velocities, planet or no planet.

It seems strange - "Where is the second rocket getting free energy from?"

It isnt - consider that the second rocket has already been given a ton of KE to get to its higher speed in the first place, it already has more energy.

 

(you can do further maths to confirm this by calculating the KE of the fuel before and after burning, but ugh, maths)

Edited April 2, 2020 by p1t1o

[magnemoe]

An tips in KSP, leave Mun or Minmus and drop your Pe down to low orbit at Kerbin before doing your burn. You keep the dV you used going to Mun or Minmus, this is only useful if you top up your tanks at location. 
use an targeting probe who has the burn plotted, this give you an insertion, you might have to spend quite some time doing extra loops to get to the time to do your burn. 
Mun is easier but refueling here is harder. 
 

[K^2]

50 minutes ago, magnemoe said:

An tips in KSP, leave Mun or Minmus and drop your Pe down to low orbit at Kerbin before doing your burn. You keep the dV you used going to Mun or Minmus, this is only useful if you top up your tanks at location. 
use an targeting probe who has the burn plotted, this give you an insertion, you might have to spend quite some time doing extra loops to get to the time to do your burn. 
Mun is easier but refueling here is harder. 
 

I might be misunderstanding something, but I'm not sure that works. Forgetting Mun/Minmus for a moment, if you're already in circular orbit around Kerbin, or any other body for that matter, direct burn to escape the system is always cheaper than performing a dive and doing your escape burn at lowest possible altitude. And if you are orbiting Mun/Minmus, the most fuel-efficient way to leave the SoI is to drift out with minimal velocity left over, leaving you in circular orbit around Kerbin. Granted, if you are doing direct ascent, you win a little bit back by burning directly to a return trajectory, making the dive a little cheaper than it would have been from circular orbit around Kerbin, but that isn't nearly enough, as the entire dive only costs you about 20m/s and burning to escape just over Kerbin's atmosphere vs directly from Mun puts you more than 200m/s in the red.

You can probably come up with a multiple gravity-assists trajectory for leaving Kerbin system from Mun/Minmus that's going to be cheaper, but it's going to be gravity assists that help you with that, not Oberth effect.

[Bill Phil]

A good formulation is characteristic energy C3:

C3 = V^2 - Vesc^2

Where C3 is characteristic energy, V is current velocity, and Vesc is the escape velocity at the current position.

A positive C3 indicates a hyperbolic orbit, a C3 of zero indicates a parabolic orbit, and a negative C3 indicates a closed orbit (either elliptical or circular). 

One interesting aspect of this equation is that a relatively small difference between V and Vesc can create a large change in C3. 

For example, in LEO the escape velocity would be around 11 km/s. Adding just 300 m/s would give a C3 of about 6.7 km^2/s^2, or a velocity at infinity of 2.6 km/s. Mars missions require a C3 of roughly 12 km^2/s^2, so to go to Mars requires 530 m/s beyond escape velocity, or just 3.73 km/s of delta-V from LEO. Meanwhile a C3 of 80 is required to reach Jupiter directly, so about 14.2 km/s of velocity - 6.4 km/s of delta-V from LEO. But this leads to an excess velocity at infinity of 8.9 km/s.

48 minutes ago, K^2 said:

I might be misunderstanding something, but I'm not sure that works. Forgetting Mun/Minmus for a moment, if you're already in circular orbit around Kerbin, or any other body for that matter, direct burn to escape the system is always cheaper than performing a dive and doing your escape burn at lowest possible altitude. And if you are orbiting Mun/Minmus, the most fuel-efficient way to leave the SoI is to drift out with minimal velocity left over, leaving you in circular orbit around Kerbin. Granted, if you are doing direct ascent, you win a little bit back by burning directly to a return trajectory, making the dive a little cheaper than it would have been from circular orbit around Kerbin, but that isn't nearly enough, as the entire dive only costs you about 20m/s and burning to escape just over Kerbin's atmosphere vs directly from Mun puts you more than 200m/s in the red.

To leave the Earth system from the Moon's altitude requires about 1.44 km/s, a delta-V of roughly 440 m/s to escape. To reach a parabolic orbit, it would be preferable to simply burn at this altitude. But what if we want to go to Mars?

A C3 of 12 would require 3.75 km/s, or about 2.75 km/s of delta-V. 

But what velocity would we have if we lowered our periapsis to LEO?

So we have two delta-Vs, the braking burn and the final escape burn. Remember that the escape only needs 530 m/s past escape velocity.

Braking burn:

Assuming the Moon's altitude is a constant 384399 km, and LEO is at 6550 km:

An orbit that goes between these two would have a semi-major axis of about 195000 km. 

Using the Vis-Viva equation the velocity after braking would be 186 m/s.

So the initial delta-V is about 813 m/s.

Escape burn:

The velocity at periapsis will be approximately 10940 m/s - once again using the Vis-Viva equation. Only 624 m/s slower than the required velocity to reach a C3 = 12.

The total delta-V would be about 1.44 km/s, more than half of which is for the braking burn. This is much less than the direct delta-V of 2.75 km/s. 

So it is cheaper for a suitably hyperbolic target trajectory.

Indeed, for a direct transfer to Jupiter, it would be much more effective to "dive" - provided you were already at the Moon's orbit when you start. 

For KSP the required C3 is much smaller, and the benefit of "diving" is much less pronounced. So it's not all that useful unless you're around a large planet like Jool (which is similar in mass to Earth) or maybe Eve. 

Another factor is that you have to actually get to the Moon's orbit - the benefit of doing this is pretty much nonexistent unless we can use resources from space as propellant. Since you'd have to escape LEO to actually do it in the first place. Maybe using electric propulsion to create in space infrastructure and then just delivering the crew later could be beneficial?

Edited April 4, 2020 by Bill Phil

[Superfluous J]

27 minutes ago, K^2 said:

I might be misunderstanding something, but I'm not sure that works.

It does, kind of. Sometimes.

Let's say you want to go to Jool from Mun. Either you build ships there with EL or you went there to refuel.

Burning from Mun Orbit to return to Kerbin takes 310m/s, and puts you in an orbit that would have cost 860m/s had you burned it from low Kerbin orbit (I'm getting these from a stock dv map). From LKO to Jool takes 1930m/s, so from your "left Mun" orbit, if you do everything perfectly, it will cost you another 1070m/s, which when you add the 310 to leave Mun costs you 1380m/s (not counting the mid course correction if any)

To get to Jool from Mun directly (again not counting the mid course correction) I just set up a node to test, and it cost me 1788m/s. So doing a deep dive saves about 400m/s.

Is it worth it? No. But it's fun and cool at least the first few times.

Duna on the other hand costs 454 from Low Mun Orbit. Instead ejecting to Kerbin for 310 and then burning the extra 220 necessary to get to Duna costs 530, so going straight to Duna from Mun is better than doing the dive.

[sevenperforce]

There have been several good explanations, plus some helpful math, but maybe this will give a way of thinking about the Oberth effect that will help conceptually.

In 1920, the New York Times published an editorial which mocked rocketeer Robert Goddard and said it was preposterous to imagine a rocket working in space, where there is no air to push against. The Times posted a retraction of that on July 17, 1969, the day after Apollo 11 lifted off from Cape Canaveral:

"Further investigation and experimentation have confirmed the findings of Isaac Newton in the 17th century and it is now definitely established that a rocket can function in a vacuum as well as in an atmosphere."

As humorous as this was, it underscores a critical element in understanding the Oberth effect, along with all other types of rocketry. Rockets are not magical force engines which produce thrust by simply flipping a switch; they are reaction issues which gain thrust by pushing against exhaust gases. The harder and faster the exhaust comes out, the farther and faster the rocket will fly. The farther you throw the exhaust, the farther it will throw you.

The Oberth effect says that the deeper you are in a gravity well, the more energy you will ultimately gain from a burn of specified length. But instead of thinking of it in terms of an engine's thrust, think of it in terms of the exhaust you are leaving behind. If you burn your engine while high in a gravity well, you're leaving your exhaust behind in its own orbit, one which has a high amount of potential energy relative to the bottom of the gravity well. The particles of exhaust will fall toward the bottom of the gravity well, their potential energy forever lost. 

By diving deeper into a gravity well, on the other hand, you convert the propellant's potential energy into kinetic energy first. A burn at periapsis leaves the exhaust behind close to the planet, rather than wasting momentum carrying it up into a high-potential-energy space to be lost.

[Laie]

2 hours ago, sevenperforce said:

The Oberth effect says that the deeper you are in a gravity well

*buzz sound* Wrong!

The only thing that matters is velocity: the faster you go, the more oberth you get.

As one tends to go faster in low orbit, this may seem like a distinction without difference. Still, I feel it's a nit worth picking.

[sevenperforce]

12 hours ago, Laie said:

*buzz sound* Wrong!

The only thing that matters is velocity: the faster you go, the more oberth you get.

As one tends to go faster in low orbit, this may seem like a distinction without difference. Still, I feel it's a nit worth picking.

I should have specified. "The Oberth effect says that for any given orbital trajectory, the deeper you are in a gravity well..."

[magnemoe]

18 hours ago, Superfluous J said:

It does, kind of. Sometimes.

Let's say you want to go to Jool from Mun. Either you build ships there with EL or you went there to refuel.

Burning from Mun Orbit to return to Kerbin takes 310m/s, and puts you in an orbit that would have cost 860m/s had you burned it from low Kerbin orbit (I'm getting these from a stock dv map). From LKO to Jool takes 1930m/s, so from your "left Mun" orbit, if you do everything perfectly, it will cost you another 1070m/s, which when you add the 310 to leave Mun costs you 1380m/s (not counting the mid course correction if any)

To get to Jool from Mun directly (again not counting the mid course correction) I just set up a node to test, and it cost me 1788m/s. So doing a deep dive saves about 400m/s.

Is it worth it? No. But it's fun and cool at least the first few times.

Duna on the other hand costs 454 from Low Mun Orbit. Instead ejecting to Kerbin for 310 and then burning the extra 220 necessary to get to Duna costs 530, so going straight to Duna from Mun is better than doing the dive.

Minmus is cheaper, mostly because its an lower eject speed. 
Your explanation also explains why I use this as much, I tend to use this to launch huge ships on 4-5 Km/s trajectories, this let me launch them pretty empty and top up the tanks at Minmus before dropping them, some times I also add some fuel for the Minmus burn in LKO.