Oberth effect

[OhioBob]

I've been recently performing some computations regarding the Oberth effect and I'd like to share the results.

I've always thought the term "Oberth effect" was a bit overused around here and somewhat misunderstood. It is not the hard and fast rule that everyone seems to thinks it is. For example, the place where I usually hear the Oberth effect evoked is when the conversation is about interplanetary ejection burns and/or orbit insertion burns. It is usually stated that performing the burn closer to the planet is always better. However, for low orbits very close to a planet, the usual thinking isn't always correct.

A transfer trajectory to another planet requires a certain hyperbolic excess velocity, denoted V_∞. This is the velocity left over after a spacecraft escapes a planet's gravity. The V_∞ required to reach a specific destination during a specific launch window is the same regardless of the orbital altitude from which the spacecraft is ejected. The equation for V_∞ is,

V_∞² = V_bo² - V_esc²

where V_bo is the burnout velocity and V_esc is the escape velocity.

V_bo is the initial orbital velocity plus the ejection Δv, i.e. V_bo = V_orb + Δv. Therefore, by substituting and rearranging the equation we get,

Δv = (V_∞² + V_esc²)^0.5 - V_orb

Of course, V_orb and V_esc are both functions of the orbital altitude. Therefore, for any given orbital altitude and V_∞, we can compute the Δv required to eject our spacecraft.

Let's take the example of a spacecraft on a transfer trajectory to Duna where V_∞ = 900 m/s. Computing Δv and plotting it on a graph, we get

Note that it initially takes less Δv the higher the orbit (significantly less in this case). It is not until we pass an altitude of 8145 km that the Δv begins to increase (i.e. the Oberth effect).

The shape of the curve varies dramatically with different values of V_∞. For instance, lets' say we are going to Jool during a launch window where V_∞ = 2800 m/s. The graph now looks like this,

Again, the Δv initially decreases with increasing altitude, though the decrease is much less than the previous example. Also note that the Δv starts to increase just past an altitude of 303 km, which is much lower than the Duna case.

What these results don't take into account is that, although it may require less Δv to eject from a higher orbit, it takes more Δv to reach the higher orbit in the first place. When we add together both the launch Δv and the ejection Δv, the lower orbit results in less total Δv.

When planning a mission, you can always use Alex Moon's Launch Window Planner and adjust the orbit altitudes to see how the ejection and insertion Δv change. However, note that when arriving at a planet, the orbit that you want to enter into will be the one that produces the lowest overall mission Δv. You don't want to enter into an orbit just because it has a low insertion Δv if its going to make subsequent maneuvers more costly.

 

Edited April 9, 2016 by OhioBob

[Norcalplanner]

OhioBob, this is great information. It shows that I haven't been completely off my rocker by putting a refueling depot at 250 km to top off interplanetary craft. One question - how do steering losses affect this? One of the reasons I've been doing higher orbits is to minimize those losses and still be able to do a transfer burn in a single orbit.

[OhioBob]

Norcalplanner said:

One question - how do steering losses affect this?

I didn't take steering losses into consideration; all my calculations assume instantaneous Δv. However, it's been discussed and demonstrated in other threads that smaller the angle that the spacecraft sweeps through during a burn, the smaller the Δv losses. Obviously the higher the orbit, the less angle a spacecraft will sweep through during a burn of given duration. A five minute burn in a 75 km will have greater losses than a 5 minute burn in a 250 km orbit. High orbits should definitely be more tolerant to long duration burns than low orbits.

I recently saw somebody post that the sweep angle during a burn should be limited to 1/6th of an orbit, i.e. 60 degrees. That's about 5 minutes in a low orbit, which sounds about right to me.
 

Edited April 9, 2016 by OhioBob

[xtoro]

While it's true that raising your AP just to take advantage of the OE is pointless since you expend dV to raise your AP in the first place, it still has advantages for situations where your departure starts near AP in a highly eccentric orbit. For example, going to Minmus where you have a refueling depot to fill your tanks with "free" fuel. Then you spend very little to escape Minmus SOI and head down steeply towards Kerbin PE. I'm sure there are scenarios in other places where this would work even more in your favor like a refueling depot on Vall, or a moon of Sarnus or Urlum etc...

[Hotaru]

it still has advantages for situations where your departure starts near AP in a highly eccentric orbit. For example, going to Minmus where you have a refueling depot to fill your tanks with "free" fuel. Then you spend very little to escape Minmus SOI and head down steeply towards Kerbin PE.

This is generally what I mean when I say "Oberth effect" or "Oberth maneuver." For instance, I had a ship departing Ike for Kerbin over the weekend using a ~200m/s burn out of low Ike orbit to lower Duna periapsis to about 70 km (as though transferring to low Duna orbit), then a ~700m/s burn at Duna periapsis to depart for Kerbin, for a total of about 900 m/s. By comparison, another rocket departing directly from low Duna orbit at the same window needed about 1300 m/s to put it on basically the same trajectory, so the savings from the Oberth departure compared to a low orbit departure was on the order of 400 m/s.

The difference was substantial enough that I'm giving serious thought to a refueling depot in Munar or Minmus orbit for departing interplanetary missions.

[GoSlash27]

OhioBob,

In addition to needing more DV to achieve the higher orbit, I also need to point out that orbital velocity is lower at higher altitudes.

The Oberth effect exists because kinetic energy is what actually gets you from place to place and DV is a nonlinear measure of it.

since Ek=1/2*M*V^2 involves the square of velocity, a change in Ek is proportional to (V2-V1)^2 where V1 is starting velocity and V2 is ending velocity.

(V2-V1)^2= (V2-V1)(V2-V1)= V2^2-2V1V2+V1^2.

Starting the burn at a higher V1 means that a change in velocity will add more kinetic energy than a burn at a a lower V1.

Best,

-Slashy

[Laie]

One question - how do steering losses affect this? One of the reasons I've been doing higher orbits is to minimize those losses and still be able to do a transfer burn in a single orbit.

Before you can begin to estimate steering losses, you need to know how you will execute a maneuver.

For example, you could point prograde all the way and spiral outward (never mind if the "spiral" doesn't even make a 1/4th of a rotation in the end). Or you could point in a certain direction relative to the fixed stars and not deviate from that direction. Both are valid methods and of no further interest, I just mention them as examples of how you may go about doing your burn.

What would probably be most interesting to us players is what happens when you follow a maneuver node. For short burns, maneuver nodes seem to follow the second method, pointing in a fixed direction. However, during longer burns the maneuver nodes update their vector to compensate for all kinds of error, and to the best of my knowledge, noone has figured out how they work.

- - - Updated - - -

Starting the burn at a higher V1 means that a change in velocity will add more kinetic energy than a burn at a a lower V1.

Yes, but that additional kinetic energy doesn't necessarily help you. What matters in the end is the kinetic energy you still have by the time you cross the SOI boundary. I can't really follow OhioBob's math, but it appears to be one of those cases where you start out with less and leave with more.

Edited October 6, 2015 by Laie

[GoSlash27]

Yes, but that additional kinetic energy doesn't necessarily help you. What matters in the end is the kinetic energy you still have by the time you cross the SOI boundary.

Laie, the amount of kinetic energy you end with will be the same regardless. What matters is how much kinetic energy you have to add to get there and thus how much kinetic energy you start with. Starting at low altitude gives you more kinetic energy to start with due to your higher orbital velocity. Starting at higher altitude gives you more potential energy, but less kinetic energy.

I can't really follow OhioBob's math, but it appears to be one of those cases where you start out with less and leave with more.

There are a couple important equations left out of the write-up. The one I suspect is in error is escape velocity as a function of altitude.

I suspect that this chart is roughly accurate *if* it involves an initial burn to lower Pe followed by the transfer burn (what Hotaru refers to as the "Oberth maneuver".

We had a really good write up on this subject a few months back. I'll see if I can dig it up.

Best,

-Slashy

Edited October 6, 2015 by GoSlash27

[Laie]

Hmm. As a simple explanation for why OhioBob might be right, I posit that a higher periapsis may reduce gravity losses on your way out. You start a bit higher up, and gravity's vector is a little more off from retrograde. That would also explain why it's less pronounced the faster you leave.

Of course, that isn't oberth's fault. The effect is stronger the faster you go and that's that. So what OhioBob is really telling us is that oberth doesn't entirely overrule everything else.

[GoSlash27]

Hmm. As a simple explanation for why OhioBob might be right, I posit that a higher periapsis may reduce gravity losses on your way out. You start a bit higher up, and gravity's vector is a little more off from retrograde. That would also explain why it's less pronounced the faster you leave.

Of course, that isn't oberth's fault. The effect is stronger the faster you go and that's that. So what OhioBob is really telling us is that oberth doesn't entirely overrule everything else.

Laie,

The proof would be in execution. If you actually place a satellite in orbit at 300km and set up a transfer burn to Jool, You'll find that it doesn't actually save DV compared to 70km unless you retroburn to reduce Pe first.

Best,

-Slashy

[Chaos_Klaus]

Hmm. As a simple explanation for why OhioBob might be right, I posit that a higher periapsis may reduce gravity losses on your way out. You start a bit higher up, and gravity's vector is a little more off from retrograde. That would also explain why it's less pronounced the faster you leave.

Well, gravity also accelerates you on the way towards PE ... there is no energy loss here ... how could there be.

Oberth actually does rule everything. It is true that it takes less delta v to depart from a circular orbit at higher altitude then at lower altitude. IF we are talking about direct ejection. The reasoning is that you could go to minmus, refuel there and have "free fuel". Ok, you had to get there in the first place, but it's kinda free. So far so good.

What most people overlook is, that when you drop your PE back towards Kerbin, you do not need to expend the same delta v at PE to do the ejection, because you already have your AP at a high alttude. Adding to that, you are going very fast at PE, which is exactly where Oberth kicks in.

If you depart from Minmus, spending the 160m/s to drop your PE towards Kerbin saves you 940m/s on your ejection burn. The reason beeing that it only takes another 20m/s to reach escape velocity.

There really is no reason at all to do direct ejections from higher orbits, because doing this "bielliptic ejection" (as one might call it) is allways more efficient.

[mhoram]

A while back there was a similar discussion.

For reference it was this post and the ones afterwards: http://forum.kerbalspaceprogram.com/threads/75542?p=1072850&viewfull=1#post1072850

It was also discussed when two burns ("lower Peri to LKO" & "ejection burn at Peri") are more efficient than a single burn.

[GoSlash27]

A while back there was a similar discussion.

For reference it was this post and the ones afterwards: http://forum.kerbalspaceprogram.com/threads/75542?p=1072850&viewfull=1#post1072850

It was also discussed when two burns ("lower Peri to LKO" & "ejection burn at Peri") are more efficient than a single burn.

MHoram,

Thank you so much! This is exactly the thread I was looking for!

Best,

-Slashy

[Laie]

Well, gravity also accelerates you on the way towards PE ... there is no energy loss here ... how could there be.

Because you're on a *escape* trajectory. You won't gain anything when you return to periapsis because you don't return. All you have is losses on the way out the door.

[OhioBob]

GoSlash27 said:

The proof would be in execution. If you actually place a satellite in orbit at 300km and set up a transfer burn to Jool, You'll find that it doesn't actually save DV compared to 70km unless you retroburn to reduce Pe first.

Mathematically I can show that it takes less Δv from the 300 km orbit than from the 70 km.

First the 70 km orbit,

V_orb = (GM/r)^0.5 = (3531600000000/670000)^0.5 = 2295.9 m/s

V_esc = (2GM/r)^0.5 = (2*3531600000000/670000)^0.5 = 3246.9 m/s

V_∞ = 2800 m/s (approximate average for Jool)

V_bo = (V_∞² + V_esc²)^0.5 = (2800² + 3246.9²)^0.5 = 4287.5 m/s

Δv = V_bo - V_orb = 4287.5 - 2295.9 = 1991.6 m/s

Now the 300 km orbit,

V_orb = (3531600000000/900000)^0.5 = 1980.9 m/s

V_esc = (2*3531600000000/900000)^0.5 = 2801.4 m/s

V_bo = (2800² + 2801.4²)^0.5 = 3960.8 m/s

Δv = 3960.8 - 1980.9 = 1979.9 m/s

Now let's show that both have the same energy post burn...

E_total = E_kinetic + E_potential = mv²/2 - GMm/r

We can divide through by m to cancel out the mass of the vehicle, giving

E_total = v²/2 - GM/r

At 70 km,

E_total = 4287.5²/2 - 3531600000000/670000 = 3,920,000 J/kg

At 300 km,

E_total = 3960.8²/2 - 3531600000000/900000 = 3,920,000 J/kg

We can see that in both cases the energy is the same, as expected. We also see that the burn at 300 km required 11.7 m/s less Δv than the burn at 70 km, which is contrary to the popular belief.

(ETA) Also note that Alex Moon's Launch Window Planner shows the same thing that my computations are showing. Find a good transfer to Jool and you will see the that ejection Δv is less at 300 km than it is at 70 km. Go much above 300 km and the Δv starts to go up, just like the graph in the OP.
 

Edited July 10, 2016 by OhioBob

[EliasDanger]

I don't understand any of the math. But I understand you guys are arguing over what's most efficient in the game. You can throw all the fancy equations at it you want, but it's all going to be hypothetical. The only way to settle the debate for sure is experimentation. Run tests, document, and share the results.

[OhioBob]

I don't understand any of the math. But I understand you guys are arguing over what's most efficient in the game. You can throw all the fancy equations at it you want, but it's all going to be hypothetical. The only way to settle the debate for sure is experimentation. Run tests, document, and share the results.

If someone where to do those tests, I recommend they do it for Duna rather than Jool. As the graphs in the OP show, the effect is much larger and more obvious with Duna. With Jool, small imperfections in the way the maneuver node is set up could result in greater error than the effect we're trying to observe.

It's also possible that the game's patched conic method could skew the results away from the purely hypothetical. Patched conics is not something I took into consideration at all when I did the math.

Edited October 6, 2015 by OhioBob

[Tourist]

I don't understand any of the math. But I understand you guys are arguing over what's most efficient in the game. You can throw all the fancy equations at it you want, but it's all going to be hypothetical. The only way to settle the debate for sure is experimentation. Run tests, document, and share the results.

Towards that end, I propose a race around the world, ending in Paris on the 4th day of... oh wait wrong bet. It would be great to see somebody demonstrate this using identical spacecraft, identical amounts of fuel etc. I've always wondering myself about the most efficient way of leaving/reentering SOIs.... after a number of, shall we say, mission planning errors, which would have lead to stranding a few Kerbals in deep space (were it not for the infinite fuel cheat, and some complicated rationalization of the use of said cheat)

[GoSlash27]

OhioBob,

I can see where you're coming from and this approach definitely has merit, but as a practical matter it doesn't quite pan out.

I was able to achieve a Jool apoapsis with less DV in both cases and the "intercept" required less DV from the lower orbit.

Something's off here, but I can't see what it is. I'll have to ponder it...

Best,

-Slashy

[cantab]

Mathematically I can show that it takes less ÃŽâ€v from the 300 km orbit than from the 70 km.

If there's no refuelling involved, though, what we care about is how much delta-V it takes from the ground.

[OhioBob]

GoSlash27 said:

I can see where you're coming from and this approach definitely has merit, but as a practical matter it doesn't quite pan out.

If there is any practical application of it, I think it is likely to be realized if we are using a refueling station or something like that. If we are simply launching from the surface of Kerbin and ejecting from orbit, then I see no practical application. The extra Δv it takes to reach the higher orbit cancels out any benefit. If we just want a short term parking orbit, then I believe the lower the better.
 

Edited April 9, 2016 by OhioBob

[More Boosters]

OhioBob,

I can see where you're coming from and this approach definitely has merit, but as a practical matter it doesn't quite pan out.

I was able to achieve a Jool apoapsis with less DV in both cases and the "intercept" required less DV from the lower orbit.

http://i52.photobucket.com/albums/g13/GoSlash27/75kXFer_zpszzpvjtvr.jpg

http://i52.photobucket.com/albums/g13/GoSlash27/300kXFer_zpsy45lsdjc.jpg

Something's off here, but I can't see what it is. I'll have to ponder it...

Best,

-Slashy

That difference is surprisingly little. I'll certainly consider that for very-low TWR vehicles as I'm sure I'd lose more to steering than a measly 20 Delta-V at a 70km orbit compared to a 300km orbit. Thanks for this!

Edit: Of course this doesn't include the Delta-V cost of getting to a 300x300 orbit as opposed to a 70x70 one, but how much would that cost anyway? 400? 500? Either way, makes rendezvous and refueling all that much easier.

Edited October 6, 2015 by More Boosters

[davidpsummers]

I think the thing is to treat the effect as an "all other things being equal" kinda thing. For example, my ion engine probes all leave from 1000 km orbit. The burns are long and that lets you not swing around too much of an orbit during the burn (and also helps you avoid having to try and do you burn on the dark side of the planet.)

[OhioBob]

If there's no refuelling involved, though, what we care about is how much delta-V it takes from the ground.

While you were posting the above, I was writing a response that agreed with you (see above post #21).

[GoSlash27]

If there is any practical application of it, I think it is likely to be realized if we are using a refueling station or something like that. If we are simply launching from the surface of Kerbin and ejecting from orbit, then I see no practical application. The extra ÃŽâ€v it takes to reach the higher orbit cancels out any benefit. If we just want a short term parking orbit, then I believe the lower the better.

OhioBob,

I always send my missions assembled and fueled in orbit.

I'm certain that the most economical (from a DV standpoint) approach is to start at a higher altitude, drop the periapsis, then hit the transfer. But launching direct from higher up? It's never worked out in my practice.

I've bookmarked this page. I'll have to ponder this...

Best,

-Slashy