It uses less dV to circuluarize at higher orbits beyond a certain point. Why?

[Kerbonaut257]

So I was playing around with orbital mechanics that I learned from scott manley. I found something interesting. Starting at a 70km Kerbin circular orbit. It costs about 1231 dv to get to a 10Mm circular orbit. However, getting to a 80Mm circular orbit is actually cheaper! It's only 1120 dV. 

I don't understand this. How is it possibly cheaper to get to a higher orbit. Can someone explain this to me? I understand that circularizing is cheaper in a higher orbit. But how is it possible that as your orbit is bigger, it actually is cheaper to get into that circular orbit.

Edited May 26, 2016 by Kerbonaut257

[p1t1o]

Possibly because you spend less of the transit time as deep in the gravity well, reducing related losses. Transit time is longer overall, but integrated over the flight, gravity losses are less. I think.

[Gaarst]

Must be a mistake. In terms of work getting to a higher orbit is always more expensive, at least from what I know in orbital mechanics.

Though maybe the speed at 80Mm being really low, circularising is very cheap, combining that with the Oberth effect that you use to get your apo to 80Mm might explain this. Would be interested to see someone do the maths.

Edited May 26, 2016 by Gaarst

[Nathair]

1 hour ago, Kerbonaut257 said:

Starting at a 70km Kerbin circular orbit. It costs about 1231 dv to get to a 10Mm circular orbit. However, getting to a 80Mm circular orbit is actually cheaper! It's only 1120 dV.

2

How did you come by those numbers?

[Gaarst]

Doing the maths gives the same result (1228m/s to 10Mm and 1118m/s to 80Mm).

I think the explanation is the difference in dV spent in LKO, where Oberth effect is strong. The higher orbit is slower (8 times higher), so the circularisation must be cheaper. For the burn in low orbit, the second manoeuvre requires a greater burn at a high orbital speed (low orbit) so more dV is saved than for the first one.

[royying]

this is the formula of attractive force between two bodies

as the distance (r) increase, the force decrease exponentially

therefore the spaceship need less delta-v to change orbit

[Gaarst]

Ah, just remembered the whole thing about bi-elliptic transfers. So this result is true as both (transfer) orbits are very elliptical and one is much higher, and slow orbital velocity and stuff...

Hohmann transfer orbits are not the most efficient in some cases. In cases even more extreme than what you described, the most efficient way can be a bi-elliptic transfer, ie: getting a higher-than-needed apo first, and then lowering your orbit is more efficient than a Hohmann transfer.

Edited May 26, 2016 by Gaarst
Typos

[Tiber9]

^this. I was really confused for a while, but check out 'worst Case's in this article.  https://en.m.wikipedia.org/wiki/Hohmann_transfer_orbit

Basically, maximum Delta v is when going to around 15.58 times the original orbit (not surface alt, but radius). 

[samstarman5]

Yeah, establishing a higher orbit is easier and cheaper. It's getting up there where the difference lies. But remember, once you make orbit, you are halfway to anywhere.

[nosirrbro]

4 hours ago, Tiber9 said:

^this. I was really confused for a while, but check out 'worst Case's in this article.  https://en.m.wikipedia.org/wiki/Hohmann_transfer_orbit

Basically, maximum Delta v is when going to around 15.58 times the original orbit (not surface alt, but radius). 

You posted this on mobile.

[Kerbonaut257]

Sadly nothing anyone has said really makes sense to me. I do understand the oberth effect, and I guess I can understand how a slightly bigger boost at the periaps is going to increase the overall energy in the system more, even though you circularize much higher, it's very cheap to circularize that high up, so the added energy from the oberth effect at the low altitude more than compensates for the cheaper circularizing, ergo the total energy in the system is... greater? Despite costing less dV? That still is hard for me to grasp.

[Tiber9]

29 minutes ago, nosirrbro said:

You posted this on mobile.

*fail*. yes, I sure did, sorry. link to wiki page

32 minutes ago, Kerbonaut257 said:

so the added energy from the oberth effect at the low altitude more than compensates for the cheaper circularizing, ergo the total energy in the system is... greater? Despite costing less dV? That still is hard for me to grasp.

I'm not sure I get it either, but I think what's going on is that the energy needed for circularizing goes down with radius faster than the energy needed for the transfer burn goes up. As you get closer to an escape, the transfer-burn delta-v gets closer to your orbital velocity times root 2, and the delta v to circularize is closer to zero. It takes more delta-v to go higher until 15-16 times the initial orbit radius, and from then on it is cheaper the higher you want to circularize. 

As for why that is, well, it is because of the oberth effect, in the sense that a burn along the velocity vector affects the energy of the orbit more when the velocity is high, so for the same delta-v you put more energy into the system. link to specific orbital energy. I hope that helps some, but really, just thanks for getting me thinking about this because it is counter-intuitive and at first just seems wrong!

 

[G'th]

A more intuitive, if potentially less accurate way of wrapping your head around it is to imagine it as a gravity issue. At that far out, the energy it takes to fight against the gravitational pull is less, so you don't have to expend as much energy to circularize as you would at lower orbit deeper in the gravity well. 

Its sort of an inverse of the Oberth effect, where the gains from burning at X location in the gravity well results in a circular trajectory rather than an elliptical one. 

[Superfluous J]

1 hour ago, Kerbonaut257 said:

Sadly nothing anyone has said really makes sense to me. I do understand the oberth effect, and I guess I can understand how a slightly bigger boost at the periaps is going to increase the overall energy in the system more, even though you circularize much higher, it's very cheap to circularize that high up, so the added energy from the oberth effect at the low altitude more than compensates for the cheaper circularizing, ergo the total energy in the system is... greater? Despite costing less dV? That still is hard for me to grasp.

You're think of this backwards, I think. It's not that the higher orbit is somehow magically creating energy, it's that the lower one is more wasteful. Both are wasteful (as thermodynamics tells us is true of everything in the universe) but the lower orbit is more so.

[Guest]

Is this the same effect as inclination changes? What my question refers to is how it's much cheaper to change or match inclination at whichever node is at the highest altitude. And yeah, my head is starting to hurt from trying to grasp this.

edit : me thinking out loud : while it costs more to bump apoapsis from 10Mm to 80 Mm, circularizing the orbit at 80Mm is so much cheaper than it is at 10Mm, that it actually makes the total dv budget less to get to the higher circular orbit, hence my origianl comparison to the inclination change thing. I need a nap now.

Edited May 27, 2016 by Otis

[qromodynmc]

Did you calculated difference between needed deltav for 70km and 80km?

[Kerbonaut257]

Basically the dV needed to get to about 10Mm from a 70km parking orbit is the MOST expensive circular orbit. After that, the larger and larger circular orbits (even calculated beyond the SOI of 84Mm) get cheaper and cheaper and approach root(2)*current velocity. Because the initial burn required to get to escape velocity (apo of infinity) is root(2)*current velocity, and the cost to circularize at infinity is zero.

 

So interestingly enough, it's more expensive than root(2)* current velocity to increase your orbit to about 10km. 

 

I do think that we're onto something thinking about it in terms of the oberth effect. Imagine it this way, EVERY m/s added to velocity at pe is MORE effective than the last, because the energy in the system uses v^2 in the calculation. So if you're squaring a number, adding a number when the number is high is MORE effective. Just like 50^2 compared to 51^2 is a bigger difference than 3^2 versus 4^2. Due to this, increasing by another 100 dV in the initial burn adds WAY more energy, thus making the circularizing at the top cheaper. It still seems pretty counter intuitive, but the numbers don't lie! 

10 hours ago, Otis said:

Is this the same effect as inclination changes? What my question refers to is how it's much cheaper to change or match inclination at whichever node is at the highest altitude. And yeah, my head is starting to hurt from trying to grasp this.

edit : me thinking out loud : while it costs more to bump apoapsis from 10Mm to 80 Mm, circularizing the orbit at 80Mm is so much cheaper than it is at 10Mm, that it actually makes the total dv budget less to get to the higher circular orbit, hence my origianl comparison to the inclination change thing. I need a nap now.

Otis, inclination changes are cheaper to do the bi-eliptic transfer if you're changing more than 45 degrees or so. (this is where you go eccentric to the TOP of the SOI, change your inclination almost for free at the top of the orbit, and then recircularize at periaps). So it's kinda a similar concept in a way. I believe below 40 degrees it's cheapest to just change your inclination at the current orbit by burning normal/anti-normal

[Snark]

Lots of good explanations in this thread, but it's not clear to me that we've quite hit the "sweet spot" for making sense to the OP. 

To @Kerbonaut257:

If I'm in a low circular orbit, and I want to go to a higher circular orbit, I have to do two things:

1. Burn when I'm in the low orbit, to raise my apoapsis up to the target altitude.
2. After I arrive at the new apoapsis, do a 2nd burn to circularize there.

Burn #1 is always bigger for a higher target orbit.  This is simple and intuitive.  It's also what's confusing you, because this is what you're mainly thinking of, and why it seems weird that a higher circular orbit would ultimately be cheaper.

However, the important thing to remember is that it's nonlinear.  The higher your Ap, the less extra dV it takes to raise your orbit further.

For example, suppose you're in LKO.  From there, it takes around 850 m/s of dV to raise your Ap up to the height of the Mun, at 12,000 kilometers.  That's a pretty big chunk of dV, right?

Now suppose you decide you're going to Minmus instead.  Minmus orbits four times higher than the Mun does.  But it takes less than 950 m/s of dV to raise your apoapsis up that high!

In other words:  Raising your apoapsis the first 12,000 kilometers takes a whopping 850 m/s of dV, but raising it an additional 33,000 km takes less than 100 m/s!

So yes, it's more expensive to boost up to Minmus' height... but it's only a tiny bit more expensive.  So getting the extra altitude only takes a little bit more for burn #1.

"Well, yes," I hear you saying, "okay, it's only a little bit more expensive.  But it's still more expensive than the lower orbit, right?"

You're right-- but don't forget about burn #2, your circularization burn.  The difference in burn #1 between "go to the Mun's altitude" versus "go to Minmus' altitude" is tiny, less than 100 m/s.  However, burn #2 changes by a lot.  Circular orbital velocity at 12,000 km is 600ish m/s, and you'll need to supply a big chunk of that because your velocity at apoapsis will be pretty small.  However, the circular velocity at 45,000 km is much smaller, only around 300 m/s.  So reaching that requires less dV.

So, what it boils down to is that when you're transferring to a higher circular orbit,

• your first burn gets a tiny bit more expensive
• your second burn gets a lot cheaper

...and that's why the higher orbit becomes cheaper.

[Fearless Son]

What @Snark said covers it pretty well.  If it helps, think of it like this: the Oberth effect works because you are adding energy to the system at the point of highest velocity (the lowest point in an orbit,) right?  But when you are circularizing, you are not actually adding much total energy to the system so much as you are taking the energy that is already in the system and evening the velocity out.  You are taking all the energy that the craft uses to go fast at that low point, and changing it to go higher-but-slower instead.  It is why a higher elliptical orbit is cheaper to circularize than a lower one: you have more potential energy at that point, and changing direction (but not speed) is cheaper when you are slower, so you can channel more of it into lateral motion.  

[tseitsei89]

3 hours ago, Kerbonaut257 said:

I do think that we're onto something thinking about it in terms of the oberth effect. Imagine it this way, EVERY m/s added to velocity at pe is MORE effective than the last, because the energy in the system uses v^2 in the calculation. So if you're squaring a number, adding a number when the number is high is MORE effective. Just like 50^2 compared to 51^2 is a bigger difference than 3^2 versus 4^2.

But... Isn't this exactly what oberth effect is about? Just in different words? Or am I understanding something wrong?

 

Also Snark has good and easy to understand explanation

[The_Rocketeer]

Really simple version:

Moving up or down a gravity well (getting more elliptical) is exponentially more work the deeper you are inside it - work per meter increases exponentially with depth.

Moving around a gravity well (getting more circular) is always the same amount of work per meter, but the further out of the well you are the more meters there are in the circle (because the circumference of your circle is bigger).

 

Edited May 27, 2016 by The_Rocketeer

[Xyphos]

all mathematics aside, and strictly from a technique point-of-view, your circulation burn depends on how you initiated your launch and how well you performed your gravity turn.

try launching the same vessel a few times differently and notice the changes in trajectory.

if you tilt your vessel 5 or 10 degrees immediately off of the launch pad, instead of going straight up, you can shave a few hundred dV off of your gravity turn maneuver, which also takes a few hundred dV off of your circulation burn.

it is also worth noting that using throttle control, it's possible to achieve orbit without needing a circulurization burn, but it'll just take slightly longer to get there and it's also a skill that's hard to master. it's just easier to do a circulation burn.

I think the least dV I've ever needed to circularize was 54 m/s, and that was at 500Km

[Cunjo Carl]

First off, it's awesome you were just playing around with orbital mechanics. Best way to learn, that!

I'm sure you're already overwhelmed with explanations, and there's quite a few excellent ones, but I figure I'll add this to the table. This is just in regards to the first burn and why you suddenly start 'saving' energy because you're going faster. To quote from Snark's excellent overview:

On 5/27/2016 at 9:24 AM, Snark said:

However, the important thing to remember is that it's nonlinear.  The higher your Ap, the less extra dV it takes to raise your orbit further.

   Here's a good way to think of this nonlinear energy saving. When doing your first burn, the higher you lift your AP the faster you go. And the faster you're going, the faster you move away from Kerbin. The faster you can pull away from Kerbin the less long you feel its gravity as you do, so the less that gravity slows you down. If you go fast enough, you can just rip straight out of Kerbin's gravity well, like ripping off a bandaid. Going slowly, it has all the time it needs to pull you back in.

  It's exactly like when you're riding your bike and need to go up over a little hill, so you speed up a bunch first. If you're going fast already you just 'whoop!' up to the top. But, if you were going slow, you'd need to crank in a ton of force to puuuuuush your way up. It only took a bit of extra force to speed up before hand, but you saved a ton when it came to the hill!

  This is thinking in terms of force over time, momentum. If we were trying to explain this same thing with energy (the Oberth effect), we'd say instead that you added your energy where your velocity was already fast (fast = before the hill or in the gravity well), so it was more effective... which is true but totally counter intuitive!

  The effects of your deltaV changing your orbit are typically most easily thought of in terms of this bicycling over a hill analogue of momentum. Your DeltaV is a measurement of your ability to change momentum after all! (with a little twist to account for your weight). Meanwhile, energy is best for steady orbits, because a thing's energy always stays the same while it's orbiting. Almost any situation can be explained with either momentum or energy, but typically one winds up being easier than the other. Here, think with momentum.

  I hope that helps!

[Nich]

So can we apply any of this to launching off the surface of kerbin?  What is the cheapest altitude to launch too?  70km? 80Mm?  Assuming going to the Mun would it be more efficient to launch higher then the Mun raise PE to the height of the Mun and have a very cheap Munar insertion burn?  Or does Munar oberth cancel this out?

[The_Rocketeer]

@Nich you would need to be more specific about the launch profile (SP vs Vertical Ascent) and definition of 'cheapest'.