Why the lowest periapsis is most efficient?

[YL3GDY]

Hi All,

Reading many manuals about interplanetar flights, often I came across a statement that "to make most efficient eject burn, you need your periapsis as low as possible". It sounds as a general rule, but I wonder which is the physical background of it?

Could anyone explain it, using strict physical proof?

 

[V8Li]

You have the highest possible velocity there. Think of it from another point of view: if you burn prograde at Ap the Pe will get up to the Ap value and from there you will start doing the interplanetary burn. So you've wasted fuel to get the Pe high for nothing.

For any other manoeuvres that alter your orbit (inclination, Pe when coming from the Moon etc) you should burn as soon as possible and as far as possible.

Also, those explains are for the most fuel-efficient travels.

 

[Urwumpe]

Hi All,

Reading many manuals about interplanetar flights, often I came across a statement that "to make most efficient eject burn, you need your periapsis as low as possible". It sounds as a general rule, but I wonder which is the physical background of it?

Could anyone explain it, using strict physical proof?

I only know that rule for inject burns, but the wisdom works only for elliptical orbits: At periapsis, your own velocity is higher. The DV for a eject burn to reach a special target is a function of local circular orbit velocity. The higher your own velocity is beyond the circular orbit velocity at periapsis radius - the less DV you need for eject.

 

[DanP]

I only know that rule for inject burns, but the wisdom works only for elliptical orbits: At periapsis, your own velocity is higher. The DV for a eject burn to reach a special target is a function of local circular orbit velocity. The higher your own velocity is beyond the circular orbit velocity at periapsis radius - the less DV you need for eject.

But, on the othere hand, wouldn't a lower periapsis mean a higher circular orbit velocity at periapsis radius thereby increasing the equired dV?

This is assuming "circular orbit velocity" means "the velocity of an object in circular orbit of a given radius".

I understand the logic behind burning at PeA instead of ApA, as V8Li explained. But why would the injection burn be shorter for a lower PeA given the same ApA?

 

[Enjo]

Have you read Duncan's "Standard Orbit, Mr. Sulu" ?

http://www.orbitermars.co.uk/

Check Old Stuff at the bottom.

The principle is that in order to raise apoapsis for example, you need to add energy to the orbit. You can only do it by adding the kinetic energy, not potential. The increase of energy in time is power, which is:

P = F * v

So the more velocity you have (the lower the periapsis), the more energy increase you'll have, for that thrust (F).

 

[V8Li]

I understand the logic behind burning at PeA instead of ApA, as V8Li explained. But why would the injection burn be shorter for a lower PeA given the same ApA?

Given 2 orbits: ApA = 200Km, PeA1 = 180Km, PeA2 = 150Km. If you would use let's say Equation MFD to calculate the velocity at Pe for the 2 orbits, you will find out that at PeA1 the velocity is less than it is at PeA2, given the same ApA (don't know the equation). That's orbital physics and there's nothing we can do about it. Consider a similar, more intuitive example: Get in a orbit af an ApA of 300Km and PeA of 150Km and your velocity is about 7500m/s. Consider another orbit in which you are returning from the Moon and your PeA is 150Km too. You probably know you will be at 10k m/s or so velocity when you're at Pe, right. Is the same exact example as the one above, only with different Ap values.

Now, if we use a special tool like TransX it will give us a burn window (an orbital point) at which we need to get an extra DV to whatever orbital velocity we will be in at that exact point. So if you need, at a certain point, a velocity of exactly 10k m/s to get us to the Moon (remember, in that orbit you will be at Pe and that's how fast you need to go to set a high Ap) and you have 2 possible orbits in which that point is at a velocity of 7000m/s and 7500m/s which one would you rather be in? The 7500m/s, of course, since you want to use less fuel to get you to 10k m/s. And TransX will tell you that by giving you a less DV. So TransX calculates your necessary projected orbit, calculates the velocity at Pe needed to get in that Orbit and tells you when to burn so that you set it just right for an intercept. That's it, it cares less if you run out of fuel.

QED

Tip: always leave from a circular 150Km orbit. It's the lowest possible Pe at which you don't encounter dynamic pressure to slow you down and alter it, the Pe position is not a decessive factor anymore since you're at a constant velocity during the entire orbit and there is not a better orbit that will give you a higher velocity. Maybe only 150Km/20Km if you really knwo what you're doing and don't need to get into a parking orbit.

 

[YL3GDY]

Thank you all very much for answers.

It looks like I've misunderstood a bit this statement. I have imagined that it was said that every orbit with low Pe (in absolute value) will be efficient for eject. As far as I understand this situation, mainly large eccentricy is needed, not always low Pe by it's absolute value.

Surely, in case of planet encounter we can't manage eccentricy a lot, but change Pe ( => gravitational potential at this point => the value of parabolical speed there ) in larger bounds. Then this rule can be used literally.

 

[V8Li]

As far as I understand this situation, mainly large eccentricy is needed, not always low Pe by it's absolute value.

It's true but only for interplanetary travels when you don't need to set a high eccentricity yourself. If you go to the Moon for example and you want to set a high eccentricity to have a high velocity at Pe you will waste fuel anyway since you can't get into that eccentricity other way than burning. It won't make any sense. Again, those theories you read in manuals assume you want to use the minimum ammount of fuel.

 

[DanP]

Thanks.

It's one of those things where you know it's true from the start but for some reason you can't get your head around it initially.

 

[Andy44]

Specific Mechanical Energy equation:

E = (v^2)/2 - mu/r

The best place to add velocity to get the most energy increase is where v is highest, because the energy is proportional to the square of the velocity:

post-burn E = ((v+deltaV)^2)/2 - mu/r

 

[Urwumpe]

Also, from looking closer at my equation: The higher the local escape velocity is, the more excess velocity you can get out of the same DV.

 

[mjessick]

Specific Mechanical Energy equation:

E = (v^2)/2 - mu/r

The best place to add velocity to get the most energy increase is where v is highest,...

Another way to put it:

Differentiate the equation above:
dE/dv = v

Energy gain per delta-V is proportional to v.

 

[Andy44]

Another way to put it:

Differentiate the equation above:
dE/dv = v

Energy gain per delta-V is proportional to v.

Beautiful! That goes into my notebook right now.

 

[synonys]

http://en.wikipedia.org/wiki/Oberth_effect

As a vehicle falls towards periapsis in any orbit (closed or escape orbits) the velocity relative to the central body increases. Burning the engine prograde at periapsis increases the velocity by the same increment as at any other time, determined by the delta-v. However, since the vehicle's kinetic energy is related to the square of its velocity, this increase in velocity has a disproportionate effect on the vehicle's kinetic energy; leaving it with higher energy than if the burn was achieved at any other time.

 

[ar81]

If you burn at a point that is not periapsis or apoapsis, you will rotate your orbit or you will be changing the altitude of periapsis/apoapsis. So you are wasting energy.

 

[Urwumpe]

If you burn at a point that is not periapsis or apoapsis, you will rotate your orbit or you will be changing the altitude of periapsis/apoapsis. So you are wasting energy.

Wrong - you only rotate the orbit when you burn out of plane. A burn at another place as the nodes can also be more effective, depending on which orbit you want to be later. For eject burns, it is ideal to burn at periapsis, but it can save you some fuel when you are doing it only close to the periapsis, if this means being closer to your final orbit before the burn.

 

[ar81]

Wrong - you only rotate the orbit when you burn out of plane. A burn at another place as the nodes can also be more effective, depending on which orbit you want to be later. For eject burns, it is ideal to burn at periapsis, but it can save you some fuel when you are doing it only close to the periapsis, if this means being closer to your final orbit before the burn.

Wrong. If you burn out of plane you are changing inclination, not really rotating the orbit shape.

 

[Enjo]

This experiment may not be a proof, answering for the "why" questions, but you can do such experiment in Energy MFD mk2, to visualise the differences in ending energies, depending on your initial velocity. The differences are dramatical:


1) Start "DeltaGlider/DG mk4 in orbit" scenario and accelerate prograde immediately (at periapsis with SimTime = 0). While accelerating, activate Energy MFD mk2
2) Keep accelerating prograde for say 200 s and note your ME at the end of burn. You'll have this ME as KE in infinity
3) Start "DeltaGlider/DG mk4 in orbit" scenario again, reach the apoapsis, and turn prograde. Restart orbiter. You'll be in apoapsis with SimTime = 0. Immediately begin accelerating prograde, and keep accelerating for the same time as before. Now note your ME. A lot smaller, despite we used the same amount of fuel, isn't it? It's like that, because you were accelerating with a smaller starting velocity, and your energy increase is proportional to your current velocity. It's "free energy".

Energy MFD mk2 can be found here:

[ame="http://www.orbithangar.com/searchid.php?ID=3448"]Energy MFD mk2 v.0.1[/ame]

 

[HiPotOk1978]

if anyone is still monitoring this thread, what your all talking about, this might shed some light for ya.

http://en.wikipedia.org/wiki/Bi-elliptic_transfer