Flight Question Orbit & Sync orbit fuel savings: WHY?

[ceauke]

Hi guys

I've been playing orbiter a few times over the last few years. I'm far from a rocket scientist or educated on extreme physics but I was hoping someone could give me some more sciency answer on fuel savings regarding adjusting your orbit to be round and aligning the orbital plane. So far all the manuals I see just say what is the most economical way of doing it. E.g. the apogee and perigee is the best places to change orbit size, and some mention that an eliptical orbit may have some points that's more fuel efficient for aligning planes.

I did some testing or playing and it seems like it's 'cheaper' in fuel to align planes at the apogee rather than perigee. Is this correct?

Then my question is why? How do you put the physics into simple english? For orbit size adjustments I guess the answer has something to do with the minimum and maximum kenetic and potential energies? I have no idea... And what about aligning planes? To me it seems like the orbit has to shift so much more at apogee, so I cannot figure out why I need a shorter burn at apogee to align plane than at perigee.

I hope some of the above made some sense

Thanks!

 

[PeterRoss]

Your velocity in apogee is always lower than in perigee. So you need less dV to change your movement vector. Just try to imagine this in vectors, you'll understand everything.

---------- Post added at 08:31 AM ---------- Previous post was at 08:24 AM ----------

Since I'm no rocket scientist either my explanation may be incomplete or even wrong. Still it works for me. I'll be glad to hear any more correct explanation.

---------- Post added at 09:13 AM ---------- Previous post was at 08:31 AM ----------

One more thing. Apogee plane alignment will be fuel-efficient only if the apogee will coincide with ascending or descending node. In any other cases - yes, it will be too much to shift.

 

[Quick_Nick]

Your velocity in apogee is always lower than in perigee. So you need less dV to change your movement vector. Just try to imagine this in vectors, you'll understand everything.

I really like that explanation. :thumbup:

Also, for perigee/apogee changes, it might be that you can just as effeciently do this at any point in your orbit, but it would involve a 'strange' orientation. It's much simpler to just wait until the opposite point in the orbit and burn perfectly prograde or retrograde.

 

[fredericva]

Also, for perigee/apogee changes, it might be that you can just as effeciently do this at any point in your orbit, but it would involve a 'strange' orientation. It's much simpler to just wait until the opposite point in the orbit and burn perfectly prograde or retrograde.

IMFD can do something like that, though it's only used for circularizing orbits to the altitude you're currently at, which I think is the second most efficient way to change ApA/PeA (with the most efficient being burns at ApA/PeA).

 

[Quick_Nick]

IMFD can do something like that, though it's only used for circularizing orbits to the altitude you're currently at, which I think is the second most efficient way to change ApA/PeA (with the most efficient being burns at ApA/PeA).

Yes, my reason for the idea was largely because when using TransX (very similar to IMFD), you can achieve the correct orbit even if you burn at the wrong time if you just burn in the direction it tells you to. I believe it may be just as effecient to follow this 'drifing' angle while you burn. You can also hold the 'average' orientation throughout the entire burn with the same result but I believe this is slightly ineffiecent. Ex: holding prograde for 2 minutes before perigee til 2 minutes after perigee will in the end only affect apogee, as is desired, but might be less effecient than 'sweeping' from a few degrees off of prograde across to a few degrees the other side of prograde. This is because with the 'sweeping' motion, all your energy goes into changing the apogee the entire time. With the 'static orientation' method, you affect your perigee as a side effect and then cancel out these effects for the second half of the burn.

EDIT: Done editting. Someone please correct me if I'm wrong on anything.

 

[TSPenguin]

I wonder if there exists an (survivable) orbit where it is more efficient to burn at Perigee due to the oberth effect.
Someone willing to do the math?

 

[tblaxland]

I wonder if there exists an (survivable) orbit where it is more efficient to burn at Perigee due to the oberth effect.
Someone willing to do the math?

This thread is discussing two types of manoeuvres: plane change and circularisation.

A plane change manoeuvre will always be cheaper (fuel volume wise) at apogee. There is no increase in the specific orbital energy in a plane change burn, therefore no help from Oberth.

For circularisation, it is always cheaper to circularise at apogee. For example, if you are in a 200 x 10,000 km orbit, you would need ~1,500 m/s dV to circularise to 200 x 200 km and ~1,200 m/s dV to circularise to 10,000 x 10,000 m/s. Circularising to perigee is reducing the specific orbital energy, so it is always more expensive by Oberth.

Does that answer the question?

 

[TSPenguin]

That makes so much sense that I will rest my head on my desk now, repeatedly.
Thanks tblaxland.

 

[Urwumpe]

I wonder if there exists an (survivable) orbit where it is more efficient to burn at Perigee due to the oberth effect.
Someone willing to do the math?

Oberth effect is for apsis changes, not plane changes. Plane changes don't change the magnitude of the orbit energy.

 

[fredericva]

A plane change manoeuvre will always be cheaper (fuel volume wise) at apogee. There is no increase in the specific orbital energy in a plane change burn, therefore no help from Oberth.

What about a plane change where there is no node coinciding with the apogee?

 

[PeterRoss]

What about a plane change where there is no node coinciding with the apogee?

The node where velocity is lowest will be the cheapest (in the sense of fuel efficiency) of two nodes. It's all about velocity.

 

[Tommy]

Circularising to perigee is reducing the specific orbital energy, so it is always more expensive by Oberth.

This holds somewhat true for circularizing an orbit from an eliptical orbit due to the way the "inverse square law" affects gravity. This means raising an orbit is a bit cheaper than lowering one, since raising the orbital mean 50k requires less change in overall energy than reducing it by the same distance. However, for an orbit insert from a hyperbolic orbit lower is better. The extent of the advantage of using the Oberth effect is determined by the distance travelled over a given time period , and this will be greater at a lower altitude (and thus a higher velocity). It doesn't matter that you are reducing velocity - subtraction is really addition (of a negative value) so the math works either way. Mind, it's cheapest to lower the PeA sooner than later in this case since it involves less deflection of the velocity vector.

 

[tblaxland]

However, for an orbit insert from a hyperbolic orbit lower is better.

Take a spacecraft on a hypothetical hyperbolic trajectory approaching Earth. Say the trajectory has a C3 of 25 km^2/s^2. To circularise to 200 km altitude would require a 4.3 km/s impulse whereas circularising to 25,000 km altitude would require a 3.5 km/s impulse. What have I missed?

 

[Tommy]

What have I missed?

Nothing, if a 25,000km altitude orbit suits the mission. If you need a low orbit, or will be landing, it should be cheaper than a high insertion and subsequent transfer to the lower orbit or re-entry (this assumes a direct re-entry or re-entry from the higher orbit isn't possible).

Also, my understanding is that the Oberth effect doesn't affect how much impulse is needed, but rather how much reaction energy is needed to create that impulse. In other words, you get more dV from each kilogram of fuel spent.

 

[Urwumpe]

Also, my understanding is that the Oberth effect doesn't affect how much impulse is needed, but rather how much reaction energy is needed to create that impulse. In other words, you get more dV from each kilogram of fuel spent.

No, it is rather change in Orbit energy vs dV. You get a bigger chance in the orbit energy, for each m/s dV.

 

[Tommy]

No, it is rather change in Orbit energy vs dV. You get a bigger chance in the orbit energy, for each m/s dV.

I stand corrected. I guess what we should be saying is "You get more energy for the same impulse." I think I got a bit confused because, although the dV provided by the impulse is the same at any altitude, the final increase of velocity is greater.

Take a spacecraft on a hypothetical hyperbolic trajectory approaching Earth. Say the trajectory has a C3 of 25 km^2/s^2. To circularise to 200 km altitude would require a 4.3 km/s impulse whereas circularising to 25,000 km altitude would require a 3.5 km/s impulse. What have I missed?

I think I did find something you missed. You are assuming that the orbit insert will be a single burn resulting in a circular orbit. Let's break that down into two burns - one to attain capture - transform from a hyperbolic orbit to an elliptical orbit, and a second to circularize. The capture burn will be more efficient at higher velocity (lower Pe). Since kinetic energy is related to the square of the velocity this can substantially increase the change in energy vs change in velocity. There may be times when it's more efficient to perform the capture burn at 200k alt, and lower the ApA to 20,000k. Then perform a circularisation burn at Ap. I suspect this would only provide any benefit if you need a substantial change in energy to attain capture (ie, coming in with a very high relative velocity to Earth). I don't have quite enough math or time to figure where that cut-off would be - if it exists. You'ld have to figure out the change in specific energy needed - not the dV needed. Then you could calculate the dV required to change the energy level that amount for each case and see which was less.

I'll offer up ten "Hail Probe"s for anyone mathy enough to calculate that, so I can find out if I'm right or wrong!

 

[Tany Jamel]

Mathematical approach to solve this question

How do you explain this through mathematical approach? I was learning the formulas and was asked to answer why it cost less at apogee. But I couldn't figure it out through formulas(such as Cost=velocity-(mu/a(1-e))*1/2). Captain here?

 

[Tommy]

How do you explain this through mathematical approach? I was learning the formulas and was asked to answer why it cost less at apogee. But I couldn't figure it out through formulas(such as Cost=velocity-(mu/a(1-e))*1/2). Captain here?

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Look up the Oberth effect on Wikipedia and it should give you the math. Its the same formula you use to calculate work, which is a function of mass, distance and time. At the higher velocity you will cover more distance therefore more work is done in the same amount of time.