some orbital mechanics / oberth thing I don't get

[Laie]

Going from Low Earth Orbit to Mars takes about 3500m/s.

A hohmann transfer from an Earth-like orbit around the Sun (say, starting from Earth's L3/4/5) to Mars requires pretty much the same dV.

It seems as if the presence/absence of Earth doesn't make much of a difference either way. Why is this so?

 

[p1t1o]

Not sure if its a complete answer but heres a thing:

If you are doing a Hohmann transfer from LEO, you can take advantage of the fact that on one side of the planet your LEO velocity is adding to your velocity around the sun, and on the other side of the planet your LEO velocity is subtracting from your velocity around the sun.

Whereas if you are simply in an earth-like heliocentric orbit your velocity is always the figure in between those values.

 

Ergo, planned properly, leaving from an Earth-present LEO gives you a dV boost (for want of a better word) when compared with leaving from an Earth-free heliocentric orbit - which offsets the dV cost (at least partly) of getting to LEO from the ground.

 

Make sense?

[Steel]

4 minutes ago, p1t1o said:

Not sure if its a complete answer but heres a thing:

If you are doing a Hohmann transfer from LEO, you can take advantage of the fact that on one side of the planet your LEO velocity is adding to your velocity around the sun, and on the other side of the planet your LEO velocity is subtracting from your velocity around the sun.

Whereas if you are simply in an earth-like heliocentric orbit your velocity is always the figure in between those values.

 

Ergo, planned properly, leaving from an Earth-present LEO gives you a dV boost (for want of a better word) when compared with leaving from an Earth-free heliocentric orbit - which offsets the dV cost (at least partly) of getting to LEO from the ground.

 

Make sense?

That's how I understand it.

 

12 minutes ago, Laie said:

Going from Low Earth Orbit to Mars takes about 3500m/s.

A hohmann transfer from an Earth-like orbit around the Sun (say, starting from Earth's L3/4/5) to Mars requires pretty much the same dV.

It seems as if the presence/absence of Earth doesn't make much of a difference either way. Why is this so?

 

How much of a difference does "pretty much the same" encompass?

[Laie]

5 minutes ago, Steel said:

How much of a difference does "pretty much the same" encompass?

I can't do the math (if I could I probably needn't ask), so I just planned maneuvers in KSP/RSS. Of my few samples, the results were very similar: 3500+-100 m/s either way. Variations between launch windows (due to eccentricity) were more important than starting from Earth or not.

[KerikBalm]

Just FYI, the same dV thing works in KSP with Duna and kerbin... The answer is that when in kerbin orbit, you're going much faster, and thus more benefit from the oberth effect.

One way of looking at it is KE=1/2 MV^2   If you take V from 10 to 20, you don't add nearly as much KE as If you take V from 1010 to 1020 (20^2-10^2 = 300, 1020^2-1010^2 = 20,300

[Steel]

41 minutes ago, KerikBalm said:

Just FYI, the same dV thing works in KSP with Duna and kerbin... The answer is that when in kerbin orbit, you're going much faster, and thus more benefit from the oberth effect.

One way of looking at it is KE=1/2 MV^2   If you take V from 10 to 20, you don't add nearly as much KE as If you take V from 1010 to 1020 (20^2-10^2 = 300, 1020^2-1010^2 = 20,300

Yeah, you've misunderstood the question, which is: why is there not a bigger difference in the situation described?

[KerikBalm]

because you're also deeper in a gravity well, which means that you have to increase your potential energy even more. Also its a bit misleading to look at absolute values like that.. proportionately, the velocity difference isn't so huge compared to solar orbit (8 vs 30 km/sec)

[PB666]

4 hours ago, p1t1o said:

Not sure if its a complete answer but heres a thing:

If you are doing a Hohmann transfer from LEO, you can take advantage of the fact that on one side of the planet your LEO velocity is adding to your velocity around the sun, and on the other side of the planet your LEO velocity is subtracting from your velocity around the sun.

Whereas if you are simply in an earth-like heliocentric orbit your velocity is always the figure in between those values.

 

Ergo, planned properly, leaving from an Earth-present LEO gives you a dV boost (for want of a better word) when compared with leaving from an Earth-free heliocentric orbit - which offsets the dV cost (at least partly) of getting to LEO from the ground.

 

Make sense?

I can put another. Try this form a low kerbol orbit, say a million km, burn to reach Jool orbit, record starting and final velocity. Next starting at the same orbit burn about half the dv of first, wait until you have lost about have of the velocity and then burn to make jool orbit. The some of the two dVs requiredis more than that required by a single burn. this is because when you a very close to a celestial,myou are going very fast, you have a square function of energy, when you add dV you add even more energy, howver there is a finite amount of exit energy required, determined by u and the periapsis radius, once you have superceded that your residual is a funciotion of 1/2v2 - energy required , the residual energy converts to velocity squar root (2 times residual energy). Remembering that you are traveling along an orbit of the celestial, so that once you are out of the SOI your residual velocity is added on top. Essentially you are exploiting the curvature of space time, which likes to hold onto to lower energetic trajectories longer, and release higher energetic trajectories, as the object is receding faster from a celestial mass it is experiencing gravity for a shorter oeriod of time, it keeps more of its velocity. 

If you are given two positions in a orbit to add velocity and you want to add dV to go away from that body, its always better to add velocity when the object is going the fastest. The closer to that fastest point you add dV the further and faster you will go away. 

 

Another example of this is kicking (repeated burns at pe for a high isp, low thrust rocket) to maximize apo or escape velocity). If the celestial is orbiting a star, then to gain orbit about the star, burn at a pe that is a sunset of the orbit (angle to prograde is greater than or equal to 180, depending on how far away you want to be) you might also consider adding normal or antinormsl intonthe burn, its cheap and it might lessen a inclination burn later.. To loose a around the celestial at or just before angle tonprograde reaches zero, dawn. If you are going to moho and for instances you can time the position correctly with regrad to the 6 degress of inclination change you need to make, you can save alot of dV) 

[fredinno]

1 hour ago, PB666 said:

I can put another. Try this form a low kerbol orbit, say a million km, burn to reach Jool orbit, record starting and final velocity. Next starting at the same orbit burn about half the dv of first, wait until you have lost about have of the velocity and then burn to make jool orbit. The some of the two dVs requiredis more than that required by a single burn. this is because when you a very close to a celestial,myou are going very fast, you have a square function of energy, when you add dV you add even more energy, howver there is a finite amount of exit energy required, determined by u and the periapsis radius, once you have superceded that your residual is a funciotion of 1/2v2 - energy required , the residual energy converts to velocity squar root (2 times residual energy). Remembering that you are traveling along an orbit of the celestial, so that once you are out of the SOI your residual velocity is added on top. Essentially you are exploiting the curvature of space time, which likes to hold onto to lower energetic trajectories longer, and release higher energetic trajectories, as the object is receding faster from a celestial mass it is experiencing gravity for a shorter oeriod of time, it keeps more of its velocity. 

You need to make eccentricity changes too if you do the burn at the half point, so it's not a very good analogy.

[PB666]

43 minutes ago, fredinno said:

You need to make eccentricity changes too if you do the burn at the half point, so it's not a very good analogy.

Right you burn a different higher pe which if you are leaving that system is unneccesary also, e doesn't really matter that much if you are leaving, its all over one, but why would you invest in a higher pe, if once you leave the system you are not going use that pe. Its sauce for the goose, because the most efficient place to burn out is always at periapse, the place were speed is fastest. BUt the most important thing is that the reason Apo increases to Escape point followed by increases of post-Escape velocities is that burning at pE increases kinetic energy, part of which is converted to potential energy as it leaves, and the rest remains kinetic energy relative to the system central mass.

The reason that Transfer energetics is not ideal in the 2 step process between orbits that start small and then increase 5 or six fold is because the second burn is velocity wise, far from the first, if it is close to the first, its efficient, and if it close to escape burn its efficient (because at escape velocity the second burn is basically 0 relative to the planet, its a limit that can never be reached in a comoving space-time system) a second burn is not needed.

Its all about the energetics, work is the addition of force over distance, if you travel into a low point around a celestial and then do work, you are moving faster relative to the celestial. Your craft has an absolute limit of thrust to weight a measure acceleration (increase in velocity) and of force/time, it does not matter if it is stationary to the celestial or moving 1/10th the speed of light. Since work is force * distance and since velocity is distance * time the force/time * distance (=velocity * time)     therefore     force/time * velocity* time  Kinetic Energy= integral (t=0 to t=end thrust)  thrust*velocity. since the moment of thrust can be converted to dV as dV = f(thrust. time) the two velocity aspects multiply by each other.

This aspect of rocket behavior can be used in the atmosphere also, conservation of motion during a launch is extremely important, for example the atmosphere aspect decreases allows increased velocity and ground referenced acceleration, loss of acceleratoin in stage changes cost efficiency for several reasons, thus its a good idea to minimize the stage losses. The more one is pushing against gravity to stay aloft the more important these are. One strategy to avoid this is to shed boosters from the side, in which the main engines continue thrusting, even if the booster engines have energy left. Once a craft has turned and has significant omega^2 r centripetal acceleration, it can relieve its stack stages without much cost.

Even though you are adding the same dV you have produced alot more energy. That celestial only took u/r potential energy coming in and will only take u/r going out, so the large ramp up of kinetic energy at the pe, is all for the space craft to keep once its leaves the SOI, if it wait halfway up the exit to add more dV it will be going slower relative to.

I think the way we can look at this is via enthalpy available in warped space-time complexes,  there because space-time is warped by energy. Bodies that enter these systems cannot gain or loose total energy and random applications of force will result in a loss of energy from the bodies (such as frictional forces causing the collapse of a nebula protosystem disk). Nature assumes the application of force is random and so enthralpy will always tend from energetic to less energetic state, application of force intelligently can favor the improved energetics of one object at the expense of something else. In this case the exhaust of gases at pE in a retrograde direction almost certain results in those gases falling into the celestial. The system assumes a continuous motion toward disorder but ordering certain aspects can be done, even randomly objects are tossed from systems into space, others are tossed into the star, and the order in a planet is sometimes smashed into collisions.

 

 

 

[GoSlash27]

Laie,

 This is an orbital mechanics question I can actually help you with

The Hohmann transfer from a Kerbin- like orbit about the sun is a straight vis-viva problem. 918.3 m/sec. Doing the same transfer from low Kerbin orbit actually takes *more* DV; 1,078.3 m/sec.

So why does it take more? Where is the benefit from our Oberth effect? Well...

The Oberth effect gives you "free" kinetic energy thanks to your velocity, but a lot of that gets traded off because you have to climb out of a gravity well that wouldn't be there if you were merely in a kerbin- like orbit about the sun. In this case, getting out of the gravity well costs more than you would save from the Oberth effect, so the total DV cost is higher.

If you were to leave the Kerbin system from a higher circular orbit, you could maximize your DV savings. For example, the DV cost to transfer from an 8 Mm altitude above Kerbin is only 649.4 m/sec. Any higher than that and the Oberth loss outstrips the gravity gain. Any lower, and the opposite happens.

But wait... there's more!

If we raise the orbit to a Munar- like orbit about Kerbin, our DV cost goes up a bit to 656.8 m/sec, but we can use the Oberth effect from a low Munar orbit to slash this even further.

Orbiting the Mun at 10 km altitude will occur at 556.9 m/sec. Ejecting to Munar escape would be 556.9*sqrt(2)= 787.6 m/sec. The excess velocity to get us to Duna from there is the aforementioned 656.8 m/sec. So the total velocity required to make the transfer is figured out Pythagoras style; sqrt(656.8^2+787.6^2) = 1,025.5 m/sec. And we already have over half of that thanks to our orbital velocity (yay Oberth!), so the DV cost to get to Duna from low Munar orbit is only 468.6 m/sec.

The general rule is that the Oberth effect benefits you for far targets and hurts you for close targets. You can get to any interplanetary destination cheaper from low Munar orbit than you can from low Kerbin orbit, but the savings are *much* higher for close targets like Duna and Eve.

TL/DR; you can save a lot of DV by refueling in low Munar orbit and leaving from there.

 

HTHs,
-Slashy

Edited April 6, 2016 by GoSlash27

[PB666]

4 minutes ago, GoSlash27 said:

Laie,

 This is an orbital mechanics question I can actually help you with

The Hohmann transfer from a Kerbin- like orbit about the sun is a straight vis-viva problem. 918.3 m/sec. Doing the same transfer from low Kerbin orbit actually takes *more* DV; 1,078.3 m/sec.

So why does it take more? Where is the benefit from our Oberth effect? Well...

The Oberth effect gives you "free" kinetic energy thanks to your velocity, but a lot of that gets traded off because you have to climb out of a gravity well that wouldn't be there if you were merely in a kerbin- like orbit about the sun. In this case, getting out of the gravity well costs more than you would save from the Oberth effect, so the total DV cost is higher.

If you were to leave the Kerbin system from a higher circular orbit, you could maximize your DV savings. For example, the DV cost to transfer from an 8 Mm altitude above Kerbin is only 649.4 m/sec. Any higher than that and the Oberth loss outstrips the gravity gain. Any lower, and the opposite happens.

But wait... there's more!

If we raise the orbit to a Munar- like orbit about Kerbin, our DV cost goes up a bit to 656.8 m/sec, but we can use the Oberth effect from a low Munar orbit to slash this even further.

Orbiting the Mun at 10 km altitude will occur at 556.9 m/sec. Ejecting to Munar escape would be 556.9*sqrt(2)= 787.6 m/sec. The excess velocity to get us to Duna from there is the aforementioned 656.8 m/sec. So the total velocity required to make the transfer is figured out Pythagoras style; sqrt(656.8^2+787.6^2) = 1,025.5 m/sec. And we already have over half of that thanks to our orbital velocity (yay Oberth!), so the DV cost to get to Duna from low Munar orbit is only 468.6 m/sec.

The general rule is that the Oberth effect benefits you for far targets and hurts you for close targets. You can get to any interplanetary destination cheaper from low Munar orbit than you can from low Kerbin orbit, but the savings are *much* higher for close targets like Duna and Eve.

TL/DR; you can save a lot of DV by refueling in low Munar orbit and leaving from there.

 

HTHs,
-Slashy

Or another way to think of it.

When you travel about the sun you have potential and kinetic energy, imagine if you could borrow temporarily some potential energy and turn it into kinetic energy, and then give it back, you could not do this because to give potential means you lower altitude.

Crossing a planets SOI does this, you give up some of your potential energy momentarily and take kinetic energy (not to the star but to the planet), but when you fire your engines you gain alot more kinetic energy because you are going faster (work = force * distance traveled), then when you leave the SOI you give back the kinetic energy loan and take back the potential energy and get to keep all the extra kinetic energy.

[Lukaszenko]

On 4/5/2016 at 4:47 PM, Laie said:

Going from Low Earth Orbit to Mars takes about 3500m/s.

A hohmann transfer from an Earth-like orbit around the Sun (say, starting from Earth's L3/4/5) to Mars requires pretty much the same dV.

It seems as if the presence/absence of Earth doesn't make much of a difference either way. Why is this so?

 

The trick is that you DON'T start at an Earth-like orbit around the Sun. You start in LEO. If you then just barely escape the Earth, and then make another burn to go to Mars (as opposed to burning directly to a Mars transfer from LEO) the difference will be much larger. 

As others have said, it takes a lot of energy to climb out of the Earth's gravity well, and it takes a lot of energy to go to Mars. Due to Oberth, you can almost buy one and get one free.

If you go even further, the discounts really start to stack up.

Edited April 7, 2016 by Lukaszenko

[sevenperforce]

On 4/6/2016 at 9:06 PM, PB666 said:

Or another way to think of it.

When you travel about the sun you have potential and kinetic energy, imagine if you could borrow temporarily some potential energy and turn it into kinetic energy, and then give it back, you could not do this because to give potential means you lower altitude.

Crossing a planets SOI does this, you give up some of your potential energy momentarily and take kinetic energy (not to the star but to the planet), but when you fire your engines you gain alot more kinetic energy because you are going faster (work = force * distance traveled), then when you leave the SOI you give back the kinetic energy loan and take back the potential energy and get to keep all the extra kinetic energy.

Which you can do because burning close to the planet leaves your exhaust with much less gravitational potential energy, allowing you to keep that extra kinetic energy.

[PB666]

54 minutes ago, sevenperforce said:

Which you can do because burning close to the planet leaves your exhaust with much less gravitational potential energy, allowing you to keep that extra kinetic energy.

Right, and in a perfect world that exhaust velocity relative to the ship is exactly equal to your forward velocity. But more to the point, that exhaust is now stuck in a gravity well. 

Edited April 7, 2016 by PB666