Question about the Oberth effect

[Snark]

The crucial difference between Oberth effect as applied to a reaction engine (like a rocket) versus a traction engine (like a car, or a guy on a bicycle):

In the bicycle example, you do work, i.e. expend energy directly. Every joule of kinetic energy that you add is coming directly out of your muscles. It takes much more energy (work) to go from 8 m/s to 10 m/s than it does to go from 2 m/s to 4 m/s, in spite of the fact that the dV is the same in both cases.

Whereas with a rocket, it's not about the energy. The vast majority of the energy is thrown away as high-speed rocket exhaust. It takes exactly the same amount of fuel burn to get 100 m/s of dV regardless of how fast you're going at the time.

In other words, with a traction engine you're spending energy; with a reaction engine you're spending velocity. The difference between the two is why Oberth works for rockets and not bicycles.

Here's another way of looking at why it works for reaction engines: If you're in LKO and do a big burn to eject to go to, say, Moho, you're leaving your fuel mass down in LKO. If you do a moderate burn to lift your Ap up to Kerbin's SoI boundary a d do a big burn there, then you just paid to lift many tons of fuel thousands of km higher than you needed to. ;) Edited November 18, 2015 by Snark

[LostOblivion]

[quote name='RocketPropelledGiraffe']Well, let me expand on this discussion and add a question of reference frame:
The gain in kinetic energy is proportional to the velocity, that's the basic idea and that's why you usually want to do pro/retro burns at periaps. So far so good.

Now we think about going interplanetary, to the outer solar system. This is easy mode, because you burn prograde in LKO, which is also prograde relative to the sun. Ideally my orbital velocity relative to Kerbin adds to the orbital velocity of Kerbin around the sun (~9200 + 2200 m/s). Do everything in one burn and you are better off, simple.

Now let's go to the inner solar system. We burn retrograde in LKO to escape Kerbin's gravity. This is done retrograde to the sun as well, which means the velocity relative to the sun is minimal, because we have to substract our velocity relative to Kerbin ( ~9200 - 2200 m/s).
What is the tradeoff here? Is it better to burn to escape Kerbin in LKO, with good relative velocity to Kerbin, and then do another burn afterwards, because we travel at the full 9200 m/s in solar orbit then?

Think about appliying 2000 m/s delta-v in two instantaneous burns of 1000 m/s each (One to escape Kerbin, one to get to the destination):
If we do everything in LKO we get 1000 m/s applied at 9200-2200 = 7000 m/s relative solar velocity, and the second one at 9200-2200-1000= 6000 m/s, which is even slower relative to the sun.
If we do 1000 m/s in LKO just to escape we apply it at the same 7000 m/s. The second 1000 m/s burn occurs after escape from Kerbin, with 9200 m/s relative solar velocity.
With regard to the Oberth effect the second one sounds better. And, at least after three wheat beers tonight, it also seems very reasonable. :huh:[/QUOTE]

Not sure if this was answered, or if his question was understood correctly, but is he right? For destinations in the inner solar system, does a sun-prograde burn out ahead of Kerbin, and then burn sun-retrograde, harness the Oberth effect better than a direct sun-retrograde burn?

[GoSlash27]

[quote name='LostOblivion']Not sure if this was answered, or if his question was understood correctly, but is he right? For destinations in the inner solar system, does a sun-prograde burn out ahead of Kerbin, and then burn sun-retrograde, harness the Oberth effect better than a direct sun-retrograde burn?[/QUOTE]

LostOblivion,
No, it doesn't. The cheap way to an inner planet is to eject solar retrograde from Kerbin in a single burn. The ideal altitude above Kerbin varies with the destination planet, but a direct ejection from LKO is still cheaper than escaping Kerbin's SoI and then burning retrograde.

Best,
-Slashy

[OhioBob]

[quote name='LostOblivion']Not sure if this was answered, or if his question was understood correctly, but is he right? For destinations in the inner solar system, does a sun-prograde burn out ahead of Kerbin, and then burn sun-retrograde, harness the Oberth effect better than a direct sun-retrograde burn?[/QUOTE]

[FONT=Verdana][COLOR=#000000]As Slashy said, no.[/COLOR][COLOR=#000000] The best way to get to any planet, inferior or superior, is to perform a single burn close to Kerbin.
[/COLOR][COLOR=#000000]
If we accelerate to exactly escape velocity, we’ll have no velocity left over, relative to Kerbin, after escaping.[/COLOR][COLOR=#000000] We would effectively be in the same orbit as Kerbin, only leading or trailing the planet by a small amount. [/COLOR][COLOR=#000000]To reach another planet we would then have to perform a second burn to either speed up relative to the Sun, to reach a superior planet, or slow down relative to the Sun, to reach an inferior planet.[/COLOR][COLOR=#000000] Instead of performing the second burn, it is better to give ourselves a little extra velocity when we perform the escape burn so that we have some velocity left over after escaping.[/COLOR][COLOR=#000000] This left over velocity is called [/COLOR][I][COLOR=#000000]hyperbolic excess velocity[/COLOR][/I][COLOR=#000000], denoted V[/COLOR][SUB][COLOR=#000000]∞[/COLOR][/SUB][COLOR=#000000]. It is measured relative to Kerbin and it doesn’t matter whether it is prograde or retrograde. [/COLOR][COLOR=#000000]We want to give V[/COLOR][SUB][COLOR=#000000]∞[/COLOR][/SUB][COLOR=#000000] just the right value that, after we escape Kerbin, we are traveling at the correct velocity relative to the Sun that we will reach the target planet without having to perform a second burn.
[/COLOR][COLOR=#000000]
Let’s say we want to reach Eve.[/COLOR][COLOR=#000000] If we performed two burns, it takes about 950 m/s to first escape Kerbin. [/COLOR][COLOR=#000000]It then takes a second burn of about 750 m/s to modify our solar orbit to reach Eve.[/COLOR][COLOR=#000000] That’s a total of 950 + 750 = 1700 m/s.
[/COLOR][COLOR=#000000]
On the other hand, let’s say we perform a single ejection burn close to Kerbin. [/COLOR][COLOR=#000000]We have to give our spacecraft enough kinetic energy to first escape, plus enough additional kinetic energy that we have some leftover after escaping. [/COLOR][COLOR=#000000]We do this by accelerating to some [/COLOR][I][COLOR=#000000]burnout velocity[/COLOR][/I][COLOR=#000000], V[/COLOR][SUB][COLOR=#000000]bo[/COLOR][/SUB][COLOR=#000000], whose kinetic energy is equal to the sum of the escape energy and the residual energy.[/COLOR][COLOR=#000000] That is,
[/COLOR][COLOR=#000000]
½ mV[/COLOR][SUB][COLOR=#000000]bo[/COLOR][/SUB][SUP][COLOR=#000000]2[/COLOR][/SUP][COLOR=#000000] = ½ mV[/COLOR][SUB][COLOR=#000000]esc[/COLOR][/SUB][SUP][COLOR=#000000]2[/COLOR][/SUP][COLOR=#000000] + ½ mV[/COLOR][SUB][COLOR=#000000]∞[/COLOR][/SUB][COLOR=#000000][SUP]2[/SUP]
[/COLOR][COLOR=#000000]
Canceling out ½ m, we have

V[/COLOR][SUB][COLOR=#000000]bo[/COLOR][/SUB][SUP][COLOR=#000000]2[/COLOR][/SUP][COLOR=#000000] = V[/COLOR][SUB][COLOR=#000000]esc[/COLOR][/SUB][SUP][COLOR=#000000]2[/COLOR][/SUP][COLOR=#000000] + V[/COLOR][SUB][COLOR=#000000]∞[/COLOR][/SUB][COLOR=#000000][SUP]2[/SUP]
[/COLOR][COLOR=#000000]
Let’s say we’re in a 75 km parking orbit, where V[/COLOR][SUB][COLOR=#000000]esc[/COLOR][/SUB][COLOR=#000000] = 3235 m/s.[/COLOR][COLOR=#000000] We set V[/COLOR][SUB][COLOR=#000000]∞[/COLOR][/SUB][COLOR=#000000] = 750 m/s, which we determined earlier is the velocity we want to be traveling relative to Kerbin in order to reach Eve. [/COLOR][COLOR=#000000]Therefore,
[/COLOR][COLOR=#000000]
V[/COLOR][SUB][COLOR=#000000]bo[/COLOR][/SUB][COLOR=#000000] = ( 3235[/COLOR][SUP][COLOR=#000000]2[/COLOR][/SUP][COLOR=#000000] + 750[/COLOR][SUP][COLOR=#000000]2[/COLOR][/SUP][COLOR=#000000] )[/COLOR][SUP][COLOR=#000000]1/2[/COLOR][/SUP][COLOR=#000000] = 3321 m/s
[/COLOR][COLOR=#000000]
Our initial orbital velocity is 2287 m/s, therefore the Δv is 3321 – 2287 = 1034 m/s. [/COLOR][COLOR=#000000]This is far less than the 1700 m/s it would take to perform the two burn strategy.[/COLOR][COLOR=#000000] Also note that the magnitude of the ejection burn is the same whether we burn prograde or retrograde. [/COLOR][COLOR=#000000]If we burn prograde, we’ll be going +750 m/s relative to Kerbin, and if we burn retrograde we’ll be traveling -750 m/s relative to Kerbin.[/COLOR][/FONT] Edited November 19, 2015 by OhioBob

[GoSlash27]

^ Trust the OhioBob. The OhioBob is wise.

Best,
-Slashy

[Goody1981]

Did you guys see my explanation? Might have gotten buried since I'm new and had to wait for the mods to approve my post

[OhioBob]

[quote name='Goody1981']Did you guys see my explanation?[/QUOTE]

Yes. I thought it was an interesting explanation. A simple illustration of the concept.

[Goody1981]

Thanks :)