Kerbin SOI versus Sun SOI ?

[Sirine]

Lets say I wanna go to Eve, from Kerbin.

Question:-

1)

A) Should I burn all the way to get the projected 'line' to touch Eve 'purple line' (Eve Planet Orbiting Sun Line).

or

Should I "Exits" Kerbin SOI 1st. Later then (in Sun SOI) burn for that Eve 'purple line'?

2) What is the difference?

Answer: A. There are tons of Delta-v save. Detail see below.

http://forum.kerbalspaceprogram.com/threads/61387-Kerbin-SOI-versus-Sun-SOI?p=834037&viewfull=1#post834037

Edited December 12, 2013 by Sirine

[astropapi1]

Oberth is your friend. The faster you're going, the more efficent a trans-planetary injection will be.

[Sirine]

Oberth is your friend. The faster you're going, the more efficent a trans-planetary injection will be.

I thought Oberth apply for Ap, and Pe node....Anyhow I still don't understand. Is A or B?

[Frederf]

You want to do A.

You want to do as much burning as possible in the deepest gravity well possible. Your orbital energy is either gravitational or kinetic so kinetic is maximized when gravity is minimized. Co-planar interplanetary transfers are most efficiently done by adding escape velocity+enough more to achieve the transfer orbit. You can think of being stuck at Kerbin's solar orbital speed while in its SOI. To do an interplanetary transfer you need to escape the SOI just to get to square one. All your velocity inside Kerbin's SOI is worthless because it's captured so you might as well use it for an Oberth effect boost to puncture through the SOI boundary and keep going.

Oberth savings are non-intuitive so you'll have to do a few math problems until the pattern becomes familiar.

[astropapi1]

Both are valid. It's just that A is more efficent than B.

[Sirine]

You want to do A.

You want to do as much burning as possible in the deepest gravity well possible. Your orbital energy is either gravitational or kinetic so kinetic is maximized when gravity is minimized. Co-planar interplanetary transfers are most efficiently done by adding escape velocity+enough more to achieve the transfer orbit. You can think of being stuck at Kerbin's solar orbital speed while in its SOI. To do an interplanetary transfer you need to escape the SOI just to get to square one. All your velocity inside Kerbin's SOI is worthless because it's captured so you might as well use it for an Oberth effect boost to puncture through the SOI boundary and keep going.

Oberth savings are non-intuitive so you'll have to do a few math problems until the pattern becomes familiar.

Thanks.

So, lets put this another way....

1) I should burn to escape Kerbin SOI, And STOP my engines once the projection shows that I'm on escape trajectory.

2) And (wait and sail until) when near the edge of Kerbin SOI, (Far from Kerbin, lower gravity, Near Minmus Altitude) And START engines to burn for Eve 'purple line'.

Are these the best approach (utilize) for Oberth effect?

Edited December 12, 2013 by Sirine

[astropapi1]

Nope. Do the entire burn as near to Kerbin as you can.

[Sirine]

Nope. Do the entire burn as near to Kerbin as you can.

Your orbital energy is either gravitational or kinetic so kinetic is maximized when gravity is minimized.

I miss-understand the above sentences?

[Superfluous J]

You "should" do whatever you're comfortable with. The dV savings for Olberth is there but it's not like you're going to save tons of dV. We're talking percentages here.

If you do B, which I personally find easier, just make sure you leave Kerbin's SOI going BACKWARDS in relation to the Sun (ie, you should burn on the day side) to get to Eve (or Moho but you said Eve). For all other planets, you want to leave Kerbin's SOI going FORWARDS.

The same goes for A, but as it's all one burn and you are able to get it to Eve I assume you already knew that

The best way to see for yourself is to do both burns. Get your ship into orbit around Kerbin and then make a quick save. Do a big burn to get all the way to Eve, and then note how much fuel you have left. Then revert to your quick save and do it the other way, with 2 burns (1 to exit Kerbin's SOI, another to intersect Eve) and look how much fuel you have left. If the differece is significant enough to be important to you, then you have your answer.

Edited December 12, 2013 by 5thHorseman

[Sirine]

You "should" do whatever you're comfortable with. The dV savings for Olberth is there but it's not like you're going to save tons of dV. We're talking percentages here.

If you do B, which I personally find easier, just make sure you leave Kerbin's SOI going BACKWARDS in relation to the Sun (ie, you should burn on the day side) to get to Eve (or Moho but you said Eve). For all other planets, you want to leave Kerbin's SOI going FORWARDS.

The same goes for A, but as it's all one burn and you are able to get it to Eve I assume you already knew that

The best way to see for yourself is to do both burns. Get your ship into orbit around Kerbin and then make a quick save. Do a big burn to get all the way to Eve, and then note how much fuel you have left. Then revert to your quick save and do it the other way, with 2 burns (1 to exit Kerbin's SOI, another to intersect Eve) and look how much fuel you have left. If the differece is significant enough to be important to you, then you have your answer.

Yeah, I think the best way to do it is to DIY.

Okay, here are the result for all.

70x70km Kerbin Orbit. I got 180 fuel, with a LV-N, small probe core, and RTG.

1) Burn Direct to Eve Purple Line....124 fuel left.

2) Burn near edge of Kerbin SOI, outer Minmus......98 fuel left.

3) Burn at Sun SOI.....97 fuel left...

Damn. The differences is huge...

Thanks again. stropapi1, Frederf and 5thHorseman.

[Kasuha]

I got 180 fuel, with a LV-N, small probe core, and RTG.

That's a bit too little fuel for that huge and heavy LV-N. I'd suggest using 48-7S instead.

The difference is 56:82 spent fuel units, i.e. about 2:3. Whether it's huge or not ... that's matter of opinion.

[Superfluous J]

I've never actually done the burn test. That is far more than I thought it would be!

[Sirine]

Yes...Imagine that if you wanna go to Moho....slingshot from Eve is.... exhilarateï¼Â

[Sirine]

That's a bit too little fuel for that huge and heavy LV-N. I'd suggest using 48-7S instead.

The difference is 56:82 spent fuel units, i.e. about 2:3. Whether it's huge or not ... that's matter of opinion.

48-7S: 119 fuel left to touch the purple line.

[wasmic]

You "should" do whatever you're comfortable with. The dV savings for Olberth is there but it's not like you're going to save tons of dV. We're talking percentages here.

If you do B, which I personally find easier, just make sure you leave Kerbin's SOI going BACKWARDS in relation to the Sun (ie, you should burn on the day side) to get to Eve (or Moho but you said Eve). For all other planets, you want to leave Kerbin's SOI going FORWARDS.

The same goes for A, but as it's all one burn and you are able to get it to Eve I assume you already knew that

The best way to see for yourself is to do both burns. Get your ship into orbit around Kerbin and then make a quick save. Do a big burn to get all the way to Eve, and then note how much fuel you have left. Then revert to your quick save and do it the other way, with 2 burns (1 to exit Kerbin's SOI, another to intersect Eve) and look how much fuel you have left. If the differece is significant enough to be important to you, then you have your answer.

An Eve capture burn from kerbin will require about 1600 m/s if i remember correctly. If you do a kerbin escape first, that costs you 950 m/s, and then you need to burn 1500 m/s to get an eve capture.

[maltesh]

Aye, the delta-V difference is significant.

The following graph assumes the spacecraft starts in a 100km orbit over Kerbin, and sets an interplanetary transfer course. The horizontal axis is semimajor axis distance from the sun in millions of kilometers, the vertical axis is delta-V requirement in kilometers per second for that departure transfer burn.

The vertical lines are, from left to right, the semi-major axes of the orbits of Moho, Eve, Kerbin,Duna,Dres, Jool, and Eeloo.

Link to the Desmos graph that created the above image: https://www.desmos.com/calculator/nrdjd8qv5x

The green curve is the delta-V requirement for doing a burn-from-LKO to the specified solar semi-major axis distance.

The red curve is the delta-V requirement for the two-step method of burning just to escape Kerbin's SOI, and then doing a second burn in solar orbit.

The closer your destination is to Kerbin, the smaller the difference is, but even with planetary destinations as close as Eve or Duna, you're still spending 50% more delta-V on the transfer departure if you use the two-step method than if you burn directly from LKO.

If you have a spacecraft design that can afford the extra losses, fine. But if you have a spacecraft whose fuel margins are thin, or if you want to shift a larger payload to the destination with a smaller spacecraft, or if you want to save delta-V to have more options when you arrive at your destinaton, or if you want to increase the chances of success when you don't know if your design is good enough, you should be burning from LKO.

[Kasuha]

There's one reason which may make exiting Kerbin SOI first more convenient:

If you don't use launch window planner, you can put a maneuver on the Sun orbit much longer ahead in time than on your LKO. Also setting up that maneuver is much easier than fiddling the LKO escape.

So if you want to be as efficient as possible, you need to start from low orbit. But if you have overengineered enough, you can afford to go the way you consider more comfortable.

[Frederf]

I miss-understand the above sentences?

The two sentences agree. One way of thinking about the Oberth effect is that you add more orbital energy when going fast (per m/s added) so burning is most efficient energy-wise when fast. You are the fastest when your kinetic energy is high. Your kinetic energy is high when your gravity energy is low. Your gravity energy is low if you are down in a gravity well (falling down speeds you up). Kinetic energy is non-relativistically 1/2mV^2. Going from V = 1 m/s to V = 2 m/s and from V = 999 m/s to V = 1000 m/s both require 1 m/s delta V. However the difference in energy in the first case is 3 units (2^2 - 1^2) and in the second case is 1999 units! (1000^2 - 999^2). By going faster you got 666x the energy from the same 1 m/s delta V!

Even if you can't hit planetary transfer exactly from LKO it's much better to guess and get close and do some fine adjustment in solar SOI.

Also that is a very very very VERY cool graph which shows the difference wonderfully.

[Yasmy]

I miss-understand the above sentences?

Gravitational potential energy is negative: E = - G M m / r,

where M is the planet mass, m is the spacecraft mass, G is Newton's constant, and r is the distance between them.

So the closer you get to the planet, the smaller r becomes, the more negative your gravitational potential energy becomes.

Gravitational potential energy is negative because it takes energy to escape the gravitation potential.

Infinitely far away, the potential energy is zero, because the gravitational force is zero.

As long as you don't spend any delta-V, your total energy, kinetic + potential, is constant: E = 1/2 m v^2 - G M m / r

Thus as you get closer to a planet, r decreases, making -1/r more negative. To keep E constant, v must increase.