Gravity Assist Challenge

[jfull]

Most people (including myself) seem to always use a direct trajectory when trying to get our missions anywhere. In real life however, scientists crunch some serious numbers to calculate gravity assists, reducing the dV necessary by huge amounts.

It seems now that some Kerbal scientists want to launch a probe to Jool, but they want to use the smallest rocket possible. Also, they don't care how long it takes to get there. The goals of this challenge are to use gravity assists as much as you can, and to use craft with as little dV as you can.

The Basic Challenge

The minimum requirements for this challenge are to launch a probe from Kerbin and gravity assist from 2 celestial bodies on your way to Jool. The same body can be used more than once, Kerbin can obviously be used, and the Mun doesn't count unless you somehow use it after leaving and re-entering Kerbin's SOI (though you'll still probably want to use it to help leave Kerbin). Also, no aerobraking unless you're landing your probe on Laythe.

All mods are allowed except those that allow warping, or other things resembling cheating. Any and all navigational aids are encouraged.

Screenshots or (if you're fancy) video must be taken of the launch and all gravity assists

Bonuses:

There isn't exactly a points system for this, but the following things will be considered

Economic Considerations: The missions with the lowest total dV will be recorded

Traveler: The missions with the highest number of SOIs entered will also be recorded

Going in Circles: gravity assist on the same celestial body multiple times

Useful Moons: get into orbit around Jool using primarily gravity assists from its moons

Style: do anything unusual and cool with your mission, or otherwise go beyond the requirements of the challenge

Runner Ups: Those who come close to completing the requirements but don't quite succeed will still get some recognition.

Hardcore Challenge

(may or may not be feasible)

The same as before, but with one major difference, it must be a manned mission carrying at least one Kerbal, and you must bring him back to Kerbin. You will need to use at least one gravity assist from a moon to help leave Jool, an then another at some point before making your final spashdown at Kerbin.

Take as much time as you need to get him there and back. We'll suppose that Kerbals have devised some kind of suspended animation for the purposes of this mission.

[Kasuha]

There's not much between Kerbin and Jool to gravity slingshot off on the way. But if you don't mind here's my entry for the Extended Vacation challenge which includes gravity slingshotting from Moho around Eve, Kerbin, Kerbin, Jool and finally to Eeloo. And since there's also gravity slingshot off Eve on the way from Kerbin to Moho it consists of total of four gravity slingshots between Kerbin and Jool.

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[jfull]

Very nice!

[ihtoit]

BOOM!

Sorry. I love the Oberth effect. Here it is on some of its glory. Well, enough to get you running, anyway:

http://www.askamathematician.com/2013/01/q-how-does-the-oberth-effect-work-and-where-does-the-extra-energy-come-from-why-is-it-better-for-a-rocket-to-fire-at-the-lowest-point-in-its-orbit/

I use this so much in KSP I've almost forgotten what a gravitational slingshot or an aerobrake manoeuvre are (is?). Bielliptical transfers? What bielliptical transfers? If I can't do it in half an orbit, I'll wait until I can. I'm a slave to the Holmann transfer.

Actually, a question: I might be able to answer this myself if I can be bothered right now to wait ten minutes for my loaded-to-the-eyeballs KSP to load (I can't). Is a bielliptic transfer (one where the higher orbit is more than 11.97 times the radius of the lower orbit, assuming both circular, and considering the BET requires three dV impulses as opposed to the HT's two) an economical way to go as far as total dV is concerned in KSP? Like say, 100km to 1200km? If so, that might answer another question that just popped into my head: should it be possible to inject a comsat into kerbalstationary orbit (100km-2868km, again assume circular) using *less* dV hence less fuel than it would to do it using a Holmann transfer by using any other method?

Edited January 30, 2014 by ihtoit

[Pds314]

Kerbin also has Minmus, does it count?

Also, from what I've seen, It is rarely very beneficial to use Gravity assists to go to Jool. The Oberth Effect and good timing are more important.

Although I guess we can always combine them by doing POWERED gravity assists.

[xoknight]

Holy crap, how is that even possible... you are a god at ksp...

[ihtoit]

::shrug:: meh, orbital mechanics 101

[Rhomphaia]

Actually, a question: I might be able to answer this myself if I can be bothered right now to wait ten minutes for my loaded-to-the-eyeballs KSP to load (I can't). Is a bielliptic transfer (one where the higher orbit is more than 11.97 times the radius of the lower orbit, assuming bother circular, and considering the BET requires three dV impulses as opposed to the HT's two) an economical way to go as far as total dV is concerned in KSP? Like say, 100km to 1200km? If so, that might answer another question that just popped into my head: should it be possible to inject a comsat into kerbalstationary orbit (100km-2868km, again assume circular) using *less* dV hence less fuel than it would to do it using a Holmann transfer by using any other method?

Remember radius of the orbit is to the center of the planet not the surface.

[ihtoit]

yep, I picked a random seed number out of my backside to make the point. And peculiar to simulated orbital mechanics in pretty much every package I've come across, gravity is a point source - in real life, this isn't the case. Example: the centre of gravity in the Solar system taken as a whole is the sum of all interacting gravitational vectors from planets, asteroids, comets, right down to the dust. It can only be approximated even on paper (because it would be simply impractical to calculate for every particle in the sphere of influence of the system), and actually lies somewhere outside of the Sun's coronasphere. In the Earth-Moon system, it lies somewhere outside the Earth's iron core. A practical demonstration of gravitational influence on a two-body system is very easy. Tie a hammer to a piece of string, hold the other end of the string at arms length and turn on the spot as fast as you can. The hammer tries to go off in a straight line, but the string (the pull of gravity) pulls it back in so the hammer describes an orbit. This interaction has another effect: it displaces the centre of rotation so it is not through your centre of mass - the centre of rotation of the two-body system, hence of the two bodies orbiting each other, are at the centre of mass of the *two* bodies, somewhere along that length of string. You rotate (orbit) around that spot at the same rate as the hammer. In KSP, to keep the physics simple, they treat every interaction as a two-body problem but with a further assumption: some of the bodies are on "rails", which means that no matter how much mass you put into orbit, the centre of mass of the system is ALWAYS a point source in the exact centre of the planet (or moon, or Kerbol). There is, somewhere, a small asteroid in the Kerbol system that *should* be possible to deflect using normal physics rules, but because even that is on rails, you cannot alter its orbit in any way, shape or form.

[Kasuha]

I think you're mixing two things. The solar system barycenter is center of mass of the whole solar system, but if you put something to that point, it will fall into the Sun. It's not a point to which gravity is attracting anything. The same goes for Earth-Moon system.

Even in solar system, in many important places, just two-body interaction is all you need: the planet you're orbiting, and your probe. With planet's mass given and probe mass negligible. Effect of all other bodies in the system is negligible at the distance and only plays role when distance to all bodies is roughly comparable, i.e. in interplanetary space.

[MilkToast]

yep, I picked a random seed number out of my backside to make the point. And peculiar to simulated orbital mechanics in pretty much every package I've come across, gravity is a point source - in real life, this isn't the case. Example: the centre of gravity in the Solar system taken as a whole is the sum of all interacting gravitational vectors from planets, asteroids, comets, right down to the dust. It can only be approximated even on paper (because it would be simply impractical to calculate for every particle in the sphere of influence of the system), and actually lies somewhere outside of the Sun's coronasphere. In the Earth-Moon system, it lies somewhere outside the Earth's iron core. A practical demonstration of gravitational influence on a two-body system is very easy. Tie a hammer to a piece of string, hold the other end of the string at arms length and turn on the spot as fast as you can. The hammer tries to go off in a straight line, but the string (the pull of gravity) pulls it back in so the hammer describes an orbit. This interaction has another effect: it displaces the centre of rotation so it is not through your centre of mass - the centre of rotation of the two-body system, hence of the two bodies orbiting each other, are at the centre of mass of the *two* bodies, somewhere along that length of string. You rotate (orbit) around that spot at the same rate as the hammer. In KSP, to keep the physics simple, they treat every interaction as a two-body problem but with a further assumption: some of the bodies are on "rails", which means that no matter how much mass you put into orbit, the centre of mass of the system is ALWAYS a point source in the exact centre of the planet (or moon, or Kerbol). There is, somewhere, a small asteroid in the Kerbol system that *should* be possible to deflect using normal physics rules, but because even that is on rails, you cannot alter its orbit in any way, shape or form.

A few notes here:

First off, gravity can be approximated as a point source (each body, that is) - for a craft in earth orbit, neglecting effect of other bodies, the orbit is exactly the same if you have earth in all it's sizeness, or just a point with the same mass. This can be shown using some exciting amounts of integration with Newton's gravitation formula, but you can find that somewhere else if you want.

As Kasuha mentioned, two-body physics is plenty in the real world for a lot of interactions - although, you'd better compensate for the moon when you're flying satellites around Earth. It does introduce a sizeable error. Once you leave the earth-moon system, pretty much everything reduces out.

It's one AM and I'm kindof gibbering, but the last thing is that for all intents and purposes, the center of mass of a two body system (that is, one that we would see set up in KSP or real world, where one body is a man-made craft), lies within a negligible distance of the center of the planet. For example, the ISS (mass 450 tons (thank you google)) and orbits roughly 250 miles up (for a total of 4250 miles, plus or minus). Earth weighs 6.58321259 × 10^21 tons.

so, a rough COM calculation (using earth as datum)

(m1*x1+m2*x2)/(m1+m2)

(6.58321259 × 10^21*0+450*4250)/(6.58321259 × 10^21+450)

2.9051166e-16 miles = 1.84068e-11 inches

For comparison, a hydrogen atom is 3.93701e-9 inches across.

So the ISS is "moving" the earth by less than a hundredth the diameter of a hydrogen atom. This, obviously, is very negligible (if even measurable) on this scale.

So, we neglect it. Asteroids would be lighter, and therefore move more:

(damn you IMGUR!) maths

(pardon my approximations]

Either way, still not moving a significant amount.

Edited January 31, 2014 by MilkToast