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How calculate ideal gravity assist


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If you arrive radially and leave tangetially, like on the diagram by TouhouTorpedo, you will get about the orbital speed of the body you're using. In the case of the Mun, about 500m/s

The idea is you're leaving with same speed you arrived in the Mun's frame of reference, but the Mun moves in Kerbin's frame, so you have to add that speed.

A nice picture is to imagine throwing a bouncy ball at a truck. Let's assume the ball moves at 10m/s and the truck at 1.

If the truck drives toward you, and you throw the ball, it is moving at 10+1=11m/s in the trucks reference frame, bounces at the same speed, so still at 11m/S for the truck, but for you, it's 10+1=12m/s.

If the truck is driving away, your ball arrives at 9m/s for the truck, and with same math, moves at only 8m/s in your reference frame.

And the use of bouncy balls is relevant. Gravity assist are technically perfectly elastic collisions, even if the force mediating it is gravity rather than electromagnetism (the force that keeps molecules binded)

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hmm..... I'm researching for the mechanism of the Gravity assist by using mathematical and physical informations of oval orbit and universal gravitition so I'll give you an mathematical sentence to you if I success.

(You can just type the information of mass and orbit speed to the letters on it when I complete to create the sentence :)

Holy mother of jeb, this thread is 5 moths old!

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It works like this:

Year 5, day 352: Leave Kerbin on a trajectory to Eve

Year 6, day 045: Eve Periapsis, correct course for Moho encounter

Year 6, day 073: Moho Encounter,

What is important is that you arrive at your Gravity-assist partner (Eve) on the exact time that the launchwindow from Eve to Moho is open. You might have to compromise your Kerbin to Eve flight to be a little suboptimal, but what you save in dV is enormous; in my case around 2000dV. By hitting Eve on the correct side, the orbit changed to one hitting Moho's orbit on Sun-Periapsis, AND the Sun-Apoapsis of that orbit was lowered to that of Eve itself. This makes the speed with which you encounter Moho lower, saving on circularization dV.

I found this particular Grav-Assist window using the Launch Planner website, by first finding all Eve> Moho transfer windows and then finding one with a reasonable "train connection" from Kerbin too.

You can have a look at the NASA flightplan for MESSENGER or Cassini for an example, This Works In Kerbal Space Program!

Cassini_interplanet_trajectory.svg

Edited by Martijn404
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  • 9 months later...
It works like this:

Year 5, day 352: Leave Kerbin on a trajectory to Eve

Year 6, day 045: Eve Periapsis, correct course for Moho encounter

Year 6, day 073: Moho Encounter,

What is important is that you arrive at your Gravity-assist partner (Eve) on the exact time that the launchwindow from Eve to Moho is open. You might have to compromise your Kerbin to Eve flight to be a little suboptimal, but what you save in dV is enormous; in my case around 2000dV. By hitting Eve on the correct side, the orbit changed to one hitting Moho's orbit on Sun-Periapsis, AND the Sun-Apoapsis of that orbit was lowered to that of Eve itself. This makes the speed with which you encounter Moho lower, saving on circularization dV.

I found this particular Grav-Assist window using the Launch Planner website, by first finding all Eve> Moho transfer windows and then finding one with a reasonable "train connection" from Kerbin too.

You can have a look at the NASA flightplan for MESSENGER or Cassini for an example, This Works In Kerbal Space Program!

http://upload.wikimedia.org/wikipedia/commons/b/bd/Cassini_interplanet_trajectory.svg

Nice explanation, but how did you find the point when eve is at the right phase angle to Moho when departing from kerbin? I still can't figure it out, so can you write an equation for that? Btw I'm good at algebra now, and learned about launch window planner.

EDIT: Does this require calculus?

Edited by Iansoreta
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