Physics help - gravity assists

[openatheclose]

Hey everyone! I'm trying to look at gravity assists. One homework problem in my physics class was the following:

A satellite (with mass negligible compared to that of the Earth) is making a fly-by of the Earth on a hyperbolic orbit. Let the velocity of the Earth be v_e. The satellite originally approaches from the negative x-direction, does a loop around the earth, and then flies off in the same direction it came from. Far from the Earth, the magnitude of the satellite's x-component of velocity is v_i when approaching and v_f when receding. For the case where v_i=4v_e, what is the final speed v_f of the satellite in terms of the speed of the earth v_e?

So, for the class, the answer was 2v_e. Simple enough. I wanted to visualize this in KSP. The first thing I did was I switched the wording from "earth" to "mun" - don't want to deal with any annoying atmospheric effects. Then I did a bunch of math to get the following HyperEdit parameters:

Inclination: 0

Eccentricity: 15.292

Semi-major axis: -855000

Lon. of asc. node: 344.657

Argument of periapsis: 216.221

Mean anomaly at epoch: -9.705

Epoch: 36643.375

These numbers should put you in a gravity assist with mun at around 9,000 m closest approach. The velocity of the mun is 542 m/s. Thus, to fit with the problem's value of 4 times the velocity of the body, we want to be going around 2168 m/s. Time warping until just before mun capture confirms this number. Once we're into the mun's SOI, we have a velocity of 1625. Once we escape the mun, we're going 1625 relative to the mun - to be expected. Once we get back out to Kerbin's SOI, we expect our velocity to be 2*542, or 1084 m/s. But we look and we're at 2165. This gravity assist only took off 3 m/s from our velocity. Nothing, really. Why is this? I think it lies in the fact that I'm not fulfilling the "does a loop around the [mun]" that existed in the original problem. But how could I possibly make it do a loop? It seems like my speed is too high to really be affected. Can anyone explain where I'm going wrong here? If there are any true saints around here, it would be excellent to get a modified set of orbital parameters, but if not, that's fine, I can do it myself. I just need to have a nudge in the right direction to understand what's wrong with my maneuver.

 

[mhoram]

Take the velocity after you get into the SOI of the Mun.
The larger this velocity is, the less the gravity of the Mun will bend your flightpath.

In order to do "a loop around the [mun]" (which I interpret as leaving the Mun's SOI in the opposite direction of entering it) you should have a minimal velocity (near to 0 after entering the Mun's SOI. For this the mentioned velocity "1625 relative to the Mun" is way to much.

Also 2v_e is the maximal velocity change, that can be reached, but it is not guaranteed that you get that in a gravity assist.

I also doubt, that it would be realistic that in the given setting the satellite would make "a loop around the earth".

As a further info: Gravity assists not only change the velocity vector, but also the altitude of where you enter Mun's SOI is different from the altitude where you exit Mun's SOI.

Edited December 3, 2015 by mhoram
typo

[openatheclose]

Hmm okay, so I'm going far too fast. Could I try a different body then? What has the best ratio of orbital speed to mass?

[mhoram]

13 minutes ago, openatheclose said:

Hmm okay, so I'm going far too fast. Could I try a different body then? What has the best ratio of orbital speed to mass?

My first guess would be Jool.

[Snark]

The issue that you've demonstrated is that the homework problem cheats.  It describes an impossible situation.

Correct me if I'm interpreting your description incorrectly, but my understanding is this:  The probe is approaching Earth from behind, traveling 4 times Earth's orbital speed relative to the sun, yes?

Therefore, its Earth-relative velocity as it approaches is 3x Earth's orbital speed.  That's when it's far from Earth; by the time it falls to a low periapsis, it'll be going even faster.

Earth orbits the sun at nearly 30 km/sec.  Triple that, you're going an utterly insane 90 km/sec, which compares to only 11 km/sec for Earth's escape velocity.  You're not even vaguely going to make a 180-degree U-turn there.  You're going to go shrieking past it like a bat out of hell, and deflect by a very shallow angle (exactly as you observed with your Mun test).  Your path will be barely affected and you will not end up changing your sun-relative velocity from 4x Earth's to 2x Earth's.  (There is nothing in the solar system that ever goes even close to that speed, since it's far higher than solar escape velocity.)

The only way to make a 180-degree U-turn around a body is if you're going right at escape velocity all the way.  This means that you have to have a very low velocity relative to the body when you're very high above it.  The change in your orbital speed around the body's primary will be 2x that relative speed, which means you'll only get a very tiny change in your orbital speed.

In other words:  a gravity assist doesn't buy you much if it's a 180-degree U-turn, because it requires that your body-relative velocity be close to zero as you approach from afar.

A gravity assist's effect approaches zero as your body-relative-velocity-from-afar approaches zero (because you only get 2x that velocity in benefit, and the fact that it's a 180-degree turn doesn't help you much).

A gravity assist's effect also approaches zero as your body-relative-velocity-from-afar approaches infinity (because then you go screaming past with hardly any deflection in your orbit, and the body has very little time to act on you.)

Therefore, the maximum benefit from a gravity assist will be somewhere in between-- there will exist some ideal deflection angle at which you get the maximum possible dV change from the maneuver.  I don't know off-hand what that angle would be, I'd have to sit down and do math.   I wouldn't be surprised if it turns out to be 90 degrees, but I'll leave that as a homework problem for you.  Simply put:

"At what deflection angle for a hyperbolic orbit past a body will you get the most dV from a gravity assist?"

Edited December 4, 2015 by Snark

[PLAD]

I think this question can be cooked, because it contains a fatal flaw which Snark touched obliquely. If it says you will leave going back the way you came after the flyby, I read that as saying the planet will turn your path exactly 180 degrees. There is only 1 type of path that will turn you exactly 180 degrees, and that is a parabola. With a parabola your v infinity is always exactly zero. Therefore the ship will end with the exact same velocity it had before the flyby, which is zero relative to the planet! Now I see the contradictory word "hyperbolic" up there, which would mean the departure velocity vector would not be exactly 180 degrees from the entry vector, but if our entry speed was, say, 1 m/s then we would end going, say, 179.9 degrees from the start direction at almost 2m/s. Now, as Snark pints out, if we keep increasing the entry speed until it equals 4 times Earth's velocity around the Sun (where in the galaxy did it come from?)  then the ship will have a v infinity relative to Earth of about 120000m/s and it will turn nowhere near 180 degrees. I think with a lowest possible flyby of 200km (avoid the air!) the ship would be turned about 0.5 degrees. In that case your final velocity would go from 4VE before the flyby to 4VE  after, in pretty much the same direction, with a little extra either towards the Sun (if you flew by outside Earth) or away from it (if you flew by inside Earth's orbit). The test maker should have postulated an Earth-mass black hole as the object to be flown by. A flyby altitude of about 2 meters above a 1-cm Earth-mass black hole from a V-infinity of 120km/s would turn you 179 degrees. Oh wait, forgot the relativistic effects...

    After some rough spots I usually came to an understanding with my physics teachers. I remember a quiz where one asked what the path of a thrown baseball was (neglecting air friction), and I answered "a little piece of an ellipse".