"Do the math? Gravity assist" Topic

Sorry, but I can't, so asking others.

The main thing that people miss about 'slingshot' is that while you accelerate towards a planet, you decelerate after you pass. I've seen it described as a net zero, that you only gain the velocity of the planet in orbit.

Is it zero? My gut feeling is that, as you start in a slower velocity, you spend a longer time in the acceleration phase than in the deceleration. Is that negated by the vectors?

Main thing is, I've seldom seen using the gravity assist as a brake, and wondered if this might have something to do with the case.

And, yes, I play with orbital mechanics in some games, but not being a rocket scientist…

Doug

that you only gain the velocity of the planet in orbit.

That's the whole point.

You can use it to brake if you come from the other direction (opposite the planet's rotation and thus lose the orbital velocity from your own).

You're doing the math?

Doug

Edit: As I made clear, that's the 'main point'. My questions is: Is it, in fact, the WHOLE point?

There are a variety of different potential options. You can gain velocity using a gravitational assist by essentially robbing the planet of some of its velocity. The planet being much much much much much more massive than a spacecraft will lose an almost infinitesimal amount of energy for a relatively large gain for the spacecraft.

The thing gravity assist is often used for is to change the vector of a spacecraft without having to eat up a bunch of fuel doing a burn. In that case the relative velocity is not so much a deal

It is not zero, not doing the math but your thought about going faster on the way "out" is the important bit. Since you spend more time being pulled toward the object on the way in(*) and the closer you are the higher the gravitational pull you gain more velocity then you lose.

BCantwell: True, but it IS used for changing velocity.

And, by vector, I'm assuming you mean direction. I'm pretty sure a vector is a 'directional velocity.'

Dynaman, but you are 'close' on the start of the way out. And the planet is 'following' you. REALLY need the math.

I know I should have taken this to a physics forum, but I don't generally join anything for 'one question.'

Thanks, all!

Doug

What If? just touched on this what-if.xkcd.com/146

Wikipedia's explanation is fine: link

> Dynaman, but you are 'close' on the start of the way out. And the planet is 'following' you. REALLY need the math.

The planet could be going in an entirely different direction once you slingshot out of it's gravity well. In a sense the spacecraft may be following the planet but going slower.

You are close at the start of the way out but the planet has less time to pull you back in and since gravity falls off somewhat rapidly that initial velocity away is very important.

That Wikipedia entry is great stuff.

To fully understand this you need to realise that velocity has two components, speed and direction. Changing either or both is possible in this situation.

Don't get me into partial derivatives though, I'm retired now so don't HAVE to be able to think like that any more.

Actually, you can't change speed without changing direction.

Okay, had it backwards; velocity is a vector. Speed and direction are not. I had the understanding of two components down pat, just confused the terminology.

The Wiki's 'simplification' just hurt my head. How did our object stay a full half orbit, THEN broke away? Still damn little math to speak of there.

No, I'll find someone that understands this is bupkus until someone isn't afraid of the math.

McKinney: WHAT?!?

Doug

GF wrote, "Changing either or both [speed and direction] is possible." You can't just change speed without changing direction. If you're falling towards a planet, you either run into it or or you pass to one side of it. If you pass to one side, the planet's gravity accelerates you laterally, changing your direction.

Dear Beasty,

What is your actual question? Please go back and look at your posts because you only ask, "Is the net speed change from a gravity slingshot zero?" And that's been answered.

I'm confused.

"How did our object stay a full half orbit, THEN broke away?"

Remem,that's a Simplified diagram! That "half-orbital" bit is really closer to a parabola.

After a complete orbit after capture it's speed has been changed but its direction (momentarily) is the same.

You can't just change speed without changing direction.

Ah, you meant within the gravity well. And you can, assuming sufficient thrust. Which I admit would mean you'd not need 'gravity assist.' But your phrasing was confusing.

Is the net speed change from a gravity slingshot zero?"

was not the original question. "Is the net speed change from a gravity slingshot zero" BEYOND the velocity of the gravitational body was the original question.

That "half-orbital" bit is really closer to a parabola.

Which means it's totally useless to my original question, which you mostly continue to ignore.

Really, let's just drop it. I can go to real scientists.

Doug

Because Doug is sounding cranky, try this page. It explains it in fairly easy terms.
http://www.mathpages.com/home/kmath114/kmath114.htm

I especially like the last paragraph where it ponders about using two black holes as a super accelerator.

I do, don't I?

Sorry, I'll look at the examples again, but I'm pretty sure GildasFacit had it right; pretty much takes Calc to get this resolved as far as I can tell. And, my Calc was about four and half decades ago. I can almost spell Calculi.., Calcuos.., oh, you know.

Thanks, all.

Doug

"Is the net speed change from a gravity slingshot zero BEYOND the velocity of the gravitational body?"
If you're talking an Oberth Effect route (parabolic approach to a single planet, not a true gravitational slingshot), you're getting "free" delta vee from the potential gravitational energy of the propellant mass: link

A gravitational slingshot (aka Gravity Assist link ) takes energy from the planet you're slinging around to change your velocity relative to the Sun.

To bring it down to a nutshell, U is the orbital velocity of the target planet. A perfect head-on approach can give you an extra 2U as the plaent pulls you with it. A perfect tail-on approach can slow you down by -2U (more with aero-braking.) Anything else is graded on a curve in between. Hitting the planet at the right angle can be a net zero. The other aspect is to change direction with a minimal amount of fuel expended.

Of course, turning those entrance and exit angles into meaningful numbers is the tricky part. The best game examples that I have seen have been gravity arrows around a planet, but that is net zero math since the planet is 'stationary'.

The math isn't particularly difficult, you can set it up in a spreadsheet. The first key is to use all metric units. The second key is everything in kilograms, meters, and seconds, and not some multiple thereof.

Then you need to doublecheck and make sure that you stay clear of the atmosphere or even the surface, because all those formulas work from the center of mass of the planet, not the surface.