# Gravitational slingshotting

1. Apr 28, 2006

### DaveC426913

Unless I misunderstand slingshotting, you are exploiting - not the gravity of the planet - but the motion of the planet in its orbit.

This would mean you cannot slingshot around a planet by moving in a retrograde direction. More importantly (to me) it would mean that you cannot use a slingshot maneuvre to gain speed perpendicular to the plane of the Solar System.

It is that last question I'm interested in the answer to. Can you end up on a trajectory outside the plane of the SS by slinghshotting around one of the inner planets?

Last edited: Apr 29, 2006
2. Apr 29, 2006

### DaveC426913

What? Did I misspell 'gravity' as 'pr0n' or something? No takers?

3. Apr 29, 2006

### Physics Monkey

Haha, I don't think so.

Of course you know that without the gravitational interaction, you couldn't slingshot around the planet. However, I think it is fair to say that the motion of the planet is where the magic comes from.

Consider a simplified example where a single planet orbits counterclockwise around a star in a circle. In the rest frame of the star, a spacecraft is flying towards the planet and the planet is flying towards the spacecraft. Now look at things in the rest frame of the planet. The spacecraft is moving with some greater velocity towards the stationary planet and is presumably in some hyperbolic orbit around the planet. The planet then scatters the spacecraft by some definite angle which depends on the details (energy of the spacecraft, impact parameter, etc.) of the situation. The spacecraft stays in the plane determined by its incident trajectory and the planet.

So what has happened? In general, the spacecraft can have a component of its velocity along the original direction (tangential) and some component in the orthogonal direction (which I leave unspecified for the moment). You now want to know what the velocity of the spacecraft is in the star's rest frame. Imagine for simplicity that the scattering angle is just right so that the spacecraft has no velocity in the tangential direction in the star's rest frame. In other words, the scattering angle is 90 degrees in the star's rest frame. With a little bit of algabra you can easily discover that the spacecraft has increased its energy.

Now I have so far left the scattering plane undetermined. Imagine the spacecraft is slightly outside the orbit of the planet in the plane of the solar system. Then the spacecraft will be scattered in the radial direction towards the star. If the spacecraft is just inside the orbit, it will be scattered out away from the star. If the spacecraft is just below the plane of the solar system, it will be scattered straight up out of the plane of the solar system. I think this is the answer you were looking for.

Hope this helps.

4. Apr 30, 2006

### pervect

Staff Emeritus
Here's a rather elegant way of looking at gravitational slingshotting for the planar case. I think it can be extended to cover the non-planar case as well, but I'm not positive about it.

We can describe an orbiting body by its orbital energy E, and it's angular momentum vector $\vec{L}$.

Without a third body, both E and $\vec{L}$ are conserved.

When we add a third body, neither E nor L is conserved, but the quantity

$$E - \vec{w} \cdot \vec{L}$$

for our small body is a conserved quantity. (Here $\vec{w}$ is the angular frequency at which the massive "third body" is orbiting.)

We can show that this quantity is conserved by converting the conserved Jacobi intergal function for the planar 3-body problem.

http://scienceworld.wolfram.com/physics/JacobiIntegral.html

from rotating to non-rotating coordinates.

Because this specific relationship was derived from the planar solution, it doesn't necessarily cover the non-planar case.

However, it does illustrate why one can gain energy via a gravitational slingshot, only if one also gains angular momentum.

One can see why this should be so from another argument. This second argument suggests that the relationship may be more general than the above specific derivation. Suppose we do a flyby of Jupiter, which we will model simplistically as being a circular orbit around the sun with an angular frequency of w.

Jupiter's orbital angular momentum will be Iw, and it's orbital energy will be Iw^2/2, I being the moment of inertia, m_juptier * r_jupiter.

When we fly by Jupiter, we actually change Jupiter's orbit very slightly.

Jupter loses an amount of angular momentum I*dw, and loses an amount of energy I*w*dw, by differentiating the above expressions for energy and momentum with respect to w.

The ratio of loss of energy to loss of angular momentum of Jupiter is fixed.

What Jupiter loses, our spacecraft gains. The spacecraft gains an amount of energy I*w*dw, and gains an amount of angular momentum I*dw - because overall, momentum and energy are conserved. Because of the relationship between the energy and momentum loss of Jupiter must be w, the ratio of the energy gained to the angular momentum gained by the "third body" is also w - w being the angular velocity of Jupiter around the sun.

Last edited: Apr 30, 2006
5. Apr 30, 2006

### DaveC426913

I. am. so. High School Math.

I once liked to think I was pretty good with Physics and a little Math - but the above leaves me in the dirt.

Um.

So, can you put yourself on a trajectory perpendicular to the plane of the Solar System by using planetary slingshotting?

6. Apr 30, 2006

### Danger

Shouldn't that be 'slingshooting'?

Sorry... I'll be slinking back to GD now...

7. Apr 30, 2006

### Physics Monkey

Absolutely. Voyager 1 went up and Voyager 2 went down. You can see some pretty clear diagrams of the motion of both probes here http://spacephysics.ucr.edu/index.php?content=v25/v0.html Remarkably, these old probes may still have something profound to tell us about the way the solar system works as they hopefully make it to the outer reaches of the Sun's influence.

The gravitational slingshot mechanism boils down to the observation that while the total energy of two bodies interacting gravitationally is conserved, the way that the energy is distributed between the bodies is not fixed. By arranging things carefully it is possible to transfer energy from a planet, say, to a spacecraft. In fact, a spacecraft will never gain energy in frame where the planet is initially at rest, but the story is quite different in a frame where the planet moves. See above for the details.

Hope this helps.

Last edited: Apr 30, 2006
8. May 1, 2006

### DaveC426913

Hm. Worth looking at closer. Thanks.

Yes. Which is the principle by which we slingshot at all. No news here.

It doesn't answer the question of whether one can use planetary motion in one direction (i.e. in the plane of the SS) to slingshot in a direction wherein the planet has no motion (i.e. perp to the plane of the SS). And that is my question.

The Voyager paths certainly suggest it is possible in principle, but it does not seem to be much of a deflection. I'm not sure that it indicates that it is practical to change a trajectory by 90 degrees out of the plane of the SS.

9. May 1, 2006

### pervect

Staff Emeritus
What's the part that's throwing you?

The point is that you can use Jupiter or whatever to help perform a plane shift maneuver of this sort, but you won't gain any energy out of the plane-shift part of the maneuver.

A lot of the details are going to depend on how close you can get to Jupiter.

10. May 1, 2006

### Physics Monkey

Hi Dave,

You are quite right that Voyager 1 was not scattered straight up. However, the fact that it was scattered up by some angle is indeed suggestive. Once again, it is possible to slingshot a spacecraft straight up.

Unfortunately, I'm not sure exactly what you're looking for anymore. I described in some detail in my first post exactly how one can compute the scattering angle in the frame where the planet is moving. As I indicated there, it is indeed possible by suitable arrangement of initial conditions to scatter the spacecraft straight up. However, you made some comment about the math being over your head so I tried to boil it down it for you, but that doesn't appear to be what you wanted either. Please let me know what, if anything, I can do to make it more clear for you.

In any event, let me be as clear as I can: it is possible to slingshot a spacecraft at 90 degress to the orbital plane.

11. May 1, 2006

### rbj

you can't gain speed when you slingshot straight up as you can when you slingshot in the same sense of revolution in the orbital plane. but you can do that straight up slingshot as the last slingshot after first using earlier planets to increase speed. but they might have wanted to use Neptune to make Voyager make an even faster departure from our solar system.

12. May 1, 2006

### turbo

Can you provide references for this claim? If you can shoot a probe very close to a massive body and provide for an appropriate exit angle, it seems like you should be able to accelerate the probe even if it does not approach or exit the massive body in its orbital plane.

13. May 2, 2006

### DaveC426913

Yes, thanks for your patience. Trouble is, I know just enough orbital mechanics, math and physics to get me into trouble, and just enough to know who to call (PF members!) for rescue.

It is true, I don't know what I want.

Well... yes I do...

What I want is to overhear someone say "You know, I've been meaning to try out this spaceflight simulator program I bought, all I need is a suitable fictional mission to plug into it, so's I can plot an interplanetary course. If only there were someone out there who could think of an interesting planetary journey!":rofl: :rofl:

This will suffice. It is plausible, therefore it can be used in a story.

14. May 2, 2006

### rbj

this is classical physics. we can figger it out. even though the frame of reference is slightly accelerated - we should use D'Alembert's principle to put the planet and the probe in an unaccelerated frame of reference, the gravitational field resulting from the sun at those great distances is small. so let's consider that planet all by itself and it is in the center of our inertial frame of reference.

now, from the perspective of the planet whether the probe is flying in (before the slingshot) or flying out (after the slingshot), because of conservation of energy, given the same distance from the planet, the velocities have to be the same. from the perspective of the planet, the probe does not gain speed after the slingshot.

now if the frame of reference is changed so that it is that of the probe long before the slingshot, you will see that the planet has some velocity and is traveling almost directly toward the probe (say from the right, moving to the left). the probe will fall toward the planet and whip around and fly out toward the left at possibly nearly twice the velocity of the planet (from the planet's POV it fell in at velocity, v, and flew out at velocity v). because energy and momentum is conserved, this had the effect of slowing down the planet very slightly, transferring some kinetic energy to the probe.

but because of conservation of momentum, the velocity gained by the probe has to be in the same direction as the velocity of the planet and if the probe is slingshoting out at a right angle to the plane of orbit, there is no velocity of the planet in that direction, so there is nothing to gain in the compoent of velocity that is perpendicular to the plane of orbit.

15. May 2, 2006

### DaveC426913

This is the very point I was making when I posed the initial question.

The source of the velocity increase is Venus' motion around the Sun - i.e. in the plane of the SS. I was not sure whether, with the right footwork (such as the right approach path and a well-timed burn), you could redirect that energy gain to modify the outgoing path.

Consensus in this thread seems to be that you can.