Slingshot Effect: Newsletter Challenge (01/03/06)

The question is a good one, because if you consider the earth by itself, you can not use it to "sling shot" yourself with any net effect. If, for the sake of argument, an object was in a non-symmetrical elliptical orbit, the part of the orbit away from the earth would lead to a slow change in velocity towards the earth, and as the object neared the earth, the pull and the change in velocity would increase, lending to the sling shot effect. However, this does not yield a net change in orbital path from the earth. Enter: The Moon If you consider the system to include the point mass of the moon, then having an orbital trajectory that would near the moon, then the effect would be a change in velocity, but sadly, does not effect the total energy (kenetic+potential) of the system. This phenomena does, however, come in handy when wishing to sling shot yourself home after a trip to visit a planet and a change in direction but not energy(speed) is required.

The sling shot effect is one where the gravity of an object (i.e., moon or planet) is used to accelerate another object. The effect is you are increasing velocity of the spacecraft by vectoring toward a planet or moon. The space ships trajectory is such that the path does not (let's hope not) impact the planet nor encounter the stratosphere.

The energy lost in the transition is only that required of the escape velocity for the planet or moon. Using another object to accelerate the space ship adds velocity. The trick is to simply set a trajectory that adds more velocity to the spaceship than is required or lost by the planet's or moon's escape velocity. You leave with a higher net velocity if it is done correctly.

I forgot an essential ingredient and I was rushed for lunch. The body that the spaceship is heading to must be in motion! The spacecraft will use that body's momentum to add to the speed of the spacecraft. Essentially, you are permanently borrowing some of the kinetic energy of the body. So the trajectory of the spacecraft must be such that the final vector for the spacecraft is at least partially aligned with the trajectory of the body.

The net result is somewhat like an elastic collision between the two objects with the smaller mass spaceship departing with its original velocity and two times the vector component velocity for the body (planet) that lies in the same direction. Conversely, the body's velocity will be slowed down a very tiny amount. However, since the mass ratio of the body versus that of the spaceship is, for all intents and purposes, zero, the amount of conserved energy lost is negligible for the body. For simplicity's sake we will ignore that and any entropy introduced in the system and the equation is:

Vf = Vi + 2*Vb

Where:

Vf = Final velocity of the spacecraft
Vi = Initial velocity of the spacecraft before encounter of the body
Vb = the vector velocity of the body that lies in the same direction as the departing spacecraft

If the speed of the spaceship is greater than the escape velocity of the body the spaceship will not return nor orbit the body. The rules of conservation of energy still apply as do the laws of Thermodynamics:

You can't win.
You can't brake even.
And you can't quit the game.

Yes, I agree with your statement above. I made my reply before I saw your update.

I disagree with your second statement "The trick is to simply set a trajectory that adds more velocity to the spaceship than is required or lost by the planet's or moon's escape velocity." It doesn't matter what path you take toward or away from the planet used for the slingshot since the gravitational potential is the symmetric. In other words, any kinetic energy you gain from the gravity of the planet, you lose as you move away from it.

I think the movement of the planet around the sun may play a role, in other words, if you can get in orbit around Jupiter, you'll be pulled along with it in it's orbit, which is pretty fast (29,303 mph)! Here's a good link that talks about it http://saturn.jpl.nasa.gov/mission/gravity-assists .cfm

Yes, I wrote in haste and on the drive home to get lunch I realized the issue. ...And 29,303 mph is a lot faster than my car!

Well stated, I forgot to take into account the relative motion effect.

The craft has the same acceleration and deceleration relative to the planet used but if it aprouches from behind the planets path it will pick up a vectored component of the planets relative speed. Also any burn of fuel at maximum velocity will be magnified.

You don't have to approach from "behind" the planet. You can approach from any angle. It is the exit trajectory relative to the trajectory of the motion of the planet that matters.

If you exit with the same trajectory as the planet you will exit with the same velocity you entered plus two times the velocity of the planet (assuming your reference is the Sun). If you exit in the opposite direction that the planet is moving you will effectively have a braking action and your final velocity will be your initial velocity minus two times the planets velocity.

For angles other than 0, 90, 180, and 270 degrees from the planets trajectory the net velocity will be a function of the exit trajectory's sin/cosine relative to the planet's trajectory, your initial velocity, and the velocity of the planet.

The actual term for this is "Gravity Assist", not slingshot. The principles involved have nothing to do with the physics of a slingshot, but closer to celestial billiards with perfectly elastic collisions, which follow the laws for the conservation of energy. As I stated earlier, Gravity Assist can be used to increase your velocity or decrease your final velocity as well as alter your final trajectory.

I like this explanation, especially the inclusion of the idea of the conservation of momentum.

It works like a parent pushing a child on a swing. While the child is going down, you give a push. The rocket engines are fired when the ship is falling toward the planet. They are shut off before it begins to leave the gravity well.

I don't believe in sling shots because you'll shoot your eye out. Having said that, I think you gain net velocity but at the cost of a change in your course vector or direction. This is just what someone wanted you to do because a straight line is not always the shortest distance between two points in the space time continuum.

The slingshot effect is very real. It does not matter about the approach or the exit. (But you do want to exit!) The increase in velocity of the space craft is compensated by the slowing down of the moon or planet. There is a huge diference in the two bodies's masses, so the planet's velocity change and displacement is for all practical purposes nothing. However, if you slingshot several billion spacecraft around Jupiter, Jupiter will be affected eventually. I don't like slingshots myself. It hurts bad when you release the wrong end. Dave Meador

Well, the exit trajectory does matter. One way will add speed and the other way will subtract speed. The latter will actually speed up the planet's velocity (ever so slightly) as the spacecraft slows down.

The Earth has been used by NASA for gravity assist a number of times for varying spacecrafts. I am surprised that environmentalists haven't gone ballistic with claims that we are destroying the Earth's orbit.

So that would be why we've been adding "leap seconds" over the past few years...the last one being just before new year and the next due in about 18 months time.

Apparantly the Japanese got quite concerned that these leap seconds might affect some of their control systems. Eventually the control engineers stopped laughing long enough to point out that the control systems worked to the nearest minute...

This is a complex question. Several factors are involved in the slingshot effect. First, both the planet or moon and the object are orbiting the sun which has a much larger mass, therefore a much larger gravitational attraction than the planet or moon. At some point the object reaches a distance at which the gravitational pull of the planet or moon will effect its motion. Since it approaches that planet an greater than escape velocity it will not attain a full orbit, but rather a hyberbolic orbit for the duration of the time that the planet's gravitational influence is greater than that of the sun. The planet's kenetic energy will change by the mathematical and goeometric opposite of the kenetic energy of the object, so the velocity of both will change on approach and on departure (in opposing manners), but since the angle will change along the hyperbolic curve the velocity of both may change as well based upon the differences in the tangents of the approach and departure vectors. But the greatest effect will be the effect imparted by the orbital velocity of the planet (which is generally very large)on the object. If it leaves the orbit in the same direction as the velocity vector of the planet then the effect will be maximized. If it leaves the orbit of the planet in the opposite direction of the planet's velocity the effect will be minimized.

The gravitational force between two objects is inversely proportional to the distance between them. That is to say that as the "ship" approaches the much larger planet, it's acceleration increases. when it reaches the point in it's vector where it is ready to "pull away" from the planet, velocity can be maintained by shifting the ships velocity so that it runs perpendicular to the gravity vector as in a satellite. The ship now "orbits the planet. From a position of orbit (now with a much higher velocity) a small amount of E (realtively) can be used to shift the ships velocity vector again, this time angling it slightly away from the planet.