
re: Voyager: The mathematics of interstellar space travel
Etmax
6/12/2013 2:38:49 AM
That is indeed interesting. Did you see the Ted talk about the 14 year old that built a fission reactor at home and now at 19 is partnering up with companies to build relatively safe(r) reactors? Another fine example of how the young developing mind should not be underestimated
Fascinating discussion gentlemen, but the part of the article I find most interesting, that I did not know before, was that the ability to use a planet to "slingshot" a spacecraft was first demonstrated by a 25 year old grad student doing numerical simulations on an ancient computer. That is pretty cool :)
re: Voyager: The mathematics of interstellar space travel
Bert22306
6/11/2013 9:38:26 PM
Oh true. I was ignoring the speed of the planet itself, just considering the gravitational attraction as if the planet were stationary.
Yes, it makes sense that the orbital speed of that planet would give the spacecraft a boost, if in the right direction. And slow down the panet by some tiny amount.
re: Voyager: The mathematics of interstellar space travel
przemek0
6/11/2013 7:44:03 PM
Bert, consider a heavy ball in free motion and a lighter ball colliding with it head on. Of course the lighter ball will gain energy after it bounces back, right?
By the way, it's quite possible to solve this type of problems by using the conservation principles; in this case, conservation of momentum, angular momentum and energy.
re: Voyager: The mathematics of interstellar space travel
Moca
6/11/2013 4:10:44 PM
Thanks for the link
It answers my question I have for so many year, and even their explanation more or less same as my gut feeling. The planet did indeed transfer its energy to voyager.
re: Voyager: The mathematics of interstellar space travel
mac_droz
6/11/2013 2:26:19 PM
Have a look at the graphs and theory Bart:
http://en.wikipedia.org/wiki/Gravity_assist
I'm not a specialist in this field but it looks like these guys did some math and it works.
Anyway it is possible to measure the speed of Voyagers...
re: Voyager: The mathematics of interstellar space travel
Bert22306
6/11/2013 7:47:32 AM
Oh, in case you're wondering why the speed would be slowing down after having swung by the planet, ask yourself this. If gravitational attraction worked to accelerate the spacecraft as it approaches the planet, why would you not expect gravitational attraction to then slow it down, as it's moving away, by exactly the same amount?
re: Voyager: The mathematics of interstellar space travel
Bert22306
6/11/2013 7:25:26 AM
The tennis racket is being swung by a person, and energy is being transferred from his muscles to the ball, via that racket. On the other hand, that planet is not losing any energy just because this object is flying by.
Think about it this way. Let's say that you measure the speed of the spacecraft X miles before it reaches its closest approach to the planet. Then you measure the speed of the spacecraft again, X miles after its closest approach. The speed will be the same. The speed will have increased as it approaches the planet and swings around that partial orbit, then it slows down again on its new course, to reach its steady state speed again.
Using the planets' gravitational attraction allows you to chart a quicker course out of the solar system. So for example, instead of having to fly toward the sun initially, and use it to slingshot Voyager out into deep space, Voyager could immediately head out toward the edges of the solar system, by doing those flybyes past other planets.
re: Voyager: The mathematics of interstellar space travel
Moca
6/11/2013 5:31:30 AM
I thought about this problem for many years, as I don't understand the fundamental question.
But I thought it indeed gain energy. The energy is transferred from the planet to voyager.
Think about a tennis ball is flying to a tennis racket. If the racket is sitting still, there would be no gain of energy. But in this case, the racket is "hitting" the tennis, and the tennis bounced back with higher energy.
The voyager flies around a moving planet. In this case, the gravitational field of the planet is the force bounce back voyager when it fly around it behind while the planet is moving forward.
re: Voyager: The mathematics of interstellar space travel
Bert22306
6/10/2013 8:17:15 PM
I believe that one equation doesn't tell the whole story, even for the twobody problem.
F = GMm/r^2 descibes the centripetal force between two bodies, i.e. attraction. What draws them together. The balancing force to achieve orbit is F = mv^2/r, socalled centrifugal force. If your mass m can acheve enough tangential speed v to balance the gravitation attraction, it will remain in orbit. Sort of like tether ball. The rope keeps the ball from escaping, but the ball needs tangential velocity to keep that rope tight.
The orbit can be circular, but it doesn't need to be, is the point. In an elliptical orbit, the speed is constantly changing, faster when the two bodies are closer (sort of sling shot efect), slower when they are further apart, so that the centrifugal and centripetal forces always balance. So it's "easier" to achieve an elliptical orbit. More generic. Less need to be so precise with speed and distance.
So now, start with that elliptical orbit and add some more speed to the rotating mass, and the mass will escape instead of staying in orbit.
So when Voyager approaches a planet at a high rate of speed, it will indeed change its direction of travel without having to use any of its own energy stores. It's basically an elliptical orbit, but with too much speed to stay in obit. But this is not the same as saying that it is getting any free energy otherwise. Had the planet NOT been there, Voyager would have continued on its journey just as well, only it would not have been able to change direction of travel for "free." It would have continued on its journey simply because it already had enough speed to escape the sun's gravitational pull.
The sling shot effect is used to change direction of travel for "free."




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