Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What is a gravitational sling-shot?

  1. Oct 3, 2013 #1
    Hi!

    I wonder how it is possible to accelerate a satellite using the gravity of a planet?

    My amateur calculations would go like this:

    Simplifying and considering a straight forward approach towards the planet where the end distance to core is r2 which of course will be grater than the planet radius which we may call R.

    Using the basic fact that speed is the integral of acceleration we may (perhaps) write

    [tex]\Delta v=-a_r\frac{1}{m} \int_\infty^{r_2} Fdr[/tex]

    and while

    [tex]F=\frac{mMG}{r^2}[/tex]

    this means

    [tex]\Delta v=-MG\int_\infty ^{r_2}\frac{1}{r^2}dr=\frac{MG}{r_2}[/tex]

    Is this right?

    I would however much more like a formula for the actual trajectory and I would very much like to try to derive it myself. But I need help.

    First, what kind of coordinate system should I use?

    I was thinking of cylindrical coordinates.

    And there will be three angles. One for the gravitational force (gamma) and one for the speed direction of the vessel (alpha) and one for the release of the vessel (beta).

    What do you think? Is it possible to derive the change of speed and beta using this?

    I am attaching a drawing of this thought.

    Take care!

    Best regards, Roger
    PS
    I have tried to use energy conservation for this problem but has failed.

    But when

    [tex]Fc=\frac{mv_2^2}{r_2}[/tex]

    equals

    [tex]Fg=\frac{mMG}{r_2^2}[/tex]

    There should be a release of vessel at

    [tex]v_2>\sqrt{\frac{MG}{r_2}}[/tex]

    Am I right?
     

    Attached Files:

  2. jcsd
  3. Oct 3, 2013 #2

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    That is not the right way to look at a gravitational sling shot maneuver. You did not account for the fact that the planet orbits the Sun. The primary purpose of a gravitational sling shot maneuver is to change the speed of the vehicle. You are not going to see that change in speed by looking at things from the perspective of the planet and vehicle. You have to look at it as a three body problem.

    Suppose a vehicle is in deep, deep space, far removed from any gravitational influence except for a single rogue planet. If the vehicle swings by this planet on a hyperbolic trajectory, the vehicle will have a different velocity vector after the encounter than it did before, but the magnitude of that velocity vector will remain unchanged. You only get a change in speed if the planet itself is undergoing acceleration.
     
  4. Oct 3, 2013 #3

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    The interaction between the planet and the satellite is the Kepler problem. It needs some analysis to solve it, but it is possible.

    That is not the main idea of a gravitational slingshot, however. The initial and final velocity relative to the planet is always the same, this follows from energy conservation. Slingshots use the fact that the planet orbits the sun - so while the speed relative to the planet stays the same, the direction changes, and therefore the speed relative to the sun can change.
     
  5. Oct 3, 2013 #4
    Thank you both for your reply!

    But I did not understand a single thing of this Wikipedia-explanation:

    "Translating this analogy into space, then, a "stationary" observer sees a planet moving left at speed U and a spaceship moving right at speed v. If the spaceship has the proper trajectory, it will pass close to the planet, moving at speed U + v relative to the planet's surface because the planet is moving in the opposite direction at speed U. When the spaceship leaves orbit, it is still moving at U + v relative to the planet's surface but in the opposite direction (to the left). Since the planet is moving left at speed U, the total velocity of the spaceship relative to the observer will be the velocity of the moving planet plus the velocity of the spaceship with respect to the planet. So the velocity will be U + ( U + v ), that is 2U + v."

    But I do understand the concept of a totally unelastic collision. Which means that the momentum change is twice the momentum on the body or

    [tex]\Delta p=2p[/tex]

    But I have a hard time seeing the analogy when it comes to a 180 degrees of trajectory change around a planet.

    It was however educational to learn that the slingshot-effect depends upon the movement of the planet and not the gravitaional pull of the planet.

    Once again the law of energy conservation says it all.

    Best regards, Roger
    PS
    Sorry for disturbing you guys with this. I could obvioulsy have tried Wikipedia first. I will do that in the future yet I find it more rewarding to be "social" and ask questions here :smile:
     
  6. Oct 3, 2013 #5

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    You mean totally elastic collision? Like when you throw a ball against a wall at v and it comes back at -v.


    It's the same thing. Direction changes by 180 degrees, while speed relative to wall / planet stays the same. If the wall / planet moves, the speed can increase. This is visualized at the begin of this video, in a different context (air masses instead of planets):

    https://www.youtube.com/watch?v=SVN-oF6tPLc
     
  7. Oct 3, 2013 #6

    Integral

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    To paraphrase what has been said already.
    The passing object picks up some of the planets ORBITAL speed as it passed through the gravity well. This is the only permanent increase in speed of the object. Any speed increase due to falling into the gravity well is lost on the way out.
     
  8. Oct 3, 2013 #7

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    And if the wall is coming toward you with speed u, you had better duck, and quickly. That ball is going to bounce elastically off the moving wall and come back at you with a speed of 2u+v in your frame of reference.
     
  9. Oct 3, 2013 #8

    rcgldr

    User Avatar
    Homework Helper

    As the satellite passes "behind" the planet, the component of gravitational force along the path of the planet results in the planet being slowed by a tiny amount (relative to the sun), and the satellite's speed being increased by a great amount (relative to the sun). During the times when the satellite is "ahead" of the planet, the satellite is much further away, and the gravitational force much less, so the net effect is to slow the planet and speed up the satellite (relative to the sun).
     
    Last edited: Oct 3, 2013
  10. Oct 4, 2013 #9
    What a great explanatory video! Thank you very much for this! Your video and the nice explanations from the other folks now makes me understand this problem completelly.

    Thank you all!

    Best regards, Roger
    PS
    To me, a stupid Swede as I am, a totally elastic collision would mean that the receiver would "absorb" the incoming object. But obviously I am real bad at english :)
     
  11. Oct 4, 2013 #10

    jbriggs444

    User Avatar
    Science Advisor

    Much better than most of us native English speakers are at Svenska, surely. For everyday English usage, this link seems apt.

    http://www.merriam-webster.com/dictionary/elastic

    The key bit is the return to original state after an interaction. That matches the sense of the word as it is used in English-speaking physics jargon.
     
  12. Oct 5, 2013 #11
    Hi!

    I have tried an energy approach to my problem and I actually think I got something :smile:

    I think we may write the total kinetic energy equations like this (skipping /2 for simplicity):

    [tex]mv_1^2+MU_1^2=mv_2^2+MU_2^2[/tex]

    Where

    v1 is the incoming speed
    v2 is the outcoming speed
    U1 is the planet speed before the approach
    U2 is the planet speed after the approach

    Putting

    [tex]v_2=v_1+2U_1[/tex]

    gives

    [tex]mv_1^2+MU_1^2=m(v1+2U1)^2+MU_2^2[/tex]

    and solving for U2 yields

    [tex]U_2=\sqrt{U_1^2(1-4\frac{m}{M})-4\frac{m}{M}U_1v_1}<U_1[/tex]

    What do you think?

    I am however a little confused because I have not used any potential energy.

    Best regards, Roger
     
    Last edited: Oct 5, 2013
  13. Oct 5, 2013 #12

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Potential energy is negligible both before and after the slingshot.
    This looks like a correct approximation for the scenario.

    Most slingshots don't include a 180°-turn, so the satellite gains less energy.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: What is a gravitational sling-shot?
  1. Tension and slings (Replies: 22)

Loading...