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Elliptical Orbits

  1. Sep 1, 2004 #1
    I was wondering if there is a simple explination to elliptical orbits. I read that if i have a satelite in circular orbit, and I then fire the engine, it will gain energy, but it will cause the new orbit to be an ellipse. How come it wont just go in a bigger circular orbit?
     
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  3. Sep 2, 2004 #2

    HallsofIvy

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    Why should it? A circle is just a very special case of an ellipse.

    In any case, if you are at a particular point in an orbit and fire your engines for a BRIEF time, you will increase your energy (and so the average distance from the earth) but not change position a great deal. The moment you turn off your engines, you are now in orbit.

    In other words, if you fire a powerful engine for a very brief time, you can alter your speed (proportional to time) greatly while not altering your position (proportional to t2) much. Your new orbit must go through your new position (close to your old one) while the average distance from the earth is increased- obviously that can't be a circle.
     
  4. Sep 2, 2004 #3
    Im afraid I diagree with you on your conculsion. You cannot increase your speed without increaseing your distance. The reason is that the orbital distance and the required velocity are dependent on eachother. Firing your rocket will NOT make you go faster with nearly the same orbit. Firing your rocket will make you go slower and obtain a higher orbit, much higher. At least thats what my physics text said about the relation between speed and velocity, I could be wrong though. hmm on second thought I think i misread what you stated, can you explain it a little more?
     
    Last edited: Sep 2, 2004
  5. Sep 2, 2004 #4

    BobG

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    This is the key point of HallsofIvy's post.

    Your maneuver point (at least the end of the manuever, where you're adding no more energy) must be part of your new orbit.

    Your average distance does increase and your average speed does decrease.

    If you started in a circular orbit and added energy, your orbit will be an ellipse with your maneuver point being perigee of the ellipse. At apogee, when your current distance is further away, the satellite will be going slower, just as you said. When you return to perigee, you will once again be going whatever speed you were going at the end of your maneuver. And, yes, your averagespeed will wind up being slower than the average speed of your circular orbit, since the average distance of the new orbit is greater.

    To get a bigger circle, it would take two maneuvers, the first to put you into a bigger, elliptical orbit, then a second at apogee to 'circularize' the orbit. After both maneuvers, the speed in the bigger circular orbit would be slower than the speed in the original circular orbit.
     
  6. Sep 2, 2004 #5
    How come the satelite would start to drift back to a lower orbit when it went around though. What is the driving force to make it go back down and speed up, and then go back out and low down. Would it not be most advantageous to just go around in uniform motion.
     
  7. Sep 2, 2004 #6

    Janus

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    Gravity and inertia are what cause this.

    After you fire your engines and have picked up speed, you will start to climb away from the body you are orbiting. As you climb away, you start to lose velocity because you are exchanging kinetic energy for potential energy. As you climb and lose velocity, you will eventually come to a point where your speed and the circular orbital speed for that distance are equal. However, your trajectory is still carrying you "upwards" and your inertia causes you to overshoot. You continue to climb and lose velocity until gravity stops your upward drift, 180 degrees around your orbit from where you started.

    By this time you are traveling much slower that the circular orbital speed for that distance, and you start to fall back, gaining velocity as you do so. Again you reach the point where your speed and orbital speed match, but your trajectory again causes you to overshoot. As you fall and gain speed your path starts to "straighten out" until when you once again retrun to your starting point, you end up in the exact same position with the exact same velocity as you had when you started, and you do the whole thing over again.
     
  8. Sep 2, 2004 #7

    HallsofIvy

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    It DOESN'T "drift back to a lower orbit". It has a new orbit some points of which are much farther from the earth but MUST (by definition of "orbit") pass through the first point at which it assumed that orbit (i.e. when the engine was turn off).
     
  9. Sep 2, 2004 #8

    HallsofIvy

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    I didn't say it would. I DID say that firing a powerful burst for a very short time would increase your speed (proportional to t) more than your distance from the earth (proportional to t2). If you burn your engine to produe an acceleration of, say, 1000000 m/s2for 0.01 second directly away from earth you will have an increase in speed of 10000 m/s and increase in distance from earth of 50 m. You shut your engine off and are now in a new orbit. You apogee must be high enough to account for that added kinetic energy changed to potential energy (and you are correct that the average speed on th e orbit will be lower- which increases the distance to the apogee even more) but the orbit will still bring the ship back to this position- only 50 m from its original height. A very elliptical orbit.
     
  10. Sep 2, 2004 #9

    pervect

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    Elliptical orbits are the general rule (for bound orbits, that is). Circular orbits are a special case of elliptical orbits.

    The shape of an orbit can be derived from to fundamental and invariant physical parameters - the energy/unit mass of the orbiting object, and the angular momentum/unit mass of the orbiting object.

    Both of these quantities are constant everywhere on the orbit of an object.

    A circular orbit occurs only when the angular momentum Lm has the correct value of GM/|2*Em| (note that E is a negative quantity for bound orbits). To be in a circular orbit requires one have exactly the right velocity at exactly the right height.

    Orbits with an inverse-square law force also have the property that they are closed - the orbit passes through it's starting point. This makes it easy to see that to transfer from one circular orbit to another circular orbit, one needs two velocity changes. The minimum-energy technique used is known as a Hohmann transfer orbit, and involves an intermediate elliptical orbit.

    http://liftoff.msfc.nasa.gov/academy/rocket_sci/satellites/hohmann.html

    BTW, I suspect that one will always need two burns to achieve a circular orbit, even for non-inverse square law forces where the orbits are not closed, but I haven't worked this out in detail.
     
  11. Sep 2, 2004 #10
    Ah now I see that makes perfect sense. I wont just happen to stop once I reach the appropriate hight /velocity. Inertia causes me to contiunally overshoot and undershoot. Therefore, If I do two burns, I will effectivly give a force great enough to cancel out this inertial effect and give me a steady circular orbit. THANKS for all your HELP guys.
    :biggrin:
    -Cyrus
     
    Last edited: Sep 2, 2004
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