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Least-energy Orbits

  1. Oct 7, 2003 #1
    How would one go about proving that a least-energy orbit of a spacecraft is actually the least-energy orbit? I don't want the answer, just a push in the right direction.
  2. jcsd
  3. Oct 7, 2003 #2


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    delta v

    Hmm, this may not be as easy as it seems at first.

    Do you have some contraints? For example, "between two neighbouring planetary orbits"? or, "excluding gravitational slingshots"?
  4. Oct 7, 2003 #3
    In context, the question is to do with sending a probe to Saturn. Not sure if that's helpful, or not. It seems to be a case of being obvious that it is the least energy orbit, but being a pain to prove it.
  5. Oct 7, 2003 #4
    The least energy orbit from Earth to another member of the solar system is an ellipse where the shuttle's perihelion, or distance where the shuttle is closest to the Sun, is the point at which the shuttle leaves the Earth's orbit. The shuttle's contact with the larger body occurs at the shuttle's aphelion or point in the elliptical orbit farthest from the Sun. It can be shown that the major axis of the resulting ellipse is found by adding the average distance in astronomical units from the Sun for each of the two bodies together: Earth and the target planet.

    Not sure if this helps, but oh well, I don't understand the question clearly enough to elaborate on mathematics. But I'm only 15, so I doubt I'd be any assistance there! I'll try though.
  6. Oct 7, 2003 #5


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    The energy of an orbit is equal to the kinetic energy of the body plus the gravitational potential energy of the the body, or

    E = mv²/2 - GMm/r

    Now for a circular orbit, v= [squ](GM/r), so

    E= GMm/2r - GMm/r = -GMm/2r

    Now for an eliptical orbit, the body swings in and out from an average distance from the sun and this average distance is equal to the semimajor axis of the orbit (a) . as it does so, its orbital velocity increases and decreases also.

    When the orbital distance is equal to 'a', the orbital velocity is equal to the velocity it would have in a circular orbit at that distance. Since the the energy of an orbit doesn't change, this means that any eliptical orbit has the same energy as a circular one of radius 'a', and we can rewrite our second equation as

    E =-GMm/2a

    As 'a' gets' larger E become less negative and thus is less.

    If you are launching a craft from orbit one to orbit two, it is obvious that if 'a' is less than the average of the distance from the sun of the two planets, then the craft won't reach the second orbit. So 'a' must at least be that large. If the outer limb of the orbit swing out further than the second orbit, then 'a' has to become larger and the energy of the orbit increases.

    Thus the energy difference between the first orbit and the transfer orbit increases (As does the energy needed to put the craft into the transfer orbit from oribt one.

    This works even if the planet you are launching to orbits closer to the sun.

    If you try and make the orbit so that it swings in closer to the sun than the target planet, then 'a' of this orbit gets smaller and the orbit has less energy. But such a tranfer orbit willhave less energy than the launch planet's orbit to start with, so decreasing its energy, increases the difference from the launch orbit, and increases the energy needed to mak the transfer.
  7. Oct 8, 2003 #6
    Yup, that answers it. Thanks Janus. Thanks for having a go, Jeebus :smile:
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