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I Minimum tangential speed for orbit

  1. Jan 28, 2017 #1
    Is there a formula describing the shape of the orbit and the size of the orbit given the tangential and perpendicular speeds? This would be based on the assumptions that the planets are point particles, everything is in a vacuum, that there are no other gravitational fields, and that one particle is much heavier than the other, so much so that it would not move noticeably at all. I would like to use it to calculate the minimum tangential velocity needed for orbit.

    Also, suppose Earth was a sphere without atmosphere (no pun intended). If the minimum tangential velocity as calculated above was achieved on the surface, would there be an orbit, or would the object just crash into the ground? If the ground was such that the object could pass through unhindered, would there be an orbit?

    Thank you for answering
  2. jcsd
  3. Jan 28, 2017 #2


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    Hello arf,

    You want to investigate Kepler orbits , or -- for a simpler math description -- circular orbits a little bit to be able to answer these questions.

    I remember a cute animation where a cannon is shot horizontally on an earth represented greatly scaled down. It the cannonball moves enough so that the falling follows teh curvature of the earth you have your circular orbit. So your question is answered with: yes.
  4. Jan 28, 2017 #3


    Staff: Mentor

    The minimum is 0. Only at 0 tangential velocity will point particles collide. At any non zero velocity they will miss and orbit.
  5. Jan 28, 2017 #4
    Thanks to BvU for the link and answer, and thanks to Dale for the answer.

    I noticed something about the eccentricity though. Its definition makes it look like it can be imaginary, with a small enough initial velocity (so that the total energy is negative, but such that the angular momentum is non-zero). It shouldn't be physically possible. Is there something I'm missing here?
    Thank you for answering
  6. Jan 30, 2017 #5


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    Do you, perhaps, mean that total energy is positive? That would apply when initial velocity exceeds escape velocity. The resulting orbit is then hyperbolic rather than elliptical.
  7. Jan 31, 2017 #6
    I meant when the total energy is negative, so that the velocity is very small. In that case the potential energy far exceeds the kinetic energy.
    But never mind I got it. If the energy is negative, the angular momentum should be very small, making their product much less than one.
  8. Feb 1, 2017 #7


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    For an elliptical or circular orbit, total energy is always negative. For a parabolic orbit, total energy is zero. For a hyperbolic orbit, total energy is positive.
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