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How do I calculate orbital height from orbital velocity?

  1. Oct 23, 2014 #1
    I'm sorry if this isn't astrophysics but I didn't know what else to post on.

    So I play this game called Kerbal Space Program, and I've recently been trying to calculate the delta v needed to launch my rockets and spaceships from all the planets and moons in the game. I learned how to calculate the orbital velocity needed to maintain a stable orbit around a body, and how much delta v I need to add my vehicle to account for atmospheric and gravitational drag. I could just use those equations for bodies with atmospheres because I know that scale heights of the atmospheres, so I can guess at what height my orbit will need to be. However for planets without atmospheres, I can't guess at what the height will be for the orbital velocity needed. So my question is how do I calculate the orbital height from my velocity? I should also mention that this is for a circular orbit.

    As I mentioned above, I'm not sure if this is astrophysics, if it isn't could you please tell me what it is?

    Thanks for your help :)
     
  2. jcsd
  3. Oct 23, 2014 #2

    mfb

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    That is not easy, and will depend on the launch profile.

    I don't understand why you can calculate something with an atmospere, but not in the much easier case of no atmosphere.

    For a circular orbit, the velocity is given by the equality of centripetal force mv^2/r and gravitational force mMG/r^2.
    deltaV calculations are a bit more tricky, as they involve an elliptical orbit in between. I'm sure there is some online calculator or some page with formulas in the internet.
     
  4. Oct 26, 2014 #3
    I misunderstood what I was asking, I've got what I was looking for now though. Thank you for your answer.
     
  5. Jun 29, 2015 #4
    To Calculate Circular Orbital Velocity, let P.E. at surface = K.E. at orbit. Thus mgh = 1/2 mvv. You see here that mass is not relevant! Thus v = square root of 2gh = approx. 17,500 mph. At this speed centripetal force = centrifugal force!
    Note: H = distance from center of earth to orbit height!!!!! or about 4200 miles for a 200 mile orbit. (Since gravity at surface is due to total earth mass concentrated at a single point of ref, e.g.,the center of earths mass)

    For moon just adjust H accordingly.

    Isn't this less complicated to get the same answer?
     
  6. Jun 29, 2015 #5

    A.T.

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    What does the surface have to do with the orbit?
     
  7. Jun 29, 2015 #6
    This doesn't look quit right, but more like the orbital velocity at zero altitude above surface.
     
  8. Jun 29, 2015 #7

    mfb

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    Not even that (the given number looks approximately right but the formula would give sqrt(2) times this value, which corresponds to the escape velocity).

    - the potential energy is not mgh in any reasonable system. It is -mgR2/r where R is the radius of earth (better: the distance where g is evaluated) and r is the distance to the center of Earth. Note: g is not constant. A better formula is -mMG/r with the gravitational constant G and the mass of Earth M. It gets rid of the detour via the radius and g at the surface.
    - the necessary kinetic energy for an orbit is not equal to the potential energy (even ignoring the opposite sign) - that would give an escape trajectory. It is equal to minus one half this value.

    I closed this thread, the original question has been answered in October 2014.
     
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