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At what altitude and velocity is an orbit geosynchronous?

  1. Jul 8, 2008 #1
    At what altitude and velocity is an orbit geosynchronous? Please include calculations. I am too old for this to be a homework problem - just exercising my mind.
  2. jcsd
  3. Jul 8, 2008 #2


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    You can use my online calculator to solve this: http://orbitsimulator.com/gravity/articles/formula55.html

    Just enter 24 hours for P (or 23 hrs 56 minutes + some seconds, Wikipedia for a more exact answer) and 1 Earth Mass for M, and you'll get your semi-major axis (same as distance for a circular orbit). For altitude, subtract the equatorial radius of Earth: 6378 km. For velocity v=d/t, or in a rotating system, v=2*pi*distance/24 hours (or 23 hours 56 min.... for extra precision)
  4. Jul 8, 2008 #3
    I should have been clearer. I am looking for an analytical solution to the problem with a circular orbit. Thanks for your assist, tony873004.
  5. Jul 9, 2008 #4
    The general equation for the orbital period of a satellite in orbit around the Earth:


    With a equal to the semi-major axis of the orbit (which is the radius r in a circular orbit). The gravitational paramter of the Earth is called muE, and it has the following value:

    muE = 3.98600441*105 km3/s2

    If you rewrite your equation for a (or r), you get:

    a = ( (muE*T2) / (4*pi2) ) 1/3

    The orbital period in geosynchronous orbit is:

    Tgeo= 86400 seconds (24 hours) ​

    Substituting all values in the equation for a (or r), you get:

    a = 42241.1 km​

    Now subtracting the radius of the Earth RE, which is equal to 6378.1 km, yields the altitude h:

    h = 35863 km​

    For the velocity, there is an equation for the circular velocity in orbit:

    Vc = sqrt(muE / r)​

    With r equal to 42241.1 km, it should be 3.1 km/s
  6. Jul 9, 2008 #5
    How about deriving the altitude of a circular, geosynchronous orbit most simply - without assuming Kepler's Laws?
  7. Jul 9, 2008 #6


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    My solution was analytic (no numerical methods used). But if you need to eliminate Kepler, this formula contains Newton, but not Kepler:

    [tex]a = \sqrt[3]{{\frac{{GMt^2 }}{{4\pi ^2 }}}} - r[/tex]

    where a is your altitude, G the gravitational constant, M the mass of the Earth, t the length of a sideral day, and r the radius of the Earth.
  8. Jul 9, 2008 #7
    Well stated, but what about deriving it?
  9. Jul 9, 2008 #8


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    Loren, here are some (hopefully) familiar formulas to assist you:

    Gravitational force:
    F = G M m / r^2
    (here M = Earth's mass, m = orbiting mass)

    Centripetal force:
    F = m w^2 r
    (here w is really "omega", the orbital frequency in radians/sec)

    Finally, w vs. period T:
    T = 2*pi / w
    w = 2*pi / T
    (where T = 1 day for geosynch. orbit)

    To derive r for a geosynch. orbit, first equate the two force expressions with each other. Solve the equation for "r" in terms of the other quantities.

    Next, figure out what w should be for T = 1 day ***, and look up G and M. Plug in the numbers to get r, the radius of the orbit. Subtract the Earth's radius if you want the altitude above the Earth's surface.



    p.s. edit:
    You'll need to work in a consistent set of units, and it's common to use meters, seconds, and kilograms. Eg., 1 day is ___ seconds? Use that number for T, rather than "1".
  10. Jul 9, 2008 #9


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    This is essentially what Redbelly posted, but I jumped straight to acceleration rather than force:

    Start with what we know:

    1. Distance = velocity * time
    2. Distance in an orbit is 2*pi*a (the circumference formula), where a is semi-major axis, same as radius for a circular orbit

    Set them equal to each other:
    (1) [tex]v_{circ} \cdot t = 2\pi a[/tex]

    Use the circular orbital velocity formula

    (2) [tex]v_{circ} = \sqrt {\frac{{GM}}{a}} [/tex]

    If you'd like to derive this, you start with formula for centrepital acceleration
    (3) [tex]\frac{{v^2 }}{a}[/tex]
    and Newton's gravitational formula for acceleration
    (4) [tex]\frac{{GM}}{{a^2 }}[/tex]

    and set them (3) & (4) equal to each other, then use algebra to isolate v

    \frac{{v^2 }}{a} = \frac{{GM}}{{a^2 }}\,\,\,\, \Rightarrow \,\,\,\, \\
    v^2 r^2 = GMr\,\,\,\, \Rightarrow \,\,\,\, \\
    v^2 r = GM\,\,\,\, \Rightarrow \,\,\,\, \\
    v^2 = \frac{{GM}}{a}\,\,\,\, \Rightarrow \,\,\,\, \\
    v = \sqrt {\frac{{GM}}{a}} \\

    Re-write formula (1), substituting the circular velocity formula for velocity
    (4) [tex]\sqrt {\frac{{GM}}{a}} \cdot t = 2\pi a[/tex]

    Use algebra to isolate a

    \sqrt {\frac{{GM}}{a}} = \frac{{2\pi a}}{t} \\
    \frac{{GM}}{a} = \left( {\frac{{2\pi a}}{t}} \right)^2 \\
    \frac{{GM}}{a} = \frac{{4\pi ^2 a^2 }}{{t^2 }} \\
    GMt^2 = 4\pi ^2 a^3 \\
    a^3 = \frac{{GMt^2 }}{{4\pi ^2 }} \\
    a = \sqrt[3]{{\frac{{GMt^2 }}{{4\pi ^2 }}}} \\

    And since you want altitude, subtract Earth's radius (r)
    (5) [tex]a = \sqrt[3]{{\frac{{GMt^2 }}{{4\pi ^2 }}}} - r[/tex]

    Plugging in numbers (using meters, kilograms, seconds)

    I used 86400 for t, which is 24 hours. You can get a little more exact if you look up the length of a sideral day (~23h 56m) in seconds
  11. Jul 10, 2008 #10
    Aha! It seems that these are what I was looking for. Thanks both.
  12. Jul 10, 2008 #11


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    Bruce too.
  13. Jul 10, 2008 #12
    Yeah, Bruce-Almight first constructed the Arc.
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