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Angle of the Sun above orbit plane on satellite-centered celestial sphere

  1. Sep 2, 2009 #1
    I'm trying to show that for certain combinations of altitude and inclination there will be periods of the year where a satellite has eclipse-free orbits. Using a satellite-centered celestial coordinate system, in which the orbit plane is the equator and the direction of Earth is fixed along the x-axis, how do I calculate the Sun's maximum angle above the orbit plane?

    I know the altitude, and thus the angular radius of the Earth disc on the sphere. In my book, to get the maximum angle of the Sun above the orbital plane, they simply take the sum of the orbit inclination and the angle between the ecliptic and the Earth's equator (23 deg). This is fine until the orbit inclination > 67 degs, for which the max sun angle would exceed 90 degrees -- which doesn't make sense (to me) if I understand the geometry in the figures of my book correctly. E.g., what if the orbit is a LEO orbit with 100 degrees inclination?
     
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  3. Sep 2, 2009 #2

    D H

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    An orbit with inclination > 90 degrees is a retrograde orbit. For example, a lot of LEO earth observation satellites are in sun synchronous orbits with inclination of about 98 degrees.
     
  4. Sep 2, 2009 #3

    D H

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    Speaking of sun synchronous satellites: If you want to be realistic, you should incorporate the nodal precession due to the Earth's oblateness into your analysis,

    [tex]\frac{\dot \Omega}{\dot m} =
    -\,\frac 3 2 J_2 \left(\frac{R_E}{a}\right)^2 \cos i[/tex]

    where
    • [tex]\dot \Omega[/tex] is the satellite's nodal precession rate,
    • [tex]\dot m = \surd(\mu_E/a^3)[/tex] is the satellite's mean motion,
    • [itex]a[/itex] is the satellite's semi-major axis,
    • [itex]i[/itex] is the satellite's inclination,
    • [itex]J_2=0.00108263[/itex] is the Earth's dynamic oblateness,
    • [itex]\mu_E[/itex] is the Earth's gravitational coefficient,
    • [itex]R_E[/itex] is the Earth's equatorial radius.
     
  5. Sep 4, 2009 #4
    Thanks for you replies DH.

    I plugged in an altitude of 700 km and inclination of 98 deg into the expression you provided, and got a nodal precession rate of 0.963 degs/day.

    I read that this is the rate of change of right ascension of the ascending node, which in this case would (almost) imply a fixed orbit plane with respect to the Earth-Sun vector (i.e. a sun-sync orbit). Is this equivalent with saying that the duration of the eclipse period for such an orbit is constant over the course of a year?

    Also, are there any good free visualization tools out there one could use to get a better feel for how different orbits behave etc?
     
  6. Sep 4, 2009 #5

    D H

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    That would be correct if the Earth's orbit about the Sun was circular. It isn't. For sun-synchronous satellites that fly roughly over local noon / local midnight, the effect is small. For those that fly roughly over the terminator, the effect is quite significant. These satellites can see the Sun all of the time for most of the year, but have a short eclipse season where the Sun is eclipsed by the Earth for part(s) of each orbit.

    STK does a good job, and the basic version is free. I don't know what the basic version's capabilities are. (A useful, non-basic is far from free. STK makes Matlab look cheap.)
     
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