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Sun-synchronous orbit: Implication on the orbit's inclination

  1. Nov 9, 2013 #1
    1. The problem statement, all variables and given/known data
    A satellite is launched into a circular sun-synchronous orbit at a height of 900km above Earth's surface. What is the implication on the orbit's inclination (in [itex]deg[/itex]) and on the change of the position of the right ascension of the ascending node per day.

    2. Relevant equations
    The change of the right ascension of the ascending node is defined as:
    [tex]\Delta\Omega = - \frac{3\pi J_2 R^2_E}{a^2(1-\epsilon^2)^2}\cos i \text{ [rad/rev]}[/tex]
    and the change of the argument of perigee is defined as
    [tex]\Delta\omega = - \frac{3\pi J_2 R^2_E}{a^2(1-\epsilon^2)^2}(4-5 \sin^2 i) \text{ [rad/rev]}[/tex][tex]R_E=6378km\\
    J_2=1082.7 \cdot 10^{-6}[/tex]
    3. The attempt at a solution
    In a previous task I already identified [itex]a = 7278 km[/itex].

    Since we have a sun-synchronous orbit, the satellite-sun vector has to be constant and equals the earth-sun vector. Hence I assume I can calculate [itex]\Delta\Omega[/itex] as following. I'm using sidereal days and the information that the Earth performs a [itex]360^\circ[/itex] rotation during 1 year.
    [tex]\frac{\Delta\Omega}{360^\circ} = \frac{23.9345h}{365d \cdot 23.9345h} \Rightarrow \Delta\Omega = 0.9863 \text{[deg/day]}[/tex]

    So now the only unknown variable in the formula for the change of the right ascension of the ascending node is [itex]i[/itex]:
    [tex]\Delta\Omega = - \frac{3\pi J_2 R^2_E}{a^2(1-\epsilon^2)^2}\cos i \\
    \Rightarrow i = \cos^{-1}(\frac{a^2(1-\epsilon^2)^2}{- 3\pi J_2 R^2_E}\cdot \Delta\Omega) = \cos^{-1}(\frac{7278^2(1-0^2)^2}{- 3\pi \cdot 1082.7 \cdot 10^{-6} 6378^2}\cdot 0.9863) = \cos^{-1}(-103.251)[/tex]

    As you can see this produces a MATH ERROR on my calculator :). Unfortunately I am not able to identify where I went wrong :). Can you please help me :)?

    Thanks in regard!
  2. jcsd
  3. Nov 9, 2013 #2


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    Staff: Mentor

    In your calculation of ΔΩ you've got the sidereal day in hours in both the numerator and denominator. They'll cancel leaving you with ΔΩ equal to 360° in 365 days. That can't be right. Instead, assume the vector rotates once in a sidereal year. So ##2\pi/yrS##, where yrS is 365.256366 day.

    Next, ΔΩ should be in radians per revolution (of the satellite), so you'll need to adjust the value accordingly.
  4. Nov 9, 2013 #3
    @gneill: Thank you very much for your help. Your hint actually helped me solving my problem. Adjusting the [deg/day] to [rad/rev] gave me a value in the range of [itex] [-1;1] [/itex]. Hence I was able to solve for i.
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