Molniya Orbit: calc angular displacement of ascending node

AI Thread Summary
The discussion focuses on calculating the angular displacement of the ascending node for a Molniya orbit with specific parameters, including an eccentricity of 0.72 and a semi-major axis of 26,554 km. The initial formula considered for this calculation was questioned for its relevance to time, leading to further clarification on its application. After applying the correct formula and calculations, the angular displacement after 100 sidereal days was determined to be approximately 24.52 degrees. The negative sign in the result indicates the direction of rotation, which was clarified as important for understanding the orbital dynamics. Overall, the calculations were validated as logically sound despite some concerns about the numerical details.
Paul Gray
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Hello there :),

Homework Statement


We have a molniya orbit with an eccentricity of e = 0.72 and a semi-major axis a = 26554 [km].

The task:
Calculate the angular displacement of the ascending node Ω after 100 sidereal days

Homework Equations


Moreover I know from Wikipedia that a molniya orbit has an inclination of i = 63.4 [deg].

The Attempt at a Solution


I considered using following formula
\Delta\Omega = - \frac{3\pi J_2 R_E^2}{a^2 (1-e^2)^2} * cos i
But I am pretty sure it is the wrong formula, because time doesn't matter in it. (Even it has an Ω at the beginning ;)).

Hence I hope on insightful input provided by you :). Thank you for your help!

Paul
 
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That formula gives you ## \Delta \Omega ## per some period of time. What is that period of time?
 
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That's a simple but clever answer. I will give it a shot as soon as possible :). Thank you ...
 
Using the hint voko provided I get the following answer

\Delta\Omega = - \frac{3\pi J_2 R_E^2}{a^2 (1-e^2)^2} \cdot \cos i <br /> = \frac{3\pi 1082.7 \cdot 10^{-6} 6378^2}{26544^2 (1-0.72^2)^2} \cdot \cos 63.4<br /> = -2.139 \cdot 10^{-3} [\frac{rad}{rev}]

One revolution takes
1 [rev] = T = 2\pi \frac{a^3}{\mu_{Earth}}^{1/2}= 11.962h

Hence the angular displacement of the ascending node after 100 sederial days (1day = 23.934h) is:
2.139 \cdot 10^{-3}\frac{23.934}{11.962}\cdot 100= 0.42798 [rad] = 24.5214[deg]

Is this a plausible result?

PS: I don't know what to do with the minus in front of -2.139 \cdot 10^{-3} :)?

Thanks for your help!
 
I did not check the numbers, but the logic is sound.
 
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Thank you very much voko. The numbers are not as important as the logic is :).

The minus in front of -2.139*10^{-3} still bugs me. How can I get rid of it? Can I neglect it?
 
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The minus gives you the direction of rotation.
 
Thank you, I got it :).
 
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