Molniya Orbit: calc angular displacement of ascending node

[Paul Gray]

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

 

[voko]

That formula gives you $\Delta \Omega$ per some period of time. What is that period of time?

 

[Paul Gray]

That's a simple but clever answer. I will give it a shot as soon as possible :). Thank you ...

 

[Paul Gray]

Using the hint voko provided I get the following answer

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

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!

 

[voko]

I did not check the numbers, but the logic is sound.

 

[Paul Gray]

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?

 

[voko]

The minus gives you the direction of rotation.

 

[Paul Gray]

Thank you, I got it :).