Register to reply

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

Share this thread:
Paul Gray
#1
Nov9-13, 10:38 AM
P: 11
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!
Phys.Org News Partner Science news on Phys.org
Experts defend operational earthquake forecasting, counter critiques
EU urged to convert TV frequencies to mobile broadband
Sierra Nevada freshwater runoff could drop 26 percent by 2100
gneill
#2
Nov9-13, 11:16 AM
Mentor
P: 11,689
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.
Paul Gray
#3
Nov9-13, 12:20 PM
P: 11
@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.


Register to reply

Related Discussions
Find mass of mars given period of moon orbit and radius of orbit Introductory Physics Homework 1
Synchronous orbit with Saturn Introductory Physics Homework 5
Synchronous orbit Introductory Physics Homework 8
A Satellite in Sun-Synchronous Orbit Introductory Physics Homework 4
Orbit Transfer to Lower Apoapsis of a Mars Orbit Introductory Physics Homework 11