5
$\begingroup$

As a continuation of my previous post regarding the use of the software GMAT (General Mission Analysis Tool) (website, YouTube) for propagating spacecraft, I have an additional question.

This time I propagated 2 spacecraft in a $J_2$ gravity field (gravity degree 2 and gravity order 0, no higher gravitational harmonics, no drag, no sun radiation pressure, no third bodies). Both spacecraft have the same initial mean orbital elements: $$[\bar{a}_0,\bar{e}_0,\bar{i}_0,\bar{\Omega}_0,\bar{\omega}_0,\bar{M}_0] = [7000\text{ km}, 0, 45 \text{ deg}, 0, 0, 0]$$ The only difference between the spacecraft is their epoch. The first spacecraft epoch is $$1/1/2005 \quad \quad 00:00:00$$ The second spacecraft epoch is $$1/1/2015 \quad \quad 00:00:00$$

Finally, I plotted their osculating inclinations in the J2000 frame (The regular Earth Centered Inertial (ECI) frame):

enter image description here

The $J_2$ induced long-period inclination oscillation (period of about 70 days) does not surprise me. What is surprising is that the year governs the oscillation amplitude.

Two points:

  • To the best of my knowledge, the utilized gravity field is axisymmetric and time-constant, so no time-varying perturbation should occur.
  • In order to verify this result, I also used STK with the same scenarios and got the same results.

I think that this phenomenon (year governing the oscillation amplitude) stems perhaps from precession and nutation effects. What do you think?

Update: Here is an output from STK where I defined some initial conditions with the aforementioned force model: enter image description here

The plots depict the osculating inclination as a function of time. We can clearly see that on long time periods, the oscillation amplitude varies. Can anyone explain this dependency?

$\endgroup$
0

1 Answer 1

3
$\begingroup$

OK, I think I figured it out. In Spacecraft Formation Flying p. 79, they write the equations of motion for the $J_2$ problem as:

enter image description here

where $X$,$Y$,$Z$ are in inertial frame ECI. In Analytical Mechanics of Space Systems pp. 379-390 there is an expression for the axi-symmetric spherical harmonic gravity perturbation $J_2 - J_6$:

enter image description here

Also here, $x$,$y$,$z$ are in inertial frame ECI.

In these examples, the expression of the coordinates in ECI is an approximation. If I would have used these expressions (in many application they are used), I would not see the phenomenon discussed above,

Because the gravity is related to the Earth and not to an arbitrary inertial coordinate frame, the coordinates should be expressed in ECEF. So, in these approximations, they assumed that $z_{ECEF} = z_{ECI}$. This is true only if you neglect nutation, precession, and polar motion.

So, an accurate expression for the gravity can be found in Fundamentals of Astrodynamics and Applications, First Edition, pp. 496-497. They write the general spherical harmonic gravity perturbation as:

enter image description here

where $U$ is the gravity potential, $\lambda$ is the longitude, $\phi_{gc}$ is the geocentric latitude, and $r_I$,$r_J$,$r_K$ are expressed in ECEF. Professional programs like STK and GMAT use the latter expression. So, once the user defines the orbital elements, they are transformed to ECI coordinates, then the gravity perturbations are calculated in ECEF and transformed to ECI, and the integration is in ECI.

So, although I disabled all the perturbations but $J_2$, there is still precession, nutation, and polar motion, which changed slightly the ECEF attitude with respect to ECI. So, precession, nutation, and polar motion vary the gravity forces over time.

$\endgroup$
2
  • 1
    $\begingroup$ Try plotting the same history in a true of date frame -- this way the frame "changes with" the orbit itself, which is closer to the original assumption of a fixed z axis. $\endgroup$
    – Chris
    Commented Jul 14, 2017 at 16:44
  • $\begingroup$ It's always okay to accept your own answer $\endgroup$
    – uhoh
    Commented Oct 10, 2019 at 7:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.