13
$\begingroup$

This is a special case follow-up to Are sun-synchronous orbits possible around any body? Reading Wikipedia on Sun-synchronous orbit a few days ago when trying to refresh my memory on orbital mechanics of SSO, I noticed this as a rather vague mention:

Sun-synchronous orbits are possible around other oblate planets, such as Mars. But Venus, for example, is too spherical to have a satellite in sun-synchronous orbit. See the article Venus where a flattening coefficient of zero for this planet is cited.

I have a problem imagining why this would be so. I would have thought (without doing actual calculations) that its long day that's longer than its year would be more problematic, e.g. for permanently sunlit dawn-to-dusk SSO but not all SSO, than it being too spherical. So what in orbital dynamics of SSO prohibits them around perfectly spherical bodies? Or is that Wikipedia quote simply inaccurate?

$\endgroup$
  • 3
    $\begingroup$ Well... Venus is so spherical because it doesn't rotate, so the two problems are related. They need the non-uniformity in the gravity field to match the Earth's precession. $\endgroup$ – Rikki-Tikki-Tavi Sep 12 '14 at 13:45
  • $\begingroup$ @Rikki-Tikki-Tavi: Venus does rotate, just very slowly. $\endgroup$ – Keith Thompson Sep 14 '14 at 0:37
20
$\begingroup$

An orbit has an angular momentum that is conserved. That angular momentum vector keeps the orbit at a fixed orientation in inertial space. As a planet orbits the Sun, the local time of the ascending node moves around the clock over that planet's year.

The only way for an orbit to be Sun-synchronous is for there to be a torque applied to the orbit to rotate the angular momentum vector of the orbit in the correct direction once per year.

It turns out rather conveniently that the gravity field of an oblate planet provides just such a torque. But only at the right orbit altitude and inclination. The first coefficient of the spherical harmonic expansion of a non-spherical gravity field is called $J_2$, and results just from the oblateness. The rotation rate of the ascending node is:

$$\dot{\Omega}=-{3\over 2}\ J_2 \left(r_e\over a\left(1-e^2\right) \right)^2\ \ \sqrt{\mu\over a^3}\ \cos{i}$$

where $r_e$ is the equatorial radius of the planet (e.g. Earth). If $J_2$ is large enough, you can select $a$, $e$ (usually chosen to be zero), and $i$ to get a rotation rate that matches the inertial rotation of a Sun-centered frame, about a degree a day. For Earth, the inclinations are around 97° to 101° (mostly polar and retrograde) for altitudes of 400 km to 1400 km respectively.

If $J_2$ is too small, you can't get sufficient torque to rotate the orbit fast enough. That is the case for Venus, whose $J_2$ is about 0.4% that of Earth's. As was pointed out by the mongoose, this lack of oblateness is due to Venus' extremely low rotation rate.

You can get an intuitive feel for where this torque comes from by considering what happens when a retrograde, mostly polar orbit approaches the equator. An equatorial bulge will accelerate the orbit a little causing it to cross the equator sooner than it would if there were no equatorial bulge. If the orbit is retrograde, then the local time below the node crossing will move a little later each orbit. If that amount is the same as the amount by which the inertial frame, e.g. the constellations, moves earlier each orbit, then the orbit is Sun-synchronous.

Due to the eccentricity of the orbit of the planet, as well as the tilt of its rotation axis to the ecliptic, a real "Sun-synchronous" orbit isn't perfect, but it's still pretty good. For Earth, the local time of the ascending node will vary about $\pm 15$ minutes, following the analemma.

$\endgroup$
  • $\begingroup$ I am not sure to understand : Are you mentioning the analemna of the sun or the analemna of the satellite ? I mean, is the local time changing along the ascending node or are you talking about the equation of time ? $\endgroup$ – radouxju Jan 6 '16 at 8:27
  • 1
    $\begingroup$ The analemma of the Sun from Earth. The ascending node crosses at a fixed mean solar time, so the apparent solar time of the ascending node varies with the equation of time. $\endgroup$ – Mark Adler Jan 6 '16 at 9:45
  • $\begingroup$ It's very handy to have that expression here - I've just used it on MOM and it seems to have worked nicely. For completeness, do you have a link or on-line reference for the expression? $\endgroup$ – uhoh Mar 17 '16 at 12:38
  • $\begingroup$ Section 10.3 here. $\endgroup$ – Mark Adler Mar 17 '16 at 14:41
4
$\begingroup$

Wikipedia is correct.

To have a sun synchronous orbit the satellite's orbital plane needs to precess by at a rate equal to the planet's mean orbital rate about the sun. The Keplerian orbital plane doesn't change. Some perturbing force is needed to make the plane precess. IN the case of sun synchronous satellites, the source of the perturbation is the planet's oblateness.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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