# Mechanics of Sun-Synchronous Orbit & North Korea's KMS-4

QUICK QUESTION: Is North Korea's KMS-4 in a proper Sun-synchronous orbit?

FOLLOW-UP QUESTION: How do I tell? I want to understand the mechanics.

RESEARCH:

This link Sun-Synchronous orbits says "Like a polar orbit, the satellite travels from the north to the south poles" I've had several other people mention the north to south thing. Why is that important? How does that aid in keeping things in sync with the sun?

The book Spacecraft Attitude Determination and Control lists critical factors in table 3.5 but doesn't have any entries below 800 kilometers. Does that mean it's not possible?

• By the way, as I recently explained here you can propagate the TLEs yourself easily with the Python package Skyfield. Don't worry, it's only Python, it won't bite!
– uhoh
Mar 2 '19 at 0:26

Is North Korea's KMS-4 in a proper Sun-synchronous orbit?

tl;dr: Based on an approximate analysis of its TLE, KMS-4 is in an imperfect Sun-synchronous orbit, and will drift slightly in sun-synchrony by about 4.8 degrees each year.

How do I tell? I want to understand the mechanics.

tl;dr: With an inclination of about 97.4 degrees at a low altitude of ~500 km it's definitely close to Sun-synchronous. The inclination is a dead giveaway.

The orbital plane of a Sun-synchronous orbit slowly precesses around the Earth at a rate of about 1 degree per day (360 degrees in ~365.2564 days). That precession is induced by the gravitational perturbation as a result of Earth's oblateness, expressed as the parameter $$J_2$$.

$$\Delta \Omega = -3 \pi \frac{J_2 R_E^2}{p^2} \cos i$$

where $$p$$ is the semi-latus rectum less useful and more useful and extremely useful:

$$p = a(1-e^2)$$

$$a, e, i$$ are the semimajor axis, eccentricity, and inclination, $$R_E$$ is the standard (and effectively the equatorial) radius of the Earth used to define the dimensionless form of $$J_2$$ (see For the mathematical relationship between J2 (km^5/s^2) and dimensionless J2 - which one is derived from the other?) and $$\Delta \Omega$$ is the precession per orbit.

If $$T$$ is the orbital period, then you'll want to multiply it by the number of orbits per day in order to compare to the 360 degrees per 365.25 days.

$$\Delta \Omega \ \frac{T_{Day}}{T} = \text{day⁻¹}$$

This link https://www.n2yo.com/satellite/?s=41332 gives the TLE which you can also get from Celestrak https://www.celestrak.com/satcat/search.php

KMS 4
1 41332U 16009A   19060.53417550  .00001024  00000-0  34521-4 0  9991
2 41332  97.3761 129.3550 0022363 328.2236 142.9344 15.32879292171098


The numbers given are not exactly Keplerian orbital elements, they are intended to be used with an SGP4 propagator (see here also), but they will be close enough for a simple test.

• orbits/day = 15.3287929 so $$T=5636.45 \text{sec}$$
• eccentricity $$e=$$0.0022363
• inclination $$i=$$97.3761
• semimajor axis $$a \approx (T^2 GM_E / 4 \pi^2)^{1/3}$$ (from here but more accurately see @Chris' answer).

So using

• $$GM_E \approx 3.986 \times 10^{14} \text{m³ s⁻²}$$ (from here)
• $$J_2 \approx 0.0010826$$ (unitless, normalized form, from here)
• $$R_E \approx 6378136.3 \text{m}$$ (from here and here)

I get

• semimajor axis $$a \approx 6845360 \text{m}$$
• semi-latus rectum $$p = 6845325.7 \text{m}$$
• $$\Delta \Omega = -3 \pi \frac{J_2 R_E^2}{p^2} \cos i \approx 0.065158 \ \text{deg/orbit} = 0.99879 \ \text{deg/day} \approx 364.81 \ \text{deg/year}$$

So what I get here is pretty close to 360 degrees per year but not exactly. I've used several approximations, though I am not sure any could result in a 1.3% error, so based on my calculation KMS-4 is in an orbit that's close to Sun-synchronous but its imperfect precession will cause its orbital plane to drift slightly in sun-synchrony by about 4.8 degrees each year.

• Thank you very much!!! Very helpful! Does the North-South thing matter at all? I've seen several people & links mention it? Is sun synchronous specific to one hemisphere only? Thank you! Mar 2 '19 at 15:38
• @Jon17 inclination of zero would be around Earth's equator, and inclination of 90 degrees exactly would be a polar orbit, meaning it would go over both the north and south poles. At 97 degrees, it's just a little past polar. At any inclination that's not exactly zero, a satellite spends half its time above the equator, and half its time below. A satellite in orbit can never say in only one hemisphere.
– uhoh
Mar 2 '19 at 15:44
• @Jon17 There are more questions here about sun-synchronous orbits, why don't you read a few and then see if you have more questions. space.stackexchange.com/questions/tagged/sun-synchronous Also check out the answers to Are sun-synchronous orbits always North to South?
– uhoh
Mar 2 '19 at 15:45