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How do the following perturbation factors affect a sun-synchronous orbit?

  1. Atmospheric drag
  2. Solar gravity
  3. Lunar gravity
  4. Solar radiation pressure
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Sun-synchronous orbit would have an altitude of between 150 and 900 km. That being said, atmospheric drag would be determined by altitude, with altitudes higher than 600 km experiencing no drag. At lower altitudes, this drag will cause the satellite to slow and lose altitude as a result (call it a change in a).
Solar radiation pressure would affect the orbit by causing the satellite to torque over time, which in combination with atmospheric drag could cause more or less drag over the surface area. This change takes a lot of time, however, because solar radiation results in a very small force.

Solar radiation perturbation can be calculated by:

$$F=\frac{F_s}{c}\cdot A_s(1+r)\cos(I)$$ where:
$F$ is the force in N
$F_s$ is the solar constant, $1358 \,\mathrm{W/m^2}$
$c$ is the speed of light, $3\cdot 10^8 \,\mathrm{m/s}$
$A_s$ is the illuminated surface area
$r$ is the surface reflectance with $1>r>0$
$I$ is the incidence angle to the Sun

This comes from: Sellers, Jerry. Understanding Space, 3rd Ed.(a book I know from my orbital mechanics class)

Unfortunately, I haven't found anything about solar or lunar gravity perturbations for sun-synch satellites. I would consider that these forces would have the minutest of minute perturbations on a satellite so close to the Earth.

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    $\begingroup$ Thanks @Tawooh. How do we know that the change in semi-major axis is caused by atmospheric drag ? $\endgroup$
    – Soumajit
    May 14 '16 at 7:02
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    $\begingroup$ If you use equations often in stackexchange you might enjoy trying Mathjax. You can add in-line equations also, even in comments! So for example if you put "c=\sqrt{a^2+b^2}" inside a pair of dollar signs, you get: $c=\sqrt{a^2+b^2}$ $\endgroup$
    – uhoh
    May 15 '16 at 2:17
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A sun synchronous orbit uses a combination of size, shape, and orientation, to exploit the gravitational perturbation of the earth's oblateness such that the orbit plane remains synchronous with the apparent motion of the sun.

Size of the orbit can be defined as the Semi-major Axis, a. Shape of the orbit is described by Eccentricity, e, a measure of non-circularity. Orientation relative to the equator is Inclination, i.

What we want is for the time rate change of the Right Ascension of the Ascending Node to match the time rate change of sun motion. Using the oblateness term of the earth's 2nd Zonal Harmonic (J2)

$d\Omega/dt=-(3/2)\sqrt{\mu/a^3}J_2(R_e/a(1-e^2))^2cos(i)$

And

Sun's apparent motion = 360 deg / 365.25 days = 0.98562628 deg/day = 1.991 rad/sec

So we set $d\Omega/dt = 1.991 rad/sec$

Examining the effects (perturbations):

Atmospheric drag lowers altitude, which changes the semi-major axis of the orbit. See above equation. Lower altitude means closer to the earth gravity perturbation, which makes the time rate change increase, moving the orbit faster eastward relative to the sun, which local time of the node earlier in the day.

Solar gravity applies a torque to the orbit that changes inclination. This is because the sun will have some elevation above or below the equator, which means over the orbit a satellite will experience a differential in gravity from one side to the other. That small difference results in a torque applied to the orbit angular momentum vector, resulting in a change in inclination. This is a very small effect that changes the inclination slightly but enough over many years (6-7 years for altitudes of 600+ km) to impact the local solar time of the node.

Lunar gravity has similar effects as solar gravity but even smaller (but with higher frequency).

Solar Radiation Pressure has negligible effect on Semi-major Axis, eccentricity, and Inclination for LEO sun-synchronous orbits (<2000 km). However it can achieve resonance effects for MEO altitudes 2000-7000 km.

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