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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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