1
$\begingroup$

I am trying to calculate the shadow time for a satellite on an orbit that is inclined by some positive amount of degrees. I am able to calculate it for a non inlined orbit but i do not know how to go about evaluation this system.

This is not a homework question, it is just for how to set mathematical foundations for physical problems.

Assumptions:

  1. Point masses.
  2. Circular orbit, of radius r, at the equator, is inclined by some amount of degrees.
  3. Light rays are perpendicular to earth.
  4. Earth casts a shadow that is a perfect circle.

When the orbit is not inclined, the stellite would travel behind the equator a certain distance $r*\theta$ that is proportional to the diameter of the circular shadow. Once the orbit is inclined, the lower end of the orbit would be now in the shadow domain. Now the satellite traverses at the lower end of the circle, so the area of the shadow covers less of the trajectory.

I am trying to calculate the time, based on the angular distance covered by the trajectory and the length of the shade at that point.

Thank you for your help.

$\endgroup$
1
$\begingroup$

Assuming that the Sun's rays are parallel near the earth and that $BC$ is a straight line

enter image description here

The satellite has to go through a distance $2BC$

$$2\sqrt{R_E^2-(R\theta)^2}\tag1$$

if $\theta$ is small

enter image description here

For an orbit of radius $R$ the speed $v$ of the satellite is from

$$\frac{v^2}{R} = \frac{GM_E}{R^2}$$

$$v= \sqrt{\frac{GM_E}{R}}\tag 2$$

where $M_E$ is the mass of the earth.

Combining (1) and (2), the time in the shadow is

$$T = 2\sqrt{\frac{R_E^2R-R^3\theta^2}{GM_E}}\tag3$$

for $R\theta \lt R_E$, otherwise there is no shadow to go through.

If $$k=\frac{R_E}{R}$$ then (3) can be written as $$T =\sqrt{\frac{4R_E^3}{GM_E}\times\frac{k^2-\theta^2}{k^3}}\tag4$$ or $$T =1610\sqrt{\frac{k^2-\theta^2}{k^3}}\tag5$$

By considering the time taken for the satellite to go through an angle $\phi$

enter image description here

An accurate formula valid for all $k$ and $\theta$ can be found

$$T = 1610 \frac{\sin^{-1}((k^2-(\sin\theta)^2)^{0.5})}{k^{3/2}}\tag6$$

There is only a small difference between (5) and (6).

$\endgroup$

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.