# Shadow time for a satellite on an inclined orbit

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.

• Are you using point masses, or do you use the first correction for the shape and mass distribution of Earth? See math.stackexchange.com/questions/2230666/… and the linked cross-postings of this question on what a difference this can make. Nov 25, 2021 at 18:43
• Please define a little closer what you want to compute. What is "shadow time", the time that the satellite has no line-of-view to the sun? What is the inclination relative to? Are ou computing with Kepler ellipses or numerical solutions? Nov 25, 2021 at 18:46
• Thanks guys for the feedback, I have edited my question. If anything isn't clear, let me know!
– RMS
Nov 25, 2021 at 19:23

Let the radius of Earth be $$R_e$$, and let the radius of the orbit of the satellite be $$R \gt R_e$$. Choose the Cartesian frame to have its origin at the center of Earth, and with $$x$$ axis pointing towards the light source. Since the light rays are parallel, they define a cylinder of light whose radius is $$R_e$$. The equation of this cylinder is

$$y^2 + z^2 = R_e^2$$

If $$r = [x, y, z]^T$$, then the above equation is

$$r^T Q r = R_e^2$$

where $$Q = \begin{bmatrix} 0 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && 1 \end{bmatrix}$$

Now the orbit of the satellite is a circle centered at the origin (the Earth center) and is spanned by two vectors $$u_1, u_2$$ which are mutually perpendicular and each having a magnitude of $$R$$. Thus the parametric equation of the satellite position is

$$p(t) = u_1 \cos t + u_2 \sin t = V u$$

where $$V = [u_1, u_2]$$ and $$u = [\cos t , \sin t ]^T$$

All we need to do is intersect this orbit with the cylinder of shadow and generate the points of intersection. Plugging the equation of the orbit into the equation of the cylinder we obtain,

$$u^T V^T Q V u = R_e^2$$

Let $$G = V^T Q V$$, then $$G$$ is a $$2 \times 2$$ matrix. Expansion of the left hand side results in,

$$G_{11} \cos^2 t + G_{22} \sin^2 t + 2 G_{12} \cos t \sin t = R_e^2$$

Using the double angle formulas this reduces to

$$\dfrac{1}{2} \left( G_{11} + G_{22} \right) + \dfrac{1}{2} (G_{11} - G_{22}) \cos 2 t + G_{12} \sin 2 t = R_e^2$$

And this equation can solved using standard methods for solving the equation

$$a \cos \phi + b \sin \phi = c$$

After finding the roots of this equation (the one involving $$\cos 2 t$$ and $$\sin 2 t$$) which can be up to $$4$$ roots, one can determine (through evaluating the position of the satellite) the shadow time.