Duration of Satellite orbit in the shadow of the Earth

Question

I want to determine how many minutes a satellite is in a circular orbit around the Earth at about $1000 km$ altitude. I assumed that the Sun-Earth vector lies exactly in the orbital plane of the satellite. Also, in this case, the Sun can be seen as a point light source and the distance to Earth is infinite. Is it possible to make an approximation of the duration that the satellite is on the 'dark' side of the Earth? Or do I need more information, like the speed of the satellite? The radius of the Earth is $6378 km$. And one orbital period is ofcourse $24$ $hours$.

Answer 1

Let's assume the light from the Sun is parallel, then the shadow of Earth looks like this:

The dotted line is the orbit of the satellite at a height $h$ (I've exaggerated the height a bit to make the diagram clearer). All we have to do is calculate the angle $\theta$, because the time the satellite is in the Earth's shadow is simply:

$$ t = \tau \frac{2\theta}{2\pi} \tag{1} $$

where $\tau$ is the period of the satellite. It should be obvious from the diagram that the distance I've labelled as $d$ is equal to the radius of the Earth, $r$, and therefore:

$$ (r + h) \sin\theta = r $$

or:

$$ \theta = \arcsin \left( \frac{r}{r + h} \right) \tag{2} $$

Finally the period of the satellite, $\tau$, is given by:

$$ \tau = 2\pi\sqrt{\frac{(r+h)^3}{GM}} \tag{3} $$

where $M$ is the mass of the Earth and $G$ is Newton's constant.

Putting all this together, for a satellite at 1000km equation 2 gives us the angle $1.044$ radians (59.8°), and equation 3 gives us the period $\tau = 105.15$ minutes. Feeding these results into equation 1 tells us that the time the satellite is in the Earth's shadow is $34.9$ minutes.