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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 $24$ $hours$.

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  • $\begingroup$ The orbital period will not be 24 hours at 1000 km altitude. Geosynchronous orbit has a radius around 42164 km. $\endgroup$ – Ross Millikan May 28 '14 at 22:45
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Let's assume the light from the Sun is parallel, then the shadow of Earth looks like this:

Shadow

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.

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  • $\begingroup$ In the past, I've been guilty of giving complete answers to homework questions, rather than nudging the OP in the direction of figuring it out. $\endgroup$ – Mike Dunlavey Apr 13 '14 at 15:27
  • $\begingroup$ I was going to sketch out the approach, but I found that doing this amounted to answering the question anyway. It's not clear to me this is set homework, though it's homework-like. If it was set homework I hereby apologise to Radda's teacher! $\endgroup$ – John Rennie Apr 13 '14 at 15:37

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