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Hello David, The answer to your question involves a little geometry and the application of simple law of orbital mechanics.

For the first part I assumed that Earth was at the June solstice. Earth's axis of rotation is titled with respect to Earth's orbit around the Sun. This is the cause of seasons on Earth. If the axis were perpendicular to the orbit there would be no seasons and everywhere on Earth woud have 12 hour days and nights. At the poles the Sun would be eternally on the horizon. But the axis is titled by 23.5° from perpendiculr. So there are times in Earth's orbit when the North Pole is in sunlight constantly and the South Pole is in darkness constantly. This is the case during the June solstice.

The Sun is so far away that its rays will be coming in essentially parallel to one another (and also the Earth's orbital plane). I used this simplification to calculate the minimum height a satellite would have to be above the Pole so that it will see sunlight (and hence be able to reflect it to the surface). The geometry of the situation is then such that you can draw a right triangle from the center of the Earth to the point above the South Pole where the satellite will be to the place on Earth where the Sun's rays will just graze by the surface (the arctic cirlce). The right angle is the one between the line connecting the arctic circle to the satellite and the center of the Earth. From this triangle I calculate the height to be about 580 kilometers.

For the second part I needed to calculate the length of time it would take to complete one orbit and what fraction of the orbit the satellite would be visible from the surface. To calculate the length of the orbit I used Kepler's 3rd law of orbits which says that there is a mathematical relationship between the period of the orbit and the size of the orbit. I assumed a simple circular orbit that goes directly over the poles with a size of the Earth's radius plus the height. As viewed from the South Pole the satellite would go in a straight line that goes directly overhead and would be visible for about 13 minutes. The entire orbit takes the satellite about 90 minutes.

I left out the details of the geometry and calculations, but I hope that this satisfies your curiosity.

-Bryan Mendez

bmendez@astro.berkeley.edu

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