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We are hovering 250m over the equator of an alien planet which is homogenous and perfectly spherical, and a satellite with a perfectly circular orbit is traveling parallel to the equator. We know that the satellite takes 800 minutes to orbit the planet, and from our position, we can see 406 minutes of that orbit, or a little over half, before the satellite is eclipsed by the planet. While standing on the ground under the spot we were hovering, we can see the satellite for exactly half of the orbit, or 400 minutes. What is the approximate diameter of the planet? Explain why the answer cannot be exact.

I don't wan't an answer, just insight into how to solve this problem, and which mathematical tools I should be looking to use. My current idea is to use the perspective shift as a form of measurement, roughly constructing the circumference of the planet into an n-sided polygon, and then summing the sides and dividing by pi. I don't exactly know how to get from the 250 ft elevation to the length of each side of this polygon, but I feel as though there'd be a way. Not sure if this intuition is correct though, so I'd be interested what those more experienced with mathematics would think. This is my first post on a forum like this, so excuse me if it's ill-placed.

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closed as off-topic by John Douma, GNUSupporter 8964民主女神 地下教會, John B, ArsenBerk, Don Thousand Nov 9 '18 at 16:50

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  • $\begingroup$ If you view the satellite from ground level, you can't possibly see it for half its orbit. So the height of your eyes above the ground is an essential parameter. I suppose this is what they mean when they say the answer can't be exact. $\endgroup$ – TonyK Nov 9 '18 at 1:48
  • $\begingroup$ That's a good point! Thank you :), I was already assuming that, but I hadn't given much thought into considering how that affects our approximation. Indeed, it appears as though it would. Also, I was imagining going away from the planet for an infinitely long time. As we go further and further away, we would see more and more of the orbit, but never all of it -- the planet will eclipse the satellite! I should think about perspective more, and how a change it can be linked to the diameter. $\endgroup$ – sasuG Nov 9 '18 at 1:51
  • $\begingroup$ @TonyK This problem does not make much sense to me. If you are supposed to see the half of the orbit from the ground level, the height of your eyes woulh have to be measured in kilometers. And if you are kilometers tall, saying that you are hovering 250m above the grand does not make any sense. The other possibility is that the whole planet is no bigger than a pumpkin. Are we sure that the numbers (especially minutes) given in the problem are correct? $\endgroup$ – Oldboy Nov 9 '18 at 11:13
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    $\begingroup$ @Oldboy: Hmmm, perhaps you are right. According to my rough calculations, on planet earth the satellite would have to be orbiting at a distance of about 8 million kilometres, which is not realistic. (If $R$ is the radius of the earth, and $h$ is the height of my eyes above the ground, then the satellite is orbiting at a distance of about $\sqrt\frac{R^3}{2h}$, to first order in $h/R$.) $\endgroup$ – TonyK Nov 9 '18 at 12:10
  • $\begingroup$ @Oldboy: ...So for a more realistic orbit of 5000 kilometres, say, the radius of the planet would have to be about 42 kilometres. Hardly pumpkin-sized; but now the planet is much too small to attract the satellite with the required gravitational force. $\endgroup$ – TonyK Nov 9 '18 at 12:19