Let's say you're in the middle of a desert, with nothing but sand. Let's also assume that you have a 20/20 vision. When you're just looking there's a point that your eye can't reach name it $M$ and let your position be modeled as a point $O$. In this case what's the best approximation for the distance $OM$. Sorry If this seems unclear but I tried my best to describe this case.


Maybe this can be a better description. Let's suppose you put an object on the floor, and you started getting back until you no longer see it, then what would be the distance walked so that can happen. We will assume the Earth is perfectly spheric

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    $\begingroup$ I don't understand "eye can't reach". We can easily see stars trillions of miles away. $\endgroup$ – Steve Byrnes Sep 11 '15 at 11:02
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    $\begingroup$ The far point of the human eye is taken to be the furthest point of distinct (i.e. clear) vision, for someone with 20/20 vision is infinity. But if you mean e.g. the furthest point on the earth then you have to consider things like the curvature of the earth etc. $\endgroup$ – Quantum spaghettification Sep 11 '15 at 11:04
  • $\begingroup$ When looking at the ocean, there's like a limiting point between sea and sky, in that case what would be $OM$ ? $\endgroup$ – Oussama Boussif Sep 11 '15 at 11:06
  • $\begingroup$ @OussamaBoussif That's due to the curvature of the earth, not any optical effects. Actually optics tends to help by allowing some light to curve along the horizon. $\endgroup$ – Blackbody Blacklight Sep 11 '15 at 11:38
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    $\begingroup$ @OussamaBoussif Fast enough to prevent you from opening your eye again, which would lead to a paradox. $\endgroup$ – Blackbody Blacklight Sep 11 '15 at 11:40

Assuming your height is $2m$, the earth is a perfect sphere of radius $6400km$, $OM$ would be $\approx 5km$. This can be achieved through basic trigonometry and approximations. Here is a very rough diagram (sorry for my pathetic paint skills). $OC = MC$ is the radius of the earth $= 6400km$. $OP= 2m$ which is greatly exaggerated here for illustration purposes. Since the line PM is tangent to the circle at M, $\angle PMC =90^{\circ}$. So $PM = PCsin\theta$, and also $cos\theta = \frac{PC}{MC}$. From these two equations $PM$ can be evaluated. Since $\theta$ is so small, we can approximate $PM$ to $OM$.

enter image description here

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    $\begingroup$ Could you show some more explaining $\endgroup$ – Oussama Boussif Sep 11 '15 at 11:28
  • $\begingroup$ This answer could use a diagram $\endgroup$ – Floris Sep 11 '15 at 12:40

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