What fraction of the earth’s surface can an astronaut see from an altitude of 400 km?

Let point O be the center of the Earth, K the cosmonaut, and G the horizon. Let’s designate the lengths of the segments: OG through R and KG through D. Then the length of the segment of the CO will be equal to R + h, where h = 400 km is the orbital height. The distance to the horizon is determined from the GOC right-angled triangle by the Pythagorean theorem: (R + h) 2 = D2 + R2, whence
D ^ 2 = 2Rh + h ^ 2 = 2Rh (1 + h / 2R)
Since h << R, the second term in this formula is much less than the first, so it can be neglected. As a result, we obtain the formula for the distance of the pre-horizon at the observer height h << R: D = √2Rh. Since D << R, the surface area of ​​the Earth that is visible to the astronaut can be calculated as the area of ​​a circle: s = πD2, since the total surface area of ​​the Earth is calculated as the area of ​​a ball: S = 4πR2. The ratio of these areas is s / S = h / 2R = 0.03 (i.e. 3%).

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