From an altitude of 254 miles (408 kilometers), what fraction of the surface of the Earth is visible to the crew of the ISS looking out the window?
I ask because I’m imagining the day when space elevators become a thing. I imagine that along the equator is ideal. So if I were 254 miles above, say, Quito, would I be able to see the lights of Bogota? Panama City? Even further north? Lima and/or La Paz (and perhaps further) to the south?
from horizon to horizon, the section of the Earth you can see at any one time is a patch about 2,000 miles across, almost enough to see the entire United States at once
This is actually a horizon calculation, if you think about it.
Doing a horizon calculation on a view height of 254 miles yields a horizon distance of 1,441.5 miles. That’s a radius, so double that for a total view disk diameter of 2,883 miles.
Maybe that is a little high? The distance to the horizon does not directly matter, only the radius of the disc (10000 km in the limit, otherwise R cos⁻¹(R/R+h) where h is the altitude and R = 6378 km
DPRK’s formula looks correct to me. I get a radius of
R \cdot cos^{-1}(\frac{R}{R+h}) = 6378 \cdot cos^{-1}(\frac{6378}{6378+408}) = 2223 km, or 1381 miles.
It is, but some horizon calculators use an approximation that’s only valid close to the surface of the Earth.
Bear in mind that the space elevator would reach out to geostationary orbit and beyond, and from the geostationary viewpoint the Earth would look quite small and distant. At geostationary orbit the Earth would occupy 20° of the sky. You’d be able to see most of the hemisphere facing you from that viewpoint.
If you descended the elevator, the Earth would appear increasingly large while (paradoxically) less and less of the surface would actually be visible. At the height of the ISS the angular size of the Earth would be about 140°, but you’d see a much smaller fraction of it.
from here
There are probably a few confounding factors when one asks how much of the Earth can be seen.
The calculation of radius is for a planar circle, whereas the surface of the Earth is a sphere, and as the size gets larger the disparity in size grows. In the limit the radius of the apparent disk is 10,000 km, but the linear extent of surface is greater by a factor of \frac \pi 2. For observations from LEO this is mostly insignificant.
The other question is how much useful surface can be seen. The edges of the disk are tangent to the line of sight. One would imagine that the usefully visible region would be smaller simply because of the angle of view. Simple horizon calculations are meant for seeing objects side on - such as seeing a ship from another ship. But watching the world from above is a different matter, and one would probably say that angles less than, say, ten degrees would be difficult to make out. Atmospheric effects would start to compound as well.
The 10000 km is what you would measure along the surface of the earth, assuming the viewpoint were far enough so that an entire hemisphere were visible. However, if the observer takes central stage it may be more appropriate to (re-)consider angular diameter rather than worry about the great-circle distance along the surface. (For an altitude of 408 km, this is about 140° in your field of view covering 0.38 steradians (3%) of the Earth’s surface as explained, whereas from the Moon, let’s call it 2 degrees.)
Circumference of the Earth is 40,000km. At least if you assume a sphere. 40,075 around the equator if you allow for the oblateness. 10,000km down the meridian through Paris from the pole to equator, by initial definition of the metre.
So I’m assuming at the limit you can see one hemisphere, so 20,000km.
I was confusing the numbers earlier and thus confused about the correction.
Yes.
Once we get past doing all the trig (correctly) to get the uneclipsed area, these effects loom large.
Just from jet airplanes’ altitudes of 30-45K ft, the geometric horizon is a long way away, as in ~200-250mi. But even on the clearest of days, the useful horizon is far closer and beyond that range the planet appears as a gray, brown, or blue featureless fog. At night distant cities are merely dim, barely perceptible glows in the otherwise invisible black thin fog.
For simplicity one could assume the atmosphere above that altitude (10-14km) is 100% vacuum-equivalently transparent. It isn’t, but what obstructions exist farther up are but a rounding error on those further down.
Which results suggest that any time your slant range through the lower atmosphere from space is above 100mi at absolute best, and more typically 50mi, the diffraction effects and generally gunkiness of the atmosphere will render seeing anything on the surface impossible. At least with the human eye and its spectrum of sensitivity.
Due to a curious measurement error, it’s actually 40,007.863 km. From this Wiki article:
In 1793, France defined the metre so as to make the polar circumference of the Earth 40,000 kilometres. In order to measure this distance accurately, the French Academy of Sciences commissioned Jean Baptiste Joseph Delambre and Pierre Méchain to lead an expedition to attempt to accurately measure the distance between a belfry in Dunkerque and Montjuïc castle in Barcelona to estimate the length of the meridian arc through Dunkerque. The length of the first prototype metre bar was based on these measurements, but it was later determined that its length was short by about 0.2 millimetres because of miscalculation of the flattening of the Earth, making the prototype about 0.02% shorter than the original proposed definition of the metre. Regardless, this length became the French standard and was progressively adopted by other countries in Europe.[25] This is why the polar circumference of the Earth is actually 40,008 kilometres, instead of 40,000.