Distance away from earth to see it as a full disk [duplicate]

This question already has an answer here:

This question is more space-related than physics-related, but here goes...

How far away the earth would I have to be in order to see the earth as a full disk? What I'm looking for is a distance in kilometers or miles. For example, when I fly in an airplane at 40,000 feet (about 12000 m), I can begin to see the curvature of the earth, but the view I have of the earth is a tiny piece of the total. Also, I know if I'm on the moon I can see the earth as a full disk. But what is the minimum distance away the earth I'd have to be to see the full disk?

The main reason I ask this question is because I am interested to know how much of the earth you can see from the International Space Station (ISS). I've seen various photo collections that supposedly show views of the earth from the ISS and some indicate that you can see the earth as a full disk. However, I am skeptical that you can see the earth as a full disk from the ISS. Below I've done some calculations to try and determine how far away from the earth you would have to be to see the earth as a full disk.

The human visual field of view is approximately 120 degrees in both the horizontal and vertical directions. If we construct a right triangle where one angle is 60 degrees (half of 120 degrees), "d" is the distance to the earth, and "r" is the radius of the earth, then d = r/tan(60) = 6371 km/1.732 = 3678 km = 2285 miles. This says that you would have to be 3678 km (2285 miles) away from the earth to see it as a full disk. Since the ISS is orbiting at an altitude of 347 km (216 miles) perigee and 360 km (224 miles) apogee (the mean is about 353 km (219 miles)), I believe that you will not be able to see the earth as a full disk from the space station.

Is this analysis correct? If not, what is the correct analysis?

Mike

marked as duplicate by ManishearthMay 12 '13 at 11:31

Someone outside a sphere, looking at the sphere, will always see a "full disk"! That's one of the properties of a sphere; it looks like a circle from every direction and distance . The two questions seem to be:

1) How large will that disk appear in the view of the person outside that sphere? Right now, I just looked straight down and noted that the circular disk of the earth that I can see is about 180 degrees wide in my field of view. Your question seems to require that the disk fit within a 120 degree field of view. In that case, the solution of Philip Gibbs, above, is correct.

2) How much of the surface of the sphere is included in that visible disk? Again, right now, looking down and from side to side, I can see the surface of the earth over a few hundreds of meters (I'm not very tall!). In the extreme, at a huge distance from the earth, I could see just essentially half the earth's surface. Even geosynchronous satellites miss the poles. Referring to the diagram in the solution of Philip Gibbs, above, note that the angle in the triangle at the center of the earth is 30 degrees. So, if you started moving straight up from the North Pole, you would see a smaller and smaller earth-disk, including more and more of the earths surface. When you reached about 1000 km, the earth would look around 120 degrees across, and you would see the polar regions down to about 60 degrees North Latitude...

How far away from earth's surface do we need to go to see its full hemispherical area? The strict mathematical answer is: no matter how far you go away from earth, you will never see its full hemisphere.

So a better question is the one that generalizes your remark "I am interested to know how much of the earth you can see from the International Space Station (ISS)":

At height $h$ above earth's surface, what fraction of earth's hemispherical area will remain invisible?

Some straightforward math (let me know via a comment in case more detail is needed) reveals an elegant answer to this question. The fraction of the hemispherical area missing equates to $\frac{r}{r+h}$, where $r$ denotes earth's radius, and $h$ observer's height above earth's surface.

So, when you observe earth ($r$ = 6.3 megameter) from ISS ($h$ = 0.35 megameter), some 95% of earth's hemispherical area will be missing from your view. Even when observing from the moon ($h$ = 360 megameter), about 1.7% of earth's hemispherical area will remain hidden from your view.

It seems to me that you have calculated the distance to the horizon rather than the altitude.

The altitude would be $\frac{r}{\sin(60^\circ)} - r = r \times 0.1547 = 986\text{ km}$. Update:

From within spacecraft the windows would restrict field of view, but in 1966 on gemini-11 mission Richard Gordon did a spacewalk at altitude of 1369km where the Earth would be 110 degrees across. The field of view of his helmet may have been sufficient to include the full width but probably not the full height of the disc. Astronauts on Apollo 8 photographed Earthrise from first lunar orbit in 1968. The Earth is 2 degrees across at that distance. In theory they could have seen Earth disk on outward journey through a window if it pointed in right direction, but 3 of 5 windows were fogged and they did not see it until they came round from far side of moon. Picture is from Gemini-11. • Which indeed proves that you would need very-wide-angle eyes/camera to see the full disc from ISS. I believe the first sight of earth like that, was from moon orbit: independent.co.uk/news/science/… – Kris Van Bael May 12 '13 at 7:46
• I think Mike is right that field of view is about 120 degrees including peripheral vision. From 1000km outwards it will look disk-like. From moon it is 2 degrees across. Highest shuttle orbit was about 620km – Philip Gibbs - inactive May 12 '13 at 8:18
• Gemini-11 reached 1369km altitude in 1966 with 2 crew. Astronaut Richard Gordon did a space-walk at that altitude. Windows on these craft would reduce angle of view to much less than 120 degrees. Even a spacesuit helmet would reduce field of view during space-walks so not sure if Gordon could get the full disk effect. I will add info to answer. – Philip Gibbs - inactive May 12 '13 at 8:34