Was the Space Shuttle and is the Crew Dragon capable of achieving a highly elliptical orbit on whose apogee you could see the entire planet Earth? There will be a touristic flight soon in which an apogee higher than that of Gemini 11 is to be reached, but I think that's still not far enough to see the entire planet. For that you'd have to go farther than 9,000 mi (14,500 km) I think. Would the Shuttle or Crew Dragon be capable of reaching such an apogee? If not by the Falcon 9, then perhaps by the Falcon Heavy? I'm talking of seeing the entire terrestrial globe from outer space, can such apogee be reached by the mentioned spacecrafts?

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    $\begingroup$ You need to be more precise in what you mean by "entire terrestrial globe"; in the limit of infinite distance from Earth, you can only see 50% of it. For a specific figure less than 50% (e.g. 49%) we can determine a specific altitude and from that a real answer to your question. $\endgroup$ – Russell Borogove Jun 28 '20 at 17:18
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    $\begingroup$ @RussellBorogove Of course I mean all of the side of Earth that's on the side you are. Or just interpret my question as whether an apogee beyond 9,000 mi above sea level can be reached through the STS, Falcon 9 or Falcon Heavy. $\endgroup$ – LoveForChrist Jun 28 '20 at 17:54
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    $\begingroup$ At 14500km you only see slightly more than 1/3 of the earth surface. $\endgroup$ – user3528438 Jun 28 '20 at 19:08
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    $\begingroup$ 9000 miles was way beyond STS capability. $\endgroup$ – Organic Marble Jun 28 '20 at 19:19
  • $\begingroup$ @OrganicMarble Then I suppose it must be beyond Falcon 9 capability too, right? $\endgroup$ – LoveForChrist Jun 29 '20 at 5:31

This is not an answer as such, but an attempt to provide the mathematical apparatus you need to think about this (and it was way too long for a comment).

There are two sensible things that might be meant by 'seeing the entire Earth':

  • one is that the Earth would be visible in the field of view of a human eye;
  • the other is that you can see a given amount of of the globe, where that amount presumably approaches $1/2$ sufficiently closely.

Both of these depend on definitions which the question doesn't provide: what is the field of view of the human eye assumed to be? how much of the Earth counts as close enough to half? Both of these questions have answers which are not obvious. For instance, the human field of view in some sense is more than $180^\circ$ horizontally: if you stand on the surface of the planet and look down then, in principle, you can see both horizons. But that's probably not a useful definition.

So without those definitions here is how to work out the values you need. If you assume that you have reached a height $h$ above the surface, and the radius of the Earth is $R$, then the angle subtended by the Earth is

$$\theta = 2 \sin^{-1}\frac{R}{R + h}$$

In other words, for a given angle of vision, $\theta$, then you need to reach a height $h$ of

$$h = R\left(\frac{1}{\sin\left(\frac{\theta}{2}\right)} - 1\right)$$

for the Earth to be in your field of view.

Wikipedia says that the binocular field of view for humans is about $114^\circ$ (with more than that vertically, so this is the critical limit). So if we assume $R = 6371\,\mathrm{km}$ then in order to be able to see the Earth, in binocular vision, you need

$$h = 1226\,\mathrm{km}$$

The other question is: at a given height, $h$, what fraction, $\alpha$, of the surface can you see? This turns out to have a remarkably simple form:

$$\alpha = \frac{1}{2}\left(1 - \frac{R}{R + h}\right)$$

Well we can rearrange this to give $h$ in terms of $\alpha$:

$$h = R\left(\frac{1}{1-2\alpha} - 1\right)$$

So now, if you specify $\alpha$ you can read off $h$.

At $h=603\,\mathrm{km}$, the first expression above gives

$$\alpha = 0.081$$

In other words, at the point where the Earth is visible as a whole to a human in binocular vision, they can see a little under 10% of the surface at a time.

  • $\begingroup$ Of course the height given by the last formula is undefined for alpha = 0.5. You can read off h only if alpha is less than 0.5. $\endgroup$ – Uwe Jun 28 '20 at 20:23
  • $\begingroup$ I mean that the entire Earth would be visible in the human's FOV (with both eyes, without having to turn). $\endgroup$ – LoveForChrist Jun 29 '20 at 5:34
  • $\begingroup$ @LoveForChrist Whenever the distance to the viewed object is huge compared to the distance between both eyes, there is no 3D view, both eyes get the same image. $\endgroup$ – Uwe Jun 29 '20 at 8:48
  • $\begingroup$ @Uwe: yes, $h\to\infty$ as $\alpha\to 0.5$, of course. $\endgroup$ – tfb Jun 29 '20 at 10:32
  • $\begingroup$ @LoveForChrist: I've added numbers assuming human binocular FOV is $114^\circ$, which comes from Wikipedia. $\endgroup$ – tfb Jun 29 '20 at 10:52

I will use a series of images of the Earth taken by the Apollo 11 crew during the flight to the Moon. I use the formulas developed by tfb to determine the percentage of the visible Earth.

enter image description here

Image AS11-36-5325 from a distance of 18,700 km showing 37.3 % using a 80 mm lens

enter image description here

Image AS11-36-5330 from a distance of 87,000 km showing 46.6 % using a 80 mm lens

enter image description here

Image AS11-36-5337 from a distance of 91,000 km showing 46.7 % using a 250 mm lens

Note the lens was changed from 80 mm to a 250 mm telephoto lens.

enter image description here

Image AS11-36-5402 from a distance of 378,000 km showing 49.2 % using a 250 mm lens

All images from the Apollo lunar surface journal.

The distances of the images are from the Apollo Flight Journal.

You don't see much more of the Earth when increasing the percentage from 37,3 to 49.2 %. You only see less details of Earth when increasing the distance by a factor of 20.

  • $\begingroup$ Why do you call it a 49% view? Isn't it 50% (if you include the night side)? $\endgroup$ – LoveForChrist Jun 29 '20 at 5:28
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    $\begingroup$ Draw a circle (the Earth) and a point (your eye in high orbit) on a sheet of paper. Draw the two tangent lines from the circle to the point with a straightedge. $\endgroup$ – Russell Borogove Jun 29 '20 at 5:50
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    $\begingroup$ @LoveForChrist I call it a 49 % view (including the night side) because a 50 % view is impossible fron a finite distance. $\endgroup$ – Uwe Jun 29 '20 at 8:04
  • $\begingroup$ So as I understand it, you can't see 50% of the Moon either from the Earth (but less)? $\endgroup$ – LoveForChrist Jun 29 '20 at 8:54
  • $\begingroup$ @LoveForChrist That is right, you cant see 50% of the Moon from the Earth. But as the Moon is smaller than Earth, the percentage is a little bit closer to 50 %. $\endgroup$ – Uwe Jun 29 '20 at 9:04

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