Virgin Galactic will take passengers aboard SpaceShipTwo as high as 65 miles above the surface of the earth. But from this altitude, passengers will only be able to see a certain segment of the curvature of the earth through windows as large as 17 inches in diameter.
How much further into space would SpaceShipTwo have to travel to give passengers a view of the entire sphere of earth through one of these windows?
As long as you don't have any mountains above the straight line between your eye and the 'ideal' horizon, you will be able to see 'the full extent', constrained only by the extent of your peripheral vision.
So isn't this question much more around what angle your peripheral vision covers? As soon as you can see to the horizon all round, when looking down, that surely meets the requirements?
Quick calculation done:
Remembered my geometry from 25 years ago:
this does assume peripheral vision is 135 degrees all round (left, right, up and down) - so feel free to update if that assumption is false
Not completely clear what you are really asking with your question - it is obviously not possible to see the entire surface of the earth as more than half of it will be on the other side. However, if you have a 17 inch diameter window, you are half way to defining a view port - and it seems to me that you are asking how far away the space ship has to be so that the earth fits inside the view port. For this we need to make an assumption about the distance of the observer to the view port.
Diagram:
For a given height above the surface of the earth we can compute the angle $\theta$ from
$$\theta = \sin^{-1}\frac{R}{R+h}$$
And in order to see all that earth through a view port of diameter $d$, you need to be closer than distance $L$ from the outer part of the view port, so
$$L < \frac{d}{2\tan\theta}$$
It should be clear from this diagram that you never even see half the earth - and that being close to the view port is essential to see enough. If the need is to see all of earth within a 135° view angle ($2\theta$ in the diagram) then we can obtain $h$ from
$$(R+h) \sin(\frac{135}{2}) = R\\ h = R(\frac{1}{\sin 67.5} - 1) = 0.082 R = 525 km$$
L<d\2tanθ"
– the_random_guy42 Jul 10 '15 at 19:59I'm going to assume that you want to see half of the Earth, as half of the Earth cannot be seen.
First of all, seeing 50% of the Earth isn't really possible, no matter how far away you get. So, I'm going to set as a goal that one can see 45% of the circumference of the Earth, as I doubt anyone would be able to tell the difference once one has gotten that far.
The size of the window doesn't really matter, as one could simply get closer to the window, and any such considerations go away. What does matter is the tangent angles seen from the observer of the Earth.
The tangent lines to the circle are at angles plus or minus $0.45\pi$. The slope of these lines will be equal to $-\cot\theta$, $x_1=r\cos\theta$, $y_1=r\sin\theta$, $x_1\times x+y_1\times y=r^2$. Solving for $y=0$, setting $r=6,371$ km, $x_1= 0.1564r$, $y_1=0.9877r$ will result in $40,735$ km. This is the distance as measured from the center of the Earth. For reference, the Geosynchronous orbit is $42,164$ km from Earth's center.
One must travel an infinite distance before 50% of the planet is visible. Any closer and the distance between your eyes will mean that less than 50% is visible - as your eyes are not the diameter of the earth apart, at any real distance it is only possible to approach the 50% boundary.
