distance from a sphere My question is how far would I have to be away from a sphere to see it in its totally. Using a view of vision of 120 degrees. radius of 3.959 miles
I am trying to work out how close an astronaut would have to be to the Moon and/or the Earth to be able to see it.
 A: You will never be far enough to see a sphere in its entirety, unless you use mirrors (or even fancier: large masses to create a gravitational lensing effect).
As an analogy: consider the Sun, which is 'pretty far' away from the Earth (when compared to the Earth's diameter). Still, at any given time, it's daytime on only half of the Earth, so the Sun 'sees' only half of Earth.
A: By "see in its totally" I assume you mean just see the edges of the disc filling your field of view (as opposed to seeing all the surface). 
We then want the edge of the field of view to be tangent to the surface of the sphere, giving us the diagram:
!http://i.imgur.com/H2FXdtS.png
With $R$ the radius of the object, and $d$ your distance from the centre of it. Now assuming we have $R$, we have $d = \frac{R}{sin(60)}$
A: You can never see the entire surface of the sphere, of course.
There's always a side of the sphere that's on the "other side" of the
center of the sphere from you.
Technically, you can never even have a direct line of sight to an
entire hemisphere (unless you consider a ray of light bent by refraction
to be "direct").
So I interpret the question to mean that you want the most extreme
"left" and "right" points for which you do have a direct line of sight
to appear no more than $120$ degrees apart when viewed from where you are.
Since the center of the sphere will appear to be midway between the
"leftmost" and "rightmost" points, you have a $60$ degree angle between
the radius to your position and the tangent line from your position 
to the "leftmost" point on the sphere.
The radius to the "leftmost" point is (of course) perpendicular to the tangent at that point, so you have a right triangle whose vertices are the
"leftmost" point, the center of the sphere, and you.
That triangle in fact is a $60$-$30$-$90$ right triangle, and the length
of its hypotenuse (the distance from you to the center of the sphere)
is $\frac{2\sqrt3}{3}$ times the length of the side opposite the
$60$-degree angle, which is the radius of the sphere.
In other words, you need to be at a distance from the center
that is equal to $\frac{2\sqrt3}{3} \approx 1.1547$ times the sphere's radius.
A: Here is a picture drawn to scale of how far above the surface of a spherical object one would have to be in order to perceive the sphere in a single view, assuming a $120^\circ$ field of view.
The astronaut would be at point $C$ in the diagram a distance $\dfrac{2\sqrt{3}}{3}r$. However, the distance above the surface would be $\left(\dfrac{2\sqrt{3}}{3}-1\right)r$.
For the earth this would be approximately $985$ km ($612$ miles).

