Okay, I’ll add my post anyway although the formula’s already been shown by JonF.
It should be noted that you can’t see “the whole diameter” at any distance either (even from the moon). The problem’s the same in both cases.
To try and visualize the problem, draw a figure I’ll try to describe :
Draw a circle. Now draw a straight tangent line from the bottom of the circle out to the right. Pick any point on this line to represent the viewer’s position. Label it V. Now draw a tangent line towards the “top” of the circle from V.
(You can see that since this is as much as you’ll see, you’ll never quite see the whole diameter.) Mark the center of the cirle as O. Now draw radii from both the bottom line and the top line. Label the bottom tangent point B and the top T.
It’ll look something like this :
_
/ \ \T
| O | \
\ / \
-B-----V
Note that OB is perpendicular to the bottom line (VB), and OT is perpendicular to the top line VT. If you aren’t convinced of this, re-orient the figure so that VT is straight along the bottom. It will look essentially the same.
Let’s call the “viewing angle” between line VB and line VT <v (angle v). Now draw a line from O straight through V. If you’ve done it right, it should look like you’ve cut <v in half. This is true (the proof is left as an exercise for the doper).
One more label, the point at which OV crosses the circle call S.
We’re going to solve for r, the radius (OB or OT equal this value). The problem can be solved whether we know distance to center or just to surface. Let’s call the distance to the surface, VS, d. Also, length to the center OV = d + r.
We have a right triangle VOB. We have a formula for the hypoteneuse length VO. We also know the angle OVB = half of <v since OV cut <v in half.
From trigonometry for right triangles, sine = opposite / hypoteneuse.
Putting in what we have,
sin (<v/2) = r/(r+d)
Solving for r gives us :
r = d* ( sin(<v/2) / (1 - sin(<v/2)) )
It should be clear that if the distance to the center (r + d) is known, the formula works the same way (subbing the value in for hypoteneuse).
Okay, for some values in your problem:
If the earth’s radius is 4000 miles, at 1000 miles <v = 106 deg. This is rather close.
As you pull back to the moon, the angle gets much smaller, and you do almost see the whole diameter. At 270,000 miles <v = 1.7 deg
Side note on applying improper observations:
BTW, ‘you’ here refers to a generic person, not to you specifically. I’m just trying to show the answers one gets when the assumptions are just a little off.
Let’s suppose you didn’t realize you can only view tangents, and assumed while standing on the moon that you could in fact see the whole diameter. So, you draw a diameter on your circle and form a triangle to where you are. Since it’s so close, you might fail to notice that you just drew your lines through the circle. You look and say, “What I have is an isosceles triangle with one angle <v, and since the other two are equal and it all adds up to 180 deg, those other angles must be (180 - <v)/2 deg.” The angle you observe is, as above, 1.7 deg so you get 89.2 deg for these.
Now then, you continue, “If I draw a line to the center of the earth, I get a right triangle with 89.2 deg in one corner and v/2 in the other. By trigonemetry, tan = opposite / adjacent, so tan (<v/2) = r/OV. Solving for r, I get r = OV * tan (v/2).” You put in the numbers, and get 4000.4 miles. So in case it wasn’t so bad at all, since it was so close.
panama jack
I saw the crescent, you saw the whole of the moon.