Nail-Tinted Glasses

Danil, did you mean equatorial Radius?

OK, help me with this.

We know the length of CB, that's the equitorial radius
We know the length of AB, that's 1/4 of the equitorial circumfrence (2 pi R) + 5 meters
we know that ABC is a right angle
we know that the formula for determinging the length of AC is sqrt (c^2-a^2)
we know that the distance off the ground is length AC - equitorial radius.

What am I missing?

who swallowed my previous followup?

I think you are missing AB, which is not 1/4 of the equatorial circumference. That's the meat of the problem, after all; finding angle ACB, which is going to be small, but non zero, and related to x in some way.

Ah. I was assuming the point at which the hypotenuse separated from the circle was 90 degrees from the lifting point, which it approaches for large enough lengths, but I don't think it ever reaches it.

Hmm. Is h = R^(1/3)[(3x)^(2/3)]/2 a general solution? If I set R at 1000m and x at 1000m, this solution tells me h = 1040m Are we to the point of losing accuracy because z isn't so small anymore?

The assumptions made to get to the tidy formula at the end presuppose that x and h are much smaller than R, so if it is general, it would be an accident.

If you examine some of the easy triangles, you'll see what I mean. Suppose θ = pi/4. AB = BC = R. AC = R^.5. In this problem, x = R(1-pi/4), roughly 1369km, which would predict h=2378km, rather sizeably off from the actual h = R(2^.5-1) = 2642km.

Similarly, if θ = pi/3, BC = R, AC = 2R, AB = R(3^.5), so x is about 4368km, which predicts h = 5154km, but the actual value is h = R (6378km).

And if x is big enough, h = x.