NavList:

Last week, Dale Lichtblau, you wrote:
"Suppose you added one yard 3 feet) to a cord stretched evenly and "tightly" around the Earth at the Equator. How far above the Earth, keeping it evenly distributed above the Earth, would it then be? (Most people would say an infinitesimal small amount...and would be surprised by the actual amount. Confusing mental pictures with mathematical reality, said Wittgenstein.)"

I think it mostly has to do with the fact that few of us deal with one-dimensional cases like this. They should be easier than three-dimensional, right? But the intuition-based "infinitesimal" reply is probably based more on our experience of adding up lumps of clay. If I have a 1000-foot wide sphere of clay, and I add a 1-foot wide additional ball to it, and then squish it into a new spherical form, what's the new radius. It's certainly not 1001 feet, and our intuition gets that right. The change is very small. Of course intuition can work against us! It does in this puzzle about a rope with an extra yard, and it does when combining spherical masses in some unusual cases. For example, suppose I have a planet ten times as massive as Jupiter, but otherwise similar in terms of composition, temperature, and so on. Now I add one more Jupiter to it. How big is the new planet? Or another extreme... A one solar-mass black hole has a radius (event horizon radius) of about 3 km. Suppose I find a huge black hole at the center of a galaxy with a mass of 10 million Suns. It has some radius... I now drop another one solar-mass star into the black hole and give it a moment to "digest". The black hole is bigger. How much bigger?

Dale, you also posted a related puzzle:
"But instead of keeping the cord spaced evenly above the (assumed perfectly spherical) Earth, you instead grasped the cord and pulled it straight up! How high, how far above the surface of the Earth would the cord then be at it's apex?"

This second rope puzzle is a good one for various reasons. As several people already explained in some detail, the problem comes down to figuring out how much of the rope no longer lies against the Earth's surface [or other circular cross-section object] and how the length of that circular portion compares to the straight length of rope now held in tension off the surface. The difference in length, d, is given by a bit of plane trig:
   d = R·(tan θ - θ)  [call this equation 1],
where R is the radius of the circle (the radius of the Earth in the original setup), θ is the angle seen at the center from the point of suspension to the point where the rope makes contact with the circle, and θ is assumed to be a "pure" number, an angle as pure ratio, or an angle "in radians" in engineering terms. Of course, we also need to relate θ and R to the height, h, of the suspension point, which also follows from a bit of plane trig:
   cos θ = R / (R + h),
or, equivalently,
   h = R·(sec θ - 1) [call this equation 2].
The solution is then easy. We pick a value for d, like 3 feet in the puzzle as given, use that in equation 1 to solve for the angle θ. Then we drop that into equation 2, and there's our value for h. Done!

Before solving the specific problem, as above, there's a nice way to verify that the two equations are valid, not missing a random factor two somewhere, no reversed signs, etc. Suppose the central angle θ is 45° (or pi/4 in "radians"). Draw it... This is an easy case. The length of the rope extending from each side of the suspension point down to the surface of the circle is exactly equal to R. So just by inspection of a diagram, we can see that the height of the suspension point is the diagonal across a square with a side of R minus the radius R or R·(sqrt(2)-1), and yes, that matches what we expect from equation 2. And the difference in length is the length of the side of the square, R, minus an eighth of the circumference of the circle, or R·(1 - pi/4), and yes, that matches what we expect from equation 1.

Back to the puzzle as stated. Suppose d is three feet and R is the radius of the Earth. Then equation 1 becomes:
   x = tan θ - θ,
where x is a very small number (about 1/7million in the puzzle as given). How do we solve that? Iteration in a spreadsheet works just fine, right? Is that an inferior approach? Isn't there some way to solve this "properly"?? Well, no, and in fact this is an example of a seemingly simple problem that is nearly identical to a still unsolved problem going back some 400 years [unsolved because it is unsolvable]. Just ask Johannes Kepler! He discovered an equation in elliptical orbit mechanics for the planets that still bears his name. Kepler's equation reads:
   M = E - e·sin E
where M, the mean anomaly, is easily calculated --a clock angle, determined from the constant rate motion of a planet around the Sun, while e is just a constant (the eccentricity of the elliptical orbit), and E, the things we're looking for, known as the eccentric anomaly, is related in a simple way to the actual position of a planet in its orbit as angle from its perihelion. We're given M and e, and we have to solve for E. It's such a deceptively simple problem, and there have been hundreds of books and articles written about it (here's a NASA paper on a "new" computer solution in 1972). After Kepler's time but before any of us were around, it was realized that this is a transcendental equation. There's no easy way out, and the iterative solution, like David Pike applied, is just fine!

Since we're looking at very small angles and differences in the case as stated, this is a great opportunity to use the second-most valuable set of equations in navigation trigonometry: the series expansions for the basic trig functions:
   sin θ = θ...
   cos θ = 1 - θ² / 2...
and, rare but we need it here,
   tan θ = θ + θ³ / 3.
All three of these apply when θ is a small angle (in practical cases, smaller than 10° is usually safe). We also benefit from a small piece of the binomial expansion which here says:
   1 / (1+x) = 1 - x... , when x is a small number.

We take the original equation 1, above, and rewrite using the expansion for tan θ. The result is just:
   d = R · θ³ / 3.
Note that this small angle approximation is now solvable without any iteration or any of the other tricks used down through the centuries to solve equations like Kepler's equation. Give me a value for "d", and if it's small enough, I can give you θ directly. With that calculated value to θ, I can then turn to the original equation 2 and produce a value for the height of the suspension point h. But here we can recognize that we're looking at another familiar problem in navigation: the angle θ here is really just the distance to the horizon as seen by an observer elevated at a height h. So we know that we can approximate this by an equation proportional to the square root of h. We get there by the expansion for cos θ, and find:
   h = R · θ² / 2.
We can solve these two very short equations and get a result connecting d and h valid so long as the angle θ is smaller than about 10°. Lars Bergman already presented that short solution in his post. A final note on the distance to the horizon here. This equation relating h to θ connects the unrefracted distance to the horizon to the height h. That's a little different from the navigational result that we have all seen many times...

Frank Reed