Ponder This Challenge - May 1998 - A belt around the earth

Ponder This Challenge:

Here's what happened... Business Week recently ran an item about a "little book of big ideas" we published for the people here at IBM Research. "Sort of like the Tao for people who think about computers all day," they wrote. Seemed harmless enough. The story highlighted some of the simple things our researchers do to get a new perspective on things, ending with this one:

Ponder something else. For example: If a belt were placed around the Earth's equator, and then had six meters of length added to it, and you grabbed it at a point and lifted it until all the slack was gone, how high above Earth's surface would you be?

In retrospect, the example should have carried a warning: For professional mathematicians only. Don't try this at home. Nonetheless, many readers were game. ("I lost the entire weekend," lamented Jan from the Bay Area.) Dozens of answers were submitted. All but one were wrong. (For what it's worth, Jan got it right.) Most interesting, though, were the different approaches people used to try and solve the problem.

One person imagined a triangle sitting tangent to the earth at its base (Hint: wrong triangle, wrong tangent point). Another tried to establish an upper limit on the height based on assuming an earth-radius of zero, then working back from that, declaring that the height must be less than 3.1416--the truncated value of pi. (Hint: try that and you'll be off by several orders of magnitude.)

Overall, though, most attempts couldn't get past the barrier of "common sense"-- if only six meters of length were added to a taut belt, wouldn't the height attained by lifting "up" on the belt have to be less than six meters, or in some way related to that limiting number? (Hint: it's not.)

Which is precisely why we posited the problem in the first place. Sometimes our most fundamental assumptions are wrong. Often, in the process of solving a seemingly unrelated problem, preexisting biases and limits to creative thinking are exposed. When that happens, the mind can open up to new ideas -- and when you're a researcher that's a good thing.

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Solution

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  First, our assumptions:

  • The radius of the Earth denoted r, is 6,400 km.
  • The additional belt length, denoted d is 6 meters, or .006 kilometers.

  If we pull the belt up at a point, we get a height h meters above the earth's surface. To determine this height , we begin by constructing a sketch of the earth with its center at c. The belt is tight against the Earth for most of its circumference, but at some point p leaves the Earth's surface, traveling up to the apex s, and returning to the Earth's surface at point q.

  • Draw the line segments cs and side cp, and let $\theta$ be the angle formed between them, expressed in radians. Also, let t denote the point at which the line segment cs intersects the earth. The arc that lies on the earth surface connecting p and t has length $r\theta$. (This is because the circumference of a circle with radius r is $2\pi r$ and an arc that spans $\theta$ radians spans $\frac{\theta}{2\pi}$ of the entire circle.)

  • Next draw the line ps connecting p and s. This line is formed by using the portion of the belt that would lie on the arc connecting p and t plus one half of the slack in the belt; thus its length is $\theta r+\frac{d}{2}$

  The length cp, is r, the radius of the earth

  The line segment cs, has length r + h (radius of the earth plus the unknown height). Since the line segment ps is tangent to the Earth's surface at point p, the three lines form a right triangle.

  We can use high school trigonometry (yes, it's really good for something) to compute the other angles and the length of the third side of the triangle.

  We first find $\theta$ by solving for $\theta$ in the following equality.

  $\tan(\theta)=\frac{\text{length ps}}{\text{length cp}}=\frac{\theta r+\frac{d}{2}}{r}$

  This can then be solved accurately by computer computation using special mathematical software. Our software found that $\theta = 0.0112$ radians.

  How does the computer find the value of $\theta$?

  Basically the computer is used to find the root of the equation

  $f(\theta)=\tan(\theta) - \frac{\theta r+\frac{d}{2}}{r}$

  Finding the root is accomplished by selecting and initial value of $\theta$, evaulating the function f, and adjusting $\theta$ if the value is not zero. The program systematically searches through values of $\theta$ until a $\theta$ for which f is zero (that is, a root) is found.

  This value of theta is then used, together with the trigonometry formula,

  $\sin(\theta)=\frac{\text{length ps}}{\text{length cs}}=\frac{\theta r+\frac{d}{2}}{r+h}$

  and some simple algebra to compute h = 402 meters.

  What if I don't have a computer to calculate the value of $\theta$?

  Calculating $\theta$ without a computer:

  To solve without mathematical software requires some freshman calculus. Since the angle $\theta$ is small for large radius r (proving this is left as an exercise for the reader) the first two terms of the Taylor series at 0 for the tan and cos functions provide a good approximation for tan $\theta$ and cos $\theta$, respectively.

  $\frac{\theta r+\frac{d}{2}}{r}=\tan(\theta)\approx\theta+\frac{\theta^3}{3}$

  Solving we get

  $\theta^3=\frac{3d}{2r}=0.00000140625$

  or

  $\theta = 0.0113$ radians.

  Substituting this value of $\theta$ into the equality

  $\frac{r}{(r+h)} = \cos(\theta)\approx 1-\frac{\theta^2}{2}+\frac{\theta^4}{24}$

  yields h = 401 meters.

  Doubting Thomas? If you don't trust our handiwork, we will prove it.

  Prove it with Pythagoras:

  Since we are working with a right triangle, the simplest way to check our answer is to plug it into the durable and ever-useful Pythagorean Theorem.

  $a^2+b^2=c^2$

  Substituting, we arrive at:

  $r^2+\left(\theta r+\frac{d}{2})\right)^2=\left(h+r\right)^2$

  $(6400)^2+(71.683)^2=(6400.401)^2$

  The left hand side is 40960000 + 5138.45 = 40965138.45 while the right hand side is 40965133, so the terms agree to 6 significant digits.

  As an alternative to using the cosine approximation, the Pythagorean Theorem could have been used to solve for h once the value of theta and the length of the side ps had been determined.

Solvers