Solution to puzzle 4: Equatorial belt

In the diagram below, the belt meets the Earth at a tangent, and therefore the angle between the belt and a radius is 90°.  Consider two lines: one joining the center of the Earth to the high point of the rope; the other joining the center of the Earth to one of the points at which the belt meets the Earth.  Let x be the angle subtended, in radians, between the two lines.  Let a be the distance from the high point of the rope to a point at which the belt meets the Earth.

The length of the belt not in contact with the Earth is 2a.  The corresponding circular arc length is r · 2x.

Let d be the extra length added to the belt: 1 meter in our example.
Then 2a = 2rx + d.
Hence a = rx + d/2, and so a/r = x + d/2r.

We also have, tan x = a/r.
Therefore tan x = x + d/2r.

Given numerical values for d and r, this equation can be solved for x to any required degree of accuracy using the Newton-Raphson Method (or Newton's Method), enabling us to calculate h.  Instead, we will pursue an approximate solution for small values of d = extra belt length.

Assume d/2r is very small, as it is for d = 1, r = 6,400,000.
Then tan x x, and so x is small.

The Maclaurin series for tan x is: x + x³/3 + ...

Hence x + x³/3 x + d/2r, and so x³ 3d/2r.

We also have h = r(sec x − 1).

At this point, we could substitute values for d and r, and calculate x³, and thereby h.  However, we will continue with an approximate solution, to determine the general expression for h in terms of r and small d.

The Maclaurin series for sec x is: 1 + x²/2 + ...
(The cube root of 3d/2r is of the order of 0.01 when d = 1, so this approximation is still acceptable.)
Therefore h rx²/2.
And so h (r/2) · (3d/2r)^2/3.

Simplifying, h k · r^1/3 · d^2/3, where k = (3/2)^2/3/2 0.65518535.
Since the Earth's radius is a constant, we see that h is proportional to d^2/3.

For d = 1, r = 6,400,000, we obtain h 121.6 meters.

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Remarks

Applying the Newton-Raphson Method to f(x) = tan x − x − d/2r = 0, yields the solution x = 0.0061654989, from which h = 121.64473 m, both correct to 8 significant figures.

The above approximate solution, h (3/2)^2/3/2 · 6,400,000^1/3 · d^2/3, is accurate to within 0.1% for extra belt lengths of up to about 2.3 km.  The table below gives further illustrations, correct to 8 significant figures.

Approximate and exact solutions

Extra belt length (m)	Approximate height (m)	Exact height (m)	Percentage difference
0.001	1.2164404	1.2164405	−0.0000056628017
0.01	5.6462162	5.6462177	−0.000026460023
0.1	26.207414	26.207446	−0.00012284293
1	121.64404	121.64473	−0.00057020334
2	193.09788	193.09962	−0.00090513744
5	355.68933	355.69526	−0.0016672651
10	564.62162	564.63656	−0.0026465916
20	896.28095	896.31860	−0.004201135
50	1650.9636	1651.0914	−0.0077382741
100	2620.7414	2621.0633	−0.012283168
200	4160.1676	4160.9789	−0.019496863
500	7663.0943	7665.8469	−0.035907446
1000	12164.404	12171.340	−0.056987112
2000	19309.788	19327.265	−0.090430158
5000	35568.933	35628.234	−0.16644292
10000	56462.162	56611.585	−0.26394511
20000	89628.095	90004.599	−0.41831634
50000	165096.36	166373.69	−0.76774618
100000	262074.14	265292.26	−1.2130464
200000	416016.76	424123.55	−1.9114198
500000	766309.43	793794.84	−3.4625335
1000000	1216440.4	1285617.5	−5.3808434
2000000	1930978.8	2104894.1	−8.2624275
5000000	3556893.3	4142952.5	−14.145932
6400000	4193186.2	5005072.4	−16.221268
10000000	5646216.2	7106259.6	−20.545878

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Further reading

1. Newton's Method Applet

Source: Pete Barnes

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