# Rope around the Earth

Rope around the Earth
Field: Geometry
Image Created By: Harrison Tasoff

Rope around the Earth

This is a puzzle about by how much a rope tied taut around the equator must be lengthened so that there is a one foot gap at all points between the rope and the Earth if the rope is made to hover. Although finding the answer requires only basic geometry, even professional mathematicians find the answer strangely counter-intuitive. There is a related problem about stretching the rope taut again where the answer is even more surprising. A question similar to this appeared in William Whiston's The Elements of Euclid circa 1702.

# Basic Description

In this puzzle, we treat the Earth as though it were a prefect sphere, even though it actually bulges toward the equator. Suppose a rope was tied taut around the Earth's equator. It would have the same circumference as the Earth (24,901.55 miles). The question is: by how much would the rope have to be lengthened so that, if made to hover, it would be one foot off the ground at all points around the Earth?

Image 1[1]. The lengthened rope lifted by a point. Image not to scale.

Despite the enormous size of the Earth, and the 1 foot gap around the entire circumference, the rope would have to be lengthened by a mere 2π feet, or roughly 6.28 feet.

In fact, 2π feet the answer regardless of the size of the ball around which the rope is wrapped.

Just as bizarre is what happens when one point on the lengthened rope is lifted up so that the rope is taut again, as in Image 1. The maximum clearance under the rope proves to be quite large. For the specific case of a rope looped around the Earth, a 2π foot extension would provide 614.771 feet of clearance if the rope were lifted. This is enough room to fit two Statues of Liberty under it, base and all. Unlike the previous question, however, this result is dependent on the size of the ball.

# A More Mathematical Explanation

Note: understanding of this explanation requires: *High-school algebra and High-school geometry

The circumference of a circle is given by the equation: $C=2\pi\,\!r$, where r is [...]

The circumference of a circle is given by the equation: $C=2\pi\,\!r$, where r is the radius.
Image 2. Image not to scale.

In Image 2:

• Lrope 2 is the length of the extended rope.
• Cearth is the Circumference of the Earth and the original length of the rope (Lrope 1).
• Rrope 2 is the radius of the circle made by the extended rope.
• Rearth is the radius of the Earth and the original radius of the rope (Rrope 1).

When the rope is taut around the globe, its length equals the circumference of the Earth.

Eq. 1        $L_{rope 1}=C_{earth}=2\pi\,\!R_{earth}$

The puzzle states that we have to lengthen the rope and made the rope hover 1 foot of the surface of the earth. Lengthening the rope so that it is 1 foot off the ground at all points simply means changing the radius of the circle it forms from:  Rrope 1= Rearth to Rrope 2= Rearth+1 ft.

Thus:

Eq. 2        $L_{rope 2}=2\pi\,\!(R_{earth}+1)$

Distributing the 2 π yields:

Eq. 3        $L_{rope 2}=2\pi\,\!R_{earth}+2\pi\,\!=C_{earth}+2\pi\,\!$

The new length of the rope is merely 2 π feet longer than the original length. Indeed, one can see that the additional 2π is a result of distributing the 2π to the 1 in Eq. 2, which yields an increase of 2π no matter what the radius of the ball is. Also, because the value of Rearth was not any number in particular throughout this proof, the answer didn't depend on the specific radius of the Earth in any way. Hence, this 2π extension would be the same for a ball, planet, or star of any size.

### Maximum Height of Rope

Note: A familiarity with trigonometry, series, and approximations is recommended for this section.

Were the lengthened rope again to be held taut by raising it at an arbitrary point (as shown in Image 3), what would be the distance from this point to the surface of the earth?

Image 3[2]. Image not to scale.

In Image 3:

• X1 is the distance between where the rope leaves the Earth's surface and the highest point on the taut rope.
• Xo is the ground distance along the curved surface of the Earth from the point where the rope leaves the globe, to the point below the apex of the rope.
• R is the radius of the globe.
• h is the height of the apex of the rope above the ground.
• θ is the angle formed between the line from the center of the Earth to the apex and the line from the center of the Earth to the point where the rope leaves the ground.

In the puzzle the slack added to the rope is 2π feet. I will be using l to represent the this slack because, in a different version of the puzzle, l can be different. Now we will develop a formula for l, the amount of slack, in terms of h, the maximum clearance the slack can provide. We start with h because it appears clearly in the diagram, where l is less visible. We hope later to invert this in-order to solve for h in terms of l. First, we will do this for the case where l=2 π.

Notice that triangle ABD is a right triangle, with (R + h) as its hypotenuse. Thus, using , we know that:

$(R+h)^2=R^2+X_1^2$

Which is equivalent to:

Eq. 4         $X_1=\sqrt{(R+h)^2-R^2}$

To find the length of Xo, we must remember what the length of an arc (Larc) is:

Eq. 5         $L_{arc}=r\theta\,\!$

Where r is the radius of the circle, and θ is the angle, in radians, formed between two radii from the center of the circle to the endpoints of the arc. For more on Arcs, see Arcs.

In the image to the right, Xo is the arc whose length we want to find; hence, it can be substituted in for Larc in Eq. 5. Also, since cos-1$\ \left (\tfrac{R}{R+h} \right )$ represents θ, the angle whose cosine is $\tfrac{R}{R+h}$, we can replace θ with cos-1$\ \left (\tfrac{R}{R+h} \right )$. For more information on cos-1 see Inverse Trig Functions.

Thus, Eq. 5 is equivalent to:

Eq. 6         $X_o=R\,\cos^{-1} \left (\frac{R}{R+h} \right )$.

Now that we have a formula for Xo, were are going to relate Xo to X1 . Because the rope is taut around most of the earth, all the slack goes to lengthening arc AC and turning it into segments AB and BC. Since we know that we lengthened the rope by 2π feet, we know that 2X1= 2Xo + 2π, because 2X1 - 2Xo is the extra slack put in to the rope. Thus, more generally:

Eq. 7         $2X_1=2X_o+ \mathit{l}$

Solving for l in Eq. 7 yields:

Eq. 8         $\mathit{l}=2 \left (X_1-X_o \right )$

Substituting Eq. 4 and Eq. 6 into Eq. 8 yields:

Eq. 9         $\mathit{l}=2 \left ( \sqrt {(R+h)^2-R^2} -R\,\cos^{-1} \left (\frac{R}{R+h} \right ) \right )$

Because h is both in the of the cos-1 and elsewhere in the problem, we can't isolate it and solve for it explicitly. Nevertheless, running Eq. 9 through a with l = 2 π and R = R earth = 20925524.9 feet one finds h ≈ 614.771 ft as the height that a 2π foot extension yields.

Eq. 9 can, however, be used to find the amount of slack needed to produce a certain height. For instance, if we set h = 35,000 feet (the average cruising altitude for commercial aircraft), Eq. 9 tells us that l = 2697.06 feet. Thus, Roughly 3,000 feet of slack must be added to the rope to that, when lifted, a plane could fly under it on its normal flight path.

If the puzzle were different, having a different amount of slack added to the rope, we could still find h if given a value for l. We can create a general formula that will find h for any value of l, in addition to when l = 2π. As mentioned above, due to the cos-1 (R / (R + h)) there is no explicit formula to find the height. Nevertheless, using series, on can obtain very good approximations. Using series approximations one can derive the following formulas for height achieved by lengthening the rope by length l; the first is a 1st order approximation, and the second, a more accurate 2nd order approximation. The approximations are more accurate the smaller the added slack is compared to the original radius of the circle.

Image 4. The series approximations for a large value of R relative to l. Image created by Harrison Tasoff

Eq. 10         $h \left (\mathit{l} \right ) \approx \frac{1}{2} \left ( \frac{3}{2} \right )^{\left ( \frac{2}{3} \right )}R^{ \left ( \frac{1}{3} \right )} \mathit{l}^{ \left ( \frac{2}{3} \right )}$

Eq. 11         $h \left (\mathit{l} \right ) \approx \frac{1}{2} \left ( \frac{3}{2} \right )^{\left ( \frac{2}{3} \right )}R^{ \left ( \frac{1}{3} \right )} \mathit{l}^{ \left ( \frac{2}{3} \right )} + \cfrac {9 \left ( \frac{3}{2} \right )^{\left ( \frac{1}{3} \right )}}{80 R^{ \left (\frac{1}{3} \right )}}\ \mathit{l}^{\left ( \frac{4}{3} \right )}$

Where l is the added slack, h(l) is the height as a function of l, and R is the initial radius of the rope circle.

Taking l = 2π feet of slack and R = R earth = 20925524.9 feet from the initial problem the first approximation yields 614.766, demonstrating the accuracy of these approximations. The second approximation yields 614.771, which is even more more accurate than the first.

Image 4 illustrates these approximations, with l on the x-axis and h(l) on the y-axis. Eq. 10 appears in red and Eq. 11 in blue. In this example, the radius R of the ball is an arbitrary number that is very large relative to l. Notice that for small values of l, Eq. 10 and Eq. 11 are indistinguishable. It is only once l starts to become a decent percentage of R that Eq. 11 becomes noticeably more accurate than Eq. 10. When R= 5000, as it does in this graph, Eq. 11 becomes appreciably more accurate than Eq. 10 when l ≈ 150, or roughly 3% of R. At this point, Eq. 10 outputs 30827.651, while the more accurate Eq. 11 outputs 30833.653, a difference of 6.002.

Here is another way to illustrate the varying degrees of accuracy in these approximations. The most accurate value the root finder yields is 614.7709915007723. Comparing this to the value found by the 1st order approximation (614.7655785603816), we find that the latter is accurate to five significant digits when rounding is accounted for. The 2nd order approximation (614.7709968720138) is accurate to seven significant figures when rounding is accounted for. Hence, though in absolute terms, both approximations are quite accurate, the 2nd order approximation is 100 times more accurate relative to the 1st order approximation.

Also, as one can see the approximations climb quickly from l = 0. This steep ascent means that even small values of l output large values for h(l), or, even small amounts of extra slack yield large increases in maximum height.

On the scale of the Earth, there is a decent margin of error in measurements: the earth is not a perfect sphere, nor perfectly smooth, ropes stretch with strain, objects expand and contract with heat. The result is that these approximations are far more accurate and precise than our measurements can ever be.

# Why It's Interesting

At first this puzzle is interesting because of its non-intuitive results. It becomes even more interesting if we can then make them intuitive after all, and indeed and we can. Though it may seem that 2π, or roughly 6.28, feet is a minuscule amount of extra rope needed to to produce such a considerable result, a look at the ratios will show otherwise.

The radius of the Earth is roughly 20,920,000 feet, though this varies because the Earth is not a perfect sphere. There is 1 foot of difference between the radius of the circle made by the lengthened rope and the radius of the Earth. This foot of difference is a mere fraction of the radius of the Earth: about five one-hundred millionths, or .000000047, of the Earth's radius. A foot doesn't seem so large anymore.

Similarly, 2 π feet is .000000047 of the circumference of the Earth (which is about 131,000,000 feet). And, unsurprisingly, the ratio of 1 foot to the Earth's radius is the same as that of 2 π feet to the Earth's circumference.

So, in this perspective, a small change in the length of the rope yields a proportionally equivalent small change in the radius of the rope circle.

# References

• Pickover, C. A. (2009). The Math Book. New York: Sterling Publishing Co.
• (2009, March 3). Roping the Earth. Message posted to: