# Rope around the Earth

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

Rope around the Earth

This is a puzzle, about a rope tied taut around the equator. How much must it be lenthened so that, if made to levitate, there is a one foot gap at all points between the rope and the Earth? This is a puzzle in that the answer is surprising. 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

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 such that, if made to hover, it would be one foot off the ground at all points around the Earth?

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, this result is independent of the size of the ball around which the rope is wrapped.

# 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.

In the image to the right:

• Lrope 2 is the length of the extended rope.
• Cearth is the Circumference of the Earth and the original length of the rope (L1).
• Rrope is the radius of the circle made by the rope.
• Rearth is the radius of the Earth and the original radius of the rope.

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

$L_{rope 1}=C_{earth}=2\pi\,\!R_{earth}$ (now this formula is correct)

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. All this can now be simplified and the sub1-2 notation eliminated

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

Distributing the 2 π yields:

$L_{rope 2}=2\pi\,\!R_{earth}+2\pi\,\!=C_earth+2\pi\,\!$

The new length of the rope is merely 2 π feet longer than the original length. Indeed, one can see that the additional 2 π is a result of extending the radius of the rope circle by one foot, an extension that will by definition be the same no matter the initial radius of the object being enclosed.

### Maximum Height of Rope

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

In the diagram to the right:

• x1 is the distance from the horizon to 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.

Now we will develop a formula for l, the amount of slack in terms of h, which we hope later to invert in-order to solve for h in terms of L. First, we will do this for the case where l=2 π, and then in general. We start with h because it appears clearly in the diagram, where l is less visible.

Using , we know that: $(R+h)^2=R^2+x_1^2$, which is equivalent to

Eq. 1         $x_1=\sqrt{(R+h)^2-R^2}$.

To find the length of xo, we must remember what the length of an arc is:

$L_{arc}=r\theta\,\!$

Where θ is the angle, in radians, formed between two radii from the center of the circle to the endpoints of the arc. Thus:

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

Where $\cos^{-1} \left (\frac{R}{R+h} \right )$ represents the angle whose cosine is $\frac{R}{R+h}$. In the picture this angle is denoted as θ.

Since we know that we lengthened the rope by 2 π feet, we know that 2x1= 2xo + 2 π, because 2x1 is the extra slack put in to the rope. Thus: x1= xo + π. The general equation, where l is the amount of added slack, is given by:

Eq. 3         $x_1=x_o+ \frac {\mathit{l}}{2}$

Substituting Eq. 1 and Eq. 2 into Eq. 3 yields:

Eq. 4         $\sqrt{(R+h)^2-R^2}=R\,\cos^{-1} \left (\frac{R}{R+h} \right )+\frac {\mathit{l}}{2}$

Solving for l in Eq. 4 yields:

Eq. 5         $l}=2 \left (\sqrt{(R+h)^2-R^2}-R\,\cos^{-1} \left (\frac{R}{R+h$

Running this through a numerical calculator, with R = R earth = 20925524.9 feet, resulted in h = 614.771 ft as the height that a l = 2π foot extension would yield.

As a result of the cos-1 (R / (R + h)), there is no explicit formula to find the height. Nevertheless, very accurate results can be achieved using Series approximations. Smaurer1My mistake to call this Taylor; that refers only to series in positive integer powers of the variable. In this way, we find the following formulas for height achieved by lengthening the rope by length l; the first is a first order approximation, and the second, a slightly more accurate second order approximation. The approximations are more accurate the smaller the added slack is compared to the original radius of the circle.

$h \left (\mathit{l} \right )= \frac{1}{2} \left ( \frac{3}{2} \right )^{\left ( \frac{2}{3} \right )} \mathit{l}^{ \left ( \frac{2}{3} \right )} R^{ \left ( \frac{1}{3} \right )}$

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

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 and precision of these approximations.First, look up these two words; they don't mean the same thing and you only mean accuracy. Second, there are two meanings of accuracy, absolute and relative. I would say that the second order approximation is much much more accurate than the first order approximation. Perhaps we need a helper page on accuracy.The second approximation yields 614.771, which is more precise than the first, but really only by a negligible amount, and simply highlights the accuracy of the first. On the scale of the Earth, there is a decent margin of error in measurements: the earth is not a perfect sphere, 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

Though it may seem that this is 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. 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 4.7 x 10-8 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: