# Difference between revisions of "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 lengthened 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.

Just as bizarre is that, if one point on the extended rope were to be lifted up, the maximum clearance under it is quite large. For this specific case it would be 614.771 feet, enough room to fit two Statues of Liberty under it, base and all. This result is, however, 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.

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 (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.

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

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.

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

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 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:

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

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

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 Formula A. Also, since θ is the angle whose cosine is $\tfrac{R}{R+h}$, we can replace θ with: cos-1$\ \left (\tfrac{R}{R+h} \right )$.

Thus, Formula A is equivalent to:

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

Keeping in mind that cos-1$\ \left (\tfrac{R}{R+h} \right )$ represents θ, the angle whose cosine is angle whose cosine is $\tfrac{R}{R+h}$. For more information on cos-1 see Inverse Trig Functions.

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

Due to the presence of h within the cos-1, we can't isolate it and solve for it explicitly. Nevertheless, running Eq. 1, Eq. 2, Eq. 3, Eq. 4, and Eq. 5 through a root finder 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.

As mentioned above, due to 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. 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 more precise second order approximation. The approximations are more accurate the smaller the added slack is compared to the original radius of the circle.

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

Eq. 6         $h \left (\mathit{l} \right )= \frac{1}{2} \left ( \frac{3}{2} \right )^{\left ( \frac{2}{3} \right )}R^{ \left ( \frac{1}{3} \right )} \mathit{l}^{ \left ( \frac{2}{3} \right )}$

Eq. 7         $h \left (\mathit{l} \right )= \frac{1}{2} \left ( \frac{3}{2} \right )^{\left ( \frac{2}{3} \right )}R^{ \left ( \frac{1}{3} \right )} \mathit{l}^{ \left ( \frac{2}{3} \right )} + \cfrac {9 \left ( \frac{3}{2} \right )^{\left ( \frac{1}{3} \right )}}{80 R^{ \left (\frac{1}{3} \right )}}\ \mathit{l}^{\left ( \frac{4}{3} \right )}$

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 and precision of these approximations. 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.

The Graph to the left illustrates these approximations, with l on the x-axis and h(l) on the y-axis. Eq. 6 appears in red and Eq. 7 in blue. In this example, the radius R of the ball is an arbitrary number that is (I stress) very large relative to l. Notice that for small values of l, Eq. 6 and Eq. 7 are indistinguishable. It is only once l starts to become a decent percentage of R that Eq. 7 becomes markedly more accurate than Eq. 6. When R= 5000, as it does in this graph, Eq. 7 becomes noticeably more accurate than Eq. 6 when l ≈ 150, or roughly 3% of R. At this point, Eq. 6 outputs 30827.651, while the more accurate Eq. 7 outputs 30833.653, a difference of 6.002. 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, 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.

# Teaching Materials

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# References

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