## Roping the Earth

### 2009/03/03

I have utterly enjoyed the very stimulating online physics book by Christoph Schiller Motion Mountain. It includes some 1700 challenges for the reader ranging from fun little puzzles to standing research problems. This is a great example of how textbooks should be written to engage the student.

Here is Challenge 70 from the book.

A rope is put around the Earth, on the Equator, as tightly as possible. The rope is then lengthened by 1 m. Can a mouse slip under the rope?

I have seen this problem previously and in both cases was dissatisfied with how understated the solution was. Schiller’s answer is:

Yes, it can. In fact, many cats can slip through as well.

The author’s reasoning must be as follows. Since the circumference of a circle is $P = 2\pi r$, then adding one meter to P adds 16 cm to the radius. Therefore, a cat could crawl under the rope. The result may be counterintuitive because of the huge proportions involved: 1 meter seems so insignificant to Earth’s circumference that it shouldn’t change anything. Yet mathematics must trump intuition’s whims.

However, this calculation assumes that the rope retains its circular shape and remains centered on Earth’s center (as in Figure a below). If one pulls the rope up as shown in Fig. b, the 1-meter slack will allow 121 meters of clearing under the rope. That’s enough to slip the Titanic under it, let alone `many cats.’

A rope around the earth

Here is the calculation as shown in the figure below. Imagine a tower h = 121.37 m high. Then the distance from the top of the tower to the horizon is $L_1 = \sqrt{(R+h)^2-R^2}=39,282.48 \text{ m}$, where $R = 6,357,000 \text{ m}$ is the Earth radius. Meanwhile, the ground distance from the bottom of the tower to the same point on the horizon is $L_0 = R\cos^{-1}\frac R {R+h} = 39,281.98 \text{ m}$, a difference of only 0.5 m. Therefore, the straight path from the horizon to the top of the tower and again to the horizon on the opposite side is 1 m longer than the ground path between the two points on the horizon $2L_1-2L_0=1 \text{ m}$.

Calculation of the height of the object that could be slipped under the rope lengthened by 1 meter.

Amazingly, adding just 1 mm of slack to the rope will make a clearing high enough for a child to walk under the rope: 1.25 m.