# Sort of a challenge

#### topsquark

##### Well-known member
MHB Math Helper
This isn't so much a challenge problem as much as it has a startling (at least I think so) answer.

Say we tie a rope around the Earth. Now we are going to cut it and add another 6 meters to it. If we pull the new rope tight (in a circle) how high is the new rope above the Earth's surface?

It's reasonably easy so if you've seen it before let someone else answer. (I first saw this problem in a "Dr. Crypton" article in "Science Digest.")

-Dan

#### Bacterius

##### Well-known member
MHB Math Helper
Well the Earth's radius is $6371 ~ \text{km}$, so the original rope has a length of $2 \pi \times 6371 ~ \text{km}$. If we add $6 ~ \text{m} = 6 \times 10^{-3} ~ \text{km}$ to this rope, its new radius is:
$$\frac{2 \pi \times 6371 + 6 \times 10^{-3}}{2 \pi} \approx 6371.0009 ~ \text{km}$$
So the rope now "floats" about $0.0009 ~ \text{km} = 90 ~ \text{cm}$ above the ground. To be exact, $95.5 ~ \text{cm}$ (this is $\frac{6}{2 \pi} ~ \text{m}$).

Wait, what? My mind is blown

#### MarkFL

Staff member
This isn't so much a challenge problem as much as it has a startling (at least I think so) answer.

Say we tie a rope around the Earth. Now we are going to cut it and add another 6 meters to it. If we pull the new rope tight (in a circle) how high is the new rope above the Earth's surface?

It's reasonably easy so if you've seen it before let someone else answer. (I first saw this problem in a "Dr. Crypton" article in "Science Digest.")

-Dan
I recall a version of this problem circulating several years ago among the performance car forums on which I used to be quite active, and there was much disbelief and dissension among the masses.

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
Well the Earth's radius is $6371 ~ \text{km}$
I take issue with this solution, which assumes that the Earth's radius is 6,371 km. This is just the mean radius. The equatorial radius is quite a bit larger at 6,378.1 km, while the polar radius is quite a bit smaller at 6,356.8 km. Clearly, the answer to the problem depends on the radius... hmm... never mind.

Here is another question: What happens if, after adding 6 m, we lift the rope by one point so that it is taut and lies flat on the surface for most of the circumference, as in the following picture?

How high above the Earth is the point by which the rope is lifted?

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#### Opalg

##### MHB Oldtimer
Staff member
I take issue with this solution, which assumes that the Earth's radius is 6,371 km. This is just the mean radius. The equatorial radius is quite larger at 6,378.1 km, while the polar radius is quite smaller at 6,356.8 km. Clearly, the answer to the problem depends on the radius... hmm... never mind.

Here is another question: What happens if, after adding 6 m, we lift the rope by one point so that it is taut and lies flat on the surface for most of the circumference, as in the following picture?

How high above the Earth is the point by which the rope is lifted?
Nice problem!

If the circle has radius $R$ and the length of the rope is increased from $2\pi R$ to $2\pi R + \delta$ then, with $\theta$ as in the diagram, $2\pi R + \delta = 2(\pi - \theta)R + 2R\tan\theta$. Thus $\tan\theta - \theta = \frac{\delta}{2R}.$ This can't be solved exactly for $\theta$, but assuming that $\delta \ll R$ we can use the approximation $\tan\theta \approx \theta + \frac13\theta^3$ (first two terms of the power series for $\tan\theta$) to get $\dfrac\delta{2R} \approx \dfrac{\theta^3}3$, so that $\theta \approx \sqrt[3]{\dfrac{3\delta}{2R}}$.

The distance $h$ of the high point of the rope from the circumference of the circle is $h = R(\sec\theta - 1)$. Again using the power series approximation, this time for $\sec\theta \approx 1 + \frac12\theta^2$, we have $$h \approx \frac R2\Bigl(\frac{3\delta}{2R}\Bigr)^{2/3} = \sqrt[3]{\frac{9\delta^2R}{32}}.$$

Since this is just a rough approximation, I'll take the radius of the Earth to be $6\times 10^6$m. If $\delta = 6$m then that formula gives $h\approx 390$m.​

#### topsquark

##### Well-known member
MHB Math Helper
I take issue with this solution, which assumes that the Earth's radius is 6,371 km. This is just the mean radius. The equatorial radius is quite larger at 6,378.1 km, while the polar radius is quite smaller at 6,356.8 km. Clearly, the answer to the problem depends on the radius... hmm... never mind.

Here is another question: What happens if, after adding 6 m, we lift the rope by one point so that it is taut and lies flat on the surface for most of the circumference, as in the following picture?

How high above the Earth is the point by which the rope is lifted?
Even more mind blowing!

-Dan

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
This problem and picture are taken from the MathForum.