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topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,135
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
Jan 26, 2012
644
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 :rolleyes:
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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. (Rofl)
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,493
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?
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,715
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
Aug 30, 2012
1,135
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! (Shake)

-Dan
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,493
This problem and picture are taken from the MathForum.