# A String Tied Around The Earth

Say you're standing on the equator and you have a string below you tied around the equator (40,075 km) that is the length of the equator + 1 meter (40,075.001 km). What is the maximum height you can you lift the string off the ground? Can you create a function of both circumference of the circle (earth) and string to output the distance between the two if pulled tight?

Assumptions:

• For illustration, the result would be pulled from a single point, making a triangle until it met with the earth, in which it would follow the curvature of the earth. Similar to a snow-cone or O>
• The string does not stretch
• The earth can be assumed to be a perfect sphere
• How tight is the string? – user328032 Apr 18 '16 at 18:50
• Do you know the formula of the circumference of a circle ? – callculus Apr 18 '16 at 18:55
• We get a transcendental equation. One can get an excellent approximation to the solution by using a power series expansion. – André Nicolas Apr 18 '16 at 19:05
• I'm far from a math major, and it took me about a day to solve, but you're correct in that the best I could do was a very good approximation. – Travis Apr 18 '16 at 19:09
• – lhf Apr 19 '16 at 0:19

## 4 Answers

We have to solve (with respect to $d$) the trascendental equation: $$2\sqrt{(R+d)^2-R^2}+2R\arcsin\frac{R}{R+d}=\pi R+1$$ or solve (with respect to $x=\frac{d}{R}$) the trascendental equation: $$2\sqrt{x^2+2x}+2\arcsin\frac{1}{1+x}=\pi+\frac{1}{R}$$ that is equivalent to: $$\sqrt{x^2+2x}-\arctan\sqrt{x^2+2x}=\frac{1}{2R}.$$ In order to solve $z-\arctan z=\frac{1}{2R}$ we may apply Newton's method, then $x=-1+\sqrt{z^2+1}$.

Since $z$ is close to zero, $z-\arctan z\approx\frac{z^3}{3}$, hence: $$d\approx \sqrt{\frac{9R}{32}}\approx \color{red}{121.5\,m}.$$

Draw a picture. Let $P$ be the peak of the stretched string, and let $C$ be the centre of the Earth. Let $G$ be the point on the Earth's surface which is on the line $PC$, and let $T$ and $T'$ be the points of tangency of the string with the Equator after the string has been pushed upwards to full height $PG$.

Let $r$ be the radius of the Earth, and let $\epsilon r$ be the excess amount of string in addition to $2\pi r$. Note that $\epsilon$, in your example, is very small.

Let $\theta=\angle TCG$ Note that $\theta$ is small. The key equation is $$\epsilon r=2r\tan\theta -2r\theta.$$ This holds because $2r\tan\theta$ is the sum of the lengths of the two tangent segments $PT$ and $PT'$, while $2 r\theta$ is the amount of string saved because it no longer covers the minor arc $TGT'$. The difference is equal to the amount $\epsilon r$ of extra string we have available.

Using the fact that $\tan\theta\approx \theta +\frac{\theta^3}{3}$, we get $$\theta\approx \left(\frac{3\epsilon}{2}\right)^{1/3}.$$

Now that we know $\theta$, we can find the height $PG$ of the peak of the string. From the diagram, we can see that $PG=r(\sec\theta-1)$. Since $\theta$ is small, this is well approximated by $\frac{\theta^2}{2}$, and we end up with $$PG\approx \frac{r}{2}\left(\frac{3\epsilon}{2}\right)^{2/3}.$$

• We end up with $$PG\approx \frac{r}{2}\left(\frac{3\delta}{2r}\right)^{2/3}?$$ – Narasimham Apr 26 '16 at 21:15
• @Narasimham: Note that I defined $\delta r$ to be the excess amount of string. If you let $\Delta$ be the excess amount, then $\Delta=\delta r$, and we get your expression, with $\frac{3\Delta}{2r}$ instead of your $\frac{3\delta}{2r}$. The reason I defined $\delta$ the way I did was "dimensional." I wanted to end up with an expression that was a constant times $r$. I will check later whether there is a mistake, but dimensionally the formula I gave has the right shape. – André Nicolas Apr 26 '16 at 21:22
• Right, my bad. I would use a symbol like $\epsilon$ Shall delete my remark. – Narasimham Apr 27 '16 at 3:05
• @Narasimham: I had hesitated between using $\delta$ and using $\epsilon$! Will take that as a suggestion and make the change. – André Nicolas Apr 27 '16 at 3:14

EDIT 1:

Let $\Delta C$ be the extra length of loose string.

$$r = \frac{C}{2 \pi} ;\ \Delta r = \frac{\Delta C}{2 \pi} = h= \frac{1000\, mm}{2 \pi} = 159. 155 \; {mm \,!}$$

The relation/ratio is the same on Mars or Saturn or any earth based large round cylinder.

In this problem you want $h$ for given input $(R, \Delta C)$

We calculate tangent length $T$ shallow horizon range using right triangle property :

$$T^2 = h ( 2 R -h ) \tag{1}$$

$$T = R \tan \theta \tag{2}$$

And eliminate $T$ to solve for $\tan \theta$

$$\tan \theta= \sqrt{ (h/a) (2-h/a) } \tag{3}$$

From the diagram the slack above involute is indicated by red line,

$$= IE =\tan \theta - \theta = \Delta C/ (2 h)$$

or,

$$\sqrt { (h/a) (2-h/a) } -\tan^{-1}{\sqrt{ (h/a) (2-h/a) } } = \Delta C/(2 R) \tag {4}$$

Plugging in numerical values to solve for $h$

$$R = 40075 * 10^3/( 2 \pi) \, meters ; \, \Delta C =1 m \rightarrow h = 121.508 m \tag{5}$$

$\because h << R$ an approximation using Madhava-Gregory-Leibnitz series

$$\tan u - \tan ^{-1}u \approx u^3/3$$

gives us $$h^3 =\frac {9}{32}\cdot {\Delta C}^2 \cdot R \tag {6}$$

giving only a bit different value

$h= 121.505 m$. Half a meter. You take the slack extra meter and lift it half a meter high.