# Extending the length of a curcumference by 1 meter

I once heard the following statement:

Take a piece of string and measure the length around a ball. Now add 1 meter to the string and stretch it out evenly around the ball. Obviously the diameter/radius will increase. Not sure by how much though.

Do the same thing for a ball the size of the earth. Add 1 meter to the string and (amazingly, at least to me) the increased radius/diameter is the same for a ball the size of the earth as for a small ball held in your hand.

Is this true? If yes, why is this true and by how much is the radius extended?

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Yes, this is true. Are you familiar with the formula $C = 2 \pi r$ relating the circumference and radius of a circle? The radius increases by $\frac{1}{2\pi}$ in both cases. – Qiaochu Yuan Aug 22 '12 at 4:13
I am familiar with the formula, but while I understand how to get $\frac{1}{2\pi}$ if the initial radius is 0, I do not understand why increasing the size makes no difference. – Michael Frey Aug 22 '12 at 4:22
@MichaelFrey: If you're at a loss for insight, just use brute force; you want to know about the difference in radii, so compute that difference. Solve for $r$ in $C=2\pi r$, solve for $R$ in $C+1=2\pi R$, and then determine $R-r$. (Afterward, look back and see if you could've anticipated the result, or at least gotten to it with less effort ... and/or fewer fractions.) – Blue Aug 22 '12 at 4:35
I heard this one in the 1970s about a kid named Kermit who eats so much that he grows to be as big as the earth itself. He still has to keep his pants on, so he wears a belt. Then he eats enough cake that he gets one foot thicker through the middle. How much longer does his belt have to get? – MJD Aug 22 '12 at 5:25

How about a proof without words?

(Of course, one should imagine the limit as the number of segments tends to infinity.)

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$C = 2 \pi r, \Delta C = 2 \pi \Delta r.$ – Narasimham Jun 15 '15 at 14:19

I really doubt any formal mathematical argument will be more "intuitive" than invoking $C = 2\pi r$. So instead I will try to appeal to intuition with a geometrical argument.

Take a small square and add a meter to the perimeter in order to form a larger square. It should be a bit more intuitive that each side of the larger square will be longer by $\frac{1}{4}$. If we interpret the "radius" of the square as the distance from the center to the side, then our radius has increased by $\frac{1}{8}$ and this value is independent of the original size of the square.

Similarly, the picture holds for shapes like pentagons, hexagons and octagons (the "radius" increase becomes more complex as the number of sides increases but it should not be too difficult to convince yourself that this increase is independent of the original size). If you view the circle as the limiting shape of these regular polygons then it's only natural that the increase in radius is independent of the original radius.

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This makes absolute sense. For any regular polygon the increase in size will be $\frac{1}{2*number-of-sides}$ – Michael Frey Aug 22 '12 at 5:19
@MichaelFrey The idea should hold but the increase is more complex than $\frac{1}{2n}$. For example, for a hexagon the "radius" increase is actually $\frac{1}{12} + \frac{\cos(30^\circ)}{6}$. – EuYu Aug 22 '12 at 5:32

Instead of adding $1$ metre to the string, let's be more generous and add $2\pi$ metres. So if the old radius is $r$, the old circumference is $2\pi r$. The new circumference is $2\pi r+2\pi$, which is $2\pi(r+1)$. This is the circumference of a circle of radius $r+1$.

So adding $2\pi$ to the string increases the radius by $1$, independently of the initial radius $r$.

In a lost valley, the unit of distance is the rad, where $2\pi$ rads happens to be $1$ metre. By the above argument, if you add $2\pi$ rads to the circumference of a circle, you add $1$ rad to its radius. In metric units, if you add $1$ metre to the circumference, then you add $\frac{1}{2\pi}$ metres to its radius.

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Let's say the Earth radius is $R_1$ and the radius of the string (after adding $1$ meter) is $R_2$.

So, the Earth circumference is $C_1 = 2\pi R_1$ and the string's circumference is $C_2=2\pi R_2$. But we know that the string circumference ($C_2$) is 1 meter longer that the Earth circumference ($C_1$), so:

\begin{align*} &C_1 + 1 = C_2\,,\\ &2 \pi R_1 + 1 = 2\pi R_2\,,\\ &2\pi R_2 - 2\pi R_1 = 1\,,\\ &2\pi (R_2-R_1) = 1\,,\text{ and}\\ &R_2-R_1 = \frac{1}{2\pi} \end{align*}

We can observe that the difference $1/2\pi$ is a constant - it is always the same and does not depend on the radius values.

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Welcome to MSE! If you format your answers using MathJax, it makes them much easier to read and review. Regards – Amzoti Jan 20 '13 at 20:09

Let’s try something new.

Notations:- V = volume; S = surface area; L = length; P = perimeter; C = circumference; R = radius.

For two similar objects (one old and one new), we already have the following formulas relating ratio of volumes (also ratio of surface areas) and ratio of corresponding lengths:- $$\frac {V_{new}}{V_{old}} = (\frac {L_{new}}{L_{old}})^3$$; and $$\frac {S_{new}}{S_{old}} = (\frac {L_{new}}{L_{old}})^2$$

Following the same logic, we should have $$\frac {P_{new}}{P_{old}} = (\frac {L_{new}}{L_{new}})^1$$, a formula that is so simple that no textbooks even bother to mention it.

In this post, two spheres are similar and the formula applies.

Thus, from $\frac {R_{new}}{R_{old}} = \frac {C_{new}}{C_{old} }$, we have $\frac {R_{new}}{R} = \frac {2 \pi R + 1}{2 \pi R}$

Then, $R_{new} = … = R + \frac {1}{2 \pi}$

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