This puzzle, also known as 'how long is a piece of string?', is based on a formula which is taught in High School; however the answer may come as a surprise to some.

**Some Information**

Assume that you have a perfectly spherical basketball, which measures exactly 75 centimetres around the circumference^{1}. If you were to magically suspend the ball in mid-air, then take a piece of string exactly one metre (100cm) long, and use it to make a perfect circle around the ball, with the centre of the circle also being the centre of the basketball, (so that the piece of string was at the same distance from the surface of the basketball all the way around), how high from the surface of the basketball would the string be?

The answer is that the string would be elevated 3.98cm above the circumference of the ball.

**The Puzzle**

Now, let us assume the Earth to be a perfect sphere (which it's not) with a radius of 6378.1 kilometres (which, at the equator, it is). Imagine that you could place a piece of string around the Earth, at the equator^{2}.

This piece of string would be 40074.78km long.

If you were now to extend the length of this piece of string by 25 centimetres (as we did with the basketball), and position it so that it is at the same distance from the surface of the Earth all the way around (as we did with the basketball), how far away would the string be from the surface of the Earth?^{3}

The answer may surprise some people and, so as not to spoil this surprise, is given in this footnote^{4}.

**Why is it so?**

The height above the surface of the sphere is only dependent on the length of the 'extension'; it bears no relationship to the size of the original sphere (or the length of the original piece of string, which is the same thing).

**Equations**

The only equation you need to validate the above is the one for the circumference of a circle, which is that C (the circumference), is equal to 2, times π, times r (the radius of the circle).

C = 2πr

After rearranging this equation, you can calculate that the radius of a circle is equal to the circumference divided by 2π.

r = c/2π

So we have two equations:

- To determine r1 (the radius of the original sphere); based on C1 (the circumference of a 'great circle' of the original sphere).
- To determine r2 (the radius of the new circle); based on C2 (the circumference of the new circle).

The height of the string above the ground is simply r2-r1.

**Determine r1 if C1 = 75**

If C1 = 75

Then, since r1 = C1/2π

r1 = 75/6.28^{5}

So r1 = 11.94268

**Determine r2 if C2 = C1 + 25**

If C2 = C1 +25,

Then r2 = (C1+25)/2π

So r2 = 100/2π

And so r2 = 15.92357

**Determine the Height of the String Above the Surface of the Basketball**

So the difference in the radii of the two circles (or the height above the surface) is:

r2-r1

which is 15.92357-11.94268 = 3.98

**Show That the Height of the String Above the Surface is not Dependant on the Value of C1**

If you leave the equation for r2 as r2 = (C1+25)/2π but, instead of doing the addition, rewrite this as:

so r2 = (C1/2π) + (25/2π)

Then express the difference between the radii as follows:

r2-r1 = (C1/2π) + (25/2π) - (C1/2π)

so (taking C1/2π away from itself to get 0, and leaving 25/2π)

r2-r1 = 25/2π

You still get 3.98 as the height above the surface, regardless of the value of C1, proving that the difference between the radii of the two circles is a function of the length of the extension (therefore height = extension/2π).

To make the calculations less messy, consider adding 6.28 metres to the length of the string, rather than 25 centimetres: no prizes if you worked out in your head that the string would be elevated one metre above the sphere (the Earth, a basketball, Neptune, a pea, or whatever).