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# Pi postcard solution: How much space does a mouse need?

Updated Monday, 25th July 2011

If you needed to lengthen some string tied round the earth for a mouse to sneak through, how much more string would you need?

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Read on to discover the solution to 'the mouse problem', which is stated on our pi postcard from The Code.

### The mouse problem

Imagine a huge circular loop of string around the Earth’s equator (assume the Earth is a sphere). You wish to lengthen the string so that it still encircles the Earth, but with a 2cm gap between the string and the Earth all the way round: just enough for a mouse to squeeze underneath. How much longer does the string need to be?

### Solution to the mouse problem

The radius of the Earth is about 637 million centimetres, but amazingly we do not need to know this to solve the mouse problem, so let’s just call the radius R. That means the circumference is 2×π×R. We usually omit the multiplication symbol × and just write 2πR. In the diagram below, the radius is labelled R and the circumference is labelled 2πR. The circumference shown is not the equator, but this doesn’t matter, because the Earth is (approximately) a sphere.

Now let us extend the length of the string so that there is a 2cm gap between the string and the Earth.

The radius of the extended loop of string is R+2, so the circumference of the extended loop of string is 2π(R+2). We can now work out how much more string is needed to create this 2cm gap. Since 2π (R+2) - 2πR = 2πR + 4π - 2πR = 4π  we see that only 4π centimetres, or about 12.5cm, of extra string is needed to create the 2cm gap all the way round!