Wrapping The Earth

Recent changes

Table of contents

Links to this page

FRONT PAGE / INDEX

Subscribe!

My latest posts can be found here:

Colins Blog

Previous blog posts:

Additionally, some earlier writings:

The "Wrapping The Earth" Problem

Our friend once again has too much to drink, and once again indulges in a little late night internet shopping. This time, however, it's fueled by curiousity, having discovered that there is a material that is infinitely stretchy, but which always retains its area. So while you can stretch it one direction, it will keep its area by shrinking in another direction.

The surface area of a sphere of radius $R$ is $4\pi R^2$. But the circumference $C$ is $2\pi R$, so $R=C/(2\pi)$. That means the area is $4\pi(C/(2\pi))^2$ which is $\frac{1}{\pi}C^2$. I find that rather neat.
So the Earth's surface area is $40^2/\pi$ (in some units), which is (very roughly) $1600/3$, or about $533$. As per the calculations in the Rope Around The Earth Refined we should then reduce that by about 5%, which is around 26 or 27, giving a more accurate answer of 506 or 507 or somewhere thereabouts.
Or reach for a calculator ...
We know the Earth's circumference is 40 million metres, because that is the original definition of the metre, so our friend purchases some 510 million million square metres (see side box for where this number comes from) of this wonderful material, expecting that it'll be about the right amount to wrap up the Earth, and goes off to bed.

(We are again in puzzle world, so we're assuming the numbers are good enough for our purposes - let's pretend they are exact.)

It takes some time for the order to arrive - next day delivery wasn't available from this vendor - but when it does arrive our friend goes to it and heroically wraps the entire Earth, only to discover that there is slightly too much material. Just one square kilometre.

One million square metres.

Never mind, we can simply prop it up the same height everywhere, as we did with the rope.

How high must it be?

It's a simple sum ...

This one's not too hard, and in truth, not too surprising. We can do it by algebra, or by calculus, both working out quite simply. We'll do it via algebra, and let the calculus wait to the next section.

We know that the area of a sphere is given by the formula $A=4\pi R^2$. Letting $E$ represent our extra area, and $h$ being the extra height we need, we have:

$A+E = 4\pi (R+h)^2$

Expanding the right-hand side we have:

$A+E = 4\pi (R^2+2hR+h^2)$

We can ignore the $h^2$, because compared with the $2hR$ term it's tiny, then substitute the formula for $A$, cancel the $4\pi R^2$ term from each side, and when the dust settles we have:

$E=8\pi hR$, so $h=E/(8\pi R)$

Remembering that $C=2\pi R$, our result is:

$h=E/4C$

With $C$ being 40 million, and $E$ being 1 million square metres, our height is ... (drum roll please) ...

6.25 millimetres.

Yup, in old money it's about a quarter of an inch.

Reflections ...

There are a few interesting differences here between this and the original "Rope around the Earth" problem. For a start, this answer, unlike the original, does depend on the size of the Earth. But that is clear from the calculus approach. In the original we have:

This is the bit where we use the language of calculus to help our thinking. Although we don't do much, it turns out to be really quite powerful.

$\frac{dC}{dR}=2\pi$

That tells us the change in circumference as a function of the change in radius is constant, and doesn't depend on the size of the Earth.

In contrast, with this problem we are going with the area, and we have:

$\frac{dA}{dR}=8\pi R$

The change in area as a function of the radius is no longer constant. We can use that to derive the answer:

$\frac{dA}{dR}=8\pi R$;
- $\frac{dA}{dR}=4C$;
  - $\frac{dR}{dA}=1/(4C)$;
    - $h=E/(4C)$;

and we have the same result as above.

The Other Wrapping the Earth Problem ...

But this was just the warm up. Next time we'll take the step of not propping up the sheet everywhere, but just in one place with a single tent pole ...

How high will that have to be?

Next time.

Acknowledgements

I believe this problem was first suggested by the Rudi Mathematici, but I'm still chasing the exact source, and will update this post when I have more information:

Update: As far as we can tell, it did originate with the Rudi Mathematici as an off-hand remark one evening, but this is relying on memory, which always has the potential to be faulty. If anyone knows of an earlier reference, do let me know. This post has been significantly improved by several small changes made by a handful of people, none of whom wish to be credited. As always, my thanks to them, and, obviously, any remaining errors are mine.

<<<< Prev <<<<

The Ring Of Steel

>>>> Next >>>>

The Other Wrapping The Earth Problem ...