Monday, May 07, 2007

Rope around the planet

Imagine a rope that fits snugly all around a planet like a ring on a person's finger.
Now imagine the rope is made one metre longer and lifted off the surface until it is taut.
What will its height be above the surface of the planet?
(For the purposes of the puzzle, assume that the planet is a perfect sphere).

4 Comments:

Blogger Jonathan said...

Well, for the degenerate case of a 0-size point-planet, the rope will be only 1 meter long, doubled back on itself, which will make the top 1/2 meter away from the planet.

So it's either always 1/2 meter or it's a function of the planet's radius. I don't actually feel like doing real math tonight, though.

May 8, 2007 at 8:57 PM  
Blogger steven said...

I disagree - a circle of 1 metre circumference (which is what you seem to be suggesting) does not have a radius of 1/2 metre.

Unless I'm misunderstanding?

May 12, 2007 at 6:06 AM  
Blogger steven said...

Let c be the circumference of the planet.
Let r be the radius of the planet.
Let h be the height of the rope above the planet's surface.

c = 2 * pi * r, obviously

Because we know the length of the rope is the circumference of the planet plus 1 metre, and the new 'radius' of the rope around the planet is the radius of the planet plus the height of the rope :-

c + 1 = 2 * pi * (r + h)

Substituting the first equation in :-

(2 * pi * r) + 1 = 2 * pi * (r + h)

Therefore,
(2 * pi * r) + 1 = (2 * pi * r) + (2 * pi * h)

Subtracting 2 * pi * r from both sides,
1 = 2 * pi * h

h = 1 / (2 * pi)

h is approximately 0.159.

What may be surprising is that this result does not depend on the radius of the planet!

May 15, 2007 at 12:29 PM  
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