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[Math] A rope around the equator...

Imagine

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A rope is tied around the equator so that it is in contact along the whole equator. (Assume that the earth is a perfectly spherical shape). You then cut the rope at some point, and add one extra meter of rope, stretching it again so that it is a perfect circle.

Could you fit a small dog through the earth and the string? (Without deforming/pulling the string?).

If so, how far off the ground would the string be?
 
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I was about to say hell no... Then I did some more calculations and yes... Yes you could fit a small dog. Now that I think about what pi means... It makes perfect sense. Oh and how far off the ground depends on how you align the new circle with the earth.
 
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I was about to say hell no... Then I did some more calculations and yes... Yes you could fit a small dog. Now that I think about what pi means... It makes perfect sense. Oh and how far off the ground depends on how you align the new circle with the earth.

Hehe, I love these types of problems. And when I mean lift it up uniformly, I mean that you have to center it at the same place the earth is centered. (concentric circles). Or I could just ask for the difference in the radii :)

(Don't post it yet, but I'm guessing you found the exact height that it's lifted off the earth?)
 
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Hehe, I love these types of problems. And when I mean lift it up uniformly, I mean that you have to center it at the same place the earth is centered. (concentric circles). Or I could just ask for the difference in the radii :)

(Don't post it yet, but I'm guessing you found the exact height that it's lifted off the earth?)

It would just be (1/pi)/2 right? (Now I'm just making shit up).
 
How small is a 'small dog'?
 
Didn't understand the question until I read the comments, strange concept. Pretty sure you could fit a dog in though after some rough calcs
 
Code:
C = 2pi*r
C + 1 = 2pi(r+n)
2pi*r + 1 = 2pi*r + 2pi*n
1 = 2pi*n
n = 1/(2pi)

yay

so for every x meters added to the rope's length, its radius will grow by x/(2pi) meters.
 
Well, then I figured it out. I got a gap of 1/(2*2*pi), and then wondered if the 7-or-so centimeters is enough to fit a dog under.

Here's something that was in my maths exam a few moons back:
The lengths of the catheti of a right triangle are 5 and 12. What is the area of the largest possible rectangle drawn inside the triangle, when one of the rectangle's sides is on the triangle's hypotenuse?
 
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