- 08 Oct '09 14:27Imagine a belt big enough to circle the entire Earth with a little slack left over. This slack can be taken up by inserting a 100 m tall stake into the Earth and letting the belt rest on top of it as it circles the Earth.

Q: If the belt were returned to its original circular shape, what would be its radius?

(Assume the Earth is a sphere with a radius of 6371 km, that the belt traces a great circle on the Earth, and that the belt does not stretch.) - 08 Oct '09 17:07 / 1 edit

My highly unchecked answer is:*Originally posted by PBE6***Never you mind about that.**

(R = radius of Earth = 6173000m, r = 100m)

1 + R/pi{sqrt[(1 + r/R)^2 - 1] - arccos(R/R+r)}

Which works out as an increase in radius of about 12cm compared to the radius of the Earth.

Plausible? Maybe. I'll try and check it later. - 08 Oct '09 17:16

That's what I got too.*Originally posted by mtthw***My highly unchecked answer is:**

(R = radius of Earth = 6173000m, r = 100m)

1 + R/pi{sqrt[(1 + r/R)^2 - 1] - arccos(R/R+r)}

Which works out as an increase in radius of about 12cm compared to the radius of the Earth.

Plausible? Maybe. I'll try and check it later. - 10 Oct '09 09:43

Inserting 100m into the belt would do that, but that's not the scenario being described. You need to work out what the length of the belt needs to be to go over the top of the mast.*Originally posted by Mephisto2***doesn't an insertion of 100m increase the radius by 100m / (2*pi) = approx. 16 m?** - 20 Oct '09 06:09

I'm you could have done this as a ratio too.*Originally posted by mtthw***My highly unchecked answer is:**

(R = radius of Earth = 6173000m, r = 100m)

1 + R/pi{sqrt[(1 + r/R)^2 - 1] - arccos(R/R+r)}

Which works out as an increase in radius of about 12cm compared to the radius of the Earth.

Plausible? Maybe. I'll try and check it later.