- 06 Jul '07 06:58OK this is an oldie but a goldie and amazing if you have not heard it before.

Imagine that the Earth is a perfect sphere and that we Earthlings (in a demonstration of 'brotherly love'/'Peace on Earth' etc) decide to put a nice piece of white ribbon around the equator.

But whoops - in some mix-up with the numbers the ribbon is a metre too long! Now if this over-long ribbon were formed into a perfect circle and magicly suspended above the Earth ... roughly how far above the Earth would it hover? - 07 Jul '07 00:41 / 2 edits

Let’s say the circumference of the earth (assuming a perfect sphere) in meters is "C" therefore using the following:*Originally posted by wolfgang59***OK this is an oldie but a goldie and amazing if you have not heard it before.**

Imagine that the Earth is a perfect sphere and that we Earthlings (in a demonstration of 'brotherly love'/'Peace on Earth' etc) decide to put a nice piece of white ribbon around the equator.

But whoops - in some mix-up with the numbers the ribbon is a metre too long! Now i ...[text shortened]... le and magicly suspended above the Earth ... roughly how far above the Earth would it hover?

C= PI x r

If the ribbon is 1 meter longer the radius will therefore be

(C+1 / PI ) - r

There is your answer. You just need to substitute the most accurate value of “C” the earth’s circumference in meters to get your answer. - 07 Jul '07 02:08 / 2 edits

It's counter intuitive but that is actually incorrect. The proof, using this example and using 6380000m for the radius of the earth is this:*Originally posted by AThousandYoung***No. The larger the object, the more distance that extra 1m will be diluted over. Therefore the larger the object the lower the ribbon will be to the surface.**

6380000*2*pi + 1 = (6380000+X)*2*pi --- the LHS is the circumference of the earth plus 1 metre. The RHS is the same measurement but calculated using the new radius (6380000+X) where X is the distance above the earth the ribbon would hover.

The rest of the proof is just rearranging the above formula:

(6380000*2*pi + 1)/(2*pi) - 6380000 = X

(63800000*2*pi)/(2*pi) + 1/(2*pi) - 6380000 = X

6380000 - 6380000 + 1/(2*pi) = X --- this shows the original radius doesn't affect the distance the ribbon would be above the surface.

1/(2*pi) = X - 07 Jul '07 14:05I knew this answer from the beginning, but I didn't want to interfere...

The radius of the Earth ( or moon or nucleus of an atom or the galaxy or whatever) doesn't matter. The answer is a constant. Yes, it is counterintuitive but nevertheless a fact.

Now another problem with the same kind of properties:

A racing track (of some kind) has (say) 5 m in width and the race goes clockwise all the way around. How much longer is the left edge of the track compared to the right edge?

( "But Fabian, is this not dependant of how the track is planned? If it’s circular or if it has a lot of left and right bends?" ) - 08 Jul '07 03:25

2*pi*(r+5) - (2*pi*r)*Originally posted by FabianFnas***I knew this answer from the beginning, but I didn't want to interfere...**

The radius of the Earth ( or moon or nucleus of an atom or the galaxy or whatever) doesn't matter. The answer is a constant. Yes, it is counterintuitive but nevertheless a fact.

Now another problem with the same kind of properties:

A racing track (of some kind) has (say) 5 m i ...[text shortened]... of how the track is planned? If it’s circular or if it has a lot of left and right bends?" )

2*pi*(r+5-r)

10*pi