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.
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:
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