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 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?
Originally posted by wolfgang59Let’s say the circumference of the earth (assuming a perfect sphere) in meters is "C" therefore using the following:
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.
Originally posted by AThousandYoungIt's counter intuitive but that is actually incorrect. The proof, using this example and using 6380000m for the radius of the earth is this:
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
I love how everyone keeps making wrong guesses after the right answer has been given ( 1/(2*pi) metres or 15.9cm)
For what it's worth though when I first thought about it I also thought it would be a really tiny distance above the surface, I was surprised by how far it actually was.
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 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?" )
Originally posted by FabianFnas2*pi*(r+5) - (2*pi*r)
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