a) assume there is a rope that goes around the equator.
if you add 1 meter to its length (and it could defy gravity), how far off of the earth would it float (in millimeters or centimeters please)?
2006-11-17 05:30:31 · 2 answers · asked by tn5421 3 in Science & Mathematics ➔ Mathematics
please show the process you used to solve this problem, as that is what i need to learn.
2006-11-17 05:34:58 · update #1
The Websense category "Non-Traditional Religions and Occult and Folklore" is filtered. from the website you listed. is there another?
2006-11-17 05:51:06 · update #2
Let L be the initial rope's length and R be the radius of the equator in metres. Then
L = 2piR
Adding 1 to the length, we get:
L+1 = 2pi(R+x)
where x is the extra radius, i.e. the height at which the rope would float. Combining these two equalities we get
2piR + 1 = 2pi(R+x)
2piR + 1 = 2piR + 2pi*x
1 = 2pi*x
x = 1/(2pi) = 0.159 or about 16 centimetres.
The cool part about this is, it's independent of the earth's radius! The result is the same whether you start with the earth, a basketball or a marble.
2006-11-17 06:21:50 · answer #1 · answered by Anonymous · 3⤊ 0⤋
less than a millimeter.
Consider two circles, one is the earth with a circumference of 40,076,000 meters. The other is the rope which has a circumference of 40,076,001 meters.
Calculate the radius of each of these circles. Subtract the smaller radius from the larger radius. This is your answer in meters. Convert this figure to milimeters.
2006-11-17 05:33:18 · answer #2 · answered by Plasmapuppy 7 · 1⤊ 2⤋