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Tue 7 Dec, 2004 08:26 am
Assume that the earth is smooth. Say a rope just went all the way around the equator of the earth (i.e. the rope just fit the earth). Now I increase the length of the rope by a mere 7 feet. Clearly the rope is now longer and so there is some room between the earth and the rope. How much is the room?
What is the room between the rope and a ping pong ball if the same procedure is performed? First get the rope snug against the ping pong ball and then add 7 feet to the rope.
Surely you'd have to assume the equator was a perfect circle as well??
If that was the case then the circumference of a circle is 2xPIxradius
You increase this by 7ft, thus the equation is as follows with r = earth radius, P = Pi and R = new radius:
2.Pi.r + 7 = 2.Pi.R
Rearrange to get:
R = (2.Pi.r + 7)/(2.Pi)
Hence the increased distance (D) is R-r or
D= [(2.Pi.r + 7)/(2.Pi)] - r
Someone else can add the exact measurements in!!
R=C/(2*PI)
R2-R1 = (C2-C1)/(2*PI) = 7/(2*PI) which is a little over a foot.
Notice that all that matters is the added length of the rope. The answer is the same for the earth and the ping pong ball.
Well approximately anyway!!!
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