Some slack (Posted on 2018-10-29)

	Submitted by Steven Lord
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Solution: About 121.6 m We use R as Earth’s radius and d as the extra length of string, which rises to a height h above ground. The angle at Earth's center to the two points on the string which just contact the ground is 2x radians. Under the raised portion, we approximate Earth as flat and of length 2xR. The string makes and isosceles triangle with the ground with sides “a” and base 2xR. The raised length is greater than the circumference under it: 2a = 2xR + d tan x = a/R = x+d/2R Using the line from Earth’s center to the top of the string, a line of length R+h, sec x = (R+h)/R h = R sec x - R There are different ways to solve this for h, e.g.: Use some small angle approximations for x: tan x = x+ x^3/3 + … sec x = 1+ x^2/2 + … with tan x = x+ d/2R ~ x +x^3/3 x^3 ~ 3d/R Using the sec approximation, h ~ (R x^2/2) This gives h ~ k R^(1/3) d^(2/3) with k = (3/2)^(2/3)/2 ~ 0.655 With R = 6400km and d = 1m, h~121.6 m Doesn't this seem counterintuitively large? Doing the solution with approximations you see it goes with radius^(1/3), which is different from a flat planet (with the string being long, say a) where the height go as a^1/2 The result for a string 2a units long with 1 unit added, tacked to flat ground is found by considering the length a+1/2 (hypotenuse) above base a and height h here h= sqrt [ (a+1/2)^2 - a^2 ]. As "2a" gets large "h" asymptotically approaches sqrt(a). It is also interesting that you can indirectly measure the circumference of a planet this way (or a person for that matter) by letting out some slack.

	Subject	Author	Date
	Solution	Kenny M	2018-10-29 21:22:39
	my solution	Charlie	2018-10-29 10:26:14