Expert: Scott A Wilson - 6/23/2010

Question
Many people know that the weight of an object varies on different planets, but did you know that the weight of an object on Earth also varies according to the elevation of the object?  In particular, the weight of an object follows the equation: w = Cr^-2, where C is a constant, and r is the distance that the object is from the center of Earth.
a.  Solve the equation w = Cr^-2 for r.
b.  Suppose that an object is 100 lbs when it is at sea level.  Find the value of C that makes the equation true. (Sea level is 3,963 miles from the center of the Earth.)
c.  Use the value of C you found in the previous question to determine how much the object would weigh in:
    i.  Death valley (282 ft below sea level).
    ii.  The top of Mount McKinley (20,320 ft above sea level).

Answer
a. Solving w = C/r² for r gives r² = C/w, so r = √(C/w).

b. If w=100 and r=3963, then the equation to solve for C would be 100 = C/3963².
To get C, multiply both sides by 3963².  That is, 3963² = 15,705,369.
Note that both have the digits 3, 6, and 9 in the last 3 digits.

ci. Now 3963 miles is (3963 miles)(5280 feet/1 mile) = 20,924,640 feet.
Since Death Valley is 282 feet below, take x = 20,924,640 - 282.
To find the weight, take 100(20,924,640/x)².

cii. To find the weight at 20,320 feet, let x = 20,924,540 + 20,320.
To find the weight, again take 100(20,924,640/x)², but this time with the new x just figured.


Trivia: The last two digits of 3963 are 63, and 63² is 3969, which is just 6 over 3963.
The number 6 is 3*2*1, or 3 factorial.   This means that 3963 - 63² = 3!.

Also, 3963 = 3*1321, and both 3 and 1321 are prime and 63 = 2^6 - 1^6.

So it can be said that the radius of the earth is 1*2^(3!) - 1 - 2*3 - that's almost as
simple as 1-2-3, except this repeats them as 1,2,3, 1,2,3 with symbols in between.