Question 366786
1. 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 this equation: , where C is a constant, and r is the distance that the object is from the center of Earth. 
a. Solve the equation for r, and then solve the equation for C. 
Multiply both sides by r^2 to get:
wr^2 = C
r^2 = C/w
r = sqrt[C/w] 
b. Suppose that an object is 160 pounds 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=160*3963^2=40205744640
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Note: This should be C = 160*3963^2 = 2512859040
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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 feet below sea level). 
Distance from center = 3963 - (282/5280) = 3962.946591 miles = 20,924,358 ft
w=Cr^-2 
w = 1570536900*3962.946591^-2= 100.0026954 miles
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Note: This should be w = 25128509040*3962.946591^(-2) = 160.004313 lbs

 
ii. The top of Mount McKinley (20,320 feet above sea level). 
3963+ (20320/5280) = 3966.848484848485 miles
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Note: w = C/r^2 = 160.311 lbs.
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2. The equation gives the distance, D, in miles that a person can see to the horizon from a height, h, in feet. 
a. Solve this equation for h. 
1.2*Sqrt (h) = D
:
Sqrt(h) = ; Divided both sides by 1.2
h = ; Squared both sides
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Note:
1.2*sqrt(h) = D
sqrt(h) = D/1.2
h = [D/1.2]^2
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b. Mauna Kea in Hawaii is 13,796 feet in elevation. How far can you see to the horizon from the top of Mauna Kea?
Note: 
D = 1.2*sqrt(h)
D = 1.2*sqrt(13796) = 140.95 miles
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Can you see Waikiki beach, Honolulu (about 198 miles away)? 
No; Waikiki is to far away.
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Cheers,
Stan H.