Hint: Pay attention to the units of measure. You may have to convert from feet to miles several times in this assignment. You can use 1 mile = 5,280 feet for your conversions.
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: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 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. 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).
ii. the top of Mount McKinley (20,320 feet above sea level).
a. To solve the equation w=Cr^-2 for r, we can rearrange the equation to isolate r:
w = C/r^2
Multiply both sides by r^2:
wr^2 = C
Divide both sides by w:
r^2 = C/w
Take the square root of both sides:
r = √(C/w)
b. Let's find the value of C that makes the equation true when the object weighs 100 pounds at sea level. Sea level is 3,963 miles from the center of the Earth.
We are given w = 100 and r = 3,963 miles. Plugging these values into the equation w = Cr^-2:
100 = C/(3,963)^2
To solve for C, multiply both sides by (3,963)^2:
100 * (3,963)^2 = C
C ≈ 157,293,369
Therefore, the value of C that makes the equation true is approximately 157,293,369.
c. Now that we have the value of C, we can use it to determine how much the object would weigh in different locations.
i. Death Valley is 282 feet below sea level. To get the distance from the center of the Earth, we need to add 282 feet to the radius of the Earth (since the radius would be negative below sea level).
Radius of Earth + 282 feet = (3,963 miles * 5,280 feet/mile) + 282 feet
Radius of Earth + 282 feet ≈ 20,924,442 feet
Using the equation w = Cr^-2, we can calculate the weight:
w = 157,293,369 / (20,924,442^2)
w ≈ 97.04 pounds
Therefore, the object would weigh approximately 97.04 pounds in Death Valley.
ii. The top of Mount McKinley is 20,320 feet above sea level. To get the distance from the center of the Earth, we need to subtract 20,320 feet from the radius of the Earth (since the radius would be greater above sea level).
Radius of Earth - 20,320 feet = (3,963 miles * 5,280 feet/mile) - 20,320 feet
Radius of Earth - 20,320 feet ≈ 20,057,120 feet
Using the equation w = Cr^-2, we can calculate the weight:
w = 157,293,369 / (20,057,120^2)
w ≈ 102.47 pounds
Therefore, the object would weigh approximately 102.47 pounds at the top of Mount McKinley.
a. To solve the equation w = Cr^(-2) for r, we need to isolate the variable r.
First, let's start by taking the reciprocal of both sides of the equation:
1/w = 1/(Cr^(-2))
Next, we can simplify the right side by multiplying the reciprocal of r^(-2), which is (r^2):
1/w = r^2/C
Now we can take the reciprocal of both sides again to solve for r:
w = C/r^2
1/w = r^2/C
Taking the square root of both sides, we get:
r = √(C/w)
Therefore, the equation w = Cr^(-2) can be solved for r as r = √(C/w).
b. Given that the object weighs 100 pounds at sea level, we can substitute w = 100 and r = 3,963 miles into the equation to solve for C:
100 = C/(3,963^(-2))
To simplify further, we need to convert 3,963 miles to feet:
3,963 miles * 5,280 feet/mile = 20,925,840 feet
Now, we substitute the values:
100 = C/(20,925,840^(-2))
To find C, we can multiply both sides by (20,925,840^2):
100 * (20,925,840^2) = C
Therefore, the value of C that makes the equation true is 4.6250402752416e+15.
c. To determine how much the object would weigh in different elevations, we can again use the equation w = Cr^(-2) with the value of C we found:
i. Death Valley is 282 feet below sea level. We need to convert this to miles:
282 feet * (1 mile/5280 feet) = 0.0534090909091 miles
Now, we can calculate the weight in Death Valley:
w = (4.6250402752416e+15) / (0.0534090909091^(-2))
w ≈ 425.32295621 pounds
ii. The top of Mount McKinley is 20,320 feet above sea level. Converting this to miles:
20,320 feet * (1 mile/5280 feet) ≈ 3.85606060606 miles
Now, we can calculate the weight at the top of Mount McKinley:
w = (4.6250402752416e+15) / (3.85606060606^(-2))
w ≈ 13,563.853594 pounds