Question 325104: Suppose an object is 100 pounds at sea level and sea level is 3,963 miles from the center of the earth and I have this equation to make this a true statement and it is correct r=3963 w=100 and w=cr^-2 100=c(3963)^-2 how would i determine the weight of an object in Death Valley which is 282 ft. below sea level and the top of Mount McKinley 120,300 feet above sea level. If you can show me which numbers to plug in where I can then figure the math. I am not sure where each number goes in the equation to get the answer.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! The equation you have is:
w = c * r^-2
This is the same as w = c / r^2
c is a constant.
r is the distance from the center of the earth.
r is given in miles, so you would need to either translate your elevation to miles or you would need to translate the distance to the center of the earth to feet.
I chose to translate the distance to the center of the earth to feet.
At sea level the object is 3963 miles from the center of the earth.
Multiply that by 5280 and the object is 20924640 feet from the center of the earth.
You are given that the weight of the object is 100 pounds at sea level.
You are also given that the equation w = c / r^2 is true at sea level.
You can use the weight of the object at sea level to calculate the value of c.
You get 100 = c / r^2 which becomes 100 = c / 20924640^2
Multiply both sides of this equation by 20924640^2 to get c = 100 * 20924640^2
That's the value of c and it will not change because it is a constant.
r will change, however, because it is the distance from the center of the earth.
To test the equation to see if it is true at sea level, then just substitute what you know and solve.
The equation at sea level is:
100 = c / r^2
Since c = 100 * 20924640^2 and r = 20924640, then you get:
100 = 100 * 20924640^2 / 20924640^2 which becomes 100 = 100 which is true.
You have confirmed the equation is true at sea level, so you can be comfortable that you calculated the value of c correctly
Your first question is how much does an object weigh that is 282 feet below sea level.
The equation is w = c / r^2
c is a constant of 100 * 20924640^2
r is now 20924640 - 282 = 20924358
Your equation becomes:
w = 100 * 20924640^2 / 20924358^2 = 100.0026954 pounds.
Your second question is how much does an object weight that is 120,300 feet above sea level.
Immediately we have a problem:
Mount McKinley is shown as 20,320 feet above sea level, so your number above sea level for Mount McKinley appears to be wrong.
I'll use 20,320 feet as the elevation of Mount McKinley above sea level and you can adjust to whatever numbers you have.
c is a constant of 100 * 20924640^2
r is now 20924640 + 20320 = 20944960
Your equation becomes:
w = 100 * 20924640^2 / 20944960^2 = 99.80606176 pounds.
If you had converted feet to miles you would have gotten the same answer.
Example:
282 feet is equivalent to .053409091 miles.
Distance to the center of the earth at sea level is 3963 miles.
Distance to the center of the earth at 282 feet below sea level is 3962.946591 miles.
Equation for 282 feet below sea level becomes:
w = 100 * 3963^2 / 3962.946591^2 = 100.0026954
Either way you get the same answer.
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