Thank you drwls.

Question

I was not reading my own question clearly. I am trying to make the equation true for an object that is 100 pounds at sea level. Find the vlaue of C that makes the equation true. wouldn't C, being the contant, be where I put my 100 pounds?

Answer 1

For the equation I gave you in the previous answer,

Weight(at r=R) = 100*[(3963)/R]^2,
the constant C in an equation
Weight (at r=R) = C/R^2
would be
C = 100*3963^2 = 1.571*10^9

You would need a different constant for objects of different mass, so that would equation would not be true for all objects. R must be expressed in miles, also, and measured from the center of the earth, not sea level.

Answer 2

That makes sense. So not so much the wieghtof the object but how high above or below sea level the object was. A mile is still 5280 I am guessing so if I have an object in Death Valley is my given example, (its not a even numbered question so I don't have to do it but I want to understand this) at 282 feet beleow sea level I need to plug this number in for r as a fraction of a mile?

Answer 3

At the -282 feet (-.053 miles) altitude at Badwater in Death Valley, the distance from the center of the Earth is
R = 3963 - .053 = 3962.947 miles, so that is what you would use in the C/R^2 equation.

The effect of the lower elevation on the weight will be very small and comparable to other corrections that should be made due to the nonspherical shape of the earth, latitude (which affects centrifugal force of Earth's rotation), and the effect of nearby mountains like Telescope Peak.

It seems to me you are beating this subject to death. I hope my explanations have been clear.