The weight of an object on earth varies according to the elevation of the object. In particular the weight of an object follows this equation w=Cr^-2, where C is the constant and R is the distance the the object is from the center of the Earth.
The first part of the problem ask for the equation w=Cr^2 for r be solved. I cam up with W=c/r^2 R=square root c/w.
The next part gives an example that the object weighs 100 pounds at sea level find the value of C (sea Level is 3963 miles from the center of earth. I can not figure out how to find the value of C.
The weight of an object is equal to the gravitational force exerted on the object by the earth.
This is usually expressed by F = W = GMm/r^2 where F/W = the attactive force of the earth on the object, or its weight, G = the universal gravitaional constant, M = the mass of the earth, m = the mass of the object and r = the distance to the object from the center of the earth.
GM = µ = 1.40766x10^16 ft^3/sec^2.
m = the mass of the object = W/g
g = the earth's gravitatyional acceleration.
The mean radius of the earth is 3963 miles = 20,924,640 ft.
In your terms, W - = C/R^2 where C = 1.40766x10^16(m)
A simple example of determining this force, or our weight, is to calculate the attractive force on the body of a 200 pound man standing on the surface of the earth. Now the man's mass is his weight divided by the acceleration due to gravity = 200/32.2 = 6.21118 lb.sec^2/ft. The radius of the surface from the center of the earth is 3963 miles x 5280 ft/mile = 20924640 feet. Thus the attractive force on his body is 1.40766x10^16(6.21118)/20924640^2 = 200 pounds. What do you know? The mans weight.
Now lets look at an astronaut inside the Space Shuttle for instance. His so called weightlessness, used to be referred to as being in a state of constant free fall. But since the term "fall" also suggests "down," while the space vehicle is moving in any direction but down, the term "zero g" was ultimately substituted for "free fall." But this clarification often leads to another misconception. The astronaut is not at all experiencing zero g as the earth's gravity is still pulling him down toward the earth. The reason he does not feel this "pull" of gravity is the fact that his velocity through the vacuum of space is creating an equal and opposite centrigugal force on his body exactly cancelling the pull of gravity on his body thus placing him in what is typically referred to as a state of weightlessness. To illustrate, lets look at our 200 pound astronaut as he is hurtling through space in a 250 mile high orbit. His velocity at this altitude is 25,155 feet per second, coincidentally the same as the Space Shuttle
.
His radius from the center of the earth is 2,224,464 feet. The pull of gravity on his body at this altitude is give by F = 1.40766x10^16(6.21118)/(22244640^2) = 176.7 pounds downward. The outward centrifugal, or inward centripital, force exerted on his body is given by F = m(V^2)/r = 6.21118(25155^2)/22244640 = 176.7 pounds. Thus he feels no feeling of weight as he would feel on the surface of the earth.
