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May 5, 2015

May 5, 2015

Posted by **tera** on Thursday, April 9, 2009 at 3:18pm.

- math -
**TchrWill**, Thursday, April 9, 2009 at 4:22pmYour weight is a measure of the force with which gravity is pulling on you and the only reason you feel it is because of the equal and opposite force being applied to your body through your feet.

The earth's gravity reaches out forever but the force of attraction on bodies at great distances would be extremely small depending on the mass of the body. The Law of Universal Gravitation states that each particle of matter attracts every other particle of matter with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Expressed mathematically,

F = GM(m)/r^2

where F is the force with which either of the particles attracts the other, M and m are the masses of two particles separated by a distance r, and G is the Universal Gravitational Constant, 1.407974x10^16 ft^3/sec^2. The product of G and, lets say, the mass of the earth, M, is sometimes denoted by µ (the greek letter pronounced meuw, as opposed to meow), the earth's gravitational constant. Thus the force of attraction exerted by the earth on any particle within, on the surface of, or above it, is

F = 1.407974x10^16 ft^3/sec^2(m)/r^2

where m is the mass of the object being attracted = W/g (W = weight), and r is the distance from the center of the earth to the center of gravity of the mass. The force of attraction which the earth exerts on our body, that is, the pull of gravity on it, is called the weight of our body, and shows how heavy our body is. Thus, our body, being pulled down by by the earth, exerts a force on the ground equal to our weight. The ground being solid and fixed, exerts an equal and opposite force upward on our body and thus we remain at rest. 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 = 20,924,640 feet. Thus the attractive force on his body is 1.407974x10^16(6.21118)/20,924,640^2 = 200 pounds. What do you know? The mans weight. Now, the attractive force on the 200 lb. man 1000 miles above the earth would only be 1.407974x10^16(6.21118)/26204640 = 127 pounds and half way to the moon, only .22 pounds.

Looking at gravity alone, the value of gravity for any body with a gravitational field varies inversely with the square of the distance from the center of the body. Put in mathematical terms, g = g(o)[r(o)^2]/r^2 where g(o) is the value of gravity at some reference point on the body, usually the outer most surface of the body, r(o) is the radius of the reference point or surface, and r is the radius of the new point where the value of gravity is of interest. Thus, for the Earth, with g(o) = ~32.15 fps^2, and r(o) = ~3963 miles = 20,924,640 ft., g = [1.407974x10^16 ft.^3/sec.^2]/r^2, or the gravitational constant of the Earth divided by the square of the altitude in question.

Thus, doubling the radius, reduces gravity to 1/4th of its surface value. Increasing the radius by three, reduces gravity to 1/9th, by four, to 1/16th, etc.

For any one body, the gravitational constant, GM , and your reference mass, m, remain constant. Therefore, you can derive your apparant weight at high altitudes from W = Wo(Ro/R)^2 where Wo = your refernce weight on the body surface, R = the altitude at which the new weight is desired and Ro = the radius of the body surface. Thus, on earth, W = Wo(3963/(3963 + h)^2 where Ro = the earth's radius and h = the height above the earth's surface, in either feet or miles..