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May 25, 2016
Posted by **Elizabeth** on Monday, October 12, 2009 at 9:37pm.

- physics -
**bobpursley**, Monday, October 12, 2009 at 9:39pmgravity= weight on Earth (re/(re+alt))^2

re is the radius of the earth. - physics -
**tchrwill**, Tuesday, October 13, 2009 at 4:06pmCan you explain wheightlessness. Do you actually weigh nothing in space or just a percentage of your Earth weight? How much would a 100 lb person weigh on each of the other planets?

An astronaut, circling the earth in the Space Shuttle, senses that his apparent weight is zero. He cannot feel the normal earthbound sensation of his feet pressing on the floor below him and the floor holding him upright. Walking in the normal sense is impossible; instead he floats from place to place. This state is often referred to as being in freefall, or in zero gravity. Since gravity in space is never zero, the phenomena that we refer to as zero g is, in reality, the human minds sensitivity to, or the feeling of, apparent weightlessness.

Weightlessnes, as perceived by astronauts in the Space Shuttle, occurs at any altitude where the spacecraft velocity is sufficient to hold the spacecraft in a circular or elliptical orbit. As long as the velocity is sufficient, the gravitational force on the astronaut is exactly balanced by the centrigugal force exerted on his body due to his circular motion about the earth.

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. The product of G and, lets say, the mass of the earth, is sometimes referred to as GM or mu (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, is F = 1.40766x10^16 ft^3/sec^2(m)/r^2 where m is the mass of the object being attracted and r is the distance from the center of the earth to 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 the body, and shows how heavy the 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 = 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 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.

Since attracting masses are present everywhere in the universe; the Sun, the fixed stars, the planets, satellites of the planets, etc., an object will everywhere and at all times be in a gravitational field. Rest and uniform motion are ficticious states and cannot exist anywhere in the universe. To avoid any misunderstanding, it must be pointed out that the gravitational field of any body is infinite. It is however true that, at great distances, the attraction becomes so small that is becomes negligible for all intensive purposes, but it cannot be said that the gravitational field of, any body, has a limit anywhere.

To summarize, gravity in space is never zero and the phenomena that we refer to as zero g is, in reality, the human minds sensitivity to, or the feeling of, apparent weightlessness. An astronaut in the space shuttle, circling the earth, senses that his apparent weight is zero.

The simplest way is to take the value of gravity on the surface of the other planet from any reference source and multiply your weight of 100 lb. by g(P)/g(E) where g(P) is the value of gravity on the planet of interest and g(E) is the value on Earth.

The value of g may also be calculated from the equation of g = GM/r^2 where GM is the Gravitational Constant of the reference body and r is the mean, or equatorial, radius of the reference body. For instance, GM for Pluto is ~1.56259x10^15ft.^3/sec.^2 and r is ~1553.5 miles. So g(P) = 1.56259x10^15/[1553.5(5280)]^2 = ~23.22 ft./sec.^2. This makes your weight of 100 lb. 100(23.22)32.15 = ~72 1/4 pounds.

I hope this is of some interest to you. - physics -
**hu**, Friday, March 18, 2011 at 4:12pmyou not answering the question!!

- physics -
**Anonymous**, Wednesday, June 12, 2013 at 6:54pmidk.