Friday

February 12, 2016
Posted by **John** on Sunday, September 11, 2011 at 8:32pm.

a) It is beyond the earth's gravitational pull.

B) The earth's gravity is balanced out by the attraction of the sun and other planets.

C) Centrifugal force keeps it up.

D) It does fall toward the earth.

- Science -
**HM**, Sunday, September 11, 2011 at 8:59pmI would say A.

What do you think?

- Science -
**tchrwill**, Monday, September 12, 2011 at 9:45amFundamentally, the sole reason that a satellite/spacecraft remains in orbit is its forward velocity. Lets explore. Assume there was some way to lift a 1000 pound satellite straight up to an altitude of 250 miles above the earth's surface. Once there, we let the satellite go, that is, drop it vertically. What happens? Of course, the gravitational pull of the earth pulls it toward the earth at an acceleration of ~32.2 feet/sec/sec. This results from the Universal Law of Gravitation which states that every particle of matter in the universe 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 gravitational masses of two particles separated by a distance r, and G is the Universal Gravitational Constant. The product of G and, lets say, M, the gravitational mass of the earth, is sometimes referred to as either 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 the earth, is F = 1.40766x10^16 ft^3/sec^2(m)/r^2 where m is the gravitational mass (weight divided by the acceleration due to gravity) 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 the satellite, that is, the pull of gravity on it, is called the weight of the satellite, and tells how heavy the satellite is.

As a satellite is circling the earth, it is constantly being pulled toward the center of the earth by gravity. However, at the same time, due to its forward velocity, it is tending to be thrown farther away from the earth due to the centrifugal force created by this forward motion on its circular path around the earth. Technically, in the motion of a body travelling around a circular path at constant speed, the body acts upon the restraining agent with a constant force directed radially outward from the center. This outward force is called the centrifugal force. Since for every action there is an equal and opposite reaction, the restraining agent exerts an equal inward force on the moving object, and this is called the centripetal force. Thus, the outward centrifugal force exerted by the satellite is exactly counterbalanced by the inward pulling centripetal force of gravity.

The simplest, and most meaningful, analogy that can be offered to illustrate this phenomena is the situation where a ball on the end of a string is being whirled overhead. The centrifugal force developed by the rotating ball is trying to tear the ball away from the string. On the other hand, the string is exerting an equal and opposite centripetal force on the ball, preventing it from flying away from the person. Obviously, an astronaut inside the Space Shuttle is not tied to a string but the reason that he does not feel this centripetal "pull" of gravity is the fact that his forward velocity, being the same as the shuttle, is creating an equal and opposite centrifugal force on his body, exactly cancelling the pull of gravity on his body and placing him, in what is typically referred to, as a state of weightlessness.

To numerically illustrate the satellite's force balance in orbit, lets look at our 1000 pound satellite as it is hurtling through space in a 250 mile high orbit. Its velocity at this altitude is 25,155 feet per second. Its mass m = W/g = 1000/32.174 = 31.08 lb.sec^2/ft. The radius from the center of the earth is 22,244,640 feet, 3963+250)5280. The pull of gravity on the satellite at this altitude is give by F = 1.40766x10^16(31.08)/(22244640^2) = ~884 pounds downward. The outward centrifugal force exerted on the satellite due to its velocity is given by F = m(V^2)/r =

31.08(25155^2)/22244640 = ~884 pounds. Therefore, gravity is pulling it down with a force of 884 pounds but centrifugal force is attempting to throw it out with an equal and opposite force of 884 pounds, and therefore the satellite remains at a constant altitude. Once the launch rocket delivers the satellite to the 300 mile high altitude, with a circular velocity of 25,155 feet per second (17,147 MPH), and a velocity direction perpendicular to the radius vector from the center of the earth, the rocket shuts down, and the satellite will remain there for an indefinite period. The constant application of rocket thrust is not at all necessary to sustain the flight of the satellite once it has achieved the required velocity. In actuality, the few molecules of atmosphere that exist at these low earth altitudes will, over time, cause the orbit to decay and the satellite's altitude will decrease at an extremely low rate. However, small thruster rockets, present on every satellite for attitude control purposes, can periodically fire to raise the satellites altitude back up to the desired 300 mile height if that is necessary.

There is another way of explaining this situation. If there was no gravitational pull on the satellite, meaning no force being applied to the satellite, it would continue on a straight line forever. However, the gravitational pull on the satellite is such that it pulls it toward the earth with an acceleration that causes it to fall toward the earth at a constant rate. But, the earth's surface just happens to be moving away from the satellite at the same rate, thus maintaining a constant distance between the satellite and the ground, or its orbit.