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Old 04-December-2006, 08:47 PM
feokidoki feokidoki is offline
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Default Centrifugal force and satellites

Hi, this seems to be a very simple question but I can't find the answer anywhere!

Here's my current (possibly flawed) understaning of centripetal/centrifugal forces:

Newton's First law states that a body remains at rest or travels in a straight line unless an unbalanced force acts on it. So to make a body travel in a curve a force must act to pull the body away from its straight line.

The force that pulls an object out of a straight line path into a circular orbit is the centripetal force (centre seeking). It is the force on the orbiting object. From Newton's Third Law if a force acts on an object then there must be an equal and opposite force acting on another body known as the centrifugal force (centre fleeing).

Stone on a string whirled around your head- the force acting on the stone is the centripetal force, the force acting on the hand is the centrifugal (centripetal and centrifugal force is the same thing but depends on your point of view-from the hand or from the stone?).

So, satellites.

An artificial satellite in orbit around the Earth experiences a centripetal force exerted by the gravitational pull of the Earth on the satellite.
What's the centrifugal force here? I've been confusing myself by thinking about the velocity of the satellite which keeps the satellite from being pulled back to the surface of the Earth but I know this doesn't make sense.

Newton's Third Law means that the centrifugal acts on the Earth. So is the centrifugal force the gravtitational pull of the satellite on the Earth?

If anyone can answer this question and/or point out the misconceptions in the above, I'd be eternally grateful!
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Old 04-December-2006, 09:18 PM
Larry Jacks Larry Jacks is offline
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According to my old physics professor, centrifugal force is really an illusion. You can easily illustrate this by your stone on a string experiment. If centrifugal force existed, when you released the string, the rock would fly directly away from you perpendicular to its motion. Instead, what happens the rock will fly off in the direction of the rock's instantaneous motion at the instant of release. So, instead of centrifugal force, think inertia. The rock's inertia would have it fly off in a straight line except for the force exerted by the string.

What's keeping the satellite in orbit is its inertia. As you said, gravity is what keeps the satellite from flying off into space. The satellite's mass and velocity combine to create its inertia.
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Old 05-December-2006, 02:14 AM
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hhEb09'1 hhEb09'1 is offline
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Quote:
Originally Posted by feokidoki View Post
Newton's Third Law means that the centrifugal acts on the Earth. So is the centrifugal force the gravtitational pull of the satellite on the Earth?
Newton's law says that the satellite pulls on the Earth just as much as the Earth pulls on the satellite.

The centrifugal force is in the reference frame of the satellite--it experiences ["feels"] no force, so the centrifugal force is balancing the gravitational force.
Quote:
Originally Posted by Larry Jacks View Post
You can easily illustrate this by your stone on a string experiment. If centrifugal force existed, when you released the string, the rock would fly directly away from you perpendicular to its motion.
In the rotating frame, that is more or less what it does do, though.
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Old 05-December-2006, 03:37 AM
Gsquare Gsquare is offline
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Quote:
Originally Posted by feokidoki View Post
.

Newton's Third Law means that the centrifugal acts on the Earth. So is the centrifugal force the gravtitational pull of the satellite on the Earth?

!
Yes, that is correct... but I wouldn't call it 'centrifugal'. The centripetal force of the earth acting on the satellite is counterbalanced by the equal and opposite centripetal force of the satellite acting upon the earth. (Just like in your rock spinning on a string). ...and as hhEb09 mentioned correctly.

No need for 'centrifugal'.

Think in terms of a larger satellite like the moon and it becomes more evident;
also when you do the equations.

G^2
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