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Circular Motion confusion

  1. Aug 21, 2009 #1

    I have some confusion with circular motions. I know centripetal force acts on the object towards the center of a circle, but where does centrifugal force (the reaction force to centripetal force) act from? Does it act from the center of the object or the center of the circle?

    Also I'm wondering if anyone can explain how something orbiting a planet escapes the orbit using circular motion concepts.

    Thanks for any help that you can provide
  2. jcsd
  3. Aug 21, 2009 #2
    The centrifugal force is not the same as the the reaction force to the centripetal force.In fact, the centrifugal force is a fictitious force that appears if you work in a rotational frame of reference while the reaction force to the centripetal force force is real force. It acts on the object that interacts with the object in circular motion.Therefore it depends on the situation.For example in a mass-string system you fave centripetal force due to string on mass- the reaction is tension on string, opposite and equal.
    I do not exactly understand your question.If you mean, why an object that has a tangential acceleration to a circular orbit changes its altitude. Consider that: [tex]a=\frac{v^2}{R}[/tex] therefore [tex]v=\sqrt{aR}[/tex]since [tex]a=\frac{Gm}{R^2}[/tex] we get [tex]v=\sqrt\frac{Gm}{{R}}[/tex]. As you can see any change in v must cause a change in R since Gm are constant.
  4. Aug 21, 2009 #3


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    The reactive centrifugal force acts at the same point of application as the centripetal force. In the case of a twirling string and rock, the string exerts a centripetal force on the rock, and the rock exerts a centrifugal force on the string, the newton third law pair of forces at work.

  5. Aug 21, 2009 #4
    Thanks for the Replies

    Then why do we use the term centrifugal force? I was taught that it is the reaction force, is this a wrong concept?

    But an object cannot orbit indefinitely if it has tangetial acceleration. Why does the force of gravity not indefinitely keep the orbit of an object as it is accelerating because theoretically based on the formulas I assume that the radius just keeps getting bigger but this is not the case in real life, the object escapes after reaching a certain velocity. How does an object escape the centripetal force/force of gravity as it circles a planet and how do we determine at what point does an object escape and its escape velocity?

    Thanks for any help that you can provide
  6. Aug 22, 2009 #5
    I am sorry i thought you were speaking of centrifugal force in this context. http://en.wikipedia.org/wiki/Centrifugal_force
    Jeff Reid's explanation and link are more pertinent. I am not used to the name of "reactive centrifugal force". As for the original question the RCF acts on the second object of the interaction, not the object in circular motion. The way in which you would draw it in a free body diagram depends on the situation. If you have string and a mass the tension vector would be drawn at the end of the string, same point as CF(even if the tension act on the whole length).If you have a planet and a small satellite the reactive force vector would be on the planet(a different point).

    Well an objects escape velocity is [tex]v=\sqrt\frac{Gm}{{R}}[/tex] meaning that for any distance R there is a velocity v such that that distance will be maintained.If the velocity changes you go up or dawn. The object never really escapes the gravitational pull(not even in real life)of the other object. if an object accelerates tangentially the distance will increase. At infinity (really really far away)the trajectory can be considered linear but the object still has not escaped the gravitational pull of the planet.
  7. Aug 22, 2009 #6


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    This occurs when the speed is so much larger than the acceleration that the path is no longer elliptical but instead a parabola if exactly at escape velocity, or a hyperbola if faster than escape velocity.

    If you have a stationary object some large distance away from a planet, then release that object so it falls towards the planet, it's possible to calculate it's speed at point close to the planet. It turns out that as the initial distance increases to infinity, the speed at the point close to the planet approaches some finite value. If the actual speed of the object close to the planet is faster than it's "falling from an infinite distance speed", the object's path will be a hyperbola.


    centrifugal force has two meanings. Physicists normally restrict it's usage to mean the apparent force obsevered in a rotating frame of reference, but a common English dictionary defines centrifugal as proceeding or acting in a direction away from a center or axis, so "reactive centrifugal force" is just a more generic usage of the word centrifugal.

    Centrifugal force is a reactive force. Other names for reactive forces are inertial force or fictitious force. The frame of reference changes the meaning. In an inertial (non-accelerating) frame of reference, the reactive force is an objects inertial reaction to being accelerated by some "real" force. In a non-inertial (accelerating) frame of reference, like a rotating frame of reference, the "fictitious" force appears to be a real one as observed in that non-inertial frame of reference.
    Last edited: Aug 22, 2009
  8. Aug 22, 2009 #7

    Doc Al

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    It is a serious error to get your physics from an English dictionary. While the root meaning of "centrifugal" does mean "away from center", the term centrifugal force has a specific meaning in physics that goes beyond its etymological origins.

    Your beloved "reactive centrifugal force" ≠ "centrifugal force". Just because something points "away from center" does not make it a centrifugal force in the standard physics sense.
  9. Aug 22, 2009 #8


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    As you have seen above there are multiple definitions of the term "centrifugal force" and it is a huge potential source for confusion. Your best bet is to not use the term at all until you have really mastered physics.

    If an object is undergoing uniform circular motion there is a centripetal force. That force will have a Newton's 3rd law "action/reaction" pair that will act as all 3rd law pairs do. Giving it a special name is not necessary (it has no special properties) and can cause confusion (it can get mixed up with the other definition of "centrifugal force").
  10. Aug 22, 2009 #9
    Ban the term Reactive Centrifugal force. Atleast in PF! :)
  11. Aug 22, 2009 #10


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    I don't see any evidence of confusion over what is going on, just a difference of opinion on the definition of a word and under what frame of reference it applies.

    The centrifugal force is the same in both cases, only the frame of reference is changing. Lots of forces get special names, thrust, drag, lift, downforce, so why not centripetal and centrifugal? Are they going to change centrifuge to centripuge? I don't see an issue or a source of confusion, since the OP included the context which defined the frame of reference, and I provided an answer to the first question in the OP in post #3.
  12. Aug 23, 2009 #11
    Are you sure? The 'reactive centrifugal force' which could be better named any other way is not a "inertial force or a fictitious force".It is in fact a very real force, the third law pair of the 'centripetal force'.
  13. Aug 23, 2009 #12


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    Both of which are very confusing for a student.

    No, the two definitions of centrifugal force are not the same at all!!!! One is a fictitious force which exists only in the rotating reference frame. The other is a real force which forms a 3rd law pair with the centripetal force and exists in both an inertial reference frame and the rotating reference frame. Note that in a rotating reference frame there are two distinct sets of forces (one real and one fictitious) which are called "centrifugal".

    Yes, you answered the question in post #3 using the meaning of "centrifugal" that the OP asked, but the confusion between the two definitions had already started by then. bp_psy's post #2 answered the question using the more typical definition of "centrifugal" that Red_CCF may not have even encountered. I stick by my assertion of confusion, and my recommendation that a new student avoid it entirely, the term "centrifugal" provides no benefit and can cause problems in learning.
    Last edited: Aug 23, 2009
  14. Aug 23, 2009 #13
    Yes that is completely correct. I was taught reactive centrifugal force but I was never made aware that there are other types of centrifugal forces until now. I will try to refrain from using the term from now on.
  15. Aug 23, 2009 #14


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    No worries. It is not an error to use it, just a big source of headaches and miscommunication.
  16. Aug 25, 2009 #15
    I didn't quite understand the wikipedia definition that escape velocity is the speed where the kinetic and gravitational potential energy equal. Does this mean that the escape velocity is changing as an object moves farther away because Ep increases with height.

    I also did not quite understand the bolded part. Are we accounting for terminal velocity of an atmosphere because how can something that is in free fall approach a finite velocity?

    I was wondering if someone can check my understanding so far. Escape velocity is basically the speed at which the eccentricity of the object's elliptical orbit is equal to 1? So the object's orbit really gets very large but not actually escaped the gravitational pull of a planet.

    I came across something that says a parabola is where the eccentricity is 1 and a hyperbola is an eccentricity greater than 1, but on the diagram they have the same shape except a hyperbola is a bit bigger. Can someone explain this? I took calculus and I can't relate the shapes of the two orbits to what I've been taught in class.

    Thanks for any help that you can provide
  17. Aug 25, 2009 #16


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    Escape speed decreases with height.

    A free falling object doesn't approach a finite velocity, instead, it's maximum velocity (before impact) versus the object's initial distance from a planet approaches a finite value as the distance approaches infinity. It would be easier to look at this the other way. If an object is moving faster than escape speed, then's it's radial component of speed away from a gravitational source approaches some finite positve value and the object never returns.

    An object never escapes the pull, but the pull diminishes faster than the speed once the speed is above escape velocity, and the object's component of speed away from the planet will always remain positive.

    If you graph a simple parabola, like y = x2, as x -> , the slope of the lines approaches vertical, but there are no asymptotes. If you graph a simple hyperbola, like y = 1 / x, you end up with two curves, which asymptotically approach the vertical line x = 0 and the horizontal line y = 0. These curves also show up as the intersection of a plane and a cone. If the plane is perpendicular to the cone, the intersection is a circle. If the plane is angled, the intersection is an ellipse. If the plane is parallel to the edge of the cone, the intersection is a parabola, and if the angle is more, the intersection is a hyperbola.


    These conic sections might help you understand another way to construct these curves, but I don't know if they help in explaining escape velocity.
    Last edited: Aug 25, 2009
  18. Aug 26, 2009 #17
    That's what I thought at first, but can you explain why escape speed is defined as when the kinetic energy and gravitational potential energy equal?

    Okay so this finite velocity changes depending on where the object is dropped

    Thanks for the response
  19. Aug 26, 2009 #18

    Doc Al

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    Note that gravitational PE, for this calculation, is zero at infinity and is negative for all finite distances from the earth's center.

    GPE = -GMm/r

    Your escape speed at any given distance will be that which makes your total energy zero:
    KE + GPE = 0
    1/2mv2 -GMm/r = 0

    (The magnitudes of KE and GPE will be equal.)
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