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I am confused about the orbital velocity

  1. Apr 8, 2010 #1
    Hello,

    We know that to calculate the orbital velocity we use the relation :

    Vo=2* pi* r /time of one cycle

    but if we want to calculate the orbital velocity when a body moves from the point a to the point b on a circular path we use the relation :

    Vo = The line segment ab / time taken (as it is written im my book)

    so i'm confused with this.
    it should be: the arc ab /time taken

    so Could u help?
    Thanks
     
  2. jcsd
  3. Apr 8, 2010 #2
    It's about the difference between speed (a scalar) and velocity (a vector)
    If you calculate the quantity in your first example, the circumference of the circle divided by the time taken, you have calculated the speed of the object. (Or more correctly, the average speed if the value is not constant)
    circle.png
    When an object moves from a to b on a circle, if you calculate the quantity arc length ab (aqb in diagram) divided by time, you have the speed of the object.
    If you calculate the quantity chord length ab divided by time (apb in diagram) you have calculated the average velocity.
    Velocity is displacement divided by time
    Speed is distance divided by time
    The distance travelled from a to b is the arc length aqb
    The displacement from a to b is the chord length apb.
     
  4. Apr 8, 2010 #3

    sophiecentaur

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    In the limit, as the interval approaches zero, the magnitude of the two quantities is the same.
    btw, is that a circle or an ellipse on post no 356? It may just be my glasses.
     
  5. Apr 8, 2010 #4

    sophiecentaur

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    In the limit, as the interval approaches zero, the magnitude of the two quantities is the same.
    btw, is that a circle or an ellipse on post no 356? It may just be my glasses.
     
  6. Apr 8, 2010 #5

    sophiecentaur

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    Wow, deja vu!
     
  7. Apr 8, 2010 #6
    It's more likely my glasses. :bugeye:
     
  8. Apr 8, 2010 #7
    the speed is constant in a circular motion .

    but speed value is equal to velocity they should have the same value
    still confused :(
    Thanks so much
     
  9. Apr 8, 2010 #8
    Often but not always.
    Certainly in most physics problems like these, involving circular motion, the speed is constant.
    Why should speed value and velocity be the same?
    The answer I gave referred to "average velocity", not instantaneous velocity.
    Unfortunately, we are not always precise with terminology. Sometimes people confuse, or are careless about, the terms speed and velocity.
    They are not the same in physics.

    Edit to add:
    If you run half way around a circle, diameter 200m, in 1 minute, you run a distance of πd/2 = 314m
    What was your average speed? Answer 314m/60s = 5.2m/s
    But your end point is 200m from your start point. Your displacement is 200m.
    What was your average velocity? Answer 200m/60s = 3.3 m/s
    Distance and displacement measure different things; and average speed and average velocity measure different things.

    When people refer to the Earth's orbital "velocity", they very often just mean the instantaneous speed at any point; the magnitude part of the velocity.
     
    Last edited: Apr 8, 2010
  10. Apr 9, 2010 #9
    Last edited: Apr 9, 2010
  11. Apr 9, 2010 #10
    What the proof does, is to say that the length of the arc and the length of the chord become nearer and nearer to equal as you make the arc smaller and smaller.
    In the limit, when you take an infinitesimally small time interval, they become equal.
    The calculus of the rate of change always uses this idea of taking a very small time interval, and looking at what happens as the interval gets smaller and smaller.
     
  12. Apr 9, 2010 #11

    sophiecentaur

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    The instantaneous speed is the same as the magnitude of the instantaneous velocity.
     
  13. Apr 9, 2010 #12
    Ahaa....I got it now
    thanks so much

    Yes
    thanks very much ,Miss Sophie
     
  14. Apr 9, 2010 #13
    MMM
    i tried it
    I dunno anything about limits but is there anything wrong about this equation?
    where Vdt is ab arc

    [PLAIN]http://img401.imageshack.us/img401/8510/limr.jpg [Broken]



    suppose we have this circle

    [PLAIN]http://img684.imageshack.us/img684/1360/circle1.jpg [Broken]

    where ab is approx. 4 cm
    ab arc = 9.4 cm

    another circle after decreasing the angle we have

    [PLAIN]http://img194.imageshack.us/img194/213/circle2i.jpg [Broken]

    ab= 3cm
    and ab arc = 3.14cm

    so they approached each other when the angle decreases (or when time interval decreases)
    till they both (the arc and the chord) equal zero when dt = zero
    right or wrong??
     
    Last edited by a moderator: May 4, 2017
  15. Apr 9, 2010 #14
    Here are the actual values (to 3 significant figures) for your circle of radius 3cm
    That's right
    As the angle gets smaller the difference between the arc and the chord gets smaller

    deg arc chord difference
    90 4.712 4.243 0.470
    80 4.189 3.857 0.332
    70 3.665 3.441 0.224
    60 3.142 3.000 0.142
    50 2.618 2.536 0.082
    40 2.094 2.052 0.042
    30 1.571 1.553 0.018
    20 1.047 1.042 0.005
    10 0.524 0.523 0.001
    00 0.000 0.000 0.000
     
  16. Apr 9, 2010 #15
    Then nothing wrong with what i have written?
     
  17. Apr 9, 2010 #16
    What you wrote is fine.
    The only mistake was the calculation of the arc for 90 degrees.
    That's why I wrote out the table of values for you.
     
  18. Apr 10, 2010 #17
    Yeah right . Actually it is 3*pi/2
    Thanks very much This really really helped me understand.
     
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