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Circumference of a circle

  1. May 5, 2009 #1
    What's the circumference of a circle of radius r, on a sphere of radius R?
     
  2. jcsd
  3. May 5, 2009 #2

    HallsofIvy

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    The circumference of a circle of radius r is [itex]2\pi r[/itex], whether it is on a sphere or not.
     
  4. May 5, 2009 #3
    The following i the context I was talking about:
    From https://www.physicsforums.com/showthread.php?t=311787 post #6,
    If r == PI*R, the circumference created by the string will be zero.
     
  5. May 5, 2009 #4

    HallsofIvy

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    So you aren't talking about radius in the usual sense- you are having the "radius" bending around the sphere. And the "center" of the circle is also on the sphere, not inside it.

    We can, without loss of generality, assume that the "center" of the circle is at the top of the circle, the "north pole". Draw a line from the north pole to the center of the sphere, then to a point on the circle. Let the angle made be [itex]\theta[/itex] ([itex]\theta[/itex] is the "co-latitude"). Dropping a perpendicular from the point on the circle to the line from north pole to center of sphere, we have a right triangle with angle [itex]\theta[/itex], hypotenuse of length R, and "opposite side" the (real!) radius of the circle, which I will call r' since we are using r for the "phony" radius. Then we have [itex]sin(\theta)= r'/R[/itex] or [itex]r'= R sin(/theta)[/itex].

    Now, the circumference is "really" [itex]2\pi r'= 2\pi R sin(\theta)[/itex]. We only need to calculate [itex]\theta[/itex] in terms of r. The spherical "distance" is measured along a great circle and a whole great circle on a sphere of radius R has circumference [itex]2\piR[/itex] and corresponds to an angle of [itex]2\pi[/itex]. We can set up the proportion
    [tex]\frac{2\pi}{2\pi R}= \frac{1}{R}= \frac{\theta}{r}[/tex]
    so [itex]\theta= r/R[/itex]

    That means our circumference formula becomes
    [tex]2\pi r'= 2\pi R sin(\theta)= 2\pi R sin(r/R)[/itex]
     
  6. May 5, 2009 #5

    DaveC426913

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    If you input values such that r > R, would you get an answer that is imaginary?
     
  7. May 6, 2009 #6
    Did I miss something?

    If r == PI*R, shouldn't the circumference created by the circle be zero?
     
  8. May 6, 2009 #7
    It should, and it is, according to that formula. ([itex]\sin \pi = 0[/itex].)
     
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