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## Main Question or Discussion Point

What's the circumference of a circle of radius r, on a sphere of radius R?

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What's the circumference of a circle of radius r, on a sphere of radius R?

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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.

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The following i the context I was talking about:The circumference of a circle of radius r is [itex]2\pi r[/itex], whether it is on a sphere or not.

From https://www.physicsforums.com/showthread.php?t=311787 post #6,

If r == PI*R, the circumference created by the string will be zero.Imagine you fix one end of a string with the length r at a point in 2d-space, and make a full circle with the other end. If you then find that the circumference of that circle is different from 2*PI*r, you conclude that the 2d-space is curved intrinsically.

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HallsofIvy

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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]

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DaveC426913

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If you input values such that r > R, would you get an answer that is imaginary?That means our circumference formula becomes

[tex]2\pi r'= 2\pi R sin(\theta)= 2\pi R sin(r/R)[/itex]

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Did I miss something?That means our circumference formula becomes

[tex]2\pi r'= 2\pi R sin(\theta)= 2\pi R sin(r/R)[/itex]

If r == PI*R, shouldn't the circumference created by the circle be zero?

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It should, and it is, according to that formula. ([itex]\sin \pi = 0[/itex].)

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