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Old 2005-02-21, 21:19   #1
jinydu
 
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I'm kind of embarrased to have to ask this here, but I just can't answer this question from my physics homework. I get this feeling that its supposed to be easy:

"What is the linear speed of a point (a) on the Equator, (b) on the Arctic Circle (latitude 66.5 degrees N), (c) at a latitude of 40.0 degrees N, due to the Earth's rotation?"

I know how I'm supposed to solve this. I find the circumference of the circle that the point traces in one day, then just divide by 86400 seconds. The problem is, I don't know how to calculate the circumference of the circle.

I guess the general way to state my problem is:

What is the circumference of a circle drawn around the Earth, parallel to the Equator, at latitude L?

I know some boundary conditions:

If C(L) represents the circumference at latitude L:

C(0) = 2*pi*R
C(90) = 0

I also know that C(L) is a strictly decreasing function (this can be seen from the geometric interpretation. As you move closer to the poles, the circles get smaller).

That is, if b > a, C(b) < C(a)

Also, since the size of the circles change continuous, C(L) is a continuous function.

But of course, there are infinitely many functions that satisfy these conditions.

Last fiddled with by jinydu on 2005-02-21 at 21:26
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Old 2005-02-21, 21:30   #2
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Quote:
Originally Posted by jinydu
I'm kind of embarrased to have to ask this here, but I just can't answer this question from my physics homework. I get this feeling that its supposed to be easy:

"What is the linear speed of a point (a) on the Equator, (b) on the Arctic Circle (latitude 66.5 degrees N), (c) at a latitude of 40.0 degrees N, due to the Earth's rotation?"

I know how I'm supposed to solve this. I find the circumference of the circle that the point traces in one day, then just divide by 86400 seconds. The problem is, I don't know how to calculate the circumference of the circle.

I guess the general way to state my problem is:

What is the circumference of a circle drawn around the Earth, parallel to the Equator, at latitude L?

I know some boundary conditions:

If C(L) represents the circumference at latitude L:

C(0) = 2*pi*R
C(90) = 0

I also know that C(L) is a strictly decreasing function (this can be seen from the geometric interpretation. As you move closer to the poles, the circles get smaller).

That is, if b > a, C(b) < C(a)

But of course, there are infinitely many functions that satisfy these conditions.

I never studied phisics, but if the earth could be considered as a sphere, and R is the radius of the sphere, the radius at latitude x should be R*cosine(x).

I'm surely wrong, but I just can't think to a better method...

Luigi
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Old 2005-02-21, 21:32   #3
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Quote:
Originally Posted by jinydu
What is the circumference of a circle drawn around the Earth, parallel to the Equator, at latitude L?
2*pi*r*cos(L) -- that is, if you don't include the flattening of the Earth at the poles.

regards, Leif.
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Old 2005-02-21, 21:38   #4
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Quote:
Originally Posted by ET_
I never studied phisics, but if the earth could be considered as a sphere, and R is the radius of the sphere, the radius at latitude x should be R*cosine(x).

I'm surely wrong, but I just can't think to a better method...

Luigi

Quote:
Originally Posted by leifbk
2*pi*r*cos(L) -- that is, if you don't include the flattening of the Earth at the poles.

regards, Leif.
Uh oh. ET_ says that its definitely not R*cos, but leifbk says it is...

Maybe a proof would help here.

Thanks

Last fiddled with by jinydu on 2005-02-21 at 21:39
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Old 2005-02-21, 21:42   #5
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I thougt that we said exactly the same thing. I've got a rule of thumb regarding trigonometrics: Cosine is 1 at 0 degrees, and 0 at 90 degrees. With the sine, it's the opposite.

regards, Leif.
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Old 2005-02-21, 21:59   #6
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here it goes just what to shoot my luck well first you need to find the point from the distence from center of the invisable axis of earth to the radius you kinda could use a model of earth to visualize it find hte point on the same plain get radius then use 3.14(2r)=circumference
r = radius

then it would be
c/(60*24*365.25)=distence per second

/end rambleing
i dont know if this can totally work hence being a freshman in highschool but who knows thats what i would do only i would proibily get it wrong gl on it jin
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Old 2005-02-21, 22:11   #7
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Once I find the radius or circumference of the circle, the rest I can do in my sleep. The problem is getting to that point.

How could I prove mathematically that

C(L) = 2pi * R * cos(L)

or equivalently

r(L) = R * cos(L)

Maybe I should have taken a non-Euclidean geometry first?

Strangely enough, Mathworld:

http://mathworld.wolfram.com/SphericalGeometry.html

says that in a sphere, there are no parallel lines. This means that longitude lines intersect?!

Last fiddled with by jinydu on 2005-02-21 at 22:14
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Old 2005-02-21, 23:01   #8
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Please? How do I prove the result?
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Old 2005-02-21, 23:04   #9
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Quote:
Originally Posted by jinydu
Strangely enough, Mathworld:

http://mathworld.wolfram.com/SphericalGeometry.html

says that in a sphere, there are no parallel lines. This means that longitude lines intersect?!
Just think to project the sphere on a plane: the longitude lines are no more parallel lines, while the latitude lines are.

Luigi

Last fiddled with by ET_ on 2005-02-21 at 23:05
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Old 2005-02-21, 23:15   #10
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Quote:
Originally Posted by jinydu
Once I find the radius or circumference of the circle, the rest I can do in my sleep. The problem is getting to that point.

How could I prove mathematically that

C(L) = 2pi * R * cos(L)

or equivalently

r(L) = R * cos(L)?!
Project a couple of radii from the center of the earth, one going straight to the equator, the other approaching the point circling at latitude L, and measure its angle, then apply trigonometry to the right triangle having the hypotenuse as the radius and the height as a vertical line from the point at latitude L. The legs are proportional by the radius to the sine and cosine of the angle.

Luigi
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Old 2005-02-22, 01:24   #11
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Points on the earth don't move in a linear fashion. Even the poles trace out the elipse of the orbit.
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