Tough Math(s) Problem | Why is the Sky Blue? | Fun Trivia Forums

Rules
Terms of Use

You are not logged in. [Log In] FunTrivia Home » Forums » The FunTrivia Community » Why is the Sky Blue? » Tough Math(s) Problem

Topic Options

#214119 - Sun Feb 22 2004 10:45 AM Tough Math(s) Problem
ozzz2002 Moderator Registered: Mon Dec 03 2001 Posts: 20912 Loc: Sydney NSW Australia	This is a question that was reecently asked of me, and I admit to being stumped. The Spin of the Earth. Obviously, the areas further from the equator will be spinning slower relative to the areas at the equator. I would assume that if the equator is eg 10000kms in circumference, then any area away from the equator would be spinning at a reduced percentage speed. So if it takes 24 hours to rotate 10000kms at a point on the equator, then at what point towards the poles it would take the earth 24 hours to spin, eg 7000kms. What happens at the poles, even though the earth is oblate? Is there a simple way of calculating the distance that a non-equatorial point would travel in 24 hours? For ease of calculation, let us assume that the earth is a perfect sphere. _________________________ The key to everything is patience. You get the chicken by hatching the egg, not smashing it. Ex-Editor, Hobbies and Sports, and Forum Moderator
Top

#214120 - Sun Feb 22 2004 11:59 AM Re: Tough Math(s) Problem
Leau Forum Champion Registered: Sun Jun 16 2002 Posts: 5337 Loc: Nijmegen/Brisbane	Quote: Is there a simple way of calculating the distance that a non-equatorial point would travel in 24 hours? I found some formulas that should/could do the trick, but I wouldn't exactly call it simple. Basically, if you want to know the distance that a point A would travel in 24 hours all you need to know is the earth's circumference at that point. Quote: As you go from 0 degrees latitude (the equator) to 90 degrees (north or south poles), the circumference of the circle defined by that latitude line will decrease in direct proportion to the cosine of the angle of latitude. Thus, the circumference of the circle is C = 2 pi r cos(x), where pi = 3.14159..., r = the earth's equatorial radius = 6378 km, and x is the angle of latitude. This formula is not entirely correct, as it assumes the earth is a perfect sphere, which it isn't. In fact, the earth bulges out a bit at the equator, as a consequence of its rotation. This effect is small, however, and the formula is accurate to within one percent. Source:www.newton.dep.anl.gov If you want to know the rotational speed for your point A, you'd have to calculate the circumference at that point and divide it by the time in a day. Edited by Leau78 (Sun Feb 22 2004 12:01 PM) _________________________ The cost of living has not affected its popularity - Loesje
Top

#214121 - Mon Feb 23 2004 05:27 AM Re: Tough Math(s) Problem
Leau Forum Champion Registered: Sun Jun 16 2002 Posts: 5337 Loc: Nijmegen/Brisbane	I tried to calculate the distance someone in Sydney (you? ) would travel in one day because of the earth's rotation, so in effect the circumference of the earth for Sydney's latitude. C = 2 pi r cos(x) The Sydney latitude is 033°52'59". In "normal" degrees - without the minutes and seconds - this is 33.88305555555556° (). The number that has to be filled in for x in the formula has to be in radians, so a conversion is necessary. Since 180°= pi radians -> 33.88° = 33.88°/180°pi radians = 0.59137088 radians. C = 2 * pi * 6378 * cos (0.59137088) = 33268.6503 So you travel 33268 km a day! (*) For clarity I've shortened this to 33.88° in the rest of the story, but the calculations have been done with the exact number. _________________________ The cost of living has not affected its popularity - Loesje
Top

#214122 - Mon Feb 23 2004 06:55 AM Re: Tough Math(s) Problem
ozzz2002 Moderator Registered: Mon Dec 03 2001 Posts: 20912 Loc: Sydney NSW Australia	Wow, thanks, Leau! I am impressed (and exhausted from covering all that distance, hehe). _________________________ The key to everything is patience. You get the chicken by hatching the egg, not smashing it. Ex-Editor, Hobbies and Sports, and Forum Moderator
Top

#214123 - Mon Feb 23 2004 01:31 PM Re: Tough Math(s) Problem
A Member Multiloquent Registered: Fri Nov 23 2001 Posts: 3082 Loc:	Am I missing something here? Assuming the Earth to be a spinning ball. If the equator is travelling at 24,000 km per second then 50 degrees latitude must be travelling at 24,000/90 x 40 (90 degrees minus the number) At 10 degrees before the pole it will be 24,000/90 x 10 so at the pole it will be 24,000/0 x 0 (ie not moving!) The logic sounds good but is it correct? _________________________
Top

#214124 - Wed Feb 25 2004 03:30 PM Re: Tough Math(s) Problem
Biggles Forum Adept Registered: Thu Jan 09 2003 Posts: 170 Loc: England	The linear relationship between latitude and speed would only be true if the bottom half of the world (or the top) was conical, with the pole at the point of the cone. As the world is spherical (well, very nearly) the distance around it at any latitude varies with the cosine. Don't ask me to prove it!
Top

View All Topics

Moderator: TabbyTom