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Physical aspects of our Solar System

  1. Jul 5, 2005 #1
    Consider the physical aspects of our Solar System that result in the Earth having seasons... how does this gives us variable hours of daylight.

    How would i calculate the shortest or longest day for a specific point on earth?

    cheers.
     
  2. jcsd
  3. Jul 5, 2005 #2

    Zurtex

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    Err, what is your level of understanding of the dynamics of the Earth? I mean what do you know about how it travels around the sun? What measurements do you have about velocity of orbit, speed or rotation, tilt of axis etc..?
     
  4. Jul 5, 2005 #3

    radou

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    ever heard of www.google.com ? :biggrin:
     
  5. Jul 5, 2005 #4
    i have absolutely no data about any of the prior. what i DO have is various cities locations and the amount of daylight they experienced on the 4th of July.

    I should be able to extend these findings to make a generalisation about the hours of daylight for any latitude and longitude.

    there should be a relationship that relates the day of the year, the hours of daylight, the latitude and the longitude.
     
  6. Jul 5, 2005 #5

    Zurtex

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    Yeah, I wouldn't imagine it's all that complex but you need to know a bit about ellipses and spheres.

    However by the sounds of it you don't really have the mathematical knowledge to construct a method to work out day light given the latitude and longitude and the date. There are calculators all over the web that will just give an output for you though, like the above poster says, you could just use Google.
     
  7. Jul 5, 2005 #6
    i formula so far is

    12+(12/67)*latitude*cos(2pi.t)=hours of sunlight

    where t=days/365

    its ALMOST right ppl...
     
  8. Jul 5, 2005 #7

    Zurtex

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    May I ask how you derived that? I may be able to help you.

    Edit: Oh and it clearly isn't right as it wouldn't work for extreme places like such as the north and south poles where there can be daylight for days.

    I speant about 5 minuites attempting to draw out a model to caclulate it when I realised quite a lot of data. Anyway here is something I found while looking it up: http://www.spsu.edu/math/stinger/101-115/109.htm

    I have a feeling that is a bit simplified but I think it might do you nicely.
     
    Last edited: Jul 5, 2005
  9. Jul 7, 2005 #8

    uart

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    I remembering figuring this one out years ago when I was a student. I'm not sure if the equation I got was exact, but I think it was pretty close. The formula I arrived at was

    daylight_hours = 12 ( 1 + 2/Pi arcsin( tan(tilt) tan(lat) sin(2 Pi * n / 365.25) ) )

    Where,
    tilt = Constant of 0.4093 radians (23.45 degrees)
    lat = Latitude of location on Earth.
    n = Number of days elapsed since the northern hemishpere spring equinox.

    Notes.

    1. I've used radian throughout, just change each literal occurance of "Pi" to 180 if you want to use degrees throughout.

    2. The equation should apply without modification to both northern and southern hemispheres provided that southern latitudes are entered as negative.
     
    Last edited: Jul 7, 2005
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