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Computing the period of a meteor using two ways.

  1. Jan 28, 2012 #1
    Computing the period of a meteor using two ways.

    I'm looking at code that computes the orbit found for a meteor. It computes the period of the meteor as:

    a = 13.07 # semi-major axis in AU
    P = m.sqrt(a*a*a) # From the program

    which yields P = 47.2512586393 Years

    If I use the "standard" approach

    P = 2*pi*m.sqrt((a*a*a)/mu),

    where mu (=GM) is the standard gravitational constant (G*M or GM).
    3986004418.0 in km**3/sec**2,

    then I get:

    P = 703462.683141 # units?

    They certainly don't look close. A source I'm looking at says 4*pi*pi/G = 1, but they are talking about solar masses.

    So why the large discrepancy? Units?
     
  2. jcsd
  3. Jan 28, 2012 #2

    D H

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    It is always a good idea to use the right constants and consistent units.

    You are using the gravitational parameter for the Earth, not the Sun. The solar gravitational parameter is 1.327124400180×1011 km3/s2. To have consistent units, you need to either convert that 13.07 AU to kilometers or convert the gravitational parameter to AU3/year2.

    With the first technique, convert that 13.07 AU to 1.955244×109 km. Then
    [itex]T = 2\pi\sqrt{a^3/\mu_{\odot}} \approx 1.49116\times10^9[/itex] seconds, or 47.25 sidereal years ( = 47.28 years).

    With the second technique, it's very handy to fold that factor of 2*pi into the square root. With this, the sun's gravitational parameter divided by 4*pi^2 is [itex]u_{\odot}' = 0.999997054[/itex] AU3/sidereal year2. Note that this is numerically equal to one to six places. This is what lets you get away with using [itex]T=\sqrt{a3}[/itex].
     
    Last edited: Jan 28, 2012
  4. Jan 28, 2012 #3
    They certainly don't look close. A source I'm looking at says 4*pi*pi/G = 1, but they are talking about solar masses.

    So why the large discrepancy? Units?[/QUOTE]

    Well, your certainly right about the units and the earth. I was using material I found on the web, and often it was unit-less. The book the computer program on never mentions how to compute P, but it's certainly in the code.

    Thanks very much.
     
  5. Jan 29, 2012 #4
    Well, not all worked out. Something is wrong. Here's a Python program and output.

    import math as m

    GMsun = 13271244018.0 # km**3/sec**-2
    Gsun = 6.67834*(10**-11)

    GMearth = 398600.4418 # km**3/sec**-2
    Rearth = 6378140.0 # radius of earth in km
    SunToEarth = 149597870.7 #km
    EarthMass = GMearth = 3986004418.0 # km**3/sec**-2
    EarthMass = 5.9742*10**24

    sidereal_year = 31558149.8 # sec
    AU = 149598000.0 # KM per AU

    pi = 3.14152965


    #
    print "Meteor Period"

    mu = GMsun/(4*pi*pi)

    a = 13.07*AU
    a1 =13.07
    P = (2.0*pi*m.sqrt((a*a*a)/mu)) / sidereal_year # Web
    Pbook = m.sqrt(a1*a1*a1) # Wray book
    print "P: ", P, " and Book: ",Pbook

    print "Check for mu_sun: ", mu

    =================
    Meteor Period
    P: 938.808534723 and Book: 47.2512586393 <---P from code is way off
    Check for mu_sun: 336178021.818
     
  6. Jan 31, 2012 #5
    To put a close to this, I have no idea why P = (2.0*pi*m.sqrt((a*a*a)/mu)) is useful. As it happens p^2/a^3 = 1 for all objects in orbit around the sun, planets anyway. So once you have a, the period is determine by the sqrt of a^3.
     
  7. Jan 31, 2012 #6

    D H

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    The expression T^2 = A^3 is valid only for orbits around the Sun, only for objects much less massive than the Sun, and only if you express time in sidereal years, distance in AU, and only if you don't care that the result is only good to a few decimal places.
     
  8. Jan 31, 2012 #7
    And the value of the other equation is what?
     
  9. Jan 31, 2012 #8

    D H

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    The other formula can be used to calculate the period of a satellite's orbit about the Earth. Or about Jupiter. Or the period of an exoplanet about some other star. And so on.
     
  10. Jan 31, 2012 #9
    Ah, thanks. Why is that true though?
     
    Last edited: Jan 31, 2012
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