Pyramid Science

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Monday, February 22, 2010

Orbit Distance With Time Relationship

The orbital distance (r) of any planet from the Sun has a definite relationship with the time it takes to circumnavigate the star. Considering the time (comparatively in P days) taken for a planet's 'year' and the distance from the central star either in millions of kilometres (mkm) or AU (Astronomical Units), a constancy is apparent.

Orbital Period (P days)

Mercury
  • P = 88, P2 = 7744
Venus
  • P = 224.7, P2 = 50490
Earth
  • P = 365.25, P2 = 133408
Mars
  • P = 686.9, P2 = 471832
Asteroid Belt
   
Jupiter
  • P = 4331.87, P2 = 18765098
Saturn
  • P = 10760.27, P2 = 115783411
Uranus
  • P = 30684.65, P2 = 941547746
Neptune
  • P = 60189.55, P2 = 3622781929
The Asteroid Belt is spread over a wide distance

Planetary distance (r) from the Sun (mkm, AU)

The respective planetary values below are for r, r3, AU, AU3

Mercury
  • 57.9, 194105, 0.387, 0.058
Venus
  • 108.2, 1266723, 0.723, 0.378
Earth
  • 149.6, 3348072, 1.000, 1.000
Mars
  • 227.9, 11836764, 1.523, 3.533
Asteroid Belt

Jupiter
  • 778.3, 471455918, 5.203, 140.852
Saturn
  • 1427, 2905841483, 9.539, 867.978
Uranus
  • 2871, 23664622310, 19.190, 7066.835
Neptune
  • 4497, 90942871470, 30.060, 27162.324

According to Kepler's third law

The respective planetary values below are for P2/r3, P2/AU3

Mercury 
  • 0.0399, 133517
Venus
  • 0.0399, 133571
Earth
  • 0.0398, 133408
Mars
  • 0.0399, 133550
Asteroid Belt                            

Jupiter
  • 0.0398, 133226
Saturn
  • 0.0398, 133394
Uranus
  • 0.0398, 133235
Neptune
  • 0.0398, 133375

Average
  • 0.0398, 133410
In both cases the constancy is clearly demonstrated and shows a relationship between a 2D quantity (function2) and a 3D quantity (function3).
    • A planet's 'year' will be completed after P days and it will travel through 360°. Relative to Mercury as the innermost planet and consequently the fastest moving, each of the other seven planets will move through less than this full circle. In the time taken for one complete orbit of Mercury (360°), the planet Venus will have moved only 88/224.7 * 360 = 140.99° and Earth by 88/365.25 * 360 = 86.735° and so on.
    Note: the ellipticity of a planet's orbit is ignored
    for the sake of this empirical argument

    Orbital Period (P days), Orbital (Circular) Angle °)

    Mercury
    • P = 88, θ = 360.00°
    Venus
    • P = 224.7, θ = 140.99°
    Earth
    • P = 365.25, θ = 86.735°
    Mars
    • P = 686.9, θ = 46.12°
    Asteroid Belt
           
    Jupiter
    • P = 4331.87, θ = 7.313°
    Saturn
    • P = 10760.27, θ = 2.944°
    Uranus
    • P = 30684.65, θ = 1.032°
    Neptune
    • P = 60189.55, θ = 0.526°
    The following values are derived from P and θ:
    • θ2 = 129600.00, AU3 = 0.058, θ2 * AU3 = 7516.80
    • θ2 = 19878.18, AU3 = 0.378, θ2 * AU3 = 7513.95
    • θ2 = 7522.96, AU3 = 1.000, θ2 * AU3 = 7522.90
    • θ2 = 2127.05, AU3 = 3.533, θ2 * AU3 = 7514.87
    • θ2 = 53.48, AU3 = 140.852, θ2 * AU3 = 7532.76
    • θ2 = 8.667, AU3 = 867.978, θ2 * AU3 = 7522.77
    • θ2 = 1.0658, AU3 = 7066.835, θ2 * AU3 = 7531.83
    • θ2 = 0.2767, AU3 = 27162.324, θ2 * AU3 = 7515.82
    In all cases the constancy is clearly demonstrated and again shows a relationship between a 2D quantity (function2) and a 3D quantity (function3).


        Sunday, February 14, 2010

        Earth: Origin Of Land Mass


        The 'mares' on the surface of the Moon can be viewed as lakes of residues from a molten mass. Unlike Earth land coastlines being well defined and the original locations easily found, the Moon surface is poorly defined. Recombining these residues creates a roughly circular area and the gravitational 'suction' pulling out a plug of molten surface matter as a slick of material is quite possible. The area of such a mass and its depth can be estimated.

        The Earth land mass volume can be approximated by calculating the surface area of the Earth and assuming an average height above the surface of 4km (the average depth - the sea floor is not flat - of the oceans is 3.79km). This allows for a global average of about 200m above sea level. The approximate surface area of the Earth is 4 * π * 6371 * 6371 = 510,064,472 square km and taking 29% as the proportion being land, this gives an area of 147,918,697 sq km. Using an average depth of the oceans as 3.79 km, an estimate of the volume of Earth's land areas can be made. Assuming the height of land above the Earth's underwater floor is 4km, a global average land mass volume is given by:

        147,918,697 * 4 = 591,674,788 cubic km

        As an average, this makes no allowance for crevices or undulations in the sea floor, but since this is a global average these are assumed to cancel out and for this reason have simply been ignored.

        Moon matter originates from the curved surface of an object much smaller than the Earth. The difference between a spherical Earth surface and a smaller spherical object (more pronounced) cannot be ignored. The apparent width of the Moon 'mares' facing the Earth compared to the diameter of the Moon itself can be estimated as a circular area of diameter approximately 2000km across. This is, however, a curved surface and as such has a greater distance: see the spherical cap. The diameter of the Moon is known (3476km) and the circumference is Dπ or 2πand equates to 10920.2km. The angle subtended to the edges of the apparent flat circle can be estimated by trigonometry from a right-angled triangle of hypotenuse (h) = 1738 (3476/2) and the opposite side (o) = 2000/2, one-half of the apparent width of the 'mares'.  This results in an angle of 70degrees. The proportion of the circumference is then provided by 70/360 * 10920.2 = 2123km (19.4%)This value provides only the curved distance in 2D. The real surface is 3D, so the area of a spherical cap of radius (r) is required: the area of a curved surface.

        In the formula 2πrh, the only value not known is that for h and is not the same quantity as the hypotenuse described above. This quantity is the height (h) above the apparent flat disc of the 'mares' and is taken as starting at the centre of the Moon (a + h = r), see the diagram: spherical cap. Introducing values for sphere radius (r = 1738km) and the chord (AB = 2000km) the value of h = 316.5km is achieved.


        Spherical cap area = 3,456,300 sq km

        This is the curved surface area of the Moon implicated in the molten splash. This area (3,456,300 sq km) would by necessity need to be over 170km deep to provide a volume of 591,674,788 cubic km.

         3,456,300sq km x 170m = 587,571,000cubic km

        It is, however, likely that the area of molten Moon mass would be shallow towards the edges and deepest in the middle. After ejection of the slick, the molten sea that remained would level itself under the influence of gravity. Around 1/6th that on Earth. More than sufficient. By comparison with conditions on the Earth, the seas are not flat, but follow the contours of Earth's curvature. The 'levelling' would not be like cutting a slice from the top of an egg, but the pool of molten sea would be curved.

        This calculation results in a very approximate value,
        but does illustrate the validity of the hypothesis
        that the land on the Earth could originate by a
        close encounter with a 'moon object'