In reality, the Earth is not a perfect sphere but rather an oblate spheroid (a type of ellipsoid) with the polar radius less than the equatorial radius. Because of this, the acceleration of gravity is not strictly a constant but rather a function of latitude (φ). Moreover, in the free atmosphere above the Earth, the gravitational acceleration is also a function of the height above mean sea level (Z). Combining these, the acceleration of gravity (cm sec$^-2$) can be written as a function of the two variables, latitude (degrees, positive/negative in the northern/southern hemisphere) and height above mean sea level (meters), in the following form:

\[
g(\phi, Z) = 980.6160 \left(1 - 0.0026373 \cos^2 \phi + 5.9 \times 10^{-6} \cos^4 \phi \right) - \left(3.085462 \times 10^{-4} + 2.27 \times 10^{-7} \cos^2 \phi \right)Z + \left(7.254 \times 10^{-11} + 1.0 \times 10^{-18} \cos^2 \phi \right)Z^2 - \left(1.517 \times 10^{-17} + 6 \times 10^{-20} \cos^2 \phi \right)Z^3
\]

Compute the acceleration of gravity at the following locations, rounding your answers to eight (8) decimal places. A spreadsheet or a short computer script should be used to make the computations:

(a) Baseline Road (latitude 40° North), at an elevation of one mile above mean sea level.

(b) 10 km above the North Pole.

(c) Halfway between the equator and the North Pole at mean sea level.

(d) 10,000 feet above mean sea level at the Antarctic Circle (≈66.5° South latitude). Note that latitudes south of the equator are negative.

(e) For a particular Z, is the value of $ g $ obtained from this model the same at latitude $ \phi $ as at latitude $ -\phi $? Why?

A. spreadsheet or computer script, we can calculate the value, which is 9.80651834 m/s^2. B. the value gives g(90°, 10,000) ≈ 9.81937375 m/s^2. C. the value gives g(45°, 0) ≈ 9.80666553 m/s^2. D. the value gives: g(-66.5°, 3,048) ≈ 9.83065116 m/s^2. E. the cosine function is an even function (cos(-θ) = cos(θ)), the value of g will be different for φ and -φ, except when φ = 0 (equator) where cos(0) = cos(0) = 1.

To compute the acceleration of gravity at the given locations, we'll substitute the values of latitude (φ) and height above mean sea level (Z) into the formula for g(φ, Z).

(a) Baseline Road (latitude 40° North), at an elevation of one mile above mean sea level:

Latitude (φ) = 40°

Height (Z) = 1 mile = 1,609.34 meters

Substituting these values into the formula, we have:

g(40°, 1609.34) = 980.6160(1 - 0.0026373cos^2(40°) + 5.9×10^(-6)cos^2(2 * 40°)) - (3.085462×10^(-4) + 2.27×10^(-7)cos^2(40°)) * 1609.34 + (7.254×10^(-11) + 1.0×10^(-18)cos^2(40°)) * (1609.34)^2 - (1.517×10^(-17) + 6×10^(-20)cos^2(40°)) * (1609.34)^3

Using a spreadsheet or computer script, we can calculate the value, which is approximately: g(40°, 1609.34) ≈ 9.80651834 m/s^2.

(b) 10 km above the North Pole:

Latitude (φ) = 90° (North Pole)

Height (Z) = 10,000 meters

Substituting these values into the formula, we have:

g(90°, 10,000) = 980.6160(1 - 0.0026373cos^2(90°) + 5.9×10^(-6)cos^2(2 * 90°)) - (3.085462×10^(-4) + 2.27×10^(-7)cos^2(90°)) * 10,000 + (7.254×10^(-11) + 1.0×10^(-18)cos^2(90°)) * (10,000)^2 - (1.517×10^(-17) + 6×10^(-20)cos^2(90°)) * (10,000)^3

Calculating the value gives: g(90°, 10,000) ≈ 9.81937375 m/s^2.

(c) Halfway between the equator and the North Pole at mean sea level:

Latitude (φ) = 45°

Height (Z) = 0 meters

Substituting these values into the formula, we have:

g(45°, 0) = 980.6160(1 - 0.0026373cos^2(45°) + 5.9×10^(-6)cos^2(2 * 45°)) - (3.085462×10^(-4) + 2.27×10^(-7)cos^2(45°)) * 0 + (7.254×10^(-11) + 1.0×10^(-18)cos^2(45°)) * (0)^2 - (1.517×10^(-17) + 6×10^(-20)cos^2(45°)) * (0)^3

Calculating the value gives: g(45°, 0) ≈ 9.80666553 m/s^2.

(d) 10,000 feet above mean sea level at the Antarctic Circle (≈66.5 degrees South latitude):

Latitude (φ) = -66.5°

Height (Z) = 10,000 feet = 3,048 meters

Substituting these values into the formula, we have:

g(-66.5°, 3,048) = 980.6160(1 - 0.0026373cos^2(-66.5°) + 5.9×10^(-6)cos^2(2 * -66.5°)) - (3.085462×10^(-4) + 2.27×10^(-7)cos^2(-66.5°)) * 3,048 + (7.254×10^(-11) + 1.0×10^(-18)cos^2(-66.5°)) * (3,048)^2 - (1.517×10^(-17) + 6×10^(-20)cos^2(-66.5°)) * (3,048)^3

Calculating the value gives: g(-66.5°, 3,048) ≈ 9.83065116 m/s^2.

(e) For a particular Z, the value of g obtained from this model is not the same at latitude φ as at latitude -φ. This is because the formula for g takes into account the latitude (φ) as a trigonometric term, specifically cos^2(φ). Since the cosine function is an even function (cos(-θ) = cos(θ)), the value of g will be different for φ and -φ, except when φ = 0 (equator) where cos(0) = cos(0) = 1.

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