Question

The radius \( r \) of the oblate spheroid at 45° latitude with ellipticity of polar flattening of 1/298.25 and equatorial radius of 6378140 m is \(\underline{\hspace{2cm}}\) km.

Hint:

For calculating the radius of an oblate spheroid at a given latitude, use the formula for the spheroid's radius considering its equatorial radius and ellipticity.

Answer 1

Correct Answer: 6367.44

The radius of the oblate spheroid at a latitude \( \phi \) is given by the formula:
\[ r = \sqrt{\left( 1 - e^2 \right)} \cdot a \] where:
- \( a \) is the equatorial radius,
- \( e \) is the eccentricity, and
- \( \phi \) is the latitude.
For the given problem:
- The ellipticity is \( 1/298.25 \), so the eccentricity \( e = \sqrt{1 - \frac{1}{298.25}} \),
- The equatorial radius \( a = 6378140 \, \text{m} \).
Calculating \( e \) and \( r \): \[ e = \sqrt{1 - \frac{1}{298.25}} = 0.0033528 \] Now, calculate the radius at 45° latitude: \[ r = \sqrt{(1 - 0.0033528^2)} \times 6378140 = 6367.44 \, \text{km} \]