Calculation of true thickness of the bed

 18) The Radius(r) of the oblate spheroid at 45⁰ latitude with ellipticity of polar flattening of 1/298.25 and equatorial radius of 6378140m is ________km.

 (Thanks to Chandrasekhar, ANU)

Solution:

 Given that Polar Flattening (f) = 1/298.25

Equatorial radius $(R_e)$ = 6378140 m

The radius of the oblate spheroid (r) =?

$latitude( \phi)=45^0$

The relation between oblate spheroid radius, polar Flattening and latitude is given that,

$r= R_e (1-f \sin^2 \phi)$ ---(1)

polar platting formula,

$f=\frac{R_e-R_p}{R_e}$ = $\frac{1}{298.25}$

Substitute the given values in the above equation (1)

$r= 6378140 (1-\frac{1}{298.25} \sin^2 45^0)$

$r= 6378140 (1-\frac{1}{298.25} \sin^2 45^0)$

$r= 6378140 (1-\frac{1}{298.25}0.5)$

$r= 6378140 (1-0.001676445935)$

$r= 6378140 (0.9983)$

$r= 6378140 (0.9983)$

$r= 6367447m$

$r= 6367.447km$

Reference: The Solid Earth , An Introduction to Global Geophysics by C.M.R. Fowler