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)
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)$
$r= 6367447m$
$r= 6367.447km$
Wrong.. it gave value for polar radius.. but the question is to find radius at 45 latitude
ReplyDeleteCan you please explain the solution?
Deleter=Re(1-f(sin∆)^2)
DeleteI'm thinking just for confusing the question he given the latitude.
ReplyDeleter=Re(1-f(sin∆)^2)
ReplyDeleteCan you please provide the reference book with page number.
ReplyDeleteIf you like this blog please follow for latest updates. Thank you.
CMR Fowler solid earth page=197
DeleteThank you so much. I will update soon.
DeleteNow, you can find the updated solution. Thank you once again.
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