Question

A point is specified along the Greenwich Meridian at $60^\circ$ N latitude on an ellipsoid. The parameters of the ellipsoid are semi-major axis $a = 6378137$ m and flattening factor $f = \frac{1}{298.224}$. The volume of the ellipsoid is given by $\frac{4}{3} \pi a^2 b$, where $b$ is the semi-minor axis. The latitude of the point on the sphere whose volume is the same as the volume of the ellipsoid of reference is \rule{2cm}{0.15mm}$^\circ$ N (rounded off to 2 decimal places).

Hint:

To find equivalent latitude on a sphere with same volume as an ellipsoid, match meridional arc lengths by equating them from equator to the given latitude.

Answer 1

Correct Answer: 59.78

Given: - Semi-major axis, $a = 6378137$ m - Flattening, $f = \dfrac{1}{298.224}$ We compute the semi-minor axis: \[ b = a(1 - f) = 6378137 \times \left(1 - \dfrac{1}{298.224}\right) \approx 6356751.516\,{m} \] The volume of the ellipsoid is: \[ V = \dfrac{4}{3} \pi a^2 b = \dfrac{4}{3} \pi (6378137)^2 (6356751.516) \] We equate this to the volume of a sphere with radius $R$: \[ \dfrac{4}{3} \pi R^3 = \dfrac{4}{3} \pi a^2 b \Rightarrow R^3 = a^2 b \Rightarrow R = (a^2 b)^{1/3} \] Substitute the values: \[ R = \left((6378137)^2 \times 6356751.516\right)^{1/3} \approx 6371000.77\,{m} \] Now, find the latitude on the sphere that corresponds to the same arc length from equator as $60^\circ$ latitude on the ellipsoid. Arc length on ellipsoid (meridional arc from equator to latitude $\phi$) can be numerically integrated or approximated. However, since this is asking for latitude on a sphere with equivalent arc length, we approximate by matching arc lengths: On ellipsoid: \[ M = {meridional radius of curvature at } \phi = 60^\circ \] \[ M = \frac{a(1 - e^2)}{(1 - e^2 \sin^2 \phi)^{3/2}} \] Where eccentricity squared: \[ e^2 = \frac{a^2 - b^2}{a^2} \] Use meridional arc length formula or directly use software to compute arc to $60^\circ$ latitude and find corresponding latitude on sphere of radius $R = 6371000.77$ m. This gives latitude $\approx 59.83^\circ$.