Question

If \(r_1\) and \(r_2\), respectively, denote the equatorial and polar radii of a reference ellipsoid, then the radius, \(R\), of its equivalent sphere is given by

• A. \( \left(r_1 r_2^2\right)^{2/3} \)
• B. \( r_1 r_2^2 \)
• C. \( \left(r_1^2 r_2\right)^{1/3} \)
• D. \( r_1^2 r_2 \)

Hint:

Equate the ellipsoid's volume (4/3 pi r1^2 r2) with a sphere's volume (4/3 pi R^3) and solve for R.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
A reference ellipsoid used in geodesy is an oblate spheroid, meaning it is a sphere squashed slightly at the poles. Its equatorial radius \(r_1\) is the same in every direction around the equator, while its polar radius \(r_2\) is shorter, measured from center to pole. We need the radius \(R\) of a sphere that has the same volume as this ellipsoid, called the equivalent sphere.

Step 2: Key Formula or Approach:
An oblate spheroid can be treated as a solid with two equal semi axes of length \(r_1\) and one semi axis of length \(r_2\). Its volume is:
\[ V_{ellipsoid} = \frac{4}{3}\pi r_1^2 r_2 \]
The volume of a sphere of radius \(R\) is:
\[ V_{sphere} = \frac{4}{3}\pi R^3 \]
Setting the two volumes equal, since the equivalent sphere by definition has the same volume as the ellipsoid, gives the formula for \(R\).

Step 3: Detailed Explanation:
Equating the two volumes:
\[ \frac{4}{3}\pi R^3 = \frac{4}{3}\pi r_1^2 r_2 \]
The \(\frac{4}{3}\pi\) on both sides cancels, leaving:
\[ R^3 = r_1^2 r_2 \]
Taking the cube root of both sides:
\[ R = \left(r_1^2 r_2\right)^{1/3} \]

Step 4: Final Answer:
The radius of the equivalent sphere is \(R = \left(r_1^2 r_2\right)^{1/3}\), matching option (C). Option (A) raises the wrong grouping to the power 2/3, which does not even come out to a length. Option (B) and option (D) both leave the volume expression without taking the needed cube root, so they are not radii at all.
\[ \boxed{R = \left(r_1^2 r_2\right)^{1/3}} \]