Understanding the Problem Statement

The question asks us to find a mathematical relationship between the radius of a sphere, denoted by $R$, and the equatorial radius ($a$) and polar radius ($c$) of an international reference ellipsoid. This relationship must hold true specifically when the volume of the sphere is equal to the volume of the ellipsoid.

Formulas for Volumes

To establish the relationship, we first need the formulas for the volume of a sphere and the volume of an ellipsoid of revolution (which is commonly used as a reference ellipsoid, like the Earth's shape).

• The volume of a sphere with radius $R$ is given by the formula:

  $$V_{sphere} = \frac{4}{3}\pi R^3$$

• The volume of an ellipsoid with semi-axes $a$, $b$, and $c$ is $V_{ellipsoid} = \frac{4}{3}\pi abc$. For an ellipsoid of revolution, typically used as a reference ellipsoid, the equatorial radii are equal ($a = b$). Therefore, the volume simplifies to:

  $$V_{ellipsoid} = \frac{4}{3}\pi a^2 c$$

Deriving the Mathematical Relationship

The problem states that the volume of the sphere is equal to the volume of the ellipsoid. We can set the two volume formulas equal to each other:

$$V_{sphere} = V_{ellipsoid}$$

$$\frac{4}{3}\pi R^3 = \frac{4}{3}\pi a^2 c$$

We can cancel out the common term $\frac{4}{3}\pi$ from both sides:

$$R^3 = a^2 c$$

This equation represents the exact relationship between $R$, $a$, and $c$ for equal volumes. However, the options provided are linear approximations. To derive such a relationship, we can use an approximation method, assuming that the radii $a$ and $c$ are close to the sphere's radius $R$. This is often the case for reference ellipsoids like the Earth's.

Let's define the deviations of $a$ and $c$ from $R$ as small quantities:

Let $a = R + \delta_a$, where $\delta_a$ is the deviation in the equatorial radius.

Let $c = R + \delta_c$, where $\delta_c$ is the deviation in the polar radius.

Substitute these into the equation $R^3 = a^2 c$:

$$R^3 = (R + \delta_a)^2 (R + \delta_c)$$

Expand the terms:

$$R^3 = (R^2 + 2R\delta_a + \delta_a^2) (R + \delta_c)$$

$$R^3 = R^3 + R^2\delta_c + 2R^2\delta_a + 2R\delta_a\delta_c + R\delta_a^2 + \delta_a^2\delta_c$$

Now, we apply the approximation by neglecting higher-order terms (terms involving products of deviations or squares of deviations, like $\delta_a^2$, $\delta_c\delta_a$, etc.), assuming $\delta_a$ and $\delta_c$ are very small compared to $R$:

$$R^3 \approx R^3 + R^2\delta_c + 2R^2\delta_a$$

Subtract $R^3$ from both sides:

$$0 \approx R^2\delta_c + 2R^2\delta_a$$

Assuming $R \neq 0$, we can divide by $R^2$:

$$0 \approx \delta_c + 2\delta_a$$

Now, substitute back $\delta_a = a - R$ and $\delta_c = c - R$:

$$0 \approx (c - R) + 2(a - R)$$

Rearrange the terms:

$$0 \approx c - R + 2a - 2R$$

Combine the constants and variables:

$$0 \approx 2a + c - 3R$$

This gives us the approximate relationship:

$$3R \approx 2a + c$$

Verifying the Solution with Options

Let's rearrange the derived approximate relationship $3R \approx 2a + c$ to see which option it matches:

Subtract $c$ from both sides:

$$3R - c \approx 2a$$

Subtract $2R$ from both sides to isolate terms involving $R$ and $a$ on one side:

$$3R - 2R - c \approx 2a - 2R$$

Simplify the equation:

$$R - c \approx 2(a - R)$$

This derived approximate relationship closely matches the third option provided in the question.

Therefore, the correct relation among the radius of sphere $R$, and equatorial and polar radii of the international reference ellipsoid $a$ and $c$ respectively, when their volumes are equal, is approximately $R-c=2(a – R)$.