On a spherical planet of radius $R$ the intensity of attraction is measured gravitational at a heigh $x$ from the surface, the same measurements is made at a depth equal to the height, achieving both situations the same result.
Find the value of $x$ that satisfies this characteristics of the gravitational field.

[i]Proposed by Francisco Soriano, Cortes[/i]
1. **Gravitational Field Above the Surface:**
   The gravitational field at a height \( x \) above the surface of a spherical planet of radius \( R \) is given by:
   \[
   g_{\text{above}} = \dfrac{G \rho \cdot \tfrac{4\pi R^3}{3}}{(R+x)^2}
   \]
   Here, \( G \) is the gravitational constant, and \( \rho \) is the density of the planet.

2. **Gravitational Field Below the Surface:**
   By Gauss's law, the gravitational field at a depth \( x \) below the surface (which is equivalent to a distance \( R-x \) from the center) is given by:
   \[
   g_{\text{below}} = \dfrac{G \rho \cdot \tfrac{4\pi (R-x)^3}{3}}{(R-x)^2}
   \]

3. **Equating the Gravitational Fields:**
   Since the gravitational field measurements are the same at both the height \( x \) above the surface and the depth \( x \) below the surface, we set the two expressions equal to each other:
   \[
   \dfrac{G \rho \cdot \tfrac{4\pi R^3}{3}}{(R+x)^2} = \dfrac{G \rho \cdot \tfrac{4\pi (R-x)^3}{3}}{(R-x)^2}
   \]

4. **Simplifying the Equation:**
   Cancel out the common factors \( G \rho \cdot \tfrac{4\pi}{3} \) from both sides:
   \[
   \dfrac{R^3}{(R+x)^2} = \dfrac{(R-x)^3}{(R-x)^2}
   \]
   Simplify the right-hand side:
   \[
   \dfrac{R^3}{(R+x)^2} = R-x
   \]

5. **Solving for \( x \):**
   Multiply both sides by \( (R+x)^2 \):
   \[
   R^3 = (R-x)(R+x)^2
   \]
   Expand the right-hand side:
   \[
   R^3 = (R-x)(R^2 + 2Rx + x^2)
   \]
   Distribute \( (R-x) \):
   \[
   R^3 = R^3 + 2R^2x + Rx^2 - R^2x - 2Rx^2 - x^3
   \]
   Combine like terms:
   \[
   R^3 = R^3 + R^2x - Rx^2 - x^3
   \]
   Subtract \( R^3 \) from both sides:
   \[
   0 = R^2x - Rx^2 - x^3
   \]
   Factor out \( x \):
   \[
   0 = x(R^2 - Rx - x^2)
   \]
   This gives us two solutions:
   \[
   x = 0 \quad \text{or} \quad R^2 - Rx - x^2 = 0
   \]

6. **Solving the Quadratic Equation:**
   Solve the quadratic equation \( R^2 - Rx - x^2 = 0 \) using the quadratic formula \( x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   x = \dfrac{R \pm \sqrt{R^2 - 4R^2}}{2}
   \]
   Simplify under the square root:
   \[
   x = \dfrac{R \pm \sqrt{5R^2}}{2}
   \]
   Simplify further:
   \[
   x = \dfrac{R(-1 + \sqrt{5})}{2} \quad \text{or} \quad x = \dfrac{R(-1 - \sqrt{5})}{2}
   \]
   Since \( x \) must be a positive distance, we discard the negative solution:
   \[
   x = R \left( \dfrac{-1 + \sqrt{5}}{2} \right)
   \]

The final answer is \( \boxed{ x = R \left( \dfrac{-1 + \sqrt{5}}{2} \right) } \).