Certainly! Let's determine the free-fall acceleration at the surface of the newly discovered planet compared to Earth, given that this planet has twice the mass and three times the radius of Earth.
The free-fall acceleration at the surface of a planet is given by the formula:
\[ g = \dfrac{GM}{R^2} \]
where:
- \( g \) is the free-fall acceleration,
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the planet, and
- \( R \) is the radius of the planet.
Given that the new planet has twice the mass (\( M_{\text{new}} = 2 \times M_{\text{Earth}} \)) and three times the radius (\( R_{\text{new}} = 3 \times R_{\text{Earth}} \)) of Earth, let's compare the free-fall acceleration at the surface of the new planet to that of Earth.
Using the formula, the ratio of the free-fall accelerations between the new planet and Earth can be expressed as:
\[ \dfrac{g_{\text{new}}}{g_{\text{Earth}}} = \dfrac{G \times 2M_{\text{Earth}} / (3R_{\text{Earth}})^2}{G \times M_{\text{Earth}} / R_{\text{Earth}}^2} \]
Simplifying this expression:
\[ \dfrac{g_{\text{new}}}{g_{\text{Earth}}} = \dfrac{2 \times M_{\text{Earth}} / (9 \times R_{\text{Earth}}^2)}{M_{\text{Earth}} / R_{\text{Earth}}^2} = \dfrac{2}{9} \]
Therefore, the ratio of the free-fall accelerations is \( \dfrac{g_{\text{new}}}{g_{\text{Earth}}} = \dfrac{2}{9} \), indicating that the free-fall acceleration at the surface of the new planet is \( \dfrac{2}{9} \) times the free-fall acceleration at the surface of Earth.
Hence, the correct answer is:
a)\( \dfrac{2}{9} \, g \)
