the value of the gravitational field strength...

# Explanation: ## Step1: Recall the inverse - square law The gravitational field strength $g$ is inversely proportional to the square of the distance $r$ from the center of the planet, i.e., $g=\frac{GM}{r^{2}}$. Let the distance from the center of the planet to location $X$ be $r_{X}$ and to location $Y$ be $r_{Y}$. Assume the distance from the center of the planet to location $X$ is $r_{X} = R$ (radius of the planet) and the distance from the center of the planet to location $Y$ is $r_{Y}=3R$ (by counting the grid - like lines in the diagram, assuming the innermost circle represents the planet's surface). ## Step2: Set up the ratio of gravitational field strengths We know that $g_{X}=\frac{GM}{r_{X}^{2}}$ and $g_{Y}=\frac{GM}{r_{Y}^{2}}$. Then $\frac{g_{Y}}{g_{X}}=\frac{r_{X}^{2}}{r_{Y}^{2}}$. ## Step3: Substitute the values Given $g_{X} = 27.0\ N/kg$, $r_{X}$ and $r_{Y}=3r_{X}$. Substituting into the ratio $\frac{g_{Y}}{g_{X}}=\frac{r_{X}^{2}}{(3r_{X})^{2}}=\frac{r_{X}^{2}}{9r_{X}^{2}}=\frac{1}{9}$. ## Step4: Solve for $g_{Y}$ $g_{Y}=\frac{g_{X}}{9}$. Since $g_{X} = 27.0\ N/kg$, then $g_{Y}=\frac{27.0}{9}=3.00\ N/kg$. # Answer: $3.00$