A planet has a core of density `rho_1` and the density of the outer shell is `rho_2`. R and 3R are the radii of the core and the planet respectively. Also, it is given that gravitational acceleration at the surface of the planet is equal to that at the surface of core. Hence, compute the ratio e ratio `(rho_1)/(rho_2)`.

Explanation

To solve the problem, we need to find the ratio of the densities of the core and the outer shell of the planet, given that the gravitational acceleration at the surface of the planet is equal to that at the surface of the core.

### Step-by-step Solution:

1. **Understanding the Problem:**
- The core of the planet has a radius \( R \) and density \( \rho_1 \).
- The outer shell of the planet has a radius \( 3R \) and density \( \rho_2 \).
- We need to find the ratio \( \frac{\rho_1}{\rho_2} \) given that the gravitational acceleration at the surface of the core is equal to that at the surface of the planet.

2. **Gravitational Acceleration at the Core:**
- The gravitational acceleration \( g_1 \) at the surface of the core can be calculated using the formula:
\[ g_1 = \frac{G M_1}{R^2} \]
- Here, \( M_1 \) is the mass of the core, which can be expressed as:
\[ M_1 = \frac{4}{3} \pi R^3 \rho_1 \]
- Substituting \( M_1 \) into the equation for \( g_1 \):
\[ g_1 = \frac{G \left( \frac{4}{3} \pi R^3 \rho_1 \right)}{R^2} = \frac{4}{3} \pi G R \rho_1 \]

3. **Gravitational Acceleration at the Surface of the Planet:**
- The gravitational acceleration \( g_2 \) at the surface of the planet can be calculated similarly:
\[ g_2 = \frac{G M_2}{(3R)^2} \]
- The mass \( M_2 \) of the entire planet can be calculated as:
\[ M_2 = M_1 + M_{shell} \]
- The mass of the shell \( M_{shell} \) is given by:
\[ M_{shell} = \frac{4}{3} \pi ( (3R)^3 - R^3 ) \rho_2 = \frac{4}{3} \pi (27R^3 - R^3) \rho_2 = \frac{4}{3} \pi (26R^3) \rho_2 \]
- Therefore, the total mass \( M_2 \) is:
\[ M_2 = \frac{4}{3} \pi R^3 \rho_1 + \frac{4}{3} \pi (26R^3) \rho_2 = \frac{4}{3} \pi R^3 \left( \rho_1 + 26 \rho_2 \right) \]
- Substituting \( M_2 \) into the equation for \( g_2 \):
\[ g_2 = \frac{G \left( \frac{4}{3} \pi R^3 \left( \rho_1 + 26 \rho_2 \right) \right)}{(3R)^2} = \frac{G \left( \frac{4}{3} \pi R^3 \left( \rho_1 + 26 \rho_2 \right) \right)}{9R^2} = \frac{4}{27} \pi G R \left( \rho_1 + 26 \rho_2 \right) \]

4. **Setting the Gravitational Accelerations Equal:**
- Since \( g_1 = g_2 \):
\[ \frac{4}{3} \pi G R \rho_1 = \frac{4}{27} \pi G R \left( \rho_1 + 26 \rho_2 \right) \]
- Canceling \( \frac{4}{\pi G R} \) from both sides:
\[ \frac{1}{3} \rho_1 = \frac{1}{27} \left( \rho_1 + 26 \rho_2 \right) \]
- Multiplying through by 27 to eliminate the fraction:
\[ 9 \rho_1 = \rho_1 + 26 \rho_2 \]
- Rearranging gives:
\[ 8 \rho_1 = 26 \rho_2 \]
- Thus, the ratio of the densities is:
\[ \frac{\rho_1}{\rho_2} = \frac{26}{8} = \frac{13}{4} \]

### Final Answer:
The ratio \( \frac{\rho_1}{\rho_2} = \frac{13}{4} \).