Three planets of same density have radii `R_(1),R_(2)` and `R_(3)` such that `R_(1) = 2R_(2) = 3R_(3)`. The gravitational field at their respective surfaces are `g_(1), g_(2)` and `g_(3)` and escape velocities from their surfeces are `upsilon_(1),upsilon_(2)` and `upsilon_(3)`, then

Options

A. `g_(1)//g_(2) = 2`
B. `g_(1)//g_(3) = 3`
C. `upsilon_(1)//upsilon_(2) = 1//4`
D. `upsilon_(1)//upsilon_(3) = 3`

Explanation

To solve the problem, we need to establish the relationships between the gravitational fields and escape velocities of the three planets based on their radii.

### Step 1: Establish the relationship between the radii
Given:
- \( R_1 = 2R_2 \)
- \( R_2 = \frac{1}{3}R_1 \)
- \( R_3 = \frac{1}{2}R_2 \)

We can express all radii in terms of \( R_3 \):
- \( R_1 = 3R_3 \)
- \( R_2 = 2R_3 \)

### Step 2: Calculate the gravitational field at the surface of each planet
The formula for gravitational field \( g \) at the surface of a planet is given by:
\[ g = \frac{GM}{R^2} \]
Where \( M \) is the mass of the planet. Since the density \( \rho \) is the same for all planets, we can express mass \( M \) in terms of density and volume:
\[ M = \rho \cdot \text{Volume} = \rho \cdot \frac{4}{3} \pi R^3 \]
Thus, the gravitational field becomes:
\[ g = \frac{G \cdot \rho \cdot \frac{4}{3} \pi R^3}{R^2} = \frac{4}{3} G \rho \cdot R \]

So we have:
- \( g_1 = \frac{4}{3} G \rho R_1 \)
- \( g_2 = \frac{4}{3} G \rho R_2 \)
- \( g_3 = \frac{4}{3} G \rho R_3 \)

### Step 3: Relate the gravitational fields
Using the relationships of the radii:
- \( g_1 = \frac{4}{3} G \rho (3R_3) = 4G \rho R_3 \)
- \( g_2 = \frac{4}{3} G \rho (2R_3) = \frac{8}{3} G \rho R_3 \)
- \( g_3 = \frac{4}{3} G \rho R_3 \)

Now we can find the ratios:
\[ \frac{g_1}{g_2} = \frac{4G \rho R_3}{\frac{8}{3} G \rho R_3} = \frac{4}{\frac{8}{3}} = \frac{4 \cdot 3}{8} = \frac{3}{2} \]
\[ \frac{g_1}{g_3} = \frac{4G \rho R_3}{\frac{4}{3} G \rho R_3} = 3 \]

### Step 4: Calculate the escape velocities
The escape velocity \( v \) from the surface of a planet is given by:
\[ v = \sqrt{2gR} \]
Thus:
- \( v_1 = \sqrt{2g_1R_1} \)
- \( v_2 = \sqrt{2g_2R_2} \)
- \( v_3 = \sqrt{2g_3R_3} \)

### Step 5: Relate the escape velocities
Using the expressions for \( g \) and \( R \):
\[ v_1 = \sqrt{2 \cdot \frac{4}{3} G \rho (3R_3) \cdot (3R_3)} = \sqrt{2 \cdot \frac{4}{3} G \rho \cdot 9R_3^2} = 3\sqrt{6G\rho} \]
\[ v_2 = \sqrt{2 \cdot \frac{8}{3} G \rho (2R_3) \cdot (2R_3)} = \sqrt{2 \cdot \frac{8}{3} G \rho \cdot 4R_3^2} = 4\sqrt{\frac{4G\rho}{3}} \]
\[ v_3 = \sqrt{2 \cdot \frac{4}{3} G \rho (R_3) \cdot (R_3)} = \sqrt{2 \cdot \frac{4}{3} G \rho \cdot R_3^2} = \sqrt{\frac{8G\rho}{3}} \]

### Step 6: Find ratios of escape velocities
\[ \frac{v_1}{v_2} = \frac{3\sqrt{6G\rho}}{4\sqrt{\frac{4G\rho}{3}}} = \frac{3\sqrt{6}}{4\cdot\frac{2}{\sqrt{3}}} = \frac{3\sqrt{18}}{8} = \frac{3\cdot3\sqrt{2}}{8} = \frac{9\sqrt{2}}{8} \]
\[ \frac{v_1}{v_3} = \frac{3\sqrt{6G\rho}}{\sqrt{\frac{8G\rho}{3}}} = \frac{3\sqrt{18}}{4} = \frac{9}{4} \]

### Final Results
- \( \frac{g_1}{g_2} = \frac{3}{2} \)
- \( \frac{g_1}{g_3} = 3 \)
- \( \frac{v_1}{v_2} = \frac{9\sqrt{2}}{8} \)
- \( \frac{v_1}{v_3} = \frac{9}{4} \)