The escape velocity of a particle on earth ( radius R and mass M ) is 11.2 `kms^(-1)` . What is the escape velocity on another planet with radius `R//2` and mass `M//4` ?

Explanation

To find the escape velocity on another planet with a radius of \( R/2 \) and mass \( M/4 \), we can use the formula for escape velocity:

\[ v_e = \sqrt{\frac{2GM}{R}} \]

where:
- \( v_e \) is the escape velocity,
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the planet,
- \( R \) is the radius of the planet.

### Step 1: Write the escape velocity for Earth
For Earth, the escape velocity \( v_e \) is given as 11.2 km/s. Using the formula:

\[ v_{e,\text{Earth}} = \sqrt{\frac{2GM}{R}} \]

### Step 2: Write the escape velocity for the other planet
For the other planet with mass \( M/4 \) and radius \( R/2 \), the escape velocity \( v_{p} \) can be expressed as:

\[ v_{p} = \sqrt{\frac{2G(M/4)}{(R/2)}} \]

### Step 3: Simplify the expression for \( v_{p} \)
Now, simplify the expression:

\[ v_{p} = \sqrt{\frac{2G(M/4)}{(R/2)}} = \sqrt{\frac{2G \cdot M/4}{R/2}} = \sqrt{\frac{2G \cdot M}{4} \cdot \frac{2}{R}} = \sqrt{\frac{2GM}{2R}} = \sqrt{\frac{GM}{R}} \]

### Step 4: Relate \( v_{p} \) to \( v_{e,\text{Earth}} \)
Now, we can relate \( v_{p} \) to the escape velocity of Earth:

\[ v_{p} = \sqrt{\frac{1}{2}} \cdot v_{e,\text{Earth}} = \frac{v_{e,\text{Earth}}}{\sqrt{2}} \]

### Step 5: Substitute the value of \( v_{e,\text{Earth}} \)
Substituting the value of \( v_{e,\text{Earth}} = 11.2 \text{ km/s} \):

\[ v_{p} = \frac{11.2 \text{ km/s}}{\sqrt{2}} \approx \frac{11.2 \text{ km/s}}{1.414} \approx 7.9 \text{ km/s} \]

### Final Answer
Thus, the escape velocity on the other planet is approximately:

\[ \boxed{7.9 \text{ km/s}} \]