What will be the escape speed from a planet having mass 16 times that of earth and diameter 8 times that of the earth ? `(v_(e)=11.2 "kms"^(-1))`

Options

A. `12.82 "kms"^(-1)`
B. `15.84 "km s"^(-1)`
C. `13.85 "km s"^(-1)`
D. `10.54 "km s"^(-1)`

Explanation

To find the escape speed from a planet with a mass 16 times that of Earth and a diameter 8 times that of Earth, we can follow these steps:

### Step 1: Understand the formula for escape velocity
The escape velocity \( v_e \) from a celestial body is given by the formula:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the body, and \( R \) is its radius.

### Step 2: Relate the planet's mass and radius to Earth's
Given:
- Mass of the planet \( M_p = 16 M_e \) (where \( M_e \) is the mass of Earth)
- Diameter of the planet \( D_p = 8 D_e \) (where \( D_e \) is the diameter of Earth)

Since radius is half of the diameter, we have:
\[ R_p = \frac{D_p}{2} = \frac{8 D_e}{2} = 4 D_e \]
Thus, the radius of the planet is:
\[ R_p = 8 R_e \]

### Step 3: Substitute the values into the escape velocity formula
Now we can substitute the values of mass and radius into the escape velocity formula:
\[ v_{e,p} = \sqrt{\frac{2G(16 M_e)}{8 R_e}} \]

### Step 4: Simplify the expression
We can simplify the expression:
\[ v_{e,p} = \sqrt{\frac{16 \cdot 2GM_e}{8 R_e}} = \sqrt{2 \cdot 2 \cdot \frac{GM_e}{R_e}} = \sqrt{4 \cdot \frac{GM_e}{R_e}} \]
Since \( \sqrt{\frac{GM_e}{R_e}} \) is the escape velocity from Earth \( v_{e,e} \):
\[ v_{e,p} = \sqrt{4} \cdot v_{e,e} = 2 v_{e,e} \]

### Step 5: Calculate the escape velocity
Given that the escape velocity from Earth \( v_{e,e} = 11.2 \, \text{km/s} \):
\[ v_{e,p} = 2 \cdot 11.2 \, \text{km/s} = 22.4 \, \text{km/s} \]

### Final Answer
The escape speed from the planet is:
\[ \boxed{22.4 \, \text{km/s}} \]