The mass of a planet is six times that of the earth. The radius of the planet is twice that of the earth. If the escape velocity from the earth is `v`, then the escape velocity from the planet is:

Options

A. `19.4 km//s`
B. `44.8 km//s`
C. `22.4 km//s`
D. `5.6 km//s`

Answer

To find the escape velocity from the planet, we can use the formula for escape velocity, which is given by:

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

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

### Step 1: Identify the parameters for Earth
Let:
- Mass of Earth = \( M \)
- Radius of Earth = \( R \)
- Escape velocity from Earth = \( v = \sqrt{\frac{2GM}{R}} \)

### Step 2: Identify the parameters for the planet
According to the problem:
- Mass of the planet = \( 6M \) (six times that of Earth)
- Radius of the planet = \( 2R \) (twice that of Earth)

### Step 3: Substitute the parameters into the escape velocity formula for the planet
Now, we can calculate the escape velocity from the planet using the modified parameters:

\[ v_{e, \text{planet}} = \sqrt{\frac{2G(6M)}{2R}} \]

### Step 4: Simplify the expression
We can simplify the expression:

\[ v_{e, \text{planet}} = \sqrt{\frac{12GM}{2R}} = \sqrt{\frac{6GM}{R}} \]

### Step 5: Relate it to the escape velocity from Earth
We know that \( v = \sqrt{\frac{2GM}{R}} \). Therefore, we can express \( \sqrt{\frac{6GM}{R}} \) in terms of \( v \):

\[ v_{e, \text{planet}} = \sqrt{3} \cdot \sqrt{\frac{2GM}{R}} = \sqrt{3} \cdot v \]

### Step 6: Substitute the value of \( v \)
Given that the escape velocity from Earth is \( v \), we can express the escape velocity from the planet as:

\[ v_{e, \text{planet}} = v \cdot \sqrt{3} \]

### Step 7: Calculate the numerical value
If the escape velocity from Earth is approximately \( 11.2 \, \text{km/s} \):

\[ v_{e, \text{planet}} = 11.2 \cdot \sqrt{3} \approx 11.2 \cdot 1.732 \approx 19.4 \, \text{km/s} \]

### Final Answer
The escape velocity from the planet is approximately \( 19.4 \, \text{km/s} \).

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