The mass of a planet is half that of the earth and the radius of the planet is one - fourth that of earth . If we plan to send an artificail satellite from the planet , the escape velocity will be ( escape velocity on earth `v_e= 11km-s^-1`)

Options

A. `11kms^(-1)`
B. `5.5 kms^(-1)`
C. `15.5 kms^(-1)`
D. `7.7 kms^(-1)`

Answer

To find the escape velocity of a planet with a mass that is half that of Earth and a radius that is one-fourth that of Earth, we can use the formula for escape velocity:

### Step 1: Write the formula for escape velocity
The escape velocity \( v_e \) is given by the formula:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
where:
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the planet,
- \( R \) is the radius of the planet.

### Step 2: Define the mass and radius of the planet
Given:
- The mass of the planet \( M_p = \frac{1}{2} M_e \) (where \( M_e \) is the mass of Earth),
- The radius of the planet \( R_p = \frac{1}{4} R_e \) (where \( R_e \) is the radius of Earth).

### Step 3: Substitute the mass and radius into the escape velocity formula
Substituting the values into the escape velocity formula:
\[ v_{e,p} = \sqrt{\frac{2G \left(\frac{1}{2} M_e\right)}{\frac{1}{4} R_e}} \]

### Step 4: Simplify the equation
Now, simplifying the equation:
\[ v_{e,p} = \sqrt{\frac{2G \cdot \frac{1}{2} M_e}{\frac{1}{4} R_e}} = \sqrt{\frac{G M_e}{\frac{1}{4} R_e}} = \sqrt{\frac{4 G M_e}{R_e}} = 2 \sqrt{\frac{G M_e}{R_e}} \]

### Step 5: Relate it to Earth's escape velocity
We know that the escape velocity on Earth \( v_{e,e} \) is:
\[ v_{e,e} = \sqrt{\frac{2GM_e}{R_e}} \]
Thus, we can express the escape velocity on the planet as:
\[ v_{e,p} = 2 \cdot \sqrt{\frac{G M_e}{R_e}} = 2 \cdot \frac{v_{e,e}}{\sqrt{2}} = \sqrt{2} \cdot v_{e,e} \]

### Step 6: Substitute the known value of Earth's escape velocity
Given that the escape velocity on Earth \( v_{e,e} = 11 \, \text{km/s} \):
\[ v_{e,p} = \sqrt{2} \cdot 11 \, \text{km/s} \approx 1.414 \cdot 11 \, \text{km/s} \approx 15.5 \, \text{km/s} \]

### Final Answer
The escape velocity from the planet is approximately \( 15.5 \, \text{km/s} \).

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