Mass and radius of a planet are two times the value of earth. What is the value of escape velocity from the surface of this planet?

Explanation

To find the escape velocity from the surface of a planet whose mass and radius are both twice that of Earth, we can follow these steps:

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

### Step 2: Identify the values for the planet
Given that the mass \( M \) of the planet is twice that of Earth (\( M = 2M_E \)) and the radius \( R \) of the planet is also twice that of Earth (\( R = 2R_E \)), we can substitute these values into the escape velocity formula.

### Step 3: Substitute the values into the formula
Substituting the values into the escape velocity formula:
\[ v_e = \sqrt{\frac{2G(2M_E)}{2R_E}} \]

### Step 4: Simplify the equation
Now, simplify the equation:
\[ v_e = \sqrt{\frac{2 \cdot 2GM_E}{2R_E}} = \sqrt{\frac{2GM_E}{R_E}} \]
Notice that the factor of 2 in the numerator and denominator cancels out.

### Step 5: Recognize the escape velocity from Earth
The expression \( \sqrt{\frac{2GM_E}{R_E}} \) is the escape velocity from the surface of Earth, which is approximately \( 11.2 \, \text{km/s} \).

### Step 6: Conclude the result
Thus, the escape velocity from the surface of the planet is the same as that from Earth:
\[ v_e = 11.2 \, \text{km/s} \]

### Final Answer
The escape velocity from the surface of the planet is \( 11.2 \, \text{km/s} \).

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