The escape velocity of a body on the surface of the earth is `V_(e)`. What is the escape velocity on a planet whose radius is thrice the radius of the earth and whose mass is double the mass of the earth?

Options

A. `sqrt((3)/(2))V_(e)`
B. `sqrt((2)/(3))Ve`
C. `(2)/(3)V_(e)`
D. `(sqrt(3))/(2)V_(e)`

Explanation

To find the escape velocity on a planet whose radius is three times the radius of the Earth and whose mass is double the mass of the Earth, 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 celestial body,
- \( R \) is the radius of the celestial body.

### Step 1: Write down the escape velocity for Earth
The escape velocity on the surface of the Earth is given as \( V_e \). We can express this as:

\[ V_e = \sqrt{\frac{2GM_e}{R_e}} \]

Where:
- \( M_e \) is the mass of the Earth,
- \( R_e \) is the radius of the Earth.

### Step 2: Define the parameters for the new planet
For the new planet:
- The radius \( R_p = 3R_e \) (three times the radius of the Earth),
- The mass \( M_p = 2M_e \) (double the mass of the Earth).

### Step 3: Write down the escape velocity for the new planet
Using the escape velocity formula for the new planet, we have:

\[ V_{ep} = \sqrt{\frac{2GM_p}{R_p}} \]

### Step 4: Substitute the values of \( M_p \) and \( R_p \)
Now, substituting \( M_p \) and \( R_p \) into the equation:

\[ V_{ep} = \sqrt{\frac{2G(2M_e)}{3R_e}} \]

### Step 5: Simplify the expression
This simplifies to:

\[ V_{ep} = \sqrt{\frac{4GM_e}{3R_e}} \]

### Step 6: Relate it to the escape velocity on Earth
Now, we can relate this to the escape velocity on Earth \( V_e \):

\[ V_{ep} = \sqrt{\frac{4}{3}} \cdot \sqrt{\frac{2GM_e}{R_e}} = \sqrt{\frac{4}{3}} \cdot V_e \]

### Final Result
Thus, the escape velocity on the new planet is:

\[ V_{ep} = \frac{2}{\sqrt{3}} V_e \]