If the escape velocity of a planet is 3 times that of the earth and its radius is 4 times that of the earth, then the mass of the planet is (Mass of the earth = `6 xx 10^24 kg`)

Options

A. `1.62 xx 10^22 kg `
B. `0.72 xx 10^22 kg `
C. `2.16 xx 10^26 kg `
D. `1.22 xx 10^22 kg `

Explanation

To find the mass of the planet given that its escape velocity is 3 times that of Earth and its radius is 4 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: Write the escape velocity for Earth and the planet
Let:
- \( v_{eE} \) be the escape velocity of Earth,
- \( M_E \) be the mass of Earth,
- \( R_E \) be the radius of Earth.

For Earth:
\[ v_{eE} = \sqrt{\frac{2GM_E}{R_E}} \]

For the planet:
- The escape velocity \( v_{eP} \) is given as \( 3v_{eE} \),
- The mass of the planet is \( M_P \),
- The radius of the planet is \( R_P = 4R_E \).

Thus, the escape velocity for the planet can be expressed as:
\[ v_{eP} = \sqrt{\frac{2GM_P}{R_P}} = \sqrt{\frac{2GM_P}{4R_E}} = \frac{1}{2} \sqrt{\frac{2GM_P}{R_E}} \]

### Step 3: Set the escape velocities equal
Since \( v_{eP} = 3v_{eE} \), we can substitute the expressions we have:
\[ 3v_{eE} = \frac{1}{2} \sqrt{\frac{2GM_P}{R_E}} \]
Substituting \( v_{eE} \):
\[ 3 \sqrt{\frac{2GM_E}{R_E}} = \frac{1}{2} \sqrt{\frac{2GM_P}{R_E}} \]

### Step 4: Square both sides
Squaring both sides to eliminate the square roots gives:
\[ 9 \left(\frac{2GM_E}{R_E}\right) = \frac{1}{4} \left(\frac{2GM_P}{R_E}\right) \]

### Step 5: Simplify the equation
Cancelling \( \frac{2G}{R_E} \) from both sides (assuming \( R_E \neq 0 \)):
\[ 9M_E = \frac{1}{4}M_P \]

### Step 6: Solve for \( M_P \)
Multiplying both sides by 4 gives:
\[ M_P = 36M_E \]

### Step 7: Substitute the mass of Earth
Given \( M_E = 6 \times 10^{24} \, \text{kg} \):
\[ M_P = 36 \times (6 \times 10^{24}) = 216 \times 10^{24} \, \text{kg} \]

### Step 8: Convert to scientific notation
\[ M_P = 2.16 \times 10^{26} \, \text{kg} \]

### Final Answer
The mass of the planet is \( 2.16 \times 10^{26} \, \text{kg} \).

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