The escape velocity at the surface of Earth is approximately 8 km/s. What is the escape velocity for a planet whose radius is 4 times and whose mass is 100 times that of Earth?

Options

A. 1.6 km/s
B. 8 km/s
C. 40 km/s
D. 200 km/s

Explanation

To find the escape velocity for a planet whose radius is 4 times and mass is 100 times that of Earth, we can use the formula for escape velocity:

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

where:
- \( v \) is the escape velocity,
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the planet,
- \( R \) is the radius of the planet.

### Step 1: Identify the escape velocity for Earth
The escape velocity at the surface of Earth is given as approximately \( 8 \, \text{km/s} \).

### Step 2: Define the parameters for the new planet
Let:
- \( M_e \) = mass of Earth
- \( R_e \) = radius of Earth

For the new planet:
- Mass \( M' = 100 M_e \)
- Radius \( R' = 4 R_e \)

### Step 3: Substitute the new parameters into the escape velocity formula
Using the escape velocity formula for the new planet:

\[ v' = \sqrt{\frac{2G(100 M_e)}{4 R_e}} \]

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

\[ v' = \sqrt{\frac{100 \cdot 2GM_e}{4R_e}} = \sqrt{\frac{100}{4} \cdot \frac{2GM_e}{R_e}} = \sqrt{25 \cdot \frac{2GM_e}{R_e}} \]

### Step 5: Relate it to Earth's escape velocity
Since \( \sqrt{\frac{2GM_e}{R_e}} \) is the escape velocity of Earth (\( v_e \)), we can write:

\[ v' = \sqrt{25} \cdot v_e = 5 v_e \]

### Step 6: Calculate the escape velocity for the new planet
Now substituting \( v_e = 8 \, \text{km/s} \):

\[ v' = 5 \cdot 8 \, \text{km/s} = 40 \, \text{km/s} \]

### Final Answer:
The escape velocity for the planet is \( 40 \, \text{km/s} \).

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