Escape velocity on earth is `11.2 "kms"^(-1)`what would be the escape velocity on a planet whose mass is 1000 times and radius is 10 times that of earth ?

Options

A. `112 "kms"^(-1)`
B. `11.2 "kms"^(-1)`
C. `1.12 "kms"^(-1)`
D. `3.7 "kms"^(-1)`

Explanation

To find the escape velocity on a planet whose mass is 1000 times that of Earth and radius is 10 times that of Earth, we can use the formula for escape velocity:

\[ V_{escape} = \sqrt{\frac{2GM}{R}} \]

Where:
- \( V_{escape} \) 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 formula
The escape velocity is given by the formula:
\[ V_{escape} = \sqrt{\frac{2GM}{R}} \]

### Step 2: Substitute the values for the new planet
Let:
- Mass of the new planet, \( M' = 1000M \) (where \( M \) is the mass of Earth),
- Radius of the new planet, \( R' = 10R \) (where \( R \) is the radius of Earth).

### Step 3: Substitute \( M' \) and \( R' \) into the escape velocity formula
Now substituting \( M' \) and \( R' \) into the escape velocity formula:
\[ V_{escape}' = \sqrt{\frac{2G(1000M)}{10R}} = \sqrt{\frac{2000GM}{10R}} = \sqrt{200 \cdot \frac{GM}{R}} \]

### Step 4: Relate it to Earth's escape velocity
We know the escape velocity on Earth is:
\[ V_{escape} = \sqrt{\frac{2GM}{R}} \]
Thus,
\[ V_{escape}' = \sqrt{200} \cdot \sqrt{\frac{2GM}{R}} = \sqrt{200} \cdot V_{escape} \]

### Step 5: Calculate \(\sqrt{200}\)
\[ \sqrt{200} = \sqrt{100 \cdot 2} = 10\sqrt{2} \]
So,
\[ V_{escape}' = 10\sqrt{2} \cdot V_{escape} \]

### Step 6: Substitute Earth's escape velocity
Given that \( V_{escape} = 11.2 \, \text{km/s} \):
\[ V_{escape}' = 10\sqrt{2} \cdot 11.2 \]

### Step 7: Calculate the final escape velocity
Calculating \( 10\sqrt{2} \):
\[ \sqrt{2} \approx 1.414 \implies 10\sqrt{2} \approx 14.14 \]
Thus,
\[ V_{escape}' \approx 14.14 \cdot 11.2 \approx 158.37 \, \text{km/s} \]

### Final Answer
The escape velocity on the new planet is approximately \( 158.37 \, \text{km/s} \).

---