A planet in a distant solar systyem is 10 times more massive than the earth and its radius is 10 times smaller. Given that the escape velocity from the earth is `11kms^(-1)`, the escape velocity from the surface of the planet would be

Options

A. `1.1 kms^(-1)`
B. `11 kms^(-1)`
C. `110 kms^(-1)`
D. `0.11 kms^(-1)`

Answer

To find the escape velocity from the surface of a planet that is 10 times more massive than Earth and has a radius that is 10 times smaller, 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: Write the escape velocity for Earth
The escape velocity from the surface of the Earth can be expressed 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.

Given that the escape velocity from the Earth is \( 11 \, \text{km/s} \), we can write:

\[ v_e = 11 \, \text{km/s} \]

### Step 2: Write the escape velocity for the planet
For the planet, which is 10 times more massive than Earth and has a radius that is 10 times smaller, we can express its mass and radius as:

\[ M = 10M_e \quad \text{and} \quad R = \frac{R_e}{10} \]

Now, substituting these values into the escape velocity formula for the planet:

\[ v_p = \sqrt{\frac{2G(10M_e)}{\frac{R_e}{10}}} \]

### Step 3: Simplify the expression
This can be simplified as follows:

\[ v_p = \sqrt{\frac{2G \cdot 10M_e \cdot 10}{R_e}} = \sqrt{\frac{100 \cdot 2GM_e}{R_e}} = 10 \sqrt{\frac{2GM_e}{R_e}} \]

### Step 4: Relate it to Earth's escape velocity
Since we know that \( \sqrt{\frac{2GM_e}{R_e}} = v_e \):

\[ v_p = 10 \cdot v_e \]

### Step 5: Substitute the known value
Now substituting the value of \( v_e \):

\[ v_p = 10 \cdot 11 \, \text{km/s} = 110 \, \text{km/s} \]

### Final Answer
The escape velocity from the surface of the planet is:

\[ \boxed{110 \, \text{km/s}} \]