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

Explanation

To find the escape velocity from the surface of the planet, we can use the formula for escape velocity, which is given by:

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

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

### Step 1: Identify the parameters for Earth
Given that the escape velocity from Earth is \( 11 \, \text{km/s} \), we can express this as:

\[ v_{e, \text{Earth}} = \sqrt{\frac{2GM_{\text{Earth}}}{R_{\text{Earth}}}} = 11 \, \text{km/s} \]

### Step 2: Define the parameters for the new planet
According to the problem:
- The mass of the new planet \( M = 10 \times M_{\text{Earth}} \)
- The radius of the new planet \( R = \frac{R_{\text{Earth}}}{10} \)

### Step 3: Substitute the new parameters into the escape velocity formula
Now we can substitute these values into the escape velocity formula for the new planet:

\[ v_{e, \text{planet}} = \sqrt{\frac{2G(10M_{\text{Earth}})}{\frac{R_{\text{Earth}}}{10}}} \]

### Step 4: Simplify the expression
This can be simplified as follows:

\[ v_{e, \text{planet}} = \sqrt{\frac{2G \cdot 10M_{\text{Earth}} \cdot 10}{R_{\text{Earth}}}} = \sqrt{\frac{100 \cdot 2GM_{\text{Earth}}}{R_{\text{Earth}}}} \]

### Step 5: Relate it to Earth's escape velocity
We know from the escape velocity of Earth that:

\[ \sqrt{\frac{2GM_{\text{Earth}}}{R_{\text{Earth}}}} = 11 \, \text{km/s} \]

Thus, we can rewrite the escape velocity for the planet as:

\[ v_{e, \text{planet}} = \sqrt{100} \cdot \sqrt{\frac{2GM_{\text{Earth}}}{R_{\text{Earth}}}} = 10 \cdot 11 \, \text{km/s} \]

### Step 6: Calculate the final escape velocity
Calculating this gives:

\[ v_{e, \text{planet}} = 110 \, \text{km/s} \]

### Final Answer:
The escape velocity from the surface of the planet would be \( 110 \, \text{km/s} \).

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