Answer: A) 110 km/s
Step 1: Understanding escape velocity
The escape velocity \( v_{\text{esc}} \) is given by the formula:
\[
v_{\text{esc}} = \sqrt{\frac{2GM}{R}}
\]
Where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is its radius.
Step 2: Scaling of mass and radius
- The planet's mass is 10 times the mass of the Earth (\( M_{\text{planet}} = 10M_{\text{earth}} \)).
- The planet's radius is 10 times smaller than Earth's radius (\( R_{\text{planet}} = \frac{R_{\text{earth}}}{10} \)).
Step 3: Relationship of escape velocity with mass and radius
For the Earth, escape velocity is \( v_{\text{esc,earth}} = 11 \, \text{km/s} \).
Escape velocity scales as:
\[
v_{\text{esc,planet}} \propto \sqrt{\frac{M}{R}}
\]
For the new planet:
\[
v_{\text{esc,planet}} = v_{\text{esc,earth}} \sqrt{\frac{M_{\text{planet}} / M_{\text{earth}}}{R_{\text{planet}} / R_{\text{earth}}}}
\]
Substitute \( M_{\text{planet}} = 10M_{\text{earth}} \) and \( R_{\text{planet}} = \frac{R_{\text{earth}}}{10} \):
\[
v_{\text{esc,planet}} = 11 \, \text{km/s} \times \sqrt{\frac{10}{\frac{1}{10}}}
\]
Step 4: Simplifying the expression
Simplify the fraction under the square root:
\[
\frac{10}{\frac{1}{10}} = 10 \times 10 = 100
\]
Thus:
\[
v_{\text{esc,planet}} = 11 \, \text{km/s} \times \sqrt{100}
\]
\[
v_{\text{esc,planet}} = 11 \, \text{km/s} \times 10 = 110 \, \text{km/s}.
\]
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