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Step-by-Step Solution
Step 1: Recall the formula for escape velocity
The escape velocity V_{es} from a spherical body of radius R and mass M is given by
V_{es} = \sqrt{\frac{2GM}{R}} .
Step 2: Identify how escape velocity changes with radius
Notice from the expression V_{es} = \sqrt{\frac{2GM}{R}} that V_{es} is inversely proportional to the square root of R , i.e.,
V_{es} \propto \frac{1}{\sqrt{R}} .
Step 3: Set up the relationship for a 10-fold increase in escape velocity
We want the new escape velocity V_{es}^\prime to be 10 times the original escape velocity V_{es} . In mathematical terms:
V_{es}^\prime = 10 \, V_{es}.
Because V_{es}^\prime \propto \frac{1}{\sqrt{R^\prime}} and V_{es} \propto \frac{1}{\sqrt{R}} , we have
\frac{V_{es}^\prime}{V_{es}} = \sqrt{\frac{R}{R^\prime}}.
Substituting V_{es}^\prime = 10 \, V_{es} gives
10 = \sqrt{\frac{R}{R^\prime}} .
Step 4: Solve for the new radius
Squaring both sides,
100 = \frac{R}{R^\prime} \Rightarrow R^\prime = \frac{R}{100}.
Given that the present radius of the Earth R = 6400\,\text{km} , the new radius R^\prime becomes
R^\prime = \frac{6400}{100} = 64\,\text{km}.
Final Answer
64 km
