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(C) The formula for escape velocity is $v = \sqrt{\frac{2GM}{R}}$.Since mass $M = 	ext{Volume} 	imes 	ext{density} = \frac{4}{3} \pi R^3 ho$,we s

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$V_{P}=2 V_{E}$

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(C) The formula for escape velocity is $v = \sqrt{\frac{2GM}{R}}$.Since mass $M = 	ext{Volume} 	imes 	ext{density} = \frac{4}{3} \pi R^3 ho$,we substitute this into the formula:$v = \sqrt{\frac{2G}{R} \cdot \frac{4}{3} \pi R^3 ho} = \sqrt{\frac{8}{3} \pi G ho R^2} = R \sqrt{\frac{8 \pi G ho}{3}}$.Thus,$v \propto R \sqrt{ho}$.Given that the average mass densities $ho$ are equal,the escape velocity is directly proportional to the radius: $v \propto R$.Therefore,$\frac{V_P}{V_E} = \frac{R_P}{R_E}$.Given $R_P = 2 R_E$,we get $\frac{V_P}{V_E} = 2$,which implies $V_P = 2 V_E$.

The radius of a planet is twice the radius of the Earth. Both have almost equal average mass densities. If $V_P$ and $V_E$ are the escape velocities of the planet and the Earth respectively,then:

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