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(C) The formula for escape velocity is $v_{e} = \sqrt{\frac{2GM}{R}}$.Let $M_{e}$ and $R_{e}$ be the mass and radius of the Earth,and $M_{m}$ and $R_{

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$4.81$

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(C) The formula for escape velocity is $v_{e} = \sqrt{\frac{2GM}{R}}$.Let $M_{e}$ and $R_{e}$ be the mass and radius of the Earth,and $M_{m}$ and $R_{m}$ be the mass and radius of the Moon.Given: $M_{e} = 81 M_{m}$ and $R_{e} = 3.5 R_{m}$.The ratio of escape velocity on Earth $(v_{e})$ to that on the Moon $(v_{m})$ is:$\frac{v_{e}}{v_{m}} = \sqrt{\frac{M_{e}}{R_{e}} 	imes \frac{R_{m}}{M_{m}}}$Substituting the given values:$\frac{v_{e}}{v_{m}} = \sqrt{\frac{81 M_{m}}{3.5 R_{m}} 	imes \frac{R_{m}}{M_{m}}}$$\frac{v_{e}}{v_{m}} = \sqrt{\frac{81}{3.5}} = \sqrt{23.14} \approx 4.81$.

The mass of the Earth is $81$ times that of the Moon and the radius of the Earth is $3.5$ times that of the Moon. The ratio of the escape velocity on the surface of the Earth to that on the surface of the Moon will be:

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