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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

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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 given by:$\frac{v_e}{v_m} = \sqrt{\frac{M_e}{R_e} \cdot \frac{R_m}{M_m}} = \sqrt{\frac{M_e}{M_m} \cdot \frac{R_m}{R_e}}$Substituting the given values:$\frac{v_e}{v_m} = \sqrt{81 \cdot \frac{1}{3.5}} = \sqrt{\frac{81}{3.5}} = \sqrt{23.14} \approx 4.81$.

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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