What multiple of the energy needed to escape from Earth givesthe energy needed to escape from (a) the Moon and (b) Jupiter?

Mass of earth,$5.95\times {\text{10}}^{\text{24}}\text{kg}$

Mass of moon,${\text{7.36×10}}^{\text{22}}\text{kg}$

Mass of Jupiter,$1.89\times {\text{10}}^{\text{27}}\text{kg}$

The radius of the earth,${\text{6.37×10}}^{\text{6}}\text{m}$

The radius of the moon,${\text{1.74×10}}^6\text{m}$

The radius of Jupiter, $\text{6.99}\times {\text{10}}^{\text{7}}\text{m}$

The amount of energy needed to escape from the surface is the same as its gravitational potential energy at the original position.

The formula is as follows:

$K=\frac{GMm}{R}$

where, M, and m are masses, R is the radius, G is the gravitational constant and K is energy.

The gravitational potential energy is written as,

$\Delta U=-\frac{GMm}{R}$

The amount of energy required to escape from the surface would be the absolute value of gravitational potential energy, so,

$K=\frac{GMm}{R}$

The ratio of energy needed to escape from the surface of the earth to energy needed to escape from the moon,

$\begin{array}{rcl}\frac{K_M}{K_E}& =& \frac{M_M{R}_E}{M_E{R}_M}\\ & =& \frac{{\text{7.36×10}}^{\text{22}}kg\times {\text{6.37×10}}^{\text{6}}m}{{\text{5.95×10}}^{\text{24}}kg\times \text{1.74×}{10}^{\text{6}}m}\\ & =& 0.045\\ & & \end{array}$

Using the values from Appendix C,

$\frac{{\text{K}}_{\text{M}}}{{\text{K}}_{\text{E}}}=0.045$

Hence, the ratio of energy needed to escape from the surface of the moon to the energy needed to escape from the earth is $0.045$.

Using the values from Appendix C,

$\begin{array}{rcl}\frac{K_J}{K_E}& =& \frac{M_J{R}_E}{M_E{R}_J} \\ & =& \frac{{\displaystyle {\text{1.89×10}}^{\text{27}}kg\times {\text{6.37×10}}^{\text{6}}m}}{{\displaystyle {\text{5.95×10}}^{\text{24}}kg\times {\text{6.99×10}}^{\text{7}}m}}\\ & =& 28.5\end{array}$

Hence, the ratio of energy needed to escape from the surface of Jupiter to energy needed to escape from the earth is $28.5$.

Therefore, the amount of energy needed to escape from the surface is the same as its gravitational potential energy at the original position. Using this concept, the given ratios can be found.