91Ó°ÊÓ

The radius of the Moon is$R_{moon}=1750\text{km}=1750×{10}^3\text{m}$.

The mass is $7×{10}^{22}\text{kg}$$7×{10}^{22}\text{kg}$.

The escape speed is defined as the speed at which the object's kinetic energy plus gravitational potential energy equals zero

$\text{m}\frac{1}{2}m{v}_{\text{escape}}^2+\left(-\frac{G{M}_{\text{moon}}m}{R_{\text{moon}}}\right)=0$

The escape velocity is

$v_{\text{escape}}=\sqrt{\frac{2G{M}_{\text{moon}}}{R_{\text{moon}}}}$

.

It's worth noting that the escape speed is defined as the speed at which the object's kinetic energy plus gravitational potential energy equals zero. Then turn this into an equation

$v_{\text{escape}}=\sqrt{\frac{2G{M}_{\text{moon}}}{R_{\text{moon}}}}$

Now determine the velocity,

$$\begin{gathered} v_{\text{escape}}=\sqrt{\frac{2G{M}_{\text{moon}}}{R_{\text{moon}}}} \\ {v}_{\text{escape}}=\sqrt{\frac{2\left(6.76×{10}^{-11}\text{N}·{\text{m}}^2{\text{/kg}}^2\right)\left(7×{10}^{22}\text{kg}\right)}{1750×{10}^3\text{m}}} \\ =2310\text{m}/\text{s} \end{gathered}$$

Reason a small rocket can lift lunar astronauts back from moon

For a small rocket, this might be a significant speed to achieve upon takeoff. However, they are able to leave the Moon because they achieve a lower escape speed at a higher altitude.

On the Moon's surface, there is no need to reach escape velocity. The third line of the equation shows that the lower the escape speed, the further away the rocket is from the Moon.

Therefore, it is easier to reach by the rocket.