Question

Calculate the escape speed from the Moon and compare with typical speeds of gas molecules. The mass of the Moon is$\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{22}}\mathbf{kg}$, and its radius is $\mathbf{1}\mathbf{.}\mathbf{75}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{}\mathbf{m}$.

Short answer

The escape velocity of an object on the Moon is $2310\mathrm{m}/\mathrm{s}$. The rms speed of nitrogen is less than half of the escape velocity of moon and the rms speed of Helium is more than half of the escape velocity of the moon.

Step-by-step solution

Identification of given data

The given data can be listed below,

• The mass of the moon is,$M=7\times {10}^{22}\mathrm{kg}$.
• The radius of the moon is,$R_{\mathrm{M}}=1.75\times {10}^6\mathrm{m}$.

Concept/Significance of escape velocity

The minimal velocity required for an object to escape a planet's gravitational field, that is, to leave the planet without ever falling back, is defined as escape velocity.

Determination of the escape speed from the Moon and compare with typical speeds of gas molecules

The formula for escape velocity of an object on moon is given by,

$v_{esc}=\sqrt{\frac{2GM}{R}}$

Here, G is the gravitational constant whose value is $6.67\times {10}^{-11}\hspace{0.17em}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2$, M is the mass of moon, R is the radius of the moon.

Substitute all the values in the above equation.

$$\begin{gathered} v_{esc}=\sqrt{\frac{2(6.67\times {10}^{-11}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2)(7\times {10}^{22}\mathrm{kg})}{1.75\times {10}^{6}m}} \\ =2310\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the escape velocity of an object on the moon is$2310\mathrm{m}/\mathrm{s}$.

The rms velocity of the Nitrogen is given by,

$v_{rms}=\sqrt{\frac{3k_{B}T}{m_N}}$

Here,$k_B$is Boltzmann constant whose value is $1.38\times {10}^{-23}\mathrm{J}/\mathrm{mol}.\mathrm{K}$, T is the temperature, $m_N$is the mass of the Nitrogen.

Substitute all the values in the above,

$$\begin{gathered} v_{\mathrm{rms}}=\sqrt{\frac{3(1.38\times {10}^{-23}\mathrm{J}/\mathrm{mol}\mathrm{K})(293\mathrm{K})}{\left({\displaystyle \frac{28\mathrm{g}/\mathrm{mol}\left({\displaystyle \frac{1\mathrm{kg}}{1000\mathrm{g}}}\right)}{6.022\times {10}^{23}/\mathrm{mol}}}\right)}} \\ =510.8m/s\left(\frac{{\mathrm{v}}_{\mathrm{rms}}}{2310\mathrm{m}/\mathrm{s}}\right) \\ =0.22v_{\mathrm{rms}} \end{gathered}$$

The rms velocity of nitrogen is less than half of the escape velocity.

The root mean square of a particle is given by,

$v_{\mathrm{rms}}=\sqrt{\frac{3k_{B}T}{m}}$

Here, $k_B$is Boltzmann constant whose value is $1.38\times {10}^{-23}\mathrm{J}/\mathrm{mol}.\mathrm{K}$,T is the temperature and m is the mass of gas particle.

Substitute values in the above for Helium atom.

$$\begin{gathered} v_{\mathrm{rms},\mathrm{He}}=\sqrt{\frac{3(1.38\times {10}^{-23}\mathrm{J}/\mathrm{mol}.\mathrm{K})(293\mathrm{K})}{6.664\times {10}^{-27}\mathrm{kg}}} \\ =1351.6\mathrm{m}/\mathrm{s} \\ =0.58{\mathrm{v}}_{\sec } \end{gathered}$$

The rms velocity of helium is more than half of the escape velocity.