Chapter 13, Problem 66Q

To determine

Escape speed on Titan and the value for limiting molecular mass of a gas, which can be retained under the effect of gravitational pull of Titan. Assume that the average atmospheric temperature on Titan is 95 K.

Answer to Problem 66Q

Solution:

$2.6\text{ km/s}$, $2.05\times {10}^{-26}\text{ kg}$

Explanation of Solution

Given data:

The average atmospheric temperature is 95 K.

Formula used:

The formula of the escape velocity is as:

$v_{escape}={\left(\frac{2GM}{R}\right)}^{\frac{1}{2}}$

Here, $v_{escape}$ is the escape velocity on Titan, M is the mass of Titan, R is the radius of Titan, and G is the universal gravitational constant having value $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$.

The requirement for the retention of gas is,

$v_{escape}>6{\left(\frac{3kT}{m}\right)}^{\frac{1}{2}}$

Here, $m$ is the atomic mass of the gas, T is the temperature of the Titan’s surface, $\mu$ is the molecular weight of the gas, and k is Boltzmann constant having value $1.38\times {10}^{-23}{\text{ m}}^{\text{2}}{\text{kg/s}}^{\text{2}}\text{K}$.

Explanation:

The diameter of Titan is 5150 km and the mass of Titan is $1.34\times {10}^{23}\text{ kg}$.

Since, radius is half of the diameter, the radius of Titan is,

$R=\frac{D}{2}$

Here, D is the diameter of Titan.

Substitute $5150\text{ km}$ for d.

$\begin{array}{c}r=\frac{5150\text{ km}}{2}\left(\frac{1000\text{ m}}{1\text{ km}}\right)\\ =2575\text{ km }\end{array}$

Recall the formula of the escape velocity.

$v_{escape}={\left(\frac{2GM}{R}\right)}^{\frac{1}{2}}$

Substitute 2575.5 km for R, $1.34\times {10}^{23}\text{ kg}$ for M and $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$ for G.

$\begin{array}{c}{v}_{escape}={\left(\frac{2\left(6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(1.34\times {10}^{23}\text{ kg}\right)}{2575\text{ km}}\right)}^{\frac{1}{2}}\\ ={\left(\frac{1.79\times {10}^{13}}{2575.5\text{ km}\left(\frac{1000\text{ m}}{1\text{ km}}\right)}\right)}^{\frac{1}{2}}\\ ={\left(6966530.771\right)}^{\frac{1}{2}}\\ =2.6\times {10}^3\text{ m/s}\end{array}$

Further solve as,

$\begin{array}{c}{v}_{escape}=2.6\times {10}^3\text{ m/s}\left(\frac{1\text{ km}}{1000\text{ m}}\right)\\ =2.6\text{ km/s}\end{array}$

Recall the requirement for the retention of gas.

$v_{escape}>6{\left(\frac{3kT}{m}\right)}^{\frac{1}{2}}$

Substitute $1.38\times {10}^{-23}{\text{ m}}^{\text{2}}\cdot {\text{kg/s}}^{\text{2}}\cdot \text{K}$ for k, 95 K for and $2.6\times {10}^3\text{ m/s}$ for $v_{escape}$.

$\begin{array}{c}2.6\times {10}^3\text{ m/s}>6{\left(\frac{3\left(1.38\times {10}^{-23}{\text{ m}}^{\text{2}}\cdot {\text{kg/s}}^{\text{2}}\cdot \text{K}\right)\left(95\text{ K}\right)}{m}\right)}^{\frac{1}{2}}\\ 2.6\times {10}^3>6{\left(\frac{3.933\times {10}^{-21}}{m}\right)}^{\frac{1}{2}}\end{array}$

Squaring both sides.

$\begin{array}{c}{\left(2.63\times {10}^3\right)}^2>36\left(\frac{3.933\times {10}^{-21}}{m}\right)\\ m>\frac{1.416\times {10}^{-19}}{6.91\times {10}^6}\\ m>2.05\times {10}^{-26}\text{ kg}\end{array}$

Conclusion:

Therefore, the gas with molecular mass greater than $2.05\times {10}^{-26}\text{ kg}$ is retained.