Question

At what temperature is ${v_{rms}}$ of ${H_2}$ molecules equal to the escape speed from earth’s surface. What is the corresponding temperature for escape of hydrogen from moon’ surface? Given
${g_{m\,\,}} = 1.6\,{m {\left/ { {m {{s^2}}}} \right. } {{s^2}}}$, ${R_e} = 6367\,km\,$ and ${R_{m\,}}\, = 1750\,km$.

Answer 1

Escape velocity is the minimum velocity required by an object to overcome the gravitational pull of a massive object
Escape velocity is given by the equation
${v_{e}} = \sqrt {\dfrac{{2GM}}{r}}$
Where $G$ is the gravitational constant $M$ is the mass of the planet or moon and $r$ is its radius.
We know $\dfrac{{GM}}{{{r^2}}} = g$ from universal law of gravitation
Where, $g$ is the acceleration due to gravity.
RMS velocity is the square root of the average square of velocity.
RMS velocity is given by the equation,
${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}}$
Where $R$ is the universal gas constant, $T$ is the temperature and $M$ is the mass
Value of universal gas constant is $R = 8.314\,J{K^{ - 1}}mo{l^{ - 1}}$
We know, Mass of ${H_2}$ is $2 \times {10^{ - 3}}\,kg$
We need to find the temperature at which escape velocity becomes equal to RMS velocity.
${v_e} = {v_{rms}}$

Complete step by step answer:
Given, Acceleration due to gravity in moon,${g_{m\,\,}} = 1.6\,{m {\left/ { {m {{s^2}}}} \right. } {{s^2}}}$
Radius of earth,
${R_e} = 6367\,km\, \\\ = 6367\, \times {10^{ - 3}}\,m \\\$
Radius of moon,
${R_{m\,}}\, = 1750\,km \\\ = 1750\, \times {10^{ - 3}}\,m \\\$
We know, Mass of ${H_2}$ is $2 \times {10^{ - 3}}\,kg$
Escape velocity is the minimum velocity required by an object to overcome the gravitational pull of a massive object
Escape velocity is given by the equation
${v_{e}} = \sqrt {\dfrac{{2GM}}{r}}$
Where $G$ is the gravitational constant $M$ is the mass of the planet or moon and $r$ is its radius.
${v_{e}} = \sqrt {\dfrac{{2GM \times r}}{{r \times r}}}$
${v_{e}} = \sqrt {2gr}$ …….(1)
Since, we know $\dfrac{{GM}}{{{r^2}}} = g$ from universal law of gravitation
Where, $g$ is the acceleration due to gravity.
RMS velocity is the square root of the average square of velocity.
RMS velocity is given by the equation,
${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}}$ ……(2)
Where $R$ is the universal gas constant, $T$ is the temperature and $M$ is the mass
Value of universal gas constant is $R = 8.314\,J{K^{ - 1}}mo{l^{ - 1}}$
We need to find the temperature at which escape velocity becomes equal to RMS velocity.
${v_e} = {v_{rms}}$
So, let us equate equation (1) and (2)
$\sqrt {2gr} = \sqrt {\dfrac{{3RT}}{M}}$
$2gr = \dfrac{{3RT}}{M}$
$T = \dfrac{{2grM}}{{3R}}$ (3)
For moon equation (3) can be written as,
${T_m} = \dfrac{{2{g_m}{R_m}M}}{{3R}}$
Substituting the given values, we get

$${T_m} = \dfrac{{2 \times 1.6\,{m {\left/ { {m {{s^2} \times }}} \right. } {{s^2} \times }}1750\, \times {{10}^3}\,m \times 2 \times {{10}^{ - 3}}\,kg}}{{3 \times 8.314}} \\\ = 449\,K \\\$$

For earth we know that the escape velocity is $11.2 \times {10^3}{{\text{m}} {\left/ { {{\text{m}} {\text{s}}}} \right. } {\text{s}}}$.
Let us equate this with root mean square velocity. Then, we get
$\sqrt {\dfrac{{3R{T_e}}}{M}} = 11.2 \times {10^3} \\\ {T_e} = \dfrac{{{{\left( {11.2 \times {{10}^3}} \right)}^2} \times M}}{{3R}} \\\$
${T_e} = \dfrac{{{{\left( {11.2 \times {{10}^3}} \right)}^2} \times 2 \times {{10}^{ - 3}}}}{{3 \times 8.314}} \\\ = 10059K \\\$
So, the temperature on earth’s surface is $10059K$ and the temperature on the moon is $449\,K$.

Note: The acceleration due to gravity and radius is different for Earth and Moon. While calculating the temperature on the Moon , substitute the value of acceleration due to gravity and radius of the moon.
The mass in the root mean square equation is the mass of hydrogen. We know the mass of hydrogen is $2\,g$. Remember to convert the value into $kg$ before substituting.