Answer
1.73
Work Step by Step
We can write the translational rms-speed $v_{rms}$ is related to the Kelvin temperature T by equation 14.6 $\frac{1}{2}mv_{rms}^{2}=\frac{3}{2}kT$. From this equation, we can get,
$$v_{rms}=\sqrt {\frac{3kT}{m}}$$
The rms-speeds in the ionosphere and near the earth's surface are,
$(v_{rms})_{ion}=\sqrt {\frac{3kT_{ion}}{m}}-(1)$ and $(v_{rms})_{es}=\sqrt {\frac{3kT_{es}}{m}}-(2)$
(1)/(2),
$\frac{(v_{rms})_{ion}}{(v_{rms})_{es}}=\frac{\sqrt {\frac{3kT_{ion}}{m}}}{\sqrt {\frac{3kT_{es}}{m}}}=\sqrt {\frac{T_{ion}}{T_{es}}}$
Given that, $T_{ion}=3T_{es}$, Therefore,
$\frac{(v_{rms})_{ion}}{(v_{rms})_{es}}=\sqrt {3}=1.73$
