Question

At what temperature the mean speed of the molecules of hydrogen gas equals the escape speed from the earth?

Answer 1

Mean speed of the molecule is given by

$$\begin{gathered} \sqrt{\frac{8\mathrm{RT}}{\mathrm{\pi M}}} \\ \mathrm{For}\mathrm{H}\mathrm{molecule},M=2\times {10}^{-3}kg \\ =\sqrt{\frac{4\mathrm{RT}\times {10}^3}{\pi }} \end{gathered}$$

For escape velocity of Earth:

$$\begin{gathered} \mathrm{Let}r\mathit{}\mathrm{be}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{Earth}. \\ v=\sqrt{\frac{2GM}{r}} \\ \mathrm{Multiplying}\mathrm{neumerator}\mathrm{and}\mathrm{denominator}\mathrm{by}\mathrm{R},\mathrm{we}\mathrm{get} \\ {v}_c=\sqrt{\frac{GM}{r^2}2r} \\ g=\frac{GM}{r^2} \\ {v}_c=\sqrt{2gr} \end{gathered}$$

$$\begin{gathered} \sqrt{\frac{4\mathrm{RT}\times {10}^3}{\pi }}=\sqrt{2gr} \\ \Rightarrow \frac{2\times 8.314\times T\times {10}^3}{3.142}=9.8\times 6.37\times {10}^6 \\ \Rightarrow T\approx 11800K \end{gathered}$$