At what temperature is $v_{r m s} o f H_{2}$ molecules equal to the escape speed from from earth's surface. What is the corresponding temperature for escape of hydrogen from moon's surface ? Given $g_{m} = 1.6 m / s^{2}, R_{e} = 6367$ km and $R_{m} = 1750$ km.

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Solution

Escape velocity is given by the formula

$v_{e} = \sqrt{\frac{2 G M}{r}} = \sqrt{2 g r}$

Thus,

$\sqrt{2 g r} = \sqrt{\frac{3 R T}{M}}$

$T = \frac{2 g M r}{3 R}$

Thus, $T_{m} = \frac{2 (1.6) (2 X 10^{- 3}) (1750 X 10^{3})}{3 (8.314)} \approx 449 K$

Also, since $T_{e} = \frac{2 (9.8) (2 X 10^{- 3}) (6367 X 10^{3})}{3 (8.314)} \approx 10059 K$

Answer is $10059 K$ and $449 K$ respectively.

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At what temperature will the hydrogen molecules escape from Earth's surface?

= 1.74 \times 10^{6} m

= 1.67 m / s^{2}

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