Question #142667
Escape velocity from the Moon is approximately 5500 miles per hour = 2.45 x 103m/s. At certain spots on the surface of the Moon, the temperature of the surface material reaches 800K. Suppose a molecule of hydrogen (m= 3.33x 0-27kg) were to hit the surface of the Moon and reach a speed equal to the root-mean-square (rms) speed for this temperature. `
What would be the velocity or root-mean-square speed (Vrms) of the Hydrogen molecule?
Will the Hydrogen molecule manage to escape?
1
Expert's answer
2020-11-09T06:48:50-0500

T=800KVescape=5500 mph=2.45×103m/smH2=3.33×1027kgVrms=?\begin{aligned} T&= 800K\\ V_{escape} &= 5500\ mph = 2.45×10^3m/s\\ m_{H_2} &= 3.33× 10^{-27} kg\\ V_{rms} &= ?\\ \end{aligned}


from, vrms=3kTmH2vrms=3×1.38×1023×8003.33×1027vrms=10,036,363.6vrms=3.168×103m/s\begin{aligned} from,\ v_{rms} &= \sqrt{\dfrac{3kT}{m_{H_2}}}\\ \\ v_{rms} &= \sqrt{\dfrac{3×1.38×10^{-23}×800}{3.33×10^{-27}}}\\ \\ v_{rms} &= \sqrt{10,036,363.6}\\ v_{rms} &= 3.168×10^3m/s \end{aligned}

Vrms>VescapeV_{rms} > V_{escape},  \ \therefore H2H_2 should escape if it hits the part of the moon at 800K.


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