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Calculate the average speed of water molecules on the surface of Enceladus and the escape velocity. Interpret the results by concluding whether water molecules can escape from the gravitational field of Enceladus. (Molecular weight of water = 18.0152 amu; mass of Enceladus = 8.4 × 1019 kg; radius of Enceladus = 250 km; T = 100 K.

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Escape speed: Ve = (2GM/R)1/2

Average speed of a molecules: Vm = (3kT/m)1/2

k = 1.38 x 10-23J/K

G = 6.67 x 10-11Nm2/kg2

Escape speed of Enceladus:

Ve = (2GM/r)1/2 = ((2 x 6.67 x 10-11 x 8.4 x 1019)/(250 x 103)1/2

Ve = (0.4482 × 105)1/2 = 211.68 m/s 

Ve = 0.2117 km/s 

Mass of one molecule of water:

m = 18.0152/(6.022 x 1023) = 2.991 x 10-23g

M = 2.991 × 10-26 kg  

Average speed of water molecules:

vm = (3KT/m)1/2 = ((3 x 1.38 x 10-23 x 100)/(2.991 x 10-26))1/2

vm = (1.384 × 105)1/2 = 371.9 m/s

vm = 0.372 km/s 

Conclusion: vm(H2O) > ve. Therefore, molecules of water can escape from Enceladus.

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