If it has enough kinetic energy, a moleculeat the surface of the Earth can escape the Earth's gravitation. The acceleration of gravity is 9.8 m/s2, and the Boltzmanns' constant is 1.38066*10-23 J/K. Using energy conservation, a) determine theminimum kinetic energy needed to escape interms of the mass of the molecule m, the free-fall acceleration at the surface g, and the radius of the Earth R

b) Calculate the temperature for which the min - imum escape energy is 11 times the average kinetic energy of an oxygen molecule.

a. mgR b. 483.6 * 10³ K

Explanation:

a. Using the law of conservation of energy,

K₁ + U₁ = K₂ + U₂ where K₁ and K₂ are the initial and final kinetic energies of the molecule and U₁ and U₂ are the initial and final potential energies of the molecule.

K₁ + - GMm/R = 0 + 0

K₁ = GMm/R since GM = gR²,

K₁ = mgR

So its minimum kinetic energy is mgR

b. The average kinetic energy energy of a gas molecule is K = 3/2kT. If K₁ = 11K, then

mgR = 11 * 3/2kT = 33/2kT.

T = 2mgR/k

For oxygen molecule, m = 32 g/6.02 * 10²³ = 5.32 * 10⁻²⁶ kg/molecule

T = 2 * 5.32 * 10⁻²⁶ kg * 9.8 m/s² 6.4 * 10⁶ m/1.38 10⁻²³ J/K = 483.6 * 10³ K