Any help on this two part problem would be greatly appreciated! If it has enough

Question

Any help on this two part problem would be greatly appreciated!

If it has enough kinetic energy, a molecule at 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 Times 10-23 J/K. Using energy conservation, determine the minimum kinetic energy needed to escape in terms of the mass of the molecule m, the free-fall acceleration at the surface g, and the radius of the Earth R. Kmin = 1/2 m g R Kmin = 2m g R Kmin = m g R Kmin = 1/ 2 = m g R Kmin = 1/3 m g R Calculate the temperature for which the minimum escape energy is 10 times the average kinetic energy of an oxygen molecule. Answer in units of K

Explanation / Answer

Take kinetic energy when the particle when it escape=0 {inorder to find the minimum kinetic energy at the surface}

apply conservation of energy,

mv2/2 -(GMm/R) =0---(1) {since when it escapes potential energy=0}

but g=GM/R2 so applying to (1)

v=sqrt(2gR).

so, Kmin=m(2gR)/2=mgR

2)from equipartition theorem in thermodynamics, average kinetic energy= 3kT/2. So, to find the T

Kmin=10*(3kT/2)=mgR; for O2 there are 6.022*1023 molecules in 32g. So, m=32*10-3/6.022*1023. Take R=6371km

T=16020.088K

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