Question

1.) 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/s
2 , and the Boltzmanns’ constant is 1.38066 ×10−23 J/K.
Using energy conservation, determine theminimum 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.
1. Kmin = 2 m g R
2. Kmin =1/2m g R
3. Kmin =1/3m g R
4. Kmin =1/√2m g R
5. Kmin = m g R

2.)Calculate the temperature for which the minimum escape energy is 9 times the average
kinetic energy of an oxygen molecule.
Answer in units of K.

Answer 1

1)

$$E=K+U=K-\frac{GmM}{R}$$

$$E=0$$

$$K_m=mgR$$

Answer: 5. Kmin= m g R 

2)

$$9\frac{3}{2}kT=mgR$$

$$\frac{27}{2}kT=m_{O_2}gR$$

$$\frac{27}{2}(1.38066\cdot 10^{-23})T=(5.31373\cdot 10^{-26})(9.8)(6.37\cdot 10^6)$$

$$T=17796.9\ K$$