# Homework Answer Key: Homework 5

1.  Calculate the "escape velocity" for Mars and Venus in units of km/sec.
•  The necessary formula is: vesc = (2GM/R)0.5
where G = 6.67 x 10-11 N m2 / kg2, M is the mass of the planet and R is the radius of the planet

for Mars, M = 6.39 x 1023 kg, and R = 3.397 x 106 m. Thus
vesc (Mars) = (2 x 6.67 x 10-11  x  6.39 x 1023 /  3.397 x 106 )0.5
vesc (Mars) = 5009 m/sec = 5.00 km/sec

for Venus, M = 4.90 x 1024 kg, and R = 6.052 x 106 m. Thus
vesc (Venus) = (2 x 6.67 x 10-11  x  4.90 x 1024 /  6.052 x 106 )0.5
vesc (Venus) = 10,400 m/sec = 10.4 km/sec

2. Calculate the average speed, vave (in units of km/sec), of an oxygen molecule (O2) in the Earth's atmosphere, assuming T = 22 C (= 295 K).  Are oxygen molecules bound to Earth?

1. The necessary formula is: vave = (3kT/m)0.5
where k = 1.38 x 10-16 gm cm2 sec-2 deg-1), T = temperature, and m = mass of the molecule or atom of gas.  Note that the mass of a hydrogen atom is 1.67 x 10-24 gm.  We need to use mass units of gm (since that is what the constant k uses and temperature in degrees K.  Our answer will be in units of cm/sec (again, since those are the units embedded in the constant k):

For an oxygen molecule, the mass is that of two O atoms, and each atom has 16 times the mass of a H atom:

vave = [ 3 x 1.38 x 10-16 x 295) / (32 x 1.67 x 10-24) ]0.5

vave = 47,800 cm/sec = 0.48 km/sec

Clearly, since 6 x vave = 2.86 km/sec is much less than  vesc (Earth) = 11.2 km/sec, oxygen atoms cannot escape from Earth.

3.  In the upper atmospheres of the terrestrial planets, the temperature increases with height.  Using the general rule of thumb (above), can hydrogen atoms that reach a height in the Earth's atmosphere such that T = 600 K escape? What about deuterium atoms (heavy hydrogen, with twice the mass as normal hydrogen)? What about helium atoms?  Which of these  three materials - hydrogen, deuterium, helium - will escape the fastest?  Why is the top of Earth's atmosphere called the 'exosphere.'

1. vave (H) = [ 3 x 1.38 x 10-16 x 600) / (1 x 1.67 x 10-24) ]0.5
vave (H) = 385,700 cm/sec = 3.86 km/sec
6 x vave (H)= 23.2 km/sec > vesc (Earth) = 11.2 so "yes," these H atoms can escape.

vave (D) = [ 3 x 1.38 x 10-16 x 600) / (2 x 1.67 x 10-24) ]0.5
vave (D) = 272,700 cm/sec = 2.73 km/sec
6 x vave (D) = 16.4 km/sec > vesc (Earth) = 11.2 so "yes," these D atoms can escape.

vave (He) = [ 3 x 1.38 x 10-16 x 600) / (4 x 1.67 x 10-24) ]0.5
vave (He) = 192,800 cm/sec = 1.93 km/sec
6 x vave (He) = 11.6 km/sec > vesc (Earth) = 11.2 so "yes," these He atoms can escape, but only barely.

H will escape fastest, D next fastest, He slowest.

The top of the Earth's atmosphere is the 'exosphere' because it is only from this layer of the atmosphere that a fast moving atom or molecule can exit, or escape.  Lower down where the air is denser, a fast moving molecule will collide with another molecule, thus preventing escape.

4. Can nitrogen atoms escape from Mars?
I forgot to assign a temperature for this calculation! As a follow-up to #3, I intended for this calculation to be done at T = 600 K:
vave (N) = [ 3 x 1.38 x 10-16 x 600) / (14 x 1.67 x 10-24) ]0.5
vave (N) = 10,300 cm/sec = 1.03 km/sec
6 x vave (N) = 6.2 km/sec > vesc (Mars) = 5.0 km/sec so "yes," these N atoms can escape.
5. What are the dominant constituents of the hydro/atmospheres of Mars, Venus and
Earth?