1. The diameter of the Moon is 3480 km. What is the surface area of the Moon? How many times larger is
the surface area of the Earth?
A  4 r 2  4  d 2    d 2 .
2
The surface area of a sphere is found by
(a)
(b)

2
AMoon   DMoon
  3.48 106 m
AEarth
AMoon

2
 3.80 1013 m2
2
2
2
 DEarth   REarth   6.38 106 m 
 DEarth


 
 
  13.4
2
6
 DMoon
D
R
1.74

10
m

 Moon   Moon  
2
2. One liter (1000 cm3) of oil is spilled onto a smooth lake. If the oil spreads out uniformly until it makes an
oil slick just one molecule thick, with adjacent molecules just touching, estimate the diameter of the oil
slick. Assume the oil molecules have a diameter of 2x10-10 meter.
The volume of the oil will be the area times the thickness.
2
The area is  r    d 2  , and so
2
3
 1m 
1000 cm 

V
2
100 cm 

V    d 2 t  d  2
2
 3  103 m .
10
t
  2  10 m 
3
3. Recent findings in astrophysics suggest that the observable Universe can be modeled as a sphere of
radius R = 13,7 x 109 light-years with an average mass density of about 1 x 10-26 kg/m3, where only about
4% of the Universe’s total mass is due to “ordinary” matter (such as protons, neutrons, and electrons). Use
this information to estimate the total mass of ordinary matter in the Observable Universe. (1 light year =
9,46 x 1015 m.)
Multiply the volume of a spherical universe times the density of matter, adjusted to ordinary
matter. The volume of a sphere is
4
3
 r 3.
3

9.46  1015 m 
26
3
9
4
m  V  1  10 kg m  3   13.7  10 ly  
  0.04 
1ly


 3.65  1051 kg  4  1051 kg