My `diameter' (height), mass (weight), and age are:

D_{Y} =
(6 ft) × (0.3048 meters/ft) =
(6 × 0.3048) meters = 1.83 meters |

M_{Y} =
(165 lb) × (0.454 kg/lb) =
(165 × 0.454) kg = 75 kg |

A_{Y} = 43 years |

Dividing the Sun's diameter, mass, and age by my diameter, mass, and age, I obtain:

D_{S} ÷ D_{Y} =
(1.4 × 10^{9} meters) ÷ (1.82 meters) =
7.7 × 10^{8} |

M_{S} ÷ M_{Y} =
(2 × 10^{30} kg) ÷ (75 kg) =
2.7 × 10^{28} |

A_{S} ÷ A_{Y} =
(4.5 × 10^{9} years) ÷ (43 years) =
1.05 × 10^{8} |

Note that the units cancel out, so the final numbers have no units
attached. These quantities are `pure' numbers, independent of the
system of units. I would have got the same final numbers with
*any* choice of units, as long as I use the *same* units for
myself and the Sun.

**1.** In which way am I least like the Sun?

- I'm least like the Sun in terms of mass; the Sun is almost 3
× 10
^{28}times my mass.

**2.** In which way are you most like the Sun?

- I'm most like the Sun in terms of age; the Sun is only
(only???) about 10
^{8}times my age.

**3.** Are you surprised by your answers to the two questions
above? Why or why not?

- The fact that the Sun is `only' about 10
^{8}times my age still seems amazing to me -- it makes me think that we can almost comprehend the vast times involved in astronomy. I'm not surprised that mass shows the largest difference; for objects of similar density, mass is proportional to volume, and volume is proportional to the cube of diameter.

Joshua E. Barnes (barnes@ifa.hawaii.edu) Last modified: September 2, 1999