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Measurement Systems
Purpose: To familiarize students with the fundamentals of different measurement
systems used in astronomy.
Goals: Upon successful completion of this lab assignment, students will be able to:
1. Understand the basis for using the metric system.
2. Convert measurements within the metric system.
3. Convert measurements between various measurement systems.
4. Use ratios as a means of comparing measurements and for setting up
proportions.
Measurements Systems
Since the late 19th century measurements of the properties of nature has been
standardized under the definitions provided within the International System of
Measurement (referred to as the ‘SI System’ due to it being established as the System
International d’Unites in France). The SI System identifies seven (7) basic properties of
nature and has established a base unit of measurement for each (Table 2.1).
Basic Physical Property
Distance
Mass
Time
Temperature
Electric Current
Amount of a Substance
Intensity of Light
Base Unit
Symbol of Base Unit
Meter
Kilogram
Second
Kelvin
Ampere
Mole
Candela
m
kg
s
K
A
mol
cd
Table 2.1: The seven basic physical properties of nature and their designated base units.
Distance, mass, time, and temperature are the basic properties that will be utilized
throughout the course, though you should be familiar with the other properties and their
units as well.
1. What base unit from Table 2.1 would you use to express the following quantities?
How far is it to the Moon?
Meters
How long does it take Venus to orbit the Sun?
Seconds
How many atoms are in a rock sample?
Moles
How heavy is the Pluto Express satellite?
Kilograms
How tall is the Martian volcano Olympus Mons?
Meters
How bright does the star Sirius appear?
Candela
What is the life expectancy of the Voyager spacecrafts?
Seconds
Dimensional Analysis
In nature there are countless physical properties beyond the base unit that require
measurement. These additional properties are a combination of the elementary properties
found in Table 2.1 and will therefore have units that are a combination of the base units.
The way these new units are determined is through a process called dimensional analysis.
By knowing how the elementary properties combine through some definition or law of
nature, we can substitute the base units to determine the appropriate unit. Below are a
couple of examples of how dimensional analysis is used to determine these new units.
Speed
The speed of an object is defined as the amount of distance covered within a
measured amount of time:
speed =
distance
time
If we wanted to know the numerical value of an object’s speed we would simply
substitute the known quantities for the distance and time (the speed of a car that goes 150
 = 50). To determine the unit of measurement for speed
miles in 3 hours; speed = 150/3
we would instead substitute the base units of the properties used in the definition for
speed:
base unit for speed =
base unit for distance
base unit for time
meters
base unit for speed =
= m ("meters - per - second" )
s
second


Acceleration
The acceleration of an object is defined as the rate at which its speed changes over a
known period of time:
acceleration =
speed
time
An example of acceleration would be a car slowing down to a stop for a red light. To
determine the unit for acceleration we substitute the base unit for speed (m/s) and the
base unit for time (s) to 
get:
base unit for acceleration
=
base unit for speed
base unit for time
m 
 
m 1
 s 
base unit
=

= m 2 ("meters - per - square - second" )
 for speed =
s
s
s
s
Force

An object is accelerated whenever a force is applied to it. The amount of force felt by
an object is defined through Newton’s 2nd Law of Motion as being equal to the mass of
an object multiplied by its acceleration (F = ma). Using dimensional analysis we find
that the base unit for force is:
base unit for force

= (base unit of mass)  (base unit of accelration)
kg  m
base unit for force = (kg)  (m 2 ) =
s
s2
The base unit for force is cumbersome to write and to read (“kilogram-meter-persquare-second). It is common to rename such units for the sake of simplicity. The base
 has been renamed the Newton (N) which is defined in the following way:
unit for force
kg  m
1 Newton (N)  1  2 
 s 

2. Some of the important physical properties we will encounter during the course
include area, volume, density and energy. In the following exercises, you will
calculate the measurements of an example of each of these properties and
determine the base unit for each by performing dimensional analysis. Be sure to
show the steps you used to arrive at your answers.
a. The area of a rectangle is determined by multiplying the length of the
rectangle by its height ( Area = length  width ). If a rectangle has a length
of 2.5 meters and a width of 3.4 meters, determine the area of the rectangle as
well as the appropriate units.
 = (2.5 m)  (3.4 m) = (2.5 x 3.4) (m  m) = 8.5 m2
Area = (l)  (w)
b. Determine the volume of a rectangular box with the dimensions listed above,
and a thickness of 1.2 meters ( Volume = length  width  height ). Be sure
to include the appropriate units.
Volume = (l)  (w)  (h) = (2.5 m)  (3.4 m)  (1.2 m)
 = (2.5 x 3.4 x 1.2) (m  m  m)
= 10.2 m3
c. The density of a substance indicates how much mass of a substance can be fit
into one cubic unit of volume. A substance’s density determines other
properties of matter like buoyancy and fluidity. Presuming that the box from
the previous example is filled with 23.5 kg of concrete, determine the density
of the concrete.
mass
Density =
Volume
Density = mass / Volume = (23.5 kg) / (10.2 m3)
= (23.5 / 10.2) (kg / m3)

= 2.30 kg/m2
d. Suppose that the box filled with concrete is dropped from the observation
deck of the Empire State Building (h = 320 meters). As the box falls toward
the ground, it gains speed and will hit the pavement with a certain amount of
energy. The amount of energy is dictated by the mass of the concrete, the
height from which it is dropped, and the rate at which it falls (g = 9.81 m/s2).
Determine the energy of the box when it hits the ground if:
Energy = mgh
Energy = (m)  (g)  (h) = (23.5 kg)  (9.81 m/s2)  (320 m)
= (23.5 x 9.81 x 320) (kg  m/s2  m)

= 7.38 x 104 kg  m2/s2
Beyond the Base Unit
There are many instances when the base unit is insufficient when describing a
physical quantity. People don’t typically refer to being alive for 662,691,456 seconds as
a birthday milestone, but that is what happens when someone turns 21 years old. In other
words, it is quite impractical to express one’s age in terms of the base unit of time
(seconds). A much larger unit, the year, being roughly 31,500,000 seconds long, is more
practical when expressing ones age. As such, each of the base units of the SI system can
be built up to describe large physical properties or cut down to describe those that are
small.
The simplest way to create new units of measurement beyond a base unit is to rely on
building up or breaking down the base unit by factors of ten. Doing so allows us to
visualize scales more easily because we are not required to memorize random conversion
factors that are found within the Imperial System of measurements used in daily
experience (e.g. 12 inches = 1 foot; 5,280 feet = 1 mile; 16 oz. = 1 pound). Making the
process easier is that each factor of ten has a unique prefix that can be added to any base
unit in question. Table 2.2 lists the prefixes of the SI system that we will utilize, and
their corresponding powers of ten.
Power of Ten
Prefix
Abbreviation
1012
109
106
103
102
101
100
10-1
10-2
10-3
10-6
10-9
10-12
TeraGigaMegaKiloHectoDeca-DeciCentiMilliMicroNanoPico-
T
G
M
k
h
da
(Base Unit)
d
c
m
*
n
p
* The symbol for micro- is the Greek letter, “mu,” because “m” is already used for milli-.
Table 2.2: SI prefixes and their powers of ten.
The prefixes listed in Table 2.2 are added to the base unit when referring to a specific
factor of the base unit. One thousand (103) meters is referred to as one kilometer. Onehundredth (10-2) of a gram is called a centigram. Similarly, to obtain the correct
abbreviation for any of these units, the prefix symbol is added before the base unit
abbreviation. One kilometer is written as 1 km while one centigram is written as 1 cg.
When presented with measurements written out completely it is useful to write the
numbers in scientific notation in order to determine the correct prefix. For example:
2,500 meters = 2.5  103 meters = 2.5 kilometers = 2.5 km

0.0041 seconds = 4.1  10 -3 seconds = 4.1 milliseconds = 4.1 ms
In practice measurements are often not written in proper scientific notation where the
decimal point would be moved so that there is a single, non-zero digit to the left. Rather,
thedecimal point is moved so that that mantissa will be followed by a factor of ten whose
power is a multiple of three (e.g. 103, 106, 10-9, etc.). To give an example, the number
56,300 grams is typically written as follows:
56,300 grams = 56.3  10 3 grams = 56.3 kilograms = 56.3 kg
3. Fill in the blank spaces of Table 2.3 below by expressing each measurement in the
desired form.

Column 1: measurements written in terms of the base unit
Column 2: measurements from Column 1 written in terms more
appropriate units by making use of prefixes
Column 3: measurements from Column 2 written with the use of the
appropriate symbol.
Base Unit Measurement
Measurement with Prefix
Measurement with Symbol
0.002 grams
0.05 meters
7.52 x 10-6 meters
84,400 seconds
1.21 x 109 Watts
0.0058 Amperes
6.5 x 106 years
16,000,000,000,000 Bytes
2.0 milligrams
5.0 centimeters
7.52 micrometers
84.4 kiloseconds
1.21 gigaWatts
5.8 milliAmpers
6.5 Megayears
16 TeraBytes
2.0 mg
5.0 cm
7.52 m
84.4 ks
1.21 GW
5.8 mA
6.5 Myrs
16 TB
Table 2.3
Sometimes when we calculate a physical quantity we end up with a value that would
be clearer if we were to express it in more convenient units. There are several methods to
go about converting measurements within the metric system, but the method described
below will always work when carried out properly:
Known Measurement
Desired Measurement
=
Desired Measurement' s Power of 10
Known Measurement' s Power of 10
Let’s consider a case where the mass of an object was calculated to be 140,000
centigrams (cg). If we wanted to express that measurement in terms of kilograms, we
 perform the conversion method from above once we define each part of the
could
proportion:
Known measurement = 140,000 cg
Desired measurement = x kg
Desired measurement’s (kg) power of ten = 103
Known measurement’s (cg) power of ten = 10-2
We begin the conversion by substituting our values into the proper places in the
proportion:
140,000 cg
10 3
=
x kg
10 -2
Next we cross-multiply:
 10 3  x kg = 140,000 cg  (10 -2 )
The we divide to get x by itself:

x kg =
(140,000 cg  10 -2 )
10 3
Calculating for x we end with the result:

x = 1.4 kg
4. Perform the following conversions using the method outlined above, or any

method that you are familiar
with. Be sure to include a page showing the steps
show how you arrived at your answer.
Convert 4 m into cm.
Convert 4 m into mm.
Convert 185 mg into cg.
4 m = 400 cm
4 m = 4000 mm
185 mg = 18.5 cg
Convert 3 x 103 ns into µs.
3 x 103 ns = 3.00 µs
Convert 0.5 MW into kW.
0.5 MW = 500 kW
Standard Astronomical Measurements
From the extraordinarily far distances to galaxies to the enormous masses of stars
astronomers contend with excessively large numbers on a regular basis. In an effort to
make these outrageous numbers more comprehensible astronomers have developed a
system of measurements that is exclusive to the scales involved in their field. Listed
below are some examples of standard astronomical measurements that are used
frequently.
Astronomical Unit (AU) – the average distance between the Earth and the Sun. It has a
measured value of:
1 AU = 1.50 x 1011 m = 1.50 x 10 8 km
The Astronomical Unit is typically used to describe the distances between objects
within our solar system, the distance an exoplanet orbits its parent star, and the distances
 make up a binary star system.
between stars that
Parsec (pc) – the distance a star would need to be in order for it to have a parallactic shift
of one arc-second. The name is derived from combination of the parallax-second. The
parsec has been determined to have a distance of:
1 pc = 3.09 x 1016 m = 3.09 x 1013 km
The parsec is the unit from which all astronomical distance equations have been
derived and is used frequently when reporting distances to objects beyond our solar
system as well
as the sizes of individual galaxies.
Light-Year (LY) – defined as the distance light travels in one year’s time. It is a
calculated distance that is found by multiplying the measured speed of light (3.00 x 108
m/s) by the number of seconds in one year (3.15 x 107 s). Thus the light-year is defined
numerically as:
1 LY = 9.46 x 1015 m = 9.46 x 1012 km
1 pc = 3.26 LY
 can be used to describe the same distance/size measurements as the
The light-year
parsec. Because the light-year is based on the speed of light, it provides an added
component of understanding.If a star were exactly one light-year away from Earth, then
it would take the light emitted exactly one year to make its trip to Earth. Similarly if a
different star were 8 LY from Earth then the light from that star would take exactly 8
years to reach Earth. Knowing the number of light-years an object is away from Earth
provides a direct measurement of how “old” the light from that object is.
Solar Mass (MSun) – defined as the amount of mass contained within our Sun. It has a
measured value of:
1 MSun = 1.99 x 10 30 kg
The solar mass is one of several solar units that are used throughout astronomy. Other
solar units include the solar luminosity (LSun), the solar radius (RSun), and the solar

lifespan (Sun). In general, solar units are useful when discussing the properties of stars
because they provide a direct comparison to the star we know most about.
Converting from standard metric measurements to standard astronomical
measurements (or vice versa) requires setting up proportions. The two key things to
remember to properly setting up the proportions are:
(1) Make sure you have the proper conversion factor.
(2) Make sure the units are the same on each side of the equal sign.
As an example, let’s determine the distance from the Sun to Jupiter (7.8 x 108 km) in
Astronomical Units (AU).
First, we write out the problem algebraically:
7.80 x 10 8 km = x AU
Next we set up the proportion using the conversion factor, 1 AU = 1.50 x 108 km, making
sure to keep the units the same on each side of the equal sign:

7.80 x 10 8 km
x AU
=
8
1.50 x 10 km
1 AU
Solving for x yields the result:

x = 5.20 AU
5. Use the Internet or your textbook, to determine the distances in each example
below. Once you are finished, convert each measurement into the units requested.

Be sure to show the steps in each conversion.
a. What is the distance from NYC to Los Angeles, in miles? Using the fact that
1 mile = 1.6 km, convert this distance to kilometers.
Distance from NYC to LA = 2784 miles = 4454 km
2784 miles
x km
=
1 mile
1.6 km

x = (1.6) x (2784) km

b. What is the equatorial diameter of the Earth, in meters? Convert the Earth’s
diameter into kilometers.
Diameter of the Earth = 12,756,000 meters = 12,756 km
2784 miles
1 mile
x km
1.6 km
=
x = (1.6) x (2784) km
c. What is the average distance from the Earth to the Moon, in kilometers?
Convert this distance into miles.


Distance from Earth to Moon = 407,000 km = 254,000 miles
x miles
1 mile
=
407,000 km
1.6 km
x = 407,000  1.6 miles
d. What is the average distance from the Sun to the Earth, in miles? Convert this
distance into kilometers.


Distance from Sun to Earth = 93 million miles = 1.5 x 108 km
9.3 x 10 7 miles
1 mile
=
x km
1.6 km
x = (1.6) x (9.3 x 10 7 ) km
e. What is the average distance from the Sun to Saturn, in kilometers? Convert
this distance into Astronomical Units(AU).

Distance from Sun to Saturn = 1.43 x 109 million km = 9.67 AU
1.43 x 10 8 km
x AU
=
8
1.50 x 10 km
1 AU
x = (1.43 x 10 9 ) x (1.50 x 10 8 ) AU
f. What is the distance from the Sun to the star, Polaris, in kilometers? Convert
(LY).
this distance into light-years
Distance from Sun to Polaris = 2.8 x 109 km = LY
4.09 x 1015 km
x LY
=
12
9.50 x 10 km
1 LY


x = (4.09 x 1015 )  (9.50 x 1012 ) LY
g. What is the diameter of the Milky Way Galaxy, in light years? Convert this
distance into kilometers.
Distance from Sun to Saturn = 1.0 x 105 LY = 9.5 x 1017 km
x km
1.0 x 10 5 LY
=
9.50 x 1012 km
1 LY
x = (9.50 x 1012 ) x (1.0 x 10 5 ) km
h. What is the distance to the Andromeda Galaxy (M31), in light-years? Convert
 this distance into parsecs (pc). 
Distance from Sun to Saturn = 2.54 x 106 LY = 7.78 x 105 pc
2.54 x 10 6 LY
x pc
=
3.26 LY
1 pc
x = (2.54 x 10 6 )  (3.26) pc
Ratios and Scale


Comparing measurements is an essential form of analysis in the sciences. A simple
glance will tell whether one measurement is larger or smaller than another, but scientists
are more interested in a more accurate analysis. To them it’s more important to know
how much one quantity is larger than another. The most effective way quantities are
compared is through the use of ratios.
Any measurement with which we are familiar can use that as a basis for comparison.
Living on in central Suffolk County on Long Island, we are most likely familiar with the
fact that a car trip to New York City takes roughly one hour. We could use this
information to estimate how long a trip to Boston would take by using ratios.
A search on MapQuest reveals that the distance form Selden to New York City is 58
miles and the distance from Selden to Boston is 252 miles. Comparing the distances
using a ratio would yield the following:
Distance to Boston
Distance to NYC
=
252
= 4.34
58
This comparison says “the distance from Selden to Boston is 4.34 times the distance
from Selden to NYC.” From there we can use the knowledge that if it’s an hour long trip
 a trip to Boston would be a little over 4.3 hours (4 hrs 18 min).
to NYC by car then
Ratios are the means for creating model scales so that accurate representations of the
physical world can be created to aid in visualization. The most common form of model
scale is the map scale. When reading a map, a scale is typically provided in the maps’
legend that lets the reader know what a unit distance on the map equates to in reality. So
a map scale of 1 cm = 25 km means, “for every centimeter’s distance that is measured on
the map, the real distance is 25 kilometers.” If the map scale is not provided then it can
be determined so long we know one real distance.
As an example, let’s assume we are presented with the map below that shows New
York City, Nassau County, and central Suffolk County and we want to determine the map
scale. Earlier we saw that the distance from Selden (indicated by the star) to NYC is 92.8
kilometers (58 miles).
Using a ruler we find a map distance of 11.6 cm directly between Selden and NYC.
The 11.6 cm distance on the map represents 92.8 km in true distance. Using the ratio of
the two measurements we can find the map scale.
True Distance
Map Distance
=
92.8 km
x km
=
11.6 cm
1 cm
The map scale is therefore, 1 cm = 8 km. This “map conversion factor” can be used to
compute the true distance between any two points on the map.

Example:
What is the true distance between Selden and Port Jefferson?
True Distance to Port Jefferson
Map Distance to Port Jefferson
=
z km
8 km
=
1.4 cm
1 cm
After cross-multiplying and dividing we find:

z =1.4  8 = 11.2 km

6. Find the true distances between the locations that are given in the examples
below. Show all the steps you use in determining your answer.
a. What is the distance from Selden to Patchogue?
Map distance between Selden and Patchogue = 1.4 cm
1.4 cm
x km
=
1 cm
8 km
x = (1.4) x (8) km
Actual distance between Selden and Patchogue = 11.2 km


b. What is the distance from Huntington to Wading River?
Map distance between Huntington and Wading River = 7.1 cm
7.1 cm
x km
=
1 cm
8 km
x = (7.1) x (8) km
Actual distance between Selden and Patchogue = 56.8 km


c. What is the distance from Manorville to Queens?
Map distance between Manorville and Queens = 12.4 cm
12.4 cm
x km
=
1 cm
8 km
x = (12.4) x (8) km
Actual distance between Selden and Patchogue = 96.0 km

