Unit Conversions - Astronomy: The Human Quest for Understanding

Astronomy is a discipline that often uses different scales of measurement for different situations. For example, the unit nanometers ($\mbox{nm}$) may be used when discussing the size of an atom, meters ($\mbox{m}$) or kilometers ($\mbox{km}$) may be used when talking about features on Earth, the Moon, or other objects in our Solar System, astronomical units ($\mbox{au}$) are often used to discuss distances between the Sun and planets, and billions of lightyears ($\mbox{Gly}$) or billions of parsecs ($\mbox{Gpc}$) may be used to describe distances characteristic of the size of the universe. All of these descriptions are measures of distance and it is frequently necessary to convert between them. Of course similar situations arise with other types of measurements as well, such as time ($\mbox{s}$ or $\mbox{y}$), mass $\left[\mbox{g},\ \mbox{kg},\ \text{mass of Earth} \left(\mbox{M}_\text{Earth}\right),\ \text{mass of the Sun}\ \left(\mbox{M}_\text{Sun}\right)\right]$, and so on. This tutorial describes a general procedure for converting between different units of measure for a given type of measurement.

In all unit conversions, the procedure is essentially identical; multiply by a fraction that has the value of one. Sometimes it is necessary for several steps to occur before getting to the desired unit, but in each step you are always multiplying by a fraction that is equal to one.

To illustrate this procedure, imagine that you make a measurement concerning how long it took for your friend to run the Boston Marathon in seconds and you want to know how many minutes it took. In your measurement you found that your friend ran the marathon in $\mbox{13,581}\ \mbox{s}$. To find the elapsed time in minutes, you know that $60\ \mbox{s} = 1\ \mbox{min}$. This means that if you divide both sides of this equality by $60\ \mbox{s}$ you will have a fraction equal to one, or

Since you can always multiply any number by $1$ and never change the answer,

The procedure started out by multiplying by $1$. Because we constructed our unit conversion fraction to be identically equal to $1$, multiplying by the fraction in the second step doesn’t change the value of the quantity.

If you reviewed the fractions tutorial, you were reminded that it is possible to reduce fractions by canceling like terms in the numerator and the denominator as long as those terms are multiplying every quantity in the numerator and every quantity in the denominator. For example,

The same is true for units! In the marathon seconds-to-minutes conversion example above, the unit of seconds (s) ended up in the numerator and in the denominator when we multiplied by the unit conversion fraction $\left(\frac{1\ \text{min}}{60\ \text{s}}\right)$. In the third step above, the seconds unit was canceled in the numerator and the denominator, leaving minutes ($\text{min}$) as the final unit.

Students who have had little practice in converting between units will sometimes get confused about how to write the unit conversion fraction. This happens because you can write the fraction two ways. In the example above, in converting from seconds to minutes the unit conversion fraction was written as

$$\frac{1\ \mbox{min}}{60\ \mbox{s}} = 1.$$

However, it could just as easily have been written as

$$\frac{60\ \mbox{s}}{1\ \mbox{min}} = 1.$$

Both fractions are equally valid since they start from the fact that (60\ \mbox{s} = 1\ \mbox{min}). So how do you know which one to use? The answer is that you want to be able to cancel the unit you start with to arrive at the final unit. This means that the unit you want to cancel must be in the denominator of your unit conversion fraction if the starting unit is in the numerator, or the unit you want to cancel must be in the numerator if the starting unit is in the denominator.

Consider another example, suppose you want to know how many seconds are in one day. To do so you will need to use several unit conversion fractions multiplied together. Again this is okay because each unit conversion fraction is just a clever way of writing the number $1$ and you can always multiply any number by $1$ without changing it. To convert from days to seconds you need to know that $1\ \mbox{day} = 24\ \mbox{hr}$, $1\ \mbox{hr} = 60\ \mbox{min}$, and $1\ \mbox{min} = 60\ \mbox{s}$. Next, think about the order in which you want to carry out the conversion; in this case you are going from days to seconds, so you need to construct your unit conversion fractions accordingly. For this example you do the following:

Notice that in every step the unit conversion fraction was constructed to cancel the prior unit, terminating when the only unit remaing was seconds. Had you wanted to know how many days corresponded to $\mbox{86,400}\ \mbox{s}$, you would have constructed your unit conversion fractions as the reciprocal of what you had before, or

Now, using what you have learned, convert $1$ mile into inches. To do so you need to know that $1\ \mbox{mi} = 5280\ \mbox{ft}$ and $\ \mbox{ft} = 12\ \mbox{in}$. You should find that $1\ \mbox{mi} = \mbox{63,360}\ \mbox{in}$. (This is a very cumbersome set of units when compared to the so-called “metric” system used by most of the rest of the world! Scientists almost always use the “metric” system, which is known as the SI system, which stands for Systeme International d’Unites.)

In some cases units show up multiplying themselves two or three times. A common example is in the measurement of area, such as the area of your home, perhaps $\mbox{15,000}$ square feet ($\mbox{ft}^2$; a quaint little cottage!). Recall that the area of a rectangle is length times width. Since you are multiplying two distances together you end up with the unit of length squared. For volumes the length unit is cubed; e.g., cubic feet ($\mbox{ft}^3$), as in “how many cubic feet of dirt do you need?”

To convert from something like square feet ($\mbox{ft}^2 = \mbox{ft} \times \mbox{ft}$) to square meters ($\mbox{m}^2 = \mbox{m} \times \mbox{m}$) you must make sure that you multiply by your unit conversion fraction enough times to cancel each factor of the unit.

Suppose you want to know the area of your $\mbox{15,000}$ square foot home in square yards, in order to purchase enough carpet to cover it completely. Since $1\ \mbox{yd} = 3\ \mbox{ft}$,

This could have also been accomplished more directly by squaring the unit conversion fractions directly:

$$\mbox{15,000}\ \mbox{ft}^2 = \mbox{15,000}\ \mbox{ft}^2 \times \left(\frac{\mbox{yd}}{3\ \mbox{ft}}\right)^2 = \mbox{15,000}\ \mbox{ft}^2 \times \left(\frac{\mbox{yd}^2}{3^2\ \mbox{ft}^2}\right) = \frac{\mbox{15,000}\ \mbox{ft}^2\times \mbox{yd}^2}{9\ \mbox{ft}^2}.$$

With $\mbox{ft}^2$ in both the numerator and the denominator, those units cancel leaving $\mbox{yd}^2$.

Quiz Yourself

Try the following exercises to make sure that you really do understand the material just presented. By the way, computers and many calculators write scientific notation in “computer-speak,” where $10$ raised to an integer (either positive or negative) replaces $\times 10^\text{exponent}$ with ${\rm E(exponent)}$ or ${\rm e(exponent)}$; in other words, $3.14159 \times 10^9$ is represented by $3.14159\text{E}9$. If your calculator has the $\Large{\hat{ }}$ symbol, then you would enter the number as $3.14159 \times 10{\Large{\hat{ }}}9$. If your calculator has a $10^x$ button then you would enter the number as $3.14159\times9\,\,10^x$. If you have having problems with entering numbers in scientific notation, read the manual or help screens, or ask your instructor.

The unit of force (such as weight) in the SI system of units ($\mbox{kg}$, $\mbox{m}$, $\mbox{s}$) is a newton ($\mbox{N}$). Suppose you weight $220$ pounds ($\mbox{lb}$), what is your weight in newtons. Note that $1\ \mbox{N} = 0.2248\ \mbox{lb}$.
The distance from the Sun to our nearest star, Proxima Centauri is $1.29\ \mbox{pc}$ (parsec). Determine the distance to Proxima Centauri in light-years ($1\ \mbox{pc} = 3.26\ \mbox{ly}$).
One light-year is the distance that light can travel in one year, which is $9.46 \times 10^{15}\ \mbox{m}$. From the previous problem, what is the distance to Proxima Centauri, measured in meters.
The Sun’s mass (symbolized by $\mbox{M}_\text{Sun}$) is typically used as the unit of measure for talking about other stars and even galaxies. In kilograms, $1\ \mbox{M}_\text{Sun} = 1.99\times 10^{30}\ \mbox{kg}$. The mass of the supermassive black hole at the center of our Galaxy is estimated to be $4.1 \times 10^6\ \mbox{M}_\text{Sun}$. What is its mass in $\mbox{kg}$?
Suppose your mass is $75\ \mbox{kg}$. Using the value of $1\ \mbox{M}_\text{Sun}$ given in the last problem, what is your mass in solar mass units?
The age of the universe is estimated to be $13.8\ \mbox{Gyr}$. How old is the universe in seconds. (If you are unfamiliar with what the prefix $\mbox{G}$ represents, please consult the prefixes tutorial.
Just for kicks let’s define a new unit, $1\ \mbox{Uyr} = 13.8\ \mbox{Gyr}$, where $1\ \mbox{Uyr}$ is the age of the universe (“universe-years”). Suppose your friend ran the Boston Marathon with a time of $\mbox{13,581}\ \mbox{s}$, convert her time into universe-years ($\mbox{Uyr}$). Your answer represents the fraction of the age of the universe your friend spent running the Boston Marathon.
The surface area of Earth is approximately $5.11 \times 10^8\ \mbox{km}^2$. Convert that surface area into square inches. The fact that $1\ \mbox{in} = 2.54\ \mbox{cm}$ will come in handy. (You may need to consult the prefixes tutorial if the prefixes are unfamiliar to you.)
The volume of the Sun is roughly $1.43\times 10^{18}\ \mbox{km}^3$. Convert the Sun’s volume to cubic parsecs ($\mbox{pc}^3$), where one light-year ($\mbox{ly}$) is $1\ \mbox{ly} = 9.46\times 10^{15}\ \mbox{m}$ and one parsec ($\mbox{pc}$) $1 \ \mbox{pc} = 3.26\ \mbox{ly}$. Your answer represents the volume that the Sun occupies relative to a sphere that is $1\ \mbox{pc}$ in radius, which is slightly less than the distance to Proxima Centauri. Most of space is very, very empty!!

Answers

$979~\text{N}$
$4.2~\text{ly}$
$3.98\times10^{16}~\text{m}$
$8.2\times10^{36}~\text{kg}$
$3.8\times10^{-29}~\text{M}_\text{Sun}$
$4.4\times10^{17}~\text{s}$
$3.1\times10^{-14}~\text{Uyr}$
$7.9\times10^{17}~\text{in}^2$
$5.2\times10^{-25}~\text{pc}^3$