Scientific Notation and Units
Steven Kaplan
Department of Physics and Astronomy, Rutgers University
Scientific Notation
A Motivation
electron mass = 0.000000000000000000000000000000911 kg
Wow, that’s a lot of zeros; thirty-one of them to be exact. Every time you
need to use the rest mass of an electron, this is the number you are going to be
using. The problem is, nobody wants to write thirty-one zeros and just three
other numbers. There has got to be a better way to write numbers like the rest
mass of an electron and others like it that have a large number of zeros (i.e.
the number in question is really small or really large) much more compactly.
Thankfully, there is such a way: scientific notation. I can write the rest mass
of an electron as 9.11 · 10−31 kg in scientific notation. I think you would agree
that this is considerably more convenient to write and is thus well worth your
time to learn how to do it well.
What Is Scientific Notation?
Basically, it is a way to write a very large or very small number in a more
compact fashion. This is accomplished by taking advantage of the powers of 10.
Let’s get reacquainted with these:
The Powers of Ten
I’ll start by listing a few of them (note that I have shown each of them multiplied
by 1.0 for reasons that will become apparent later on):
1
1.0 · 10−3 = 0.001
1.0 · 10−2 = 0.01
1.0 · 10−1 = 0.1
1.0 · 100 = 1.0
1.0 · 101 = 10.0
1.0 · 102 = 100.0
1.0 · 103 = 1000.0
Table 1: Some Powers of Ten
Let us note the correlation between the position of the decimal point and the
power of ten that we are dealing with. Using 1.0 · 10−3 = 0.001 as an example,
we notice that the decimal point on the right hand side is three numbers to the
left of where it was on the left hand side. This is to say, multiplying 1.0 by
10−3 just moved the decimal point over three numbers to the left. The exponent of 10 was −3, and the decimal point moved three spaces to the left. Hmm...
Now, let’s look at 1.0 · 102 = 100.0 . Multiplying 1.0 by 102 just moved the
decimal point two points to the right. The exponent of 10 was 2, and the decimal point moved two spaces to the right. Hmm...it looks like there is a HUGE
connection between the number in the exponent and the number of spaces the
decimal point moves (i.e. the two are exactly equal)!
This is the backbone of scientific notation. We multiply a number by 10another number .
The power that 10 is raised to is negative if I want the decimal point to be moved
to the left (i.e. to make the number smaller), and it is positive if I want the
decimal point moved to the right (i.e. to make the number bigger). In these
examples, the number we are making larger and smaller is just 1, but scientific
notation can be (and is) used in this capacity for any number.
Examples
1. Let’s try writing the number 0.0023 in scientific notation:
Ok, let’s first pick out the number that is going to be multiplying 10.
The most straightforward choice is 2.3. Alright, now we just need to find
the right power for 10. First, do we want to make 2.3 smaller or bigger?
Well, 0.0023 is smaller than 2.3, so I am going to have to pick my power
of 10 such that it makes 2.3 smaller. Like we have seen above, this means
that the power must be negative. Also, note that if I were to move the
decimal point in 2.3 three numbers to the left, I would have exactly 0.0023.
Now, I can write 0.0023 in scientific notation!
Namely:
0.0023 = 2.3 · 10−3
2
2. How about 345000000?
Ok, first pick your number. Let’s pick 3.45.
Note that I could have also picked 34.5 or 345 (or really, even something like 0.00000345, but that isn’t at all practical and defeats the whole
purpose of using scientific notation). The power of 10 that we need to
use depends on what we pick for the number we are multiplying it by.
This is obviously true because the only difference between any of numbers
we would pick is where the decimal point is. Since the exponent is the
difference between where we want the decimal point and where it is before
we multiply by a power of ten, the number we pick has a direct effect on
what the power of ten is.
Since 345000000 is larger than 3.45 (i.e. we need to move the decimal
point to the right), the power of ten needs to be positive. How many
spaces to the right does the decimal point need to move? It looks like 8 to
me. How did I know this if there was no decimal point shown? Well, any
number is the same number if I put a .0 after it, right?. Like, for example,
550 = 550.0. Just the same, 345000000 = 345000000.0. You can use this
if you need to visualize the decimal point. Ok, after our analysis, we see
that:
345000000 = 3.45 · 108
Say however, that we picked 345 instead of 3.45 as the number to which
we multiply the power of ten. We reasoned above that the only difference
in how I write the number in scientific notation (other than using that
number instead, of course) is the power of ten I use. Since 345 is larger
than 3.45, it stands to reason that I need to use a smaller power of 10
(in this case, 106 ). I can see in three different ways that we can write
345000000 = 345 · 106 .
(a) Through the same way we have done our examples, I notice that
the decimal point just needs to be shifted by 6 numbers to the right
(hence, the power of ten needed is 6)
(b) A slightly more crafty method: I know that 3.45 = 345 · 10−2 (can
you figure out why?). Using the fact that 345000000 = 3.45 · 108 :
345000000 = 3.45 · 108
345000000 = (345 · 10−2 ) · 108
345000000 = 345 · 106
(c) 106 = 1000000, that is to say, one million. We know that 345 · one
million = 345 million, so if one million is the same thing as 106 , I
can write 345 million as 345 · 106 . This method is especially useful
in physics. We will see this later on in the section on units.
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3. Let’s try going the other way now. Write 6.67 · 10−11 in decimal notation.
(Note: You will be using this number, Newton’s Universal Gravitational
Constant, extensively in your first semester).
Ok, The power is negative, so I know the number we come out with should
be smaller than 6.67. Also, the -11 tells me that the decimal point should
be moved 11 spaces to the left. Therefore, I should arrive at the following
result:
6.67 · 10−11 = 0.0000000000667
Again, we see the power (pun intended) that using scientific notation
affords us.
4. One more going the other way: 8.99 · 109 . (Note: This is a constant that
appears in many places in electricity. One such place you will see it is the
electric field generated by a point charge like a proton or electron.)
Since the exponent is positive 9, the decimal point should be moved 9
spaces to the right (hence, a bigger number). Therefore:
8.99 · 109 = 8990000000
Note that, along the lines of example 2, we also could have noticed that 109
= one billion. So, 8.99 · 109 = 8.99 billion. That translates to 8990000000.
Units
This section is probably one that is least understood by students. In my opinion,
not enough emphasis is placed on these ideas. Hopefully, this section will make
the concept of units and how they are used clear.
Definition and Background Information
An Example: Length
What is a unit anyway? Let’s learn by an example. Your height is measurement
of length, right? When you are standing straight, it is the distance between the
ground and the top of your head. How tall are you? I’m about 5 feet and 10
inches (5.83 feet). When I say that my height is a certain number of feet, that
implies that there is a reference I am comparing it to; namely, one foot. There
is a certain amount of length that is considered one foot, and my height is 5.83
times that length. Just the same, I could say that my height is 1.78 meters.
One meter is defined to be a certain length, and my height is 1.78 times that
length. We can see that one meter is longer than one foot by the fact that it
takes 5.83 feet to span my height, but it only takes 1.78 meters.
4
Feet and meters are examples of units of length. Length units are (literally, in this case) a measuring stick to quantify the length of an object. Just
as well, I could define a length measurement called the Kaplan that is defined
to be exactly my height. Then, of course, my height would be quantified as one
1
Kaplans. Can you
Kaplan. If an object is one meter long, it is equivalently 1.78
figure out why this is the case? If not, I will use this as an example when we
get to unit conversions.
Mass
Just as, say, one meter can be a measuring stick for quantifying the length of an
object, there exist mass ”measuring sticks” that quantify the mass of an object.
Note that while I gave the example of feet when we discussed length units, I
will not be using pounds as an example of mass here. Why not? Because the
pound is simply not a unit of mass; it is a unit of weight. As much as I would
like to, this forum isn’t the place for a discussion of the difference between the
two.
I do, however, owe you an explanation of what mass is. The definition of mass
is a rather tricky one to put into words, but I will do my best. Mass is a certain
quantity that every object has that describes the degree of effort required to
move it from its original position. I am assuming here that the object isn’t resting on anything. Of course, you could have even a very massive object that is
very delicately balanced at the top of a very steep hill (on Earth, where there is
gravity). All it would take is a finger flick to get it to rolling. That is not what
I am talking about. Imagine that the object is at rest in space, with nothing
else (not even gravity) interfering. The more massive the object is, the harder
it is to get it to change its position.
Describing a unit of mass is much like describing a unit of length. There is a
certain amount of mass that is designated to be one unit of mass. For example,
let us take the kilogram. A kilogram, like any other unit of mass, is designated
to be a certain amount of mass. If an object has twice that designated amount
1
of mass, the object is said to have a mass of 2 kilograms. If an object has 1000
1
of that designated amount of mass, the object is said to have a mass of 1000
kilograms (or one gram, if you are familiar with the metric system, which you
should be).
Time
Time, again, can be tricky to define; however, I think you already have a feel
for what it is. Just like length and mass, time also has units. These units,
however, are more familiar to you. Take the time unit of one second: currently
defined to be ”The second is the duration of 9192631770 periods of the radiation
corresponding to the transition between the two hyperfine levels of the ground
state of the caesium [sic] 133 atom.” (source=http://bit.ly/aReTzE) The idea
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of a time unit is still the same as the other types of units: there is a certain
duration of time that is defined to be one second, and all other time durations
are measured relative to this special time duration. If there is a time duration
that is three times as long as the time duration defined to be one second, then
that time duration is said to be three seconds long. If there is a time duration
that is sixty times as long as the time duration defined to be one second, then
that time duration is said to be sixty seconds long (or one minute).
Systems of Units
This is a very important concept, and it is also one that many of my students
did not give enough thought to. To use a particular system of units is basically a
decision to always choose the same units for length, mass, time, electric charge,
and temperature. In physics, the system that is used by basically everyone is
the SI system which chooses the length unit to be the meter, the mass unit to be
the kilogram, the time unit to be the second, the electric charge unit to be the
Coulomb, and the temperature unit to be the Kelvin. (used wikipedia). While
we didn’t discuss the Coulomb and the Kelvin explicitly, they are defined using
the same idea as how the meter, kilogram, and second were defined. Why the
kilogram as opposed to the gram some of you might ask? Not sure, I’ll have to
look into it.
Who Cares? YOU Should
Seriously! Why can’t I just pick whatever units I feel like using that day? Why
limit myself? The answer is simple: if you stick to a certain system of units,
you are guaranteed to always have a unit in that system at the end of the day.
Solving physics problems can involve a number of calculations. If you are not
consistent with your units, your answer may end up with a really weird unit. I
suppose that doesn’t make your answer wrong, but it makes it weird. Nobody
will know what you are talking about. What’s even worse, you may even think
your answer is in a certain unit when it isn’t! If you stay in the SI system, for
example, ANY answer you have for length will automatically be in meters. Any
answer you have for a temperature will automatically be in Kelvins. The same
is true for any other physical quantity you’ll be solving for. You don’t even need
to think about it.
Unit Conversion
In my time as a TA, I would say that the number one source of errors that
can easily be corrected are those having to do with units: in particular, unit
conversions. As far as complexity is concerned, unit conversion is not among
the top when considering all of the other concepts one sees in an introductory
physics course. To have your success hampered down by something so easily
corrected is probably not the situation you want to be in.
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What Is Unit Conversion?
Simple: Say that you have a quantity measured in one particular unit and that
quantity would be more useful to you in another unit, you can change from the
current unit to the one desired by the process of unit conversion.
Let us start with an example we are all familiar with: ordering food. You’re
staring at the menu with a few of your friends and you decide that, collectively,
you want three burgers, two orders of fries, and three sodas. Great, but now
you need to pay for it. The burgers, fries, and the coke each have a certain
price. To figure out the total price, we must ”convert” the food to its monetary
value. We all know how to do this. Each item has its price. If you get more
than one item, you need to multiply the price per item by the number of items
to get the total price. Say for example that each burger costs $3.00, each order
of fries $2.00, and each coke $1.50. Let’s total the bill up.
Burgers
3 burgers ·
$3.00
= $9.00
1 burger
Fries
2 orders of fries ·
$2.00
= $4.00
1 order of fries
Sodas
3 sodas ·
$1.50
= $4.50
1 soda
So, the total price = $9.00 + $4.00 + $4.50 = $17.50. What is written above here
is exactly the process described above, but written perhaps a bit different from
what you imagined. Let’s speak out one of the calculations to gain some insight
into what’s going on here. Three burgers times $3.00 per burger comes out to
$9.00. In other words, I originally start with an amount of burgers, and since I
know how much one burger costs, I can translate the number of burgers I have
to the amount of money those burgers cost. The key information is the rate:
in this case, the price of one burger. Since I know the price of one burger, I can
tell you how much any number of burgers would cost. I would just multiply that
number by the price just as I did above. Also, notice that I did NOT write the
calculations in the following way:
3 burgers ·
1 burger
=?
$3.00
In this case, the rate (again, the price per burger) is flipped the wrong way. How
can I tell? Well, what would the resultant unit be in this case? Dollars? Not
in this case; and this is absolutely critical: Just like the numbers multiply
in this calculation, we also ”multiply”, so to speak, the units as well.
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With the rate the way we have it shown, we get the right hand side to be
2
1 burgers
dollar . That makes absolutely no sense whatsoever. This, believe it or not,
is one of the most common mistakes that students make when converting units:
the rate is flipped the wrong way. Let’s compare this to the correct way to do
the calculation:
· $3.00 = $9.00
3
burgers
1
burger
Now, we can see very clearly that the burgers ”cancel out” since we have one in
the numerator and one in the denominator, and we are left with dollars in the
end. Since we are trying to figure out how much the burgers cost, this is what
we want. As you can see, the correct way to convert means that you need to
multiply by the correct rate. It did me absolutely no good to multiply 3 burgers
by 1 burger per $3.00; however, when I multiplied by $3.00 per burger, it made
sense. Keep this very crucial point in mind when you see the following examples
and when you do the practice problems.
Conversion Within One Particular Unit Of The Metric System
This is a trick of sorts that will serve you very well in your physics course. We
spoke before about systems of units. In particular, we discussed the SI system
which makes use of the metric system of units. Along with this comes the use
of the prefixes of the metric system. This is to say that when there is a prefix
before a particular unit, like the meter, that signifies that we are talking about
a different amount than the original unit itself. Below is a table of the prefixes
used in the Metric System.
Prefix
femto
pico
nano
micro
mili
centi
kilo
mega
Multiply Original Unit By Either
Associated Power of Ten Fractional Or Decimal Form
1
10−15
1,000,000,000,000,000
1
−12
10
1,000,000,000,000
1
−9
10
1,000,000,000
1
10−6
1,000,000
1
−3
10
1,000
1
−2
10
100
3
10
1,000
106
1,000,000
Table 2: Commonly Used Metric System Prefixes And Their Associated Mutiplication Factors
An example we all know: the centimeter. The ’centi’ prefix in front of the
meter denotes that we take the length that is defined to be 1 meter and divide
1
that length by 100. That new length (i.e. 100
of a meter) is called a centimeter.
8
Now, let’s look at doing the easy conversion between centimeters and meters.
How about, let’s convert 67 centimeters into meters.
67 cm = 67 · (10−2 m) = 67 · 10−2 m = 0.67 m
I simply substituted the centimeter unit with its equivalent in meters and just
multiplied by the rate. In reality, this is not any different from the burger example or the examples we’ll see further on, but it is a slightly different way of
writing down the conversion that is taking place. (i.e. how many meters are in
a centimeter in this case).
Let’s see one more example: 55 kilograms converted into grams
55 kg = 55 · (103 g) = 55 · 103 g = 55000 g
Again, from knowledge of the metric system, I replaced kilograms with its equivalent in grams and just did the multipication.
Examples
I think the best way to learn all of these concepts well is to see examples.
1. Convert 120 seconds into minutes.
Ok, the answer here is obvious: two minutes; however, going through the
formal process of conversion to do an obvious example will help to solidify
your understanding of the process itself. Plus, you can tell pretty easily if
you messed up since you know what the answer should be. Following the
burgers example:
120 seconds ·
120
1 minute
=
minutes = 2 minutes
60 seconds
60
Note that I had my rate such that the units I started with (i.e. seconds)
canceled out, and I was left with the units that I wanted at the end
(i.e. minutes). You might say, ”Ok, saying that there is 1 minute per
60 seconds is the exact same thing as saying that there are 60 seconds
in one minute, so what difference does it make which one I use?” You’re
certainly right in saying that the statements are equivalent; however, as we
saw in our burger example, using the rate incorrectly gives us something
absolutely non-sensical. While the rate and its inverse mean the same
thing, they each give us something different when using them to convert.
One product makes sense and is the correct answer, the other is nonsensical and absolutely wrong.
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2. Convert 19 inches into centimeters.
Ok, same as the first example. We all know from grade school that there
are about 2.54 centimeters in one inch, we just need to display this rate
properly in our calculation. This is to say, we have the choice between
using the following two calculations (again, one is right and the other
terribly wrong):
(a) 19 inches ·
OR
(b) 19 inches ·
1 inch
2.54 centimeters
2.54 centimeters
1 inch
Which one should we pick? Well, like every other example, we want the
units of what we started with to cancel out, and the units of what we want
to convert to to remain. With choice ’a’, nothing cancels out, and we are
inch2
left with a weird unit of centimeter
. This is obviously wrong, so choice ’b’
is what we want to use:
· 2.54 centimeters = 48.26 centimeters
19 inches
1
inch
I showed the inch units as being crossed out to drive home the point that
this was the correct way to convert from inches to centimeters.
The question should arise: would we ever use the rate that was in choice
a? Yes, when we are converting from centimeters to inches. That way, the
centimeters will cancel.
3. Convert 100 yards into meters. Note: 1 meter ≈ 1.09 yards
Again, we must position the rate such that yards cancel out and we are
left with meters. Therefore, it looks like this:
· 1 meter ≈ 91.74 meters
100 yards
yards
1.09 4. A certain carmaker has advertised that their new car has a 350 kilowatt
(abbreviated kW) engine. Since you live in the US, you have no idea if
this is powerful or not. Convert this to horsepower. Is this a powerful
engine (as cars go)?
Note: 1 horsepower (hp) = 745.7 watts (W)
Ok, first we need to use what we have already learned to convert from
kilowatts to watts. We look at this in a couple of ways:
(a) Knowledge of the metric system tells us that there are 1000 watts in
a kilowatt (just as there are 1000 meters in 1 kilometer). Therefore:
350 kW ·
1000 W
= 350000 W
1 kW
10
(b) There is an easier way to look at this. Instead of physically writing
the rate down when you are just switching between things like meters
and kilometers or watts and kilowatts (i.e you are just going to be
multiplying or dividing by a power of 10), it is easier to write it
explicitly as follows:
350 kW = 350 · (103 W) = 350 · 103 W = 350000 W
Note that I just took the ’kW’ part of the expression and I just
substituted its value in ’W’. All that was left was to multiply the
numbers. While this is of course the same thing we did in part a,
it really comes in handy in more complex situations. You can see
that each prefix in the metric system has an associated power of ten
(centi is 10−2 , micro is 10−6 , kilo is 103 , nano is 10−9 , etc.). When
you convert within a particular unit within the metric system, you
just need to know the relative power of 10.
Ok. Now that we’ve got our engine power in watts, we can convert to
horsepower using the rate given in the beginning of the problem:
·
350000 W
1 hp
= 469.4 hp
745.7 W
This looks like a powerful engine to me!
5. If a serving of food has an energy of 30 kcal (not to burst your bubble,
but when the package says Calories, it means kilocalories), what is this in
Joules?
Note: The calorie (abbreviated cal) is more or less defined as the amount
of energy it takes to raise the temperature of 1 gram of water by 1◦ C.
(used wikipedia). In the metric system, this is measured as 4.184 Joules
(abbreviated J). This is to say, 4.184 J = 1 cal.
I’ll start out by writing the energy in cal instead of kcal by using my
trick from example three. The power of 10 associated with ’kilo’ is 103 ;
therefore, 30 kcal = 30 · (103 cal) = 30 · 103 cal. Now, let’s convert from
calories to joules.
· 4.184 J = 1.25 · 105 J
30 · 103 cal
1
cal
An Interesting Fact: There is a similar concept of a ’calorie’ in the english
system of units: the BTU (british thermal unit). It is defined as the
amount of energy required to raise the temperature of 1 lb of water by
1◦ F. This comes out to about 252 calories.
6. Convert from 18 MJ (megajoules) to kWh (kilowatt hours).
Note: 1 kWh = 3.6 · 106 J.
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First, let’s convert from MJ to J. Note that the prefix mega means ”a
million”; therefore, there are one million Joules in one Megajoule:
18 MJ = 18 · (106 J) = 18 · 106 J
Now from J to kWh:
18 · 106 J
·
1 kWh
= 5 kWh
3.6 · 106 J
mi
7. Convert from 100 km
h to h (miles per hour).
Note: 1 kilometer = 0.621 miles
At first, this may seem more tricky because we have two units involved
now, but this is just as simple as the previous problems. Since the unit
in the denominator of both velocity units is hours in both cases, it just
stays along for the ride. This makes sense, because a velocity is just a
statement that an object traveling at that velocity will go that far in a
certain amount of time. For example, an object travelling at 100 km will
do just that: go 100 km in 1 hour. The question on the table is how many
miles is that? It is the same exact distance that is being travelled in that
hour, but instead of measuring that distance in km, I want it measured in
miles.
0.621 mi
mi
km
·
100
≈ 62.1 h
h
1
km
Again, I have shown the original units as being cancelled out to prove the
correctness of the solution. Notice in particular that since the hours unit
was never cancelled, it is still there at the end.
8. The speed of light in a vacuum as measured in the SI system is approximately 3 · 108 ms . How fast is this in mi
h ?
Wow, now this seems tough. In the previous example, we didn’t have
to worry about the time unit in the denominator because it was the same
in what was given and in what was converted to; however, in this example,
they differ. Do we need to do anything other than what we already know
to convert this? Absolutely not. Let’s first do it in two steps and then I’ll
show it being done all together. First, let’s convert from ms to mi
s using
the same method as the previous example.
3 · 108
m 0.621 · 10−3 mi
mi
·
= 1.863 · 105
s
1
m
s
12
How did I know this conversion rate you ask? Well, we know from before
that 0.621 mi = 1 km, right? Also, 1 km = 103 m. So...
0.621 mi = 1 km
0.621 mi = 1 · 103 m
Dividing each side by 103 :
0.621 · 10−3 mi = 1 m
Again, this type of algebraic manipulation with these units will prove very
mi
useful for you in physics. Ok, let’s do part two (from mi
s to h ).
1.863 · 105
mi 60 s 60 min
mi
·
≈ 6.71 · 108
·
s 1 1h
h
min
What I did here with the time was to show two steps at once. First, I converted from miles per second to miles per minute. Then, I converted from
miles per minute to miles per hour. We have seen many times that converting is just multiplying by the conversion rate, so successive conversions
can be done by just multiplying the rates together.
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Practice Problems
Note that if a conversion rate isn’t given anywhere in the chapter, it is very easy
to look it up on the web. Answers can also be checked either with a calculator
or on an internet converting site.
1. Write 0.000345 in scientific notation.
2. Write 0.000000000345 in scientific notation.
3. Write 34500000000 in scientific notation.
4. Write 2.34 · 10−3 in regular notation.
5. Write 2.34 · 10−7 in regular notation.
6. Write 2.34 · 102 in regular notation.
7. Write 2.34 · 106 in regular notation.
8. Convert 7.2 · 10−2 minutes into seconds.
9. Convert 210 miles to kilometers.
10. Convert 200
mi
h
to
km
h .
11. Convert 760
mi
h
to
m
s .
12. Convert 526 kW to hp.
13. Convert 125 · 103 BTU into Joules.
14. Convert 123 mm2 into m2 . Try to do this by using my trick rather than
just looking up the conversion rate.
15. Convert 522 m3 into cm3 . Again, try to do this using my trick.
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