Astronomy and astrophysics involve working with the enormously large (the size of the entire universe) and the very small (the size of a nucleus of an atom). Clearly writing out such numbers is impractical, error prone, and very hard to read. For example, the size of the nucleus of an atom is roughly \(0.000\,000\,000\,000\,001\) m while the mass of our Sun is about \(\mbox{2,000,000,000,000,000,000,000,000,000,000}\) kg! Not only do you not want to count all of those zeros, but misreading or miswriting them is very easy to do. It is much more straightforward to simply express the number of zeros explicitly for the reader. It also turns out that doing calculations with numbers is much easier if the zeros are expressed explicitly. This is the motivation behind writing large and small numbers in scientific notion.
Sections
- For Numbers Greater Than One
- For Numbers Less Than One
- CAUTION: Notation with Scientific Calculators and Computers
- Summary Table
- Significant Figures (optional)
- Quiz Yourself
For Numbers Greater Than One
First, recall a simple rule about exponents (see the exponents tutorial for more information), integer exponents are simply a shorthand for indicating how many times a number (or variable) is to be multiplied by itself. This means that if you encounter the value \(2^2\) (”two squared”), that is really saying multiply two by itself, or
$$2^2 = 2 \times 2 = 4.$$
Similarly, \(2^3\) (”two cubed”) is shorthand for
$$2^3 = 2 \times 2 \times 2 = 8,$$
and so on. This simple rule is the entire basis for scientific notation, only instead of using \(2\) as the base of the exponent, we use \(10\) instead.
Since we usually do our mathematics in base-\(10\) (unless you are a computer or a computer programmer, in which case you would use base-\(2\), or base-\(8\), or even base-\(16\)!), using \(10\) as the base for exponents greatly simplifies the writing of very large or small numbers. For example:
$$10^{\color{red}3} = 10 \times 10 \times 10 = 1\color{red}0\color{red}0\color{red}0;$$
the exponent \(\color{red}3\) ends up representing the number of zeros in the number \(1000\). Similarly, \(10^9\) is
$$10^{\color{red}9} = 10 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10 = 1,\!\color{red}0\color{red}0\color{red}0,\!\color{red}0\color{red}0\color{red}0,\!\color{red}0\color{red}0\color{red}0.$$
which is one billion, a \(1\) with \(\color{red}9\) zeros after it.
Taking the process one more step, what do you need to express a large number that is more complicated than just \(1\) with a bunch of zeros after it? Suppose you were amazingly fortunate and found a backpack with \(\mbox{30,000}\) dollars in it! Before you do the right thing (of course) and turn it all in to the authorities, you inexplicably want to write the amount in scientific notation. First of all, you know that \(\mbox{30,000}\) is just \(\mbox{10,000} + \mbox{10,000} + \mbox{10,000}\) (\(\mbox{10,000}\) added together \(3\) times), or \(3 \times \mbox{10,000}\). But you know that since \(\mbox{10,000}\) is a \(1\) with \(4\) trailing zeros, that is the same thing as \(10^4\). As a result
$$\$\color{red}3\color{blue}0,\!{\color{blue}000} = \$\mbox{10,000} + \$\mbox{10,000} + \$\mbox{10,000} = 3\times \$\mbox{10,000} = {\color{red}3 \times 10}^{\color{blue}4}\ \mbox{dollars.}$$
All you needed to do is look at the leading number (\(3\) in this case) and the number of trailing zeros (\(4\)) and write the number as \({3 \times 10}^4\) dollars.
The procedure is only slightly more difficult if not all of the numbers after the first one are zero. In this case you need to imagine putting a decimal point after the first number, and write the remaining numbers after the decimal point. Then it is a matter of figuring out how many times you need to multiply the number by \(10\) to get the correct result.
It turns out that the number of seconds in one year is approximately \(\mbox{31,560,000}\) s. Following the procedure described in the last paragraph,
$$\mbox{31,560,000}\ \mbox{s} = 3.156\,0000 \color{red}{\times 10 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10}\ \mbox{s} = 3.156 \times 10^{\color{red}7} \mbox{s}.$$
Each time \(3.156\) is multiplied by \(10\), the decimal point moves to the right one place: \(3.156 \times 10 = 31.56\), \(31.56 \times 10 = 315.6\), \(315.6 \times 10 = 3156\), \(3156 \times 10 = \mbox{31,560}\), and so on until you reach \(\mbox{31,560,000}\). Adding up the number of times you multiplied by \(10\) you will find that it is \(7\).
The quick way to figure out how many powers of \(10\) you need once you have written the number with the decimal point after the first digit is to count how many times you need to move the decimal point from its original position until it is located after the first digit. For \(\mbox{31,560,000}\), the decimal point is actually located after the number (we just didn’t bother to write it down), so in reality we want \(\mbox{31,560,000}\color{red}.\) as the number to start with. If you move the decimal point \(7\) places to the left you will arrive at \(3.156\,0000 \times {10}^7\). Since eliminating all of the trailing zeros at this point doesn’t change the answer, it is customary to write the result as \(3.156 \times {10}^7\).
As another example to test your understanding: How do you write \(\mbox{85,500,000,000,000}\) in scientific notation? The answer is \(8.55 \times {10}^{13}\). You need to move the decimal point \(13\) places to the left in order to end up after the first number.
For Numbers Less Than One
For very small numbers (numbers whose absolute value is less than one), the process is very similar, except that decimal points need to move to the right until they are after the first non-zero number. To understand why, it is important to realize two additional points about exponents:
- When any number (except \(0\) is raised to the zero power the answer is always \(1\). For base-\(10\) this means \({10}^0 = 1\). (Technically \(0^0\) is undefined.)
- When a number is raised to a negative power, that is the same thing as dividing one by the number raised to the positive power.
As an example of the second point,
$${10}^{-2} = \frac{1}{{10}^2} = \frac{1}{100} = 0.01.$$
In this case the decimal point would need to be moved to the right \(2\) places in order for the decimal point to follow the first non-zero number (hence the minus sign in the exponent).
To illustrate how very small numbers are written, write the number \(0.000\,000\,004\,56\) in scientific notation. In order to have the decimal point after the first non-zero number (in this case between the \(4\) and the \(5\)), the decimal point needs to be moved \(9\) places to the right, meaning that the number is represented as \(4.56 \times {10}^{-9}\).
CAUTION: Notation With Scientific Calculators and Computers
Calculators and computers can automatically display large and small numbers in scientific notation, but for most calculators the display is not in the formal scientific notation form. For example, the number displayed may look like
$$3.1415\mbox{E}9\ \ \ \mbox{or perhaps}\ \ \ 3.1415\mbox{EE}9\ \ \ \mbox{or perhaps}\ \ \ 3.1415\mbox{E}009.$$
Numbers much smaller than one might be displayed as
$$3.1415\mbox{E}-9\ \ \ \mbox{or perhaps}\ \ \ 3.1415\mbox{EE}-9\ \ \ \mbox{or perhaps}\ \ \ 3.1415\mbox{E}-009.$$
Even more confusing, some calculators represent very large and very small numbers without the E, or simply as
$$3.1415\ \ 09\ \ \ \mbox{or even}\ \ \ 3.1415{\Large{\hat{ }}}{09}.$$
You should never write these forms yourself, nor are you likely to see them in textbooks; they are reserved entirely for calculators and computers. Even though it is not displayed traditionally, the same information is present. In the large-number example, you must translate the value to be
$$3.1415 \times {10}^9.$$
Be careful, the meaning of your display IS NOT \(3.14159^9\). Those are very different numbers! Compare the values of the two represenations:
$$3.1415 \times {10}^9 = \mbox{3,141,500,000}$$
while
| \(3.1415^9\) | \(=\) | \(3.1415 \times 3.1415 \times 3.1415 \times 3.1415 \times 3.1415\) |
| \(\times\ 3.1415 \times 3.1415 \times 3.1415 \times 3.1415\) | ||
| \(=\) | \(\mbox{29,801}.187\,948\,655\,318\,249\,871\,346\,761\,148\,833\,984\,375\) |
Many students fall into this trap, not realizing that scientific notation really is just an application of exponents involving the base of \(10\), no more and no less. Getting marked wrong for using incorrect scientific notation is not just your instructor being picky, you really are representing a number that does not equal the correct answer.
If you are not sure how to read the display on your calculator, be sure to ask your instructor or someone else familiar with using scientific calculators!
Summary Table
To summarize what we have been saying above:
| Scientific Notation | Standard Notation | |
|---|---|---|
| \({10}^{6\ \ }\) | \(=\) | \(\mbox{1,000,000}\) |
| \({10}^{5\ \ }\) | \(=\) | \(\mbox{100,000}\) |
| \({10}^{4\ \ }\) | \(=\) | \(\mbox{10,000}\) |
| \({10}^{3\ \ }\) | \(=\) | \(1000\) |
| \({10}^{2\ \ }\) | \(=\) | \(100\) |
| \({10}^{1\ \ }\) | \(=\) | \(10\) |
| \({10}^{0\ \ }\) | \(=\) | \(1\) |
| \({10}^{-1}\) | \(=\) | \(0.1\) |
| \({10}^{-2}\) | \(=\) | \(0.01\) |
| \({10}^{-3}\) | \(=\) | \(0.001\) |
| \({10}^{-4}\) | \(=\) | \(0.0001\) |
| \({10}^{-5}\) | \(=\) | \(0.000\,01\) |
| \({10}^{-6}\) | \(=\) | \(0.000\,001\) |
Scientific notation always starts with shifting the decimal point immediately to the right of the first non-zero number. Therefore \(\mbox{314,159} = 3.14159 \times {10}^5\) (the decimal point was shifted \(5\) places to the left) and \(0.000\,031\,4159 = 3.141\,59 \times {10}^{-5}\) (the decimal point was shifted \(5\) places to the right).
Significant Figures
[Note: Your author will not emphasize the material in this final section although your instructor may rightfully choose to do so.]
Technically there are reasons to include additional zeros to the right of the decimal point following the last non-zero number. The total number of digits, both zeros and non-zero digits, are used to inform the reader of how precise a measurement is. For example, it may be the case that a measurement is known to \(7\) significant figures, even though the last \(3\) might actually be the digit \(0\), such as \(3.142\,000\).
It should be noted that there can be some ambiguity when all of the zeros are to the left of the decimal point, such as for a measurement that gives \(\mbox{3,142,000}\). How many significant figures are there? In other words, how precise was the measurement? Was it accurate to \(4\) digits? \(5\)? \(6\)? \(7\)? From the way the number is written there is no way to know! But if you write the number in scientific notation, the answer becomes clear. If written as \(3.142\,00 \times {10}^6\) then it is obvious that there are \(6\) significant figures in the measurement.
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.
- At the top of this tutorial the mass of the Sun was written in kilograms. How is this number expressed in scientific notation?
- The radius of a hydrogen atom is \(0.000\,000\,000\,053\) m. Express the atom’s radius in scientific notation.
- Earth is \(\mbox{150,000,000,000}\) m from our Sun. What is this distance, written in scientific notation?
- The mass of a neutron is \(0.000\,000\,000\,000\,000\,000\,000\,000\,001\,67\) kg. Write this number in scientific notation. (Have fun counting all of the zeros!)
- The radium of Earth is \(6.38 \times {10}^7\) m. In this problem please write this number in standard mathematical notation rather than in scientific notation.
Answers
- \(2\times 10^{30}\) kg
- \(5.3\times 10^{-11}\) m
- \(1.5\times 10^{11}\) m
- \(1.67\times 10^{-27}\) kg
- \(\mbox{63,800,000}\) m