OK, you can pass my courses without knowing any math beyond arithmetic. But I hope you will want to learn a little more math so you can approach quantitative astronomy problems with confidence and success.

Here are some links to help you with the math in your astronomy course.  Red stars indicate issues with which students often have the most difficulty.

*** Scientific Notation    how to deal with 93,000,000 and what 10¹² means ***

More Scientific Notation Practice    using it in astronomy

*** Canceling Units   useful help from Purplemath ***

Metric Units   centimeter, meter, kilometer, etc help from Math League

Exponents1   for help evaluating things like (x⁵)(x³)

Exponents2   for help evaluating things like 5³

Exponents3   for help evaluating things like (x³)⁵

Exponents4  for help evaluating things like x^-4

Positive and negative numbers  x+-y and the like from Math League

***  Multiplication Property   make your equation balance by multiplying on both sides from Dr. Brennan***

Graphing1   practice graphing from James Brennan

Graphing2    practice graphing from Purplemath

Slopes of lines1    ASTR 14 students may use this in research experiments (from James Brennan)

Slopes of lines2    more slopes from Purplemath

Equation Solver    it's automatic!  Use it to check your algebra.

Here are some rules of math you might use in astronomy. If you understand these rules already, you know enough math to do any problem I give you. If you don't understand these mathematical rules, you have the choice of learning them (I'll use and explain them several times in class over the semester) or not bothering (if you decide not to learn the math, you can still do well in class as long as you score well on non-quantitative problems). These rules are used and explained in the two examples I give below them.

EXAMPLE 1

As an example of the math you might need to understand in astronomy class, consider the equation relating the wavelength and frequency of electromagnetic radiation,

where is wavelength, c is the speed of light, and f is frequency.

A. First, examine the equation. It says that wavelength is equal to the speed of light divided by the frequency. Because the wavelength is at the top of one side of the equation and the frequency is at the bottom of the other side, wavelength and frequency are said to be inversely proportional to one another. This means that when the wavelength is big, the frequency is small and vice versa. More on this later!

B. Now, can you manipulate this equation so that the symbol for frequency is by itself and at the top rather than the bottom? You will want to use the multiplication property above.

Here’s what to do:

1. Get the f from the bottom to the top by multiplying both sides by f



2. On the right hand side, you have an f on the top and an f on the bottom so they cancel to give:



3. Now get the away from the f by dividing both sides by



4. Now we have a on the top and a

    and you’ve done it.

For practice, try getting the a alone in this equation for force:

C. One possibility is that you will be asked to calculate wavelength, given a certain frequency. Frequency comes in units of Hertz, abbreviated Hz, and equivalent to 1/seconds or cycles/second.

Suppose you are given a problem in which the frequency is

and you know the speed of light is always

.

Then

D. Another possibility is that you are asked to compare things, in this case, two types of light. Here is an example.

Red light has a wavelength of 700nm and violet light has a wavelength of 300nm. Which has the highest frequency?

1. Write down the information you are given using sensible symbols.

GIVEN:

Red light wavelength

Violet light wavelength

Notice that the subscript r above refers to the red light and the subscript v to the violet light.

2. Write down what you are supposed to find.

FIND:

Comparing frequencies – or any other property – usually means finding the ratio for the two things you are comparing. In this case, you want to compare the frequencies of red and violet light. That means you need to determine either , that is, the ratio of the frequency of red light to the frequency of violet light or the ratio of the frequency of violet light to the frequency of red light.

3. Next find an equation that will help you solve your problem or draw a picture if that might help you understand the problem. In this case, you might want look at or copy the figure of the electromagnetic spectrum from the textbook to see if it helps you understand the problem. But you’ve probably already figured out which equation to use.

EQUATION:



4. There are two main ways to go about solving this problem. I’ll show you the long way first, then show you a shortcut.

SOLVE:

For the red light, ; and for violet light, .

The ratio will be



so you have just shown that



The only two algebraic properties you had to remember are:

*** 1. If you have the same symbol or quantity on the top of the dividing line and on the bottom of the dividing line, you get to cancel that symbol out! ***

That’s how I got rid of the speed of light, c. There was a c on the "top of the top" part of the equation and another c on the "top of the bottom" part of the equation. I got to cancel them out.

*** ***

That’s how I flipped everything in the 3^rd step of manipulating the ratio. The a/b part is the and the c/d part is the .

5. Here’s the shortcut.

Well, we know that the wavelength of the red light is 2.3 times bigger than the wavelength of the violet light because  

Because the red light wavelength is 2.3 times bigger than the violet light wavelength, the equation tells us the red light frequency is 2.3 times smaller than the violet light frequency.

And you’ve just told me that the frequency of red light is 2.3 times smaller than the frequency of violet light or, just as did the long way, the short way gives

Let's try another equation for practice.

Newton's Law of Universal Gravitation tells us that the force of gravity depends in direct proportionality upon the masses of the objects being considered and in inverse proportionality to the square of the distance between their centers

The direct proportionality between force and mass is due to mass being on the top on the right hand side of the equation and force being on the top of the left hand side. The inverse relationship between force and distance is due to distance being on the bottom of the right hand side of the equation and force being on the top of the left hand side.

If one or both of the masses in this equation increase, the force increases:

If one or both of the masses in this equation decreases, the force decreases:

If the distance increases, the force decreases:

If the distance decreases, the force increases:

Here is a sample problem:

The planet Jupiter is about 300 times more massive than Earth; Jupiter’s radius is about 10 times bigger than Earth’s. Will you weigh more or less on Jupiter than you do on Earth?

1) Write down the information you are given about the problem. Use sensible mathematic symbols, like M for mass for instance. Label the symbols with subscripts, say using J for Jupiter and E for Earth, so you can tell them apart.

2) See if you have enough information to determine what equation you will use to calculate your answer. Think about the quantity you are asked to find and decide whether the information you have been given is sufficient to allow a calculation of that quantity. Here you are asked to find weight. Well, maybe you would like to use the symbol W for weight but I hope you’ll remember that the gravitational force between the mass of your body and the planet you are standing on is your weight. So you will use Newton’s Law of Universal Gravitation to calculate your weight.

EQUATION:

Have you been given enough information to make this calculation? You know the ratio of the mass of Jupiter and the mass of Earth, and the ratio of the radius of Jupiter and the radius of Earth.

You need to know some masses and some distances or sizes to calculate the force of gravity in the equation. Now decide if you have the necessary ones.

3) Try drawing a picture of what you are trying to calculate. You want to compare your weight when you are standing on Earth and when you are standing on Jupiter.

The line between you and the center of the planet is the radius of the planet. When you calculate your weight on a planet, the radius is the distance you need in the gravitational force equation. So, on Earth, in order to calculate your weight, you need information about your mass, Earth’s mass, and the radius of Earth. On Jupiter, in order to calculate your weight, you need information about your mass, Jupiter’s mass, and the radius of Jupiter.

You have the ratios of the masses and radii so let’s make it easy on ourselves and use a ratio to solve the problem.

4) You could look up the gravitational constant, G, and the masses and radii of Earth and Jupiter in a book, and you could determine your mass then put all that information in the gravitational force equation. Let’s not. It is much easier to use a ratio to solve this problem.

Here are the equations you could use to calculate your weight on Earth and your weight on Jupiter separately. Each equation includes the mass of the planet, your mass, and the radius of the planet.

Now we can see the power of ratios.

5) The ratio of the force of gravity on Jupiter to the force of gravity on Earth gives you the relationship between your weight on Jupiter and your weight on Earth. This is what the problem asked you to find. You will be able to determine if you weigh more on Earth or on Jupiter. Try to follow the math below step by step.

1. Here’s the shortcut:

Your weight on Earth can be calculated using the equation below. Now let’s see what happens if you move to Jupiter.

Your mass stays the same no matter where you go (as long as you don’t lose a limb or diet extensively on the way to Jupiter). But Jupiter’s mass is 300 times larger than Earth’s mass and Jupiter’s radius is 10 times as big as Earth’s radius:

Remember these rules?

That the mass gets bigger, means that the force on the other side of the equation gets bigger. But that the radius gets bigger means that the force on the other side of the equation gets smaller.

So if you move to Jupiter your weight increases by a factor of 300 due to the mass but decreases by a factor of 100 due to the radius. Because 300/100=3, overall your weight increases by a factor of 3. Hey, that’s exactly what you calculated using the long way above!