NAME___________________________
Review of Some Mathematical Concepts for Space Science
College of Staten Island - CUNY
Objective: Space Science (including Astronomy and Astrophysics) is a measurement science. In order to draw
conclusions about measurements we usually organize them in tables or graphs or formulas. These give rise to
the discovery of basic natural principles or laws with enormous predictive power (which gives science an edge
over other human endeavors in interpreting existence).
In the laboratory part of the course you will carry out from time to time various measurements and will be
asked to draw conclusions. In the course you will be asked to solve problems for homework of an analytic
nature. To help you do these activities you will need to understand the following concepts and/or techniques of
analytical thinking.
Your work will be judged on quality which usually mean time commitment. Your mathematical skills are
only improved with practice. . this exercise is a review of mathematical concepts you will use, master
these and you will not find this course difficult. Whatever is not finished in the laboratory session should
be finished at home and this entire exercise handed in next laboratory period.
I. Metric measurement.
Study the Metric Ruler as you do this exercise.
The measure of length is now universally based on the meter.
Examine a meter stick and note that 1 meter = 100 centimeters and further
note that 1 cm. = 10 millimeter.
Measurements are expressed in decimal fashion,
for example: 1.57 meters, 12.55 cm. 345.5 mm 3 kilometers
TASK 1. How many millimeters in one meter. ______________________
TASK 2. 25.5 cm = ? mm
______________________
TASK 3. 1.579 meters = ? cm = ? mm
______________________
The metric system was developed around 1775 by French scientists. It is convenient to use because its units are
related by powers of ten. A standardized system now exists worldwide. It is referred to as the International
System (SI) of weights and measures.
Traditionally in the study of physics, hence astrophysics, two systems of measure have been used. They are
referred to as the MKS and the CGS systems.
MKS system meter - kilogram - second
energy is joule temperature is Kelvin
CGS system centimeter - gram - second energy is erg
temperature is Kelvin
Many textbooks emphasize the MKS system but ours uses mostly the CGS system
The CGS Metric System: some basic fundamental units of measure are displayed.
Length: base unit = centimeter (cm)
Mass: base unit = gram (g)
Time: base unit = second (s)
Energy: base unit=erg
Temperature: base unit=Kelvin (K)
Derived units from the fundamental
Speed (v)= cm/sec
Acceleration (a) = cm/sec2
Volume(V): = liter (derived unit = 1000 ml or =1000 cm^3(cubic centimeter sometimes called a “cc”)
Force: =dynes from F=ma
1 dyne= 1 g x 1cm/sec2
LOOK OVER APPENDIX C UNITS AND CONVERSION TO SEE OTHER UNITS WE
USE IN OUR COURSE…
TASK 3b NAME SEVERAL ENERGY UNITS NOT MENTIONED ABOVE?
LOOK OVER APPENDIX B Physical and astronomical constants to see how some of the
units are used
TASK 3c write down here three constants, all with different units…
Note sometimes I used ^ for exponent!
Units of measure are divided into two categories called the Fundamental unit and the Derived Unit.
Fundamental Unit: These are units of measure obtained directly from measurement observations using SI units
and cannot be obtained indirectly from any simpler or more basic units of measure. Mass, length, and time are
examples of fundamental units.
Derived Unit: These are units of measure which may be able to be observed through direct measurement (i.e.
volume can be observed directly by visibly observing the reading off of the side of a graduate cylinder.), but
also can be obtained by derivation using simpler fundamental units of measure. As an example consider that
volume can be obtained from length by the formula V = L x W x H.
Example: Length of course is a length measurement (L), but so are width (W) and height (H) length
measurements simply oriented in different directions. And it follows then that volume, which is a multiplication
product of the three, is derived from length.
Metric Prefixes - Units of Measure base units may be inappropriate for a given measurement or
calculations so we use multipliers of the base units with prefix names, as follows
Prefix Symbol Fractional Equivalent Example (using the meter)
Kilo K 1000 x base unit kilometer (Km)
hecto h 100 x base unit hectometer (hm)
deka da 10 x base unit dekameter (dam)
___ ___ 1 x base unit meter (m)
deci d 1/10 x base unit decimeter (dm)
centi c 1/100 x base unit centimeter (cm)
milli m 1/1000 x base unit millimeter (mm)
------------Additional Units:
Prefix (Symbol) Fractional Equivalent Example (using the meter)
giga (G) 1 x 10^9 x base unit gigameter (Gm)
mega (M) 1 x 10^6 x base unit megameter (Mm)
micro () 1/1 x 10^-6 x base unit micrometer (m ) greek letter is called mu
nano (n) 1/1 x 10^-9 x base unit nanometer (nm) ..we use this a lot!
pico (p) 1/1 x 10^-12 x base unit picometer (pm)
------------Common Metric - English Equivalents:
1 m = 39.37 in 1 cm = 0.394 in 1 in = 2.54 cm
1 kg = 2.21 lb 1 lb = 454 g 1 L = 1.057 qt
1 oz = 28.4 g 1 in^3 = 16.5 cm^3 1 gal = 3.78 L
Metric Conversion Conversions between metric units:you started this exercise by doing just that!
The metric system is very easily manipulated as can be seen by the following two examples. Converting
between two units of the same category of measurement can be done simply by moving the position of the
decimal place at the same time the prefix is changed. As the unit of measure is increased the decimal is moved
to the left making the numerical value smaller proportionally. As the unit of measure is decreased the decimal is
moved to the right making the numerical value larger proportionally. Look at the following two examples of
this concept.
Example 1: Consider the measurement 2.5 m and the variations of the same length measurement expressed in
other metric units of length.
km
hm
dam
m
dm
cm
mm
0.0025
0.025
0.25
2.5
25.0
250.
2,500.
Can you see all these ….study this!
Example 2: Consider the measurement 4,200 m and the variations of the same length measurement expressed
in other metric units of length.
km
hm
dam
4.2
42.
420.
m
4,200.
dm
cm
42,000.
420,000.
mm
4,200,000.
TASK 4 Convert each of the example values to nm
1_______________________________________________ 2 ________________________________________
Working with multiple dimensional conversions.
Student who study doing conversions between units within a measurement system such as Metric to Metric or
between measuring systems such as the Metric to English and English to Metric conversions usually can handle
conversion factors that are one dimensional. However, a problem often arises when conversion factors are
needed for two and three dimensional problems. The following information compares handling units in one,
two, and three dimensions where conversions are involved.
------------1. One Dimensional conversion.
Given a length of 100 inches, convert this to centimeters.
The one dimensionsl (linear) conversion factor is 1 in = 2.54 cm.
The calculation involves 100 in x 2.54 cm / in = 254 cm.
- - - - - - - - - - - - - You did some of this at the start!
2. Two Dimensional conversion.
Given a surface area of 500 square inches (500 in^2), convert this to square centimeters.
The two dimensional (Area) conversion factor is (1 in)^2 = (2.54 cm)^2 or 1 in^2 = 6.4516 cm^2
The calculation involves 500 in^2 x 6.4516 cm^2 / in^2 = 3,225.8 cm^2
------------3. Three Dimensional conversion.
Given a volume of 1200 cubic inches (1200 in^3), convert this to cubic centimeters.
The three dimesnional (Volume) conversion factor is (1 in)^3 = (2.54 cm)^3 or 1 in^3 = 16.387064 cm^3
The calculation involves 1200 in^3 x 16.387064 cm^2 / in^3 = 19,664.4768 cm^3
Dimensional Analysis, also called Factor Labeling.
Dimensional analysis focuses on the units of measurement, which is an area of mathematics that many students
overlook. Dimensional analysis focuses on the use of measurements in calculations. Below are some examples
of using units in calculations. By doing this when working out a formula one can avoid calculations errors since
the final dimension must be what one is looking for..
------------1. Calculating the surface area of a shelf board.
Given a 24 inch by 48 inch shelf board calculate the amount of surface area available to store objects on.
Equation to be used: Area = length x width
Substitution:
Area = 48 inches x 24 inches
Answer:
Area = 1,152 square inches or 1,152 in^2
------------2. Converting the surface area determined in the above problem into square feet.
The shelf board above has a surface area of 1,152 in^2
There are two ways to go about this. One is to convert the original dimensions to feet and then substitute into
the equation and solve. The other way is to solve for the answer by doing a conversion calculation.
The 1st way: The length is 48 in x 1 ft / 12 in = 4 ft
The width is 24 in x 1 ft / 12 in = 2 ft
Calculation of Area:
A = L x w = 4 ft x 2 ft = 8 square feet or 8 ft^2
The 2nd way: The area is 1,152 in^2
The conversion factor between the inch and the foot is 12 in = 1 ft
The conversion factor between the square inch and the square foot involves squaring both sides of the above
conversion factor. This gives us:
Conversion factor in^2 --> to ft^2 is (12 in)^2 = (1 ft)^2
Using this conversion factor: 1,152 in^2 x (1 ft^2 / 144 in^2)
Answer:
8 ft^2
------------3. The same approach is used in volume problems.
Calculate the volume of a box having the following dimensions.
48 inches long, 24 inches wide, 36 inches deep (or high)
Equation to be used:
Volume = length x width x depth (or height)
Substitution:
Volume = 48 in x 24 in x 36 in
Answer:
Volume = 41,472 in^3
4. Determining the volume in cubic feet would involve the following process.
Again we could approach this in two different ways.
The 1st way:
The length is 48 in x 1 ft / 12 in = 4 ft
The width is 24 in x 1 ft / 12 in = 2 ft
The depth is 36 in x 1 ft / 12 in = 3 ft (also height)
Calculation of Volume:
The 2nd way:
V = L x W x H = 4 ft x 2 ft x 3 ft = 24 cubic feet or 24 ft^3
The area is 41,472 in^3
The conversion factor between the inch and the foot is 12 in = 1 ft
The conversion factor between the cubic inch and the cubic foot involves cubing both sides of the above
conversion factor. This gives us:
Conversion factor in^3 --> to ft^3 is (12 in)^3 = (1 ft)^3
Using this conversion factor: 41,472 in^2 x (1 ft^2 / 1,728 in^2)
Answer:
24 ft^3
------------5. Determining the Density of a substance.
The mass of a rectangular solid piece of metal is 100 g
Its dimensions are: Length is 5 cm, Width is 4 cm, and thickness (height) is 2 cm.
Equation used:
Volume is:
Substitution:
Density = Mass / Volume
V = L x W x H = 5 cm x 4 cm x 2 cm = 40 cm^3
D = M / V = 100 g / 40 cm^3 = 2.5 g/cm^3
Note: Density is a derived unit with a complex unit of measure.
Angles and Protractor.
An angle is a measure of the space between the intersection of two lines. It has come down to us from
ancient times that a circle can be divided into 360 parts each of which subtends an angle of 1 degree at the
center. Recall the class discussion the 360 was the number of days for the Sun to go completely around the
Zodiac or the length of a Year as measured in ancient times.. now we know it is ~ 365.25 days
The protractor is used to measure the angle. It usually is half a circle and, hence, contains 180 degrees
marked in various convenient fashions. You must center the intersection point of the lines you are measuring or
constructing at the center point (instructor will demonstrate) of the protractor and also match one of the lines to
the base line of the protractor.
IF A LINE IS NOT LONG ENOUGH TO REACH THE ANGLE SCALE ON THE PROTRACTOR IT IS
CUSTOMARY TO CAREFULLY ENLARGE THE LINE WITH THE STRAIGHT EDGE OF THE
PROTRACTOR.
TASK 5. With one arm outstretched and pointing to the horizon and the other to the point over your head, (Do
not be bashful- of course you look silly)
approximately what angle are you constructing on the sky = _______________.
TASK 6. The Zodiac Constellations lie along the path of the sun which may be thought of as a great circle all
around us. If there are twelve constellations all evenly spaced, then what angle describes their "size" on the sky
=_______.
TASK 7. Define another unit of angle known as the “Radian” what is the relationship to degrees!
Use the space below.
TASK 8 Draw a these angles carefully with your protractor 38, 112, 195 on a sheet of paper or on the
back of this exercise.
Right angle triangle Trigonometry: these functions are extremely useful in all fields of science
Tangent
The three Trigonometry functions of Sine, Cosine, and Tangent are very important. All three functions are
defined as ratios of lengths of sides in a right triangle (a triangle with one angle being 90 degrees such that two
of the sides of the triangle meet as lines perpendicular to each other). In such a situation the some of the other
two of the three angles present in the right triangle add up to 90 degrees and are called complementary angles.
------------The Sine function is described as the ratio of the length of the side opposite a defined angle (other than the 90
degree angle) to the length of the hypotenuse.
In equation form it is written as:
Sine of the angle = Opposite side / Hypotenuse
OR
Sin  = OPP / HYP
For example: In a triangle having one of the angles equaling 30 degrees the ratio of the length of the side
opposite to the angle to the length of the hypotenuse is 0.5000
------------The Cosine function is described as the ratio of the length of the side adjacent to a defined angle (other than the
90 degree angle) to the length of the hypotenuse.
In equation form it is written as:
Cosine of the angle = Adjacent side / Hypotenuse
OR
Cos = ADJ / HYP
For example: In a triangle having one of the angles equaling 30 degrees the ratio of the length of the side
adjacent to the angle to the length of the hypotenuse is 0.8660
The Tangent function is described as the ratio of the length of the opposite side to a defined angle (other than
the 90 degree angle) to the length of the side adjacent the defined angle.
In equation form it is written as:
Tangent of the angle = Opposite side / Adjacent side
OR
Tan  = OPP / ADJ
For example: In a triangle having one of the angles equaling 30 degrees the the ratio of the length of the side
opposite to the angle to the length of the side adjacent the angle is 0.5774



TASK 9 cos (46)= ?
Tan (55)= ?
if the tan ? If the sin 
CALCULATOR NOTE: Turn a computer on and choose the windows option. Under the Astronomy Group is a
Scientific Calculator. Use this calculator by clicking the mouse in doing your calculations. Try a simple
calculation like 2 x 4 and 8 / 2 to be sure you understand the calculators use. Who will you ask if you do not?
II Scientific Notation (Exponential Notation)
In scientific and engineering work very large and very small numbers are encountered. In order to make
calculations easier the use of the fact that our number system is based on powers of 10 is invoked.
For example: a number written as 5,280 is short for adding 5 1000's, 2 100's and 8 10's. We can write large
and small numbers using the fact that powers of ten can be written as exponents and some of them have names
in the metric system. Study the following table.
10
100
1,000
10,000
100,000
1,000,000
= 10
= 101 = 10
= 10 x 10
= 102
= 10 x 10 x 10
= 103 = kilo
=10 x 10 x 10 x 10
= 104
= 10 x 10 x 10 x 10 x 10
= 105
= 10 x 10 x 10 x 10 x 10 x 10 = 106 = mega
The exponents used here are just the number of zero's in the original number, or, how many places to the
right one moves the understood decimal point from the one.
Small numbers use negative exponents to represent the reciprocal of the positive values, as in the following
table.
a tenth
= 0.1
= 1/10
= 1/101
= 10-1
a hundredth = 0.01
= 1/100
= 1/102
= 10-2 = centi
3
a thousandth = 0.001
= 1/1000
= 1/10
= 10-3 = milli
a millionth
= 0.000001 = 1/1000000 = 1/106 = 10-6 = micro
10-9 = nano etc
The exponent used in the negative powers are the number of places to the left of the one to move the decimal
place to get the original number.
With these exponents we can represent all numbers and especially cumbersome ones. Study the following
illustrated values for Scientific numbers. We note that we round all numbers to an accuracy of two digits to
the right of the decimal point and we keep only a single digit to the left of the decimal point. The numbers are
in the format of d.dd x 10power with d being a digit.
5,000 = 5.00 x 103
8,000,000 = 8.00 x 106
1,231,780,000,000,000 = 1.23 x 1015
.05 = 5.00 x 10-2
.00007 = 7.00 x 10-5
.0000000000000083752 = 8.38 x 10-19
This form of notation is a way of writing numbers using exponents to show how large or small a number is in
terms of powers of ten.
Powers of ten are a means of writing what we might think of as everyday numbers as powers of ten such as 234
= 2.34 x 10^2, which can be read as two point three four times ten raised to the second power or ten squared.
The powers of ten can also be written using the "carrot" symbol to indicate the exponent of the power of ten, as
in one time ten to the third power is written as 1 x 10^3. or if I take the time as above as an exponent 103
Example 1: Simple powers of ten in this notation
1 x 10^9 = 1,000,000,000
1 x 10^6 = 1,000,000
1 x 10^3 = 1,000
1 x 10^1 = 10
1 x 10^0 = 1
1 x 10^-1 = 1 / 10
1 x 10^-3 = 1 / 1000
1 x 10^-6 = 1 / 1,000,000
1 x 10^-9 = 1 / 1,000,000,000
Example 2: Converting numbers into scientific notation..a more detailed explanation.
The numbers must be written according to a simple rule. The none power of ten portion of the number may only
have one digit before the decimal. i.e. 22.65 x 10^3 would be incorrect while 2.265 x 10^4 is correct
314 = 3.14 x 10^2
7,183 = 7.183 x 10^3
230,100 = 2.301 x 10^5
0.0204 = 2.04 x 10^-2
0.0004607 = 4.067 x 10^-4
0.00000198 = 1.98 x 10^-6
Example 3: Converting from scientific notation
6.9702 x 10^5 = 69,702
4.599 x 10^3 = 4,599
2.0023 x 10^1 = 20.023
4.205 x 10^-2 = 0.04205
7.45 x 10^-4 = 0.000745
3.97 x 10^-5 = 0.0000397
1-3 Adding and Subtracting in Scientific Notation
Though calculators can do this for us, it is always important to see the math process behind the operation.
The fundamental rule when adding and subtracting numbers in scientific notation (exponential notation) is to be
sure that the powers of ten for each number being added (or subtracted) is the same.
----------
Adding numbers in scientific notation
Add the two numbers 4.55 x 10^2 plus 3.77 x 10^3
1st change one of the values so that it has the same power of ten
i.e. 0.455 x 10^3 plus 3.77 x 10^3
2nd add the none exponent values 0.455 + 3.77 = 4.225
3rd write the sum followed by the power of ten
Answer: 4.225 x 10^3
Example 2: Subtracting numbers in scientific notation
Subtract the two numbers 7.65 x 10^4 minus 4.32 x 10^3
1st change one of the values so that it has the same power of ten
i.e. 7.65 x 10^4 minus 0.432 x 10^4
2nd subtract the none exponent values 7.65 - 0.432 = 7.218
3rd write the difference followed by the power of ten
Answer: 7.218 x 10^4
---------Multiplying and Dividing in Scientific Notation
Though calculators can do this for us, it is always important to see the math process behind the operation.
The process of multiplying and dividing numbers written in scientific notation involves focusing on the
exponents themselves. In the process of multiplication the exponents are added while in the process of division
the exponents are
subtracted.
Example 1: Multiplying numbers in scientific notation
Multiply the two numbers 3.0 x 10^4 times 6.0 x 10^5
1st multiply the none powers of ten 3.0 x 6.0 = 18
2nd add the exponents of the powers of ten 4 + 5 = 9
3rd recombine the information as a single answer
Answer: 18 x 10^9 which is correctly written as 1.8 x 10^10
Example 2: Dividing numbers in scientific notation
Divide the two numbers 8.0 x 10^6 divided by 2.0 x 10^3
1st divide the none powers of ten 8.0 / 2.0 = 4
2nd subtract the exponents of the powers of ten 6 - 3 = 3
3rd recombine the information as a single answer
Answer: 4 x 10^3 which is correctly written
Originally calculations by hand involving multiplication and division were easier in this notation because rules
of adding exponents for multiplication and subtracting exponents in division were used. The scientific
calculator permits entry of exponential numbers by the use of the "EXP" key. If you haven’t already turn on the
computer and choose WINDOWS from the menu. Use the mouse to pick the scientific calculator and try the
next example.
TASK 10. Using the Windows or other scientific calculator compute the following (enter all numbers with at
the most 2 decimal digit accuracy as scientific numbers (with powers) and use the appropriate OPERATORS (
* or /).
I) 5,000 x 120,000
=
ii) 345,000 / 9,000
=
iii) 54,000 x 0.00006
=
iv) 3,400 / 56,000,030
=
v) 7,567,345 / 0.00567
=
vi) 45,678 x 0.0000000978 / 3.45 =
vii) 89,000,000,001 / 0.0000034
viii) 111,980 / 76,567,890,345
=
=
III: Solving Algebraic equations:
A great deal of the HW is solving equations
We start with the basic rules and advance from there (some might find the discussion too
simple).
1. Solving for a single variable in a first degree equation.
In an equation such as a = b / c being able to solve for a, b, or c correctly is
extremely important. The most common mistake in this situation is thinking that the
solution in the case of solving for c is not recognizing that c = b / a . Way too often
students think that the value for c is equal to a / b. The most basic rule for solving
algebraic equations of any size or complexity is that:
"You must always do the same thing to both sides of an equation, regardless of what
operation you are applying."
When a student thinks that c = a / b in the above equation rather than c = b / a, it is
because they did NOT apply this rule.
To begin consider the ratio and proportion equation a / b = c / d.
Step 1. When solving for any variable in this type of equation first cross multiply.
This means that a is multiplied times d, and b is multiplied times c.
This yields a d = b c
Step 2. To solve for any one of the four variables, the variable being solved for must be
isolated by the use of division.
For Example, if the variable being solved for was d, both sides of the equation would be
divided by a.
This yields d = b c / a
As a similar example, consider the following.
The equation
Step 1.
a = b / c
Multiply both sides of the equation by c
This yields
c a = c b / c
This simplifies to
Step 2.
can be solved for c by the following two steps.
c a = b
Divide both sides of this equation by c
This yields
c a / c
= b / c
This simplifies to
a = b / c
Thus solving for c
in the expression (equation)
results in the expression (equation)
a = b / c
c = b / a
[Note: For convenience a could be written over 1, as in a / 1 = b / c, and c could be
solved for using a ratio and proportion approach.]
Additional examples of solving for variables:
As stated above the most important rule to focus on in solving any equation is to avoid
destroying an equality by doing some operation to one side that you do not do to the
other side.
To see the significance of this statement look at the following examples.
Example 1: solve for "v" in the equation v = d / t
The "d" is a numerator (a value in the top of a fraction)
The "t" is a denominator (a value in the bottom of a fraction)
The "v" is equal to "d" divided by "t"
Example 2: solve for "d" in the equation v = d / t
The "d" is a numerator
The "t" is a denominator
"v" is equal to "d" divided by "t"
"d" can be solved for by multiplying both sides by "t"
This gives us v t = d or d = v t
Example 3: solve for t in the equation v = d / t
The "d" is a numerator
The "t" is a denominator
The "v" is equal to "d" divided by "t"
"t" can be solved for by
1st multiplying both sides by "t" which gives
v t = d
2nd dividing both sides by "v" which gives
t = d / v
The second rule is to be patient when solving equations which will require that you use
several steps.
Example 4: Solve for "t" in the equation v(f) = v(i) + at
There are no fractions here but there is an addition
of unlike variables on the right side of the equation.
"t" can be solved for by
1st subtracting "v(i)" from both sides which gives
v(f) - v(i) = at
2nd divide both sides by "a" which gives
a = v(f) - v(i) / a = t
Similarly you learned to manipulate more advanced equations
TASK 11 given the following equations, solve for the indicated unknown.
(a) 50 = 10 + 5a : find a
(b) vf2 = vo2 + (1/2)at2 : find a if vf = 0, vo = 20, and t = 4
(c) F = mv2/r : find r if F = 455, m = 94, and v = 22
(d) 3t2 = 5t + 2: find t
Formulas in your text! Be sure to check with me if you cannot do these!
(e) b1/b2 =100(m2 –m1)/5 if m2= 18 and m1=3 the ratio of b’s is =?
If m2=9 and m1=3 the ratio of b’s is = ?
(f)m2-m1= 2.5 log10 (b1/b2)
b ratio is 109 then was is the value of m2-m1 ?
(g)c/if c= speed of light and  = 600 nm then what is  (the frequency)… be sure units used are ok!
For c check out the appendix for constants. See page 11 after you do it..
(h) luminosity of the sun L =(4R2 )(T4) with called sigma) is the Stefan-Bolzmann constant (gee
where do I get this professor?). If R the radius of the sun is 7 x 105 km then what is L….be careful to
put all units in the proper values…ie matching in the calculation…..see page 15 after you do it….
(i) Solve the equation in (g) for in terms of the other variables
(j) Solve the equation in (h) for T= in terms of all the other variables.
Additional useful review exercises to do for HW or what is the formula?
a. What is the area of a circle with a diameter of 8 cm?
b. What is the circumference of the circle above?
c. What is the area of a sphere with a radius of 5 cm - express the answer in square meters.
d. What is the volume of a cube with an edge length of 2 cm?
e. What is the surface area of the above cube?
f. What is the volume of a sphere with diameter of 1.5 m