Model of the Earth-Moon System<\/p>\n
ASTR 1010L<\/p>\n
The goal of this lab is twofold \u2013 first, to help you get a feel for the the relative sizes of the Earth and moon and just how far apart they are from one another (and how much EMPTY SPACE is out there!!, and second, to review and practice mathematical skills you will need in later labs. We can work on this together \u2013 join a WebEx session during any tutoring hours or schedule a time that works for you!<\/p>\n
This lab has lots of math in it \u2013 this is not like the majority of our labs this semester. Most will have some small calculations here and there (though two will be more math intensive), so do not fear the rest of the semester if this is difficult for you. Definitely take a break if you are getting frustrated, and remember you can send me an e-mail or text any time. Make sure to give me guidance as to what you need \u2013 just saying \u201cI\u2019m stuck on question 12\u201d doesn\u2019t help me help you!<\/p>\n
To follow up on all the amazing pictures in the first lab, let\u2019s start looking at the actual sizes of objects in the solar system. The Earth\u2019s average diameter is 12,756 km. Because it is difficult to have a feel for the size of this numbers like these, we will often use ratios, or the number of times one number contains the other, to express sizes and distances.<\/p>\n
You may have seen ratios written like 16:9 (the aspect ratio of HDTVs and movie screens), indicating that the screen is 16 units wide and 9 units (any unit!) high. Instead of this notation with a colon, we will simply divide the first number by the second, and say that the ratio of the width to height of an HDTV is 16\/9 = 1.8. In other words, the screen is 1.8 times wider than it is tall.<\/p>\n
The Moon\u2019s average diameter is 3,476 km. How many times larger is the Earth than the moon?<\/p>\n
12,756 km<\/p>\n
137160065405003,476 km = 27.25%Km<\/p>\n
If you\u2019re not used to dealing with ratios like this, two questions may arise. First, what happens to the units? Second, your calculator may say something like 3.669735327963. What do you actually write down!?<\/p>\n
First, the units. Almost any time that you are presented with a measurement in astronomy, it will have some sort of unit attached. After all, 12,756 km is a very different size than 12,756 inches. Remember from algebra that if you have the same variable in the numerator and the denominator of a fraction, you can cancel it? For example, 4ab\/a just equals 4b, right? You can do the same with units. In the case of ratios, you will have the same units on the top and bottom, and so they cancel out, leaving an answer with NO UNITS \u2013 it\u2019s just the amount of times one number goes into another. We\u2019ll talk about other cases in a bit when we discuss converting between units.<\/p>\n
Second, how many digits do you write down when taking an answer from your calculator? The key is this \u2013 your calculator may be faster than you are at doing math, but you\u2019re the one with the brain! Almost all of the numbers you will be seeing this semester are measurements that have been made about objects in the solar system. This makes them different than numbers in math. Numbers in<\/p>\n
math are typically exact numbers \u2013 if you have 3 dozen eggs, you know you have exactly 36 eggs. You count 1 \u2013 2 \u2013 3 dozen and know that a dozen contains 12, and 3 x 12 = 36.<\/p>\n
Not so with measurements! If I tell you that I measured a piece of paper to be 12 cm wide, that just means it was closer to 12 cm than it was to 11 cm or 13 cm. If really I measured the piece of paper to be really close to 12 cm \u2013 closer to 12.0 cm than 11.9 cm or 12.1 cm, then I write my measurement as 12.0 cm. If the paper was really, really close to 12 cm, like right on the line of the ruler, I would say it was 12.00 cm. The extra digits tell us about the precision of the measurement. The more precise your measurement, then more digits you are justified in writing down.<\/p>\n
Take a few minutes watch the significant figures videos in the Introduction to the Solar System folder, then determine the number of significant digits (also known as \u201csig figs\u201d!) in each of these measurements, underline which digits are significant, and briefly explain why. HINT: The answers are NOT all the same \u2013 whether or not the zeroes are significant depends on where they are!<\/p>\n
Significant figures are also discussed on p. xiii of the Mathematical Review document in iCollege.<\/p>\n
12 cm#2 of SigFigs:Explanation: All non-zero numbers are significant.<\/p>\n
12.0 cm#3 of SigFigs:Explanation: All trailing zeros that are placeholders are not significant<\/p>\n
12.00 cm#4 of SigFigs:Explanation: All trailing zeros that are placeholders are not significant<\/p>\n
120 cm#2 of SigFigs:Explanation: Zeros at the end of numbers which are not significant but are not removed.<\/p>\n
1200 cm#2 of SigFigs:Explanation: Zeros at the end of numbers which are not significant but are not removed<\/p>\n
102 cm#3 of SigFigs:Explanation: Zeros between non-zero numbers are significant.<\/p>\n
102.0 cm#4 of SigFigs:Explanation: All trailing zeros that are placeholders are not<\/p>\n
0.012 cm#2of SigFigs:Explanation: The zero to the left of the decimal value less than 1 is not significant.<\/p>\n
When you are calculating with measurements, just keep in mind the adage that \u201cA chain is no stronger than its weakest link.\u201d If you are multiplying or dividing \u2013 which we will do more often than adding and subtracting since we\u2019ll often be dealing with ratios \u2013 then your answer should have no more significant digits than your least precise (i.e. weakest) measurement.<\/p>\n
In the case of the diameters given above, how many significant digits does 12,756 km have? <\/p>\n
5 significant digits<\/p>\n
How many significant digits does 3,476 km have?<\/p>\n
4 significant digits<\/p>\n
So how many significant digits should your answer have?<\/p>\n
The answer should have a total of 9 significant digits. <\/p>\n
So you take that number that was given to you by your calculator (3.669735\u2026) and round it off to 4 significant digits. In this case, we have to think carefully. If we just truncated the answer, we\u2019d get a ratio of 3.669. However, because the next digit is greater than 5, we need to round up. This makes the ratio 3.670. You need that zero at the end to show that you really do know the answer precisely! Just saying 3.67 means your calculated ratio is between 3.665 and 3.675 \u2013 but the numbers you were given for the Earth\u2019s and Moon\u2019s radii allow you to know better than that.<\/p>\n
I know this can be a bit confusing, and we\u2019re not going to get overly obsessed with significant figures, but it is important that you know how to decide to write down off your calculator based on the numbers you use in a calculation.<\/p>\n
Let\u2019s try it out on a few calculations.<\/p>\n
Jupiter\u2019s diameter (measured at its equator; it\u2019s not perfectly spherical\u2026) is 143,000 km. How many times larger than Earth\u2019s diameter is Jupiter\u2019s diameter?<\/p>\n
143000 km Answer equals to: 11.222 times therefore, Jupiter\u2019s <\/p>\n
12,742 km diameter is 11 times larger than earth\u2019s diameter.<\/p>\n
How many sig figs should your answer have?Why?<\/p>\n
11.222 has 2 significant digits because all non-zero numbers are significant.<\/p>\n
What units does your answer have? Why?<\/p>\n
The answer has 5 units because all non-zero numbers after the decimal point are significant hence they ought to be counted.<\/p>\n
The Moon\u2019s average distance from the Earth is 384,400 km. How many times the Earth\u2019s diameter is this? Make sure to show your work like in part a!!)<\/p>\n
384,400 <\/p>\n
12,742 = 30.1679<\/p>\n
Therefore, the Moon\u2019s average distance from the Earth is 30 times the diameter of the Earth. <\/p>\n
Ok, now it\u2019s about time to start using this information! On the next page, I want you to draw a \u201cscale model\u201d of the Earth and Moon. I\u2019ve started you off with a circle in the upper right corner for the Moon.<\/p>\n
Next, we need to measure the diameter of the Moon circle in centimeters. It\u2019s small, but you should be able to measure to hundredths of cm (e.g. 0.33 cm or 1.55 cm or 3.02 cm) by estimating between tick marks on a centimeter ruler. If you don\u2019t have a centimeter ruler, then you can print the one posted in iCollege \u2013 just make sure to print it at 100%. On a typical centimeter ruler, the millimeters are the little tick marks between the centimeter marks (remember that 10 mm = 1 cm). In order to be as precise as possible about the size of Earth in our model, we need to be precise in our measurements; because the Moon circle is so small, by estimating between tick marks you should be able to make a measurement with 2 significant figures (0._ _ cm; remember that the leading 0 is not significant). If the measurement falls exactly on one of the tick marks, then the final (estimated) digit will be a 0 (e.g. 0.50 cm).<\/p>\n
So, what is the diameter of the model Moon (the circle on the next page) in cm with 2 significant digits? <\/p>\n
The diameter of the model moon is 420500cm. In this case, the figure has 4 significant digits. The zero between units 2 and 5 is significant because it appears between two significant digits. <\/p>\n
Now to the Earth\u2026 You figured out that the Earth\u2019s diameter is 3.670 times the Moon\u2019s diameter. The point of a scale model is that if the real Earth has a diameter is 3.670 times the Moon\u2019s real diameter, then the diameter of Earth in your model is 3.670 times the Moon\u2019s diameter in the model.<\/p>\n
So, in a few minutes, when you calculate where to draw the Earth, how large will you draw it (what will be its diameter)? Don\u2019t forget to show your units in your work and in your answer. (Hint \u2013 you should get an answer smaller than the width of your fingertip, and the units of your answer should be cm because 3.670 is a ratio and doesn\u2019t have a unit and you measured the model Moon in cm.)<\/p>\n
Earth\u2019s diameter= 12742km<\/p>\n
Moon\u2019s diameter 3.670 times smaller than 12742km Earth\u2019s diameter<\/p>\n
Therefore, Moon\u2019s diameter is 12742 divided by 3.670 times<\/p>\n
12742 = 3,471.93 km rounded off to 3,472km<\/p>\n
3.670<\/p>\n
3,742km equals to 3.472e+8cm<\/p>\n
Answer= moon\u2019s diameter is 3.472e+8cm.<\/p>\n
You also figured out the ratio between the Earth-Moon distance and the Earth\u2019s diameter. How many cm apart will your circles representing the Earth and the Moon be drawn?<\/p>\n
I didn\u2019t say it, but you did show your work, right? The distance you find should span a large portion of the page but still fit on the page. Check in with me if you\u2019re not sure you got the correct number.<\/p>\n
Earth-Moon distance is 384,400Kkm<\/p>\n
Earth\u2019s diameter is 12742km<\/p>\n
384,400 km = 3.844e+10<\/p>\n
12742 = 1.2742e+9<\/p>\n
3.844e+10 – 1.2742e+9 = 3.71658e+15<\/p>\n
Answer = 3.71658e+15cm<\/p>\n
The distances between objects in astronomy is almost always given as the distances between the CENTERS of the objects, not the distances between their surfaces. (For relatively little things far apart from one another, this doesn\u2019t matter too much, but if you\u2019re dealing with relatively large things close together, it can be very important.)<\/p>\n
Draw a line from the center of the Moon circle to where the center of your Earth circle will be.<\/p>\n
187642549593500The model Moon \uf022<\/p>\n
Because you know where the center of the Earth will be in your model, it will be easier to draw the Earth if you know its radius, the distance from its center to its surface, rather than its diameter.<\/p>\n
The Earth\u2019s diameter is two times its radius. A note about sig figs here\u2026 In this case, 2 is an exact number \u2013 the diameter is defined by mathematics to be exactly two times the radius. So, because it\u2019s exact, you don\u2019t have to worry about how many significant digits it has \u2013 it\u2019s \u201cperfect\u201d. Your answer for the Earth\u2019s will should have 4 sig figs because 6,378 km has 4 sig figs, and it\u2019s the \u201cweak link\u201d since 2 is exact.<\/p>\n
Using your value for the diameter of the model Earth in cm from part b, what is the model Earth\u2019s radius (in cm)?<\/p>\n
Earth\u2019s diameter 12742km which is equivalent to 1.2742e+9cm<\/p>\n
Radius is 2 times smaller than earths diameter<\/p>\n
Therefore 12742 = 6371km<\/p>\n
2 <\/p>\n
6371km = 6.371e+8cm<\/p>\n
Answer = 6.371e+8cm<\/p>\n
Next, draw a circle the appropriate size for the model Earth using the radius you calculated in part e. The center of the circle should be the end of your line. (You don\u2019t need a compass or anything special, just do your best to draw a shape that\u2019s close to a circle instead of something random and lumpy.)<\/p>\n
Model Earth<\/p>\n
Label your model Earth.<\/p>\n
Congratulations! You\u2019ve constructed a scale model of the Earth-Moon system!<\/p>\n
We\u2019re not quite done with the lab, though. Keep pressing!<\/p>\n
Any time you use a model or look at a picture, there is some sort of scale to the model or the picture. The scale is what allows you to convert between measurements you make on the model or picture and the real sizes of the objects in the model or picture. In your Earth-Moon model on the previous page, the scale was 1 cm = 17,500 km. This isn\u2019t a conversion factor like there are 10 mm in 1 cm, but it tells you that 1 cm on the model (your drawing on the previous page) is 17,500 km in real life. However, the math that you use to convert between units is the same as you would use to convert between sizes in the model and sizes in real life.<\/p>\n
Before we proceed, review the \u201cHow to Convert Units of Measurement\u201d resource in the Introduction to the Solar System folder.<\/p>\n
What is \u201cThe Big Secret\u201d? <\/p>\n
Numbers with units, like 16.2\u00a0meters or 32\u00a0ft\/sec\u00b2, are treated\u00a0exactly\u00a0the same as coefficients with variables, like 16.2x\u00a0or 32y\/z\u00b2<\/p>\n
Give an example, different than one in the \u201cHow to Convert Units\u201d webpage, of using the principle of multiplying by 1 to convert one unit to another.<\/p>\n
60 minutes = 1<\/p>\n
1 hour<\/p>\n
Let\u2019s say we want to convert 5 hours into minutes, we multiply it by 60 minutes which is equal to 1 hour.<\/p>\n
60 minutes x 5 hours = 300 minutes<\/p>\n
I hour<\/p>\n
What are the two steps to \u201cHow to Pick a \u20181\u2019\u201d?<\/p>\n
Construct a fraction that is equal to 1 and multiply the original measurement by that fraction.<\/p>\n
Now let\u2019s use this method to confirm the scale given above. Fill in the blank below with the length of your line between the centers of the Earth and Moon (the distance you found in part c of #5), then complete the calculation. Because in the model 1 cm = 17,500 km, the fraction<\/p>\n
11430001435100017,500 km equals 1 and so is a valid conversion factor.<\/p>\n
1 cm<\/p>\n
3,742km \u00d7<\/p>\n
17,500 km<\/p>\n
224028065405001 cm=<\/p>\n
What units cancel?<\/p>\n
Kilometers (Km) cancel in the calculation. <\/p>\n
What unit stays in your answer?<\/p>\n
Since we are looking for the distance in centimeters, the centimeters units will stay in the answer<\/p>\n
Does your answer roughly agree with the actual distance between the Earth and Moon? (This value was given to you in part b of #5.) If it\u2019s FAR off, then check your math on the last few pages, but if it\u2019s pretty close, then the difference is due to rounding and imperfect measurements<\/p>\n
\u2013 remember you were only able to measure the size of the model Moon to 2 significant figures.<\/p>\n
To see how the Earth-Moon system is different from Jupiter and its moons, let\u2019s use the same scale (from the Earth-Moon model in Part 1) to calculate values that you can use to construct a model of Jupiter and its moons.<\/p>\n
This time, you are starting with actual sizes in km and need to convert them to cm in your model. Therefore, you need to cancel km out, and so your fraction will be reversed. Fill in the table on the following page using the equation below. Make sure that you show the units of the model distances and diameters in the table. Show the work for a few of your calculations below.<\/p>\n
509651018605500Each conversion now will look like this:km \u00d71 cm=<\/p>\n
17,500 km<\/p>\n
Object Actual<\/p>\n
Diameter Diameter in Model Actual Distance<\/p>\n
from Jupiter Distance from<\/p>\n
Jupiter in Model<\/p>\n
Jupiter 143,000 km 143.0e+10cm N\/A N\/A<\/p>\n
Io 3,640 km 3.640e+15cm 421,800 km 421.8e+ 12cm<\/p>\n
Europa 3,120 km 3.120e+8cm 671,100 km 671.1e+10cm<\/p>\n
Ganymede 5,260 km 5.260e+12cm 1,070,000 km 107.0e+20cm<\/p>\n
Callisto 4,820 km 4.820e+15cm 1,883,000 km 1883.0e+8cm<\/p>\n
Whoa, there\u2019s a problem here, isn\u2019t there! You can\u2019t fit that on one (regular-sized) page! This is why pictures in astronomy can often be laid out in a way that is really deceiving! Look at the \u201cJupiter and its Moons\u201d picture.<\/p>\n
Based on your numbers above, explain what is meant by the following sentence: This isn\u2019t a real picture, though it is made from real pictures.<\/p>\n
Although the figures appear as a true representation of the measurements, they are figures that have been rounded off to the nearest digit.<\/p>\n
To at least get a feel for this model, and see the relative sizes of Jupiter and Earth, draw a circle representing Jupiter at the bottom right of this page and then \u2013 with the correct spacing between the centers of the circles \u2013 add Jupiter\u2019s closest \u201cbig\u201d moon Io to your model (Jupiter has LOTS of moons, but only the four listed above are more than a hundred miles across). It should fit on this page \u2013 just find a place that doesn\u2019t have text on it that is the correct distance from Jupiter.<\/p>\n
Does your drawing of Jupiter seem like it\u2019s about as many times across as your drawing of Jupiter as you calculated in part a of #5?<\/p>\n
Yes, the model looks quite as many times as the drawing. The Jupiter drawing is actually larger than the Jupiter model.<\/p>\n
If so, congratulations and good job! If not, then there\u2019s been a calculation error somewhere, and you need to find it\u2026 Remember that you can consult classmates or Dr. Skelton!<\/p>\n
Io is MUCH smaller than Jupiter, but is Io similar in size, much larger than, or much smaller than the Earth\u2019s Moon?<\/p>\n
Io\u00a0is the fifth\u00a0moon\u00a0from\u00a0Jupiter. Its average distance is about 262,000 miles (422,000 km) from earth.\u00a0Io\u00a0is locked such that the same side always faces\u00a0Jupiter. Io\u00a0has a mean radius of 1,131.7 miles hence making it slightly\u00a0larger than Earth’s moon.<\/p>\n
A few more subtopics before we\u2019re done with reviewing math\u2026 First, prefixes for metric units.<\/p>\n
The basic unit of length for the metric system is the meter. For most adults a meter is roughly the distance from the tip of your nose to the end of your outstretched arm. This distance is useful for measuring the sizes of cars and buildings, but not the size of a cell or the distance to Pluto.<\/p>\n
The basic unit of mass is the gram. A regular paperclip is about half a gram; one of the big paperclips is about a gram and a half. A nickel has a mass of 5 g (actually 5.000 g). Useful for measuring the mass of a bug, but not the mass of an ingredient in medication or the mass of a person.<\/p>\n
So, we add prefixes to the basic units that indicate a multiplier that makes the unit either bigger or smaller. In words, what do the following prefixes mean? (You may research this online if you need to.) What symbol (all are regular letters except for micro-, which is a Greek letter\u2026) do we use to represent the prefix?<\/p>\n
I\u2019ll get you started with the first one; note that this means that a nanometer (nm) is one billionth of meter, a nanosecond (ns) is one billionth of a second, and a nanogram (ng) is one billionth of a gram!<\/p>\n
nano-one billionthn micro- one millionth \u03bcm<\/p>\n
milli- one thousandth mm <\/p>\n
Using the prefixes does help us more easily represent objects that are very large and very small, but honestly, in astronomy, we are dealing with objects that are very, very massive and very, very far away. We deal with these very, very large numbers in different ways. For masses, we use scientific notation or use ratios to compare an object\u2019s mass to the Earth\u2019s mass or to the Sun\u2019s mass. (You will see this in the Moons of Jupiter lab.)<\/p>\n
For distances, however, we define two new units. For distances in the solar system, we typically use astronomical units. One astronomical unit (abbreviated AU) is equal to the average distance between the Earth and the Sun.<\/p>\n
Jupiter is, on average, 778,600,000 km from the Sun. To find out how many AU this is, fill in the conversion below. Put Jupiter\u2019s distance in the first blank, then complete the conversion factor with the average distance between the Earth and Sun in km (149,600,000 km, often rounded to 150,000,000 km). Cross out units that cancel, and make sure use the correct number of sig figs and the correct units in your answer.<\/p>\n
149,600,000km<\/p>\n
24809458636000778,600,000 km \u00d7 <\/p>\n
150,000,000km = 7,765,237,333.3<\/p>\n
That\u2019s a lot easier number to get your head around, isn\u2019t it? Even Pluto at its farthest is 49 AU, or 49 times the Earth\u2019s distance, from the Sun.<\/p>\n
For even farther distances, we use a unit called a light year, defined simply as the distance light travels in one year. Light travels remarkably fast \u2013 299,800,000 meters every second! Even so, distances in space are vast \u2013 it takes light 500 seconds (8 minutes and 20 seconds) to travel from the Sun to the Earth. Light from nearby stars takes years to reach us; light from distant objects may take millions or even billions of years to get to us.<\/p>\n
In order to figure out how large a light year is, we need two more pieces of information. One is the relationship between distance travelled, speed, and time. You\u2019ve seen this before, I\u2019m sure: ???????? = ????????????????.<\/p>\n
The second thing we need is the number of seconds in a year. This is relatively easy using the conversion method you discussed in previous questions. Fill in the missing information in the conversion below, then cross out units that appear on both the bottom and the top, and then calculate. Your answer should have 5 sig figs because 365.24 days in a year is a measurement rather than a definition. (The number of hours in a day, minutes in an hour, and seconds in a minute are definitions and therefore are exact numbers, so we don\u2019t have to worry about them when deciding what the weakest link is.) Don\u2019t forget that your answer should have units!<\/p>\n
1 yr \u00d7<\/p>\n
365.24 days<\/p>\n
180721065405001 yr\u00d7<\/p>\n
hrs<\/p>\n
277050565405001 day\u00d7<\/p>\n
min hr\u00d7<\/p>\n
s min =<\/p>\n
3529330-300355004361815-30035500To figure out how big a light year is, you take that number of seconds and multiply it by the speed of light. Again, remember that your answer needs to have units and the correct number of sig figs.<\/p>\n
252984014287500287655019621500???????? = ???????????????? = 299,800 km \u00d7 365.24s = 109,498,952 km<\/p>\n
s<\/p>\n
That\u2019s a big number! Did your calculator tell you (after using your own brain to figure out the correct number of significant digits and units\u2026) 9.460 x 1012 km or 9,460,000,000,000 km? These are the same amount, just represented in different ways. In either case, a light year is over nine trillion kilometers \u2013 this is close to six trillion miles!!! And the closest star to us is over four light years away!<\/p>\n
Watch the two scientific notation videos on iCollege. Scientific notation is also discussed on pages viii and ix of the Mathematical Review.<\/p>\n
What is the point\/purpose of scientific notation?<\/p>\n
The purpose of scientific notation is to represent numbers that are either too small or too large to be captured.<\/p>\n
Numbers in scientific notation are written in the form a x 10b, where a is called the coefficient and b<\/p>\n
is the exponent.<\/p>\n
What is the rule for what kind of number a must be?<\/p>\n
\u00a0The\u00a0whole numbers<\/p>\n
What is the rule for what kind of number b must be?<\/p>\n
\u00a0The\u00a0integers\u00a0including; the positive whole numbers, their negative counterparts, and zero.<\/p>\n
In some sciences, there are very, very small numbers (not negative, just much smaller than 1) that need to be represented by scientific notation. We use negative exponents for these very small numbers \u2013 a negative exponent just means it belongs in the denominator of a fraction instead of the numerator. For example, 6 x 10-6 is the same as 6\/106 or \u201csix millionths\u201d, whereas 6 x 106 is \u201csix million\u201d. In astronomy, though, we\u2019re almost always dealing with very large numbers so we usually don\u2019t need to deal with negative exponents.<\/p>\n
Write the following numbers in scientific notation.<\/p>\n
1,400,000 km (the Sun’s diameter) = 1.4 \u00d7 106<\/p>\n
b. 5,980,000,000,000,000,000,000,000 kg (the Earth\u2019s mass) = 5.98 x 10\u00b2\u2074<\/p>\n
c. 4,560,000,000 yr (the age of the solar system) = 4.56\u00d7108\u00a0<\/p>\n
Write the following in regular, non-scientific-notation and then in words as you might say them (e.g. “120 trillion”).<\/p>\n
Time of the last ice age: 1.8 x 104 years ago<\/p>\n
187.2 which is 187 years, 2 months ago<\/p>\n
Time when dinosaurs became extinct: 6.5 x 107 years ago<\/p>\n
695.5 Which is 695 years and 5 months ago<\/p>\n
Age of the universe: 1.38 x 1010 years<\/p>\n
1393.8 years which is rounded off to 1394 years<\/p>\n
Distance to the Moon: 3.84 x 105 km = 384\u00a0000 kilometers<\/p>\n
Average distance between the Sun and Pluto: 5.91 x 109 km<\/p>\n
5910000000 kilometers<\/p>\n
Mass of the Sun: 2.0 x 1030 kg<\/p>\n
2e+30 grams<\/p>\n
(That last one is kind of mean, isn\u2019t it\u2026 See what you can figure out without spending too much time \u2013 but it helps make the point why scientific notation is pretty useful, doesn\u2019t it?)<\/p>\n
One time you will see negative exponents is in the Moons of Jupiter lab, so a question to think about (and hopefully remember in a couple of weeks!):<\/p>\n
Consider the numbers 1.05 x 10-3 and 9.89 x 10-4. Are these numbers fairly close to one another, or quite different? Explain your conclusion.<\/p>\n
Hint: write them in decimal (normal, non-scientific) notation and the round them to one significant figure\u2026<\/p>\n
9.89 x 10-4 = 0.000989<\/p>\n
1.05 x 10-3 = 0.00105<\/p>\n
1.05 x 10-3 is larger than 9.89 x 10-4 based on the number of zeros after the decimal point and the number raised to the figure. -3 is greater than -4.<\/p>\n
One final scientific notation note \u2013 when you have to do calculations with numbers in scientific notation, it is important to make sure you enter them into your calculator correctly. Your calculator has a specific way of entering scientific notation \u2013 and it really is better to use the button designed for scientific notation rather than \u201c*10^\u201d and then the exponent!! (The reason has to do with the way that the calculator deals with the number and order of operations.)<\/p>\n
Usually there is either a CAPITAL E (not the lower case one \u2013 it\u2019s something entirely different!), capital EE, or a capital EXP marked on or above one of the calculator keys. On many TI calculators the E is above the comma, so you have to press \u201c2nd\u201d then \u201c,\u201d. If you can\u2019t find a button that seems right, e-mail Dr. Skelton a picture of your calculator or look at the calculator documentation (you can search your model online).<\/p>\n
The E or EXP means \u201ctimes ten to the\u201d, so you would enter 2×104 into your calculator as 2E4 without any multiplication signs, 10s, or exponents. Your calculator now knows this is a single number rather than a calculation it\u2019s supposed to do. For negative exponents, you will probably need to use the (-) button or the +\/- button to let the calculator know you\u2019re not subtracting.<\/p>\n
What button (or sequence of buttons) does your calculator have to enter scientific notation?<\/p>\n
MODE and Sci (Short form for Scientific mode) buttons <\/p>\n
If your calculator returns an answer in scientific notation to you, it may not look just like we\u2019ve been writing \u2013 instead, it will have an E (which, recall, means \u201ctimes ten to the\u201d) or just a small superscript number at the far right of the display. If your calculator says 6.05E7, what should YOU write down as your answer? Hint: It shouldn\u2019t have an E in it\u2026<\/p>\n
Answer: 9.67e7<\/p>\n
To finish, I would like you to complete a couple more unit conversion problems then do a graphing exercise.<\/p>\n
You can use the same method of converting between units that was used even when there are more than one unit \u2013 for example, if you are travelling at 55 miles per hour, what is your speed in meters per second? You just need to concentrate on changing one unit at a time \u2013 miles to m and hours to s. First, fill in the blanks in the conversion below, cancel any units that appear on both the top and bottom, then calculate the answer. You will need to figure out what goes into the completely blank numerators based on what units need to cancel and then looking up the conversion factor.<\/p>\n
55 mi<\/p>\n
1371600654050019202406540500hr\u00d7<\/p>\n
0.9hr <\/p>\n
5min \u00d7<\/p>\n
55min<\/p>\n
271907086360003300s \u00d7<\/p>\n
17kmm \u00d7<\/p>\n
1700m 17km = <\/p>\n
Answer = 100km<\/p>\n
3398520-266700004285615-26670000Now try one by yourself, making sure to show all your steps like in the previous problem.<\/p>\n
The Earth travels around the Sun at a speed of 29.8 km\/s. How fast is this in miles per hour?<\/p>\n
While its speed averages out to about 29.8 km\/s (18.5 mps) or 107,000 km\/h (66487 mph), it actually ranges by a full km per second during the course of the year \u2013 between 30.29 km\/s and 29.29 km\/s (109,044 \u2013 105,444 km\/h; 67,756.8 \u2013 65,519.864 mph)<\/p>\n
Almost there! Just a little more about graphing!<\/p>\n
Look at the graph of the temperature structure of Venus\u2019 atmosphere.<\/p>\n
Based on this figure, describe how the temperature changes with altitude on Venus.<\/p>\n
4572000476250As sunlight passes through the atmosphere, it heats up the surface of Venus.The greenhouse effect on Venus causes the temperatures at its surface to reach 864 degrees Fahrenheit (462 degrees Celsius)<\/p>\n
Based on this figure, what is the temperature at the surface of Venus? Explain or show mathematically how you arrived at your answer using the information in the graph.<\/p>\n
I multiplied the highest middle and lower temperatures<\/p>\n
How much does the temperature drop from the surface to an altitude of 100 km (just over 60 miles)?<\/p>\n
Above 60 miles (100 km) from Earth’s surface the chemical composition of air becomes strongly dependent on altitude and the atmosphere becomes enriched with lighter gases (atomic oxygen, helium and hydrogen).\u00a0<\/p>\n
The final exercise (yes, really!!) helps demonstrate why graphs can often demonstrate data much more clearly<\/p>\n","protected":false},"excerpt":{"rendered":"
Model of the Earth-Moon System ASTR 1010L The goal of this lab is twofold \u2013 first, to help you get<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[],"tags":[],"class_list":["post-52850","post","type-post","status-publish","format-standard","hentry"],"yoast_head":"\n