ASTRO 1050 Scientific Notation, Model Scales, and Calculations The simple truth is, interstellar distances will not fit into the human imagination. - Douglas Adams Materials: Ping pong balls, meter sticks, protractors 1. Scientific Notation (12pts) Describing the universe requires some very big (and some very small) numbers. Such numbers are tough to write in long decimal notation, so we ll be using scientific notation. Scientific notation is written as a power of 10 in the form m x 10 e where m is the mantissa and e is the exponent. The mantissa is a decimal number between 1.0 and 9.999 and the exponent is an integer. To write numbers in scientific notation, move the decimal until only one digit appears to the left of the decimal. Count the number of places the decimal was moved and place that number in the exponent. For example, 540,000 = 5.4x10 5 or, in many calculators and computer programs this is written: 5.4E5 meaning 5.4 with the decimal moved 5 places to the right. Similarly 314.15 = 3.1415x10 2 and 0.00042 = 4.2x10-4 and 234.5x10 2 = 2.345x10 4 You get the idea. Now try it. Convert the following to scientific notation. Decimal Scientific Decimal Scientific 2345.4578 2.3x10 3 (1pt) 0.000005 5.0x10-6 (1pt) 356,000,000,000 3.5x10 11 (1pt) 0.0345 3.45x10-2 (1pt) 111x10 5 1.11x10 7 (1pt) 2345x10-8 2.34x10-5 (1pt) 2. Arithmetic in Scientific Notation To multiply numbers in scientific notation, first multiply the mantissas and then add the exponents. For example, 2.5x10 6 x 2.0x10 4 = (2.5x2.0) x10 6+4 = 5.0x10 10. To divide, divide the mantissas and then subtract the exponents. For example, 6.4x10 5 / 3.2x10 2 = (6.4 / 3.2) x 10 5-2 = 2.0x10 3 Now try the following. 4.52x10 12 x 1.5x10 16 = 6.78x10 28 (1pt) 9.9x10 7 x 8.0x10 2 = 7.92x10 10 (1pt) 1.5x10-3 x 1.5x10 2 = 2.25x10-1 (1pt) 8.1x10-5 x 1.5x10-6 = 1.21x10-10 (1pt) 1.5x10 32 / 3.0x10 2 = 5.0x10 29 (1pt) 8.0x10-5 / 2.0x10-6 = 40 (1pt) Be careful if you need to add or subtract numbers in scientific notation. 4.0x10 6 + 2.0x10 5 = 4.2x10 6 since 4.0x10 6 = 4,000,000 +2.0x10 5 = + 200,000 4.2x10 6 = 4,200,000
3. Converting Units (13pts) Often we make a measurement in one unit (such as meters) but some other unit is desired for a computation or answer (such as kilometers). Example: You have 2,340,000,000,000 meters. How many kilometers is this? How many astronomical units? There are 1000 m/km. Because kilometers are larger than meters, we need fewer of them to specify the same distance, so divide the number of meters by the number of meters per kilometer and notice how the units cancel out and leave you with the desired result. 2.34x10 12 m 1000 m = 2.34 10 9 km Another way to think about this operation, is that you want fewer km than m, km so just move the decimal place three to the left since there are 10 3 m per kilometer. Or, if the new desired unit is smaller, and you expect more of them, then multiply. For example, how many cm are there in 42 km? 42 km 10 5 cm km = 42 105 cm Use the information in the appendix of your text to convert the following. Write these in Scientific Notation. 50 km = 5x10 4 m (1pt) 3x10 6 m = 3x10 8 cm (1pt) 52,600,000,000 km = 5.2x10 13 m (1pt) 300,000,000,000 km = 2.0x10 3 AU (1pt) 6.0x10 18 m = 6.3x10 2 ly (1pt) 4500 parsecs = 1.4x10 17 km (1pt) 5.2x10 12 kg = 5.2x10 15 g (1pt) 4.0x10-5 sun masses = 7.6x10 25 kg (1pt) 3450 seconds = 5.7x10 1 minute (1pt) 99 minutes = 1.65 hr (1pt) 600 hours = 2.5x10 1 days (1pt) 1200 days = 3.28 yr (1pt) 1 year = 3.15x10 7 s (1pt) 4. Scale Models (12pts) The Universe is such a big place, scale models are almost required to help visualize it and help it fit into our imagination. If we are going to represent the earth by a ping long ball (diameter=4.0 cm), how large should be the planet Venus? To make the scale model we need to look up the real size of Earth and Venus and use the method of ratios to find the model size of Venus. 12756 km real Earth diameter = 4 cm model size 12101 km real Venus diameter = x cm model size Set up the ratio, and then solve for x (cross multiply and divide) 12756 km 12101km = 4cm x cm therefore, x cm= 4cm 12101km 12756km = 3.79cm Now do the following: Real Size (km) Model size (cm) The moon (diameter) 3.4748x10 3 (1pt) 1.08 (1pt) The distance to the moon 3.84403x10 5 (1pt) 1.2x10 2 (1pt)
The distance from sun to earth 1.49x10 8 (1pt) 4.6x10 4 (1pt) The sun (diameter) 1.3x10 6 (1pt) 4.3x10 2 (1pt) The size of Jupiter (diameter) 1.42x10 5 (1pt) 4.3x10 1 (1pt) The radius of Pluto s orbit (semi-major axis) 5.9x10 9 (1pt) 1.8x10 6 (1pt) Use the space below for calculations. 5. Time to travel (14pts) Suppose you could travel at the speed of the fastest jets, about 1500 miles per hour, or about 2250 km/hr. The distance, D, traveled in time, t, and speed, v, is D(km) = v(km/hr) x t(hr) D(km) = v(km/s) x t(s) D(m) = v(m/s) x t(s) Notice that the units of time in v and in t must be the same, as must the units of distance in the v and in D, otherwise, we ll get a nonsensical answer! Traveling at the speed of a fast jet, how long would it take to get from earth to: Distance (km) Time The moon: 3.84403x10 5 (1pt) 6.1x10 3 sec (1pt) The sun: 1.49x10 8 (1pt) 6.6x10 4 hours (1pt) Mars (closest to Earth) 5.4x10 7 (1pt) 1.0x10 3 days (1pt)
Alpha Centauri: 4.01x10 13 (1pt) 7.4x10 8 days (1pt) Now suppose you could travel at the speed of light (3x10 8 meters/sec). How long would it take to get to: Distance (km) Time The sun: 1.49x10 8 (1pt) 8.30 minutes (1pt) Pluto (closest to Earth): 4.18x10 9 (1pt) 3.87 hours (1pt) Alpha Centauri: 4.01x10 13 (1pt) 4.24 years (1pt) 6. Angles (12pts) Often in astronomy we measure the size of things in the sky, or the motion of things across the sky, in terms of angles. In one circle there are 360 degrees. In one-half of a circle there are 180 degrees, etc. 180 90 30 Make estimation of the following angles. Then use a protractor to measure them. a rough ~45 degrees (1pt) ~135 degrees (1pt) ~5 degrees (1pt) Degrees are further subdivided into minutes of arc or arcminutes. There are 60 arcminutes in one degree. Acrminutes are further subdivided into seconds of arc or arcseconds. As you might imagine, there are 60 arcseconds in one arcminute. Convert the following. 120 arcminutes = 2 degrees (1pt) 900 arcseconds = 1.5x10 1 arcmin (1pt) 12 arcminutes = 2x10-1 degrees (1pt) 6 arcseconds = 1.6x10-3 degrees (1pt) 12 degrees = 7.2x10 2 arcminutes (1pt) 55 arcmin = 3.3x10 3 arcsec (1pt) 360 degrees = 2.16x10 4 arcmin (1pt) 360 degrees = 1.3x10 6 arcsec (1pt) 57.29 degrees = 2.06x10 5 arcsec (1pt) This last one is special. This number will show up a lot
this semester. And here s where it comes from. There is also a special unit of angular measure called a radian. 1 radian = 57.3 degrees. We won t use it explicitly this semester, but we ll be using it implicitly, so we thought you might like to know. 1 rad = 57.3 deg 7. Measurements and Graphing (11pts) Measure the height and shoe length (we really mean foot length, but we don t want everyone taking off their shoes in lab...) and the length of fingernails of your lab mates, then record data for everyone else in your lab. (2pts) Height Shoe Length Nail Length Height Shoe Length Nail Length (cm) (cm) (mm) (cm) (cm) (mm) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. Do you expect to see a correlation between height and shoe length? Between shoe length and nail length? Why or why not?(3pts) *As long as all three questions are answered, thoughtfully, points can be given at your discretion.
Now plot these data on the graphs provided. You ll need to add the axis numbers. (4pts) Nail length (mm) Shoe length (cm) 140 145 150 155 160 165 170 175 180 185 190 140 145 150 155 160 165 170 175 180 185 190 Height (cm) Height (cm) Which variables show correlations? (1pt) The shoe length and height variables show correlations, the height and nail length do not. What is the explanation for this correlation?(1pt) As a person is taller they need more area to stand on to balance (therefore taller people need longer feet), whereas nail length can be grown and cut to lengths determined by a person s personal preference.