Answers to HW01
1) Briefly describe the Solar System and its component parts. What units are used to describe sizes and distances?
The Sun, a typical star, dominates the solar system in size and mass. It is the gravitational "anchor" of the solar system -- all other objects orbit around the Sun, bound by gravity. Nuclear fusion inside the Sun is the main souce of heat and light throughout the solar system. [If you like to learn by song...enjoy!]
The four planets orbiting closest to the Sun (Mercury, Venus, Earth and Mars) are similar in size, and much smaller than the outer four planets (Jupiter, Saturn, Uranus and Neptune) [For Your Info: the differences run deeper, and we refer to the two groups of planets as "Terrestrial" (Earth-like) and "Jovian" (Jupiter-like); stay tuned for more!]. Between these groups of planets lies the asteroid belt, a ring of orbiting fragments [mostly rock and metal]. Outside the four outer planets is a ring of orbiting icy fragments, of which Pluto is a member.
Planet diameters are usually measured in kilometers (km) or in Earth-radii (Rearth, the circle-with-a-plus symbol); 1 Rearth = 6378 km. Planetary masses are expressed in kilograms (kg) or Earth-masses (1 M = 5.97x1024 kg!!). Distances between planets are usually measured in km or Astrononomical Units (AU, the average distance between the Earth and the Sun); 1 AU = 1.495978706 x 108 km or about 1.5 x 108 km.
2) What are stars? Galaxies? Describe the size of the Universe in comparison to our Solar System.
Stars are large spheres of gas inside which nuclear fusion releases energy. Stars come in different sizes, 0.1 to 100 times the mass of our Sun, and the size and color/temperature of the surface of a star are related to the star's mass. Our Sun is one of several hundred billion (2x1011) stars that orbit around a disk-shaped structure we call the Milky Way galaxy, which also contains large clouds of gas and dust (and the enigmatic "dark matter"). The size of our Galaxy is so large (about 100 million times the diameter of our soloar system, or 100,000 light-years in diameter) that the stars (and any systems of planets that may orbit them) are surrounded by lots of empty space [and they rarely collide, whew!].
Millions of other galaxies exist in space outside our Milky Way galaxy. Galaxies tend to cluster together to form galaxy groups, clusters, and superclusters. We can see light coming from over 13 billion light-years away, so the Universe must be at least that large. This is enormous compared to our solar system -- you might do a size-ratio calculation here to provide some sense of how much larger it is. For example, using the rough size scales from the Powers of 10 video, (Diam of Universe) / (Diam of Solar System) = (1024) / (1013) = 1011. That is, you would have to lay 100 billion solar systems side-by-side to span the Universe.
Note: people commonly confuse "solar system", "galaxy" and "universe". Hopefully this problem helps you keep them straight, and recognize that in our course "The Solar System" we will be studying a tiny fraction of the Universe, albeit the one we can study in the most detail. Take ASTR 201 "Modern Astronomy" if you are interested in "extra-solar" topics, or come to astronomy lunch/coffee with Andy Fridays Time TBA.
3) Write out the following distances in units of meters using both "regular" and scientific notations:
Note: To answer these questions, you can draw information from the text, from Table 2.1, or even Fig. 3.2. These sources will give you approximate values, accurate to the nearest power of ten or better. More precise values can be derived using data found in the Appendix. In many cases in astronomy and life in general, knowing about how large (or distant, or old, or expensive, etc) something is can be sufficient for your purposes. In the answers below, I've included both (i) an estimation -- the method I'd intended you to take-- and (ii) a precise answer -- preferable if you require a detailed answer. Notice the difference in difficulty and accuracy beetween the two methods. Actually, there are lots of ways you can answer these questions, and your answer may vary a bit depending on the source of your facts.
(a) The diameter of the Earth:
(i) Table 2.1 states that the Earth's approximate radius is 6400 km. The diameter is twice the radius, so we get 12,800 km. There are 1000 meters in a kilometer, so multiply: 12,800 km X 1000 m/km = 12,800,000 meters = 1.28x107 meters = nearly 13 million meters.
(ii) Another way to answer this question is to head for the appendix of the textbook. Table 5 on page A-3 lists the Earth's Equatorial Radius to be 6378 km. Following the steps above, this gives 12,756,000 meters= 1.2756 x 107 meters.
(b) The distance from the Earth to the Sun:
(i) In Table 2.1, the distance from Earth to Sun is listed as 150 million km. Multiply by 1000 m/km to get 150,000,000,000 meters = 1.5x1011 meters.(ii) Table 6 on page A-4 lists this distance as 149.6 in units of "106 km" or millions of kilometers. So that would be 149,600,000 km, or 149,600,000,000 meters = 1.496 x 1011 meters.
(c) The distance from the Sun to nearby stars:
(i) In Table 2.1, the distance to the nearest star is listed as 4.2 light-years. We need to convert to units of meters using the fact that 1 light-year equals about 10 trillion km (see text on p.16), or 10,000,000,000,000 = 1x1013 km = 1x1016 meters. Multiply 4.2 ly X 1x1016 m/ly = 4.2x1016 m = 42,000,000,000,000,000 m.
(ii) Table 10 on page A-7 lists the distance to the star Proxima Centauri as 4.23 light-years. This time, let's use the more precise value for the number of meters in a light year given in Table 1 on p. A-2. So, multiply: (4.23 ly) X (9.4605 x 1015 m/ly) = 4.00 x 1016 meters = 40,000,000,000,000,000 m.
Note: notice that these numbers don't fit on the display of your calculator if you use regular notation. To enter astronomical numbers into your calculator, you almost have to use scientific notation.
(d) The diameter of the Milky Way galaxy:
From Table 2.1, we find an approximate radius of 50,000 ly, so the diameter is about 100,000 ly. Multiply by 1x1016 m/ly = 1x1021 m = 1,000,000,000,000,000,000,000 m. Since the Milky Way has no abrupt edge (the density of stars falls off gradually), we can't really define the diameter more precisely than 100,000 ly, so I've included no "precise" calculation here.
(e) How many "powers of ten" larger is (b) compared to (a):
Using a size ratio: (sun-earth distance) / (diameter of earth) = (1.5x1011 meters) / (1.28x107 meters) = 1.2x104. This means the sun-earth distance is twelve thousand times (about 4 powers of ten) larger than the diameter of the Earth. Another way to think of this quesiton is that you would need to lay about 12,000 Earths side by side to reach from the Earth to the Sun. There is a lot of empty space out there! Also, remember that size ratios are unitless -- (meters/meters cancels out ... leaving one thing so many times larger than another).
(e) How many "powers of ten" larger is (c) compared to (a):
Using a size ratio: (sun-star distance) / (diameter of earth) = (4.2x1016 meters) / (1.28x107 meters) = 3.3x109. This means the distance between the Earth and Proxima Centauri is about three billion times (9-10 powers of ten) larger than the diameter of the Earth. You would need to lay about three billion Earths side by side to reach from Earth to Proxima Cen. There is a whole lot of empty space out there!!!
Note: The Appendix of the textbook describes Scientific Notation and reviews techniques for solving math problems -- remember them as the topics crop up in class. If you want practice, you can try these practice problems to hone your skills.
4) Repeat #3 (a-d) using units of Astronomical Units (AUs) instead of meters:
(a) The diameter of the Earth:
Each of these problems requires a unit conversion from distances in meters to AUs. They key is the conversion factor 1.496 x 1011 meters/AU from problem 3(b) above. Divide the diameter of the Earth by this factor: (1.2756 x 107 meters) / (1.496 x 1011 meters/AU) = 8.56 x 10-5 AU. It makes sense that Earth's diameter is a very small number of AUs, because AUs are a "big yardskick."
(b) The distance from the Earth to the Sun:
This one is easy -- by definition, the Earth-Sun distance is 1.0 AU.
(c) The distance from the Sun to nearby stars:
(4.00 x 1016 meters) / (1.496 x 1011 meters/AU) = 2.67 x 105 AU (over 25 thousand AUs). Watch the unit analysis: meters/(meters/AU) = AU.
(d) The diameter of the Milky Way galaxy:
(1x1021 m) / (1.496 x 1011 meters/AU) = 6.7 x 109 AU (over six billionAUs).