|Astronomy 110||PRINT Name   __________________________|
|Fall 2005   Section 006|| |
Homework 1 : How Many ?
(Due Thursday, Sep 1, 2005)
Use scientific (exponential) notation wherever possible.
There are about 1057 atoms in the Sun. It turns out that the Sun is a pretty average sort of star, and there are about a hundred billion stars in our galaxy, and about a hundred billion galaxies in the "observable" Universe. Can you put all these numbers together to estimate the number of atoms we can observe? (3 pts.)
In this problem we need to multiply the number of atoms in a star, by the number of stars in a galaxy, and again by the number of galaxies in the universe. We know the number of atoms in a star: 1057, the number of stars in a galaxy is one hundred billion: 100,000,000,000 more easily written as 1011, and with 1011 galaxies in the universe we have: 1057*1011*1011=1079. So we can observe 1079 (one followed by 79 zeros) atoms in the universe.
At a rate of one digit per second, how long would it take you to write down this number in its long form (with all the zeroes)? Could light travel between the Sun and the Earth in this time? (Check with Figure 1.3 on page 6 of the textbook.)(3 pts.)
If 1079 is one followed by 79 zeros, that means that this number has 80 digits, so if we write one digit per second (that's pretty slow) it will take 80 sec., or 1 min. 20 sec. It takes light about 8 min. to reach Earth from the Sun, so we would be done before light could make the trip.
At a rate of one per second, how long would it take you to count all these atoms? Express this in terms of the lifetime of the Universe (14 billion years). Could light travel across the observable Universe in this time? Explain. (4 pts.)
If we count atoms at one per second it will take us 1079 sec. to finish. How long is that? It will make more sense if we convert 1079 sec. into years. We can figure out how many seconds are in a year by multiplying numbers that everyone should know, 60 seconds per minute, times 60 minutes per hour, times 24 hours per day, times 365 days per year: 60*60*24*365=31536000 or about 3.2*107. So 1079/(3.2*107)=3.2*1071 years. The lifetime of the universe is 14 billion or 1.4*1010 years, so (3.2*1071)/(1.4*1010)=2.3*1061 lifetimes of the universe. The radius of the observable universe is defined by the distance light could travel from the beginning of the universe until now, which we know is about 14 billion years. To cross the entire universe we need to go twice the radius (to get the diameter), so light could travel across the observable universe in 2.8*1010 years or 2 lifetimes, much much less time than it would take to count all the atoms.
Express a "typical" human lifetime in years in scientific (exponential) notation, to the nearest power of ten. We think the Big Bang happened about 14 billion years ago. How many human lifetimes is this? (4 pts.)
Let's say that a human lifetime is about 100 or 102 years. If the universe is 14 billion or 1.4*1010 years old, then there are (1.4*1010)/(102)=1.4*108 lifetimes.
Humans have been around for about a million years (very approximately). If we represent all the time since the Big Bang as a year, how long have humans been around? (6 pts.)
If humans have been around for about 1 million or 106 years, then the proportion of the lifetime of the universe that we've existed is: (106)/(1.4*1010)=7.1*10-5 or 0.000071 the lifetime of the universe. So, if the universe is one year, then humans have been around for 7.1*10-5 of a year. How long is that? Recall that there are about 3.2*107 seconds in a year, so (7.1*10-5)*(3.2*107)=2300 sec. Let's make that a little more manageable by converting to minutes: 2300/60=38 min. So if the universe is a year, humans have been around for about 38 min.
Study Figure 1.3 on page 6 of the textbook. Calculate in the most direct way how many times our galaxy, the Milky Way, would fit in the distance between the Milky Way and its nearest neighbor galaxy, Andromeda. (3 pts.)
Page 6 tells us that the radius of the Milky Way is about 1*105 light-years, and that the distance between the Milky Way and Andromeda is 2.2*106 light-years. If we want to know how many Milky Ways would fit between the two, we need the Milky Way's diameter (twice the radius): 2*105 light-years. Dividing 2.2*106 by 2*105, we find that it's a little more than 10 (we don't round up to 11 because we are interested in whole galaxies). So ten Milky Ways could fit in the distance between it and the Andromeda Galaxy.
Now find the radius of the Sun in kilometers in the textbook (close to the beginning of Chapter 13). Calculate in the most direct way how many times the Sun would fit in the distance between the Sun and its nearest neighbor, Proxima Centauri. (What extra step do you have to make, compared to the previous calculation?) (5 pts.)
We find that the radius of the Sun is about 7*105 km, and the distance between the Sun and Proxima Centauri is 4.2 light-years. As with the Milky Way, we will use the diameter of the Sun: 2*7*105= 1.4*106 km. In order to compare this with the distance between the two stars we need to make their units the same. The appendix of our textbook on page A-8 tells us that there are about 9.5*1015 meters in a light-year, which means there are 9.5*1012 kilometers in a light-year. So if Proxima Centauri is 4.2 light-years away, that's about 4*1013 km. That distance divided by the diameter of the Sun, gives us our answer: (4*1013)/(1.4*106)=2.9*107. This tells us that about 30 million suns could fit between it and our nearest neighbor star Proxima Centauri.
Relative to their size, which are more closely packed, stars or galaxies? (2 pts.)
Given that galaxies are spread about ten apart, while our stars are spread about 30 million apart, galaxies are packed much closer together than stars (proportionally to their size).