Astronomy 110 | PRINT Name __________________________ |

Fall 2005 Section 006 | |

Homework 1 : How Many ? |
(Due Thursday, Sep 1, 2005) |

**Use scientific (exponential) notation wherever possible.**

- How many atoms ?
There are about 10

^{57}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: 10**^{57}, the number of stars in a galaxy is one hundred billion: 100,000,000,000 more easily written as 10^{11}, and with 10^{11}galaxies in the universe we have: 10^{57}*10^{11}*10^{11}=10^{79}. So we can observe 10^{79}(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 10**^{79}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 10**^{79}sec. to finish. How long is that? It will make more sense if we convert 10^{79}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*10^{7}. So 10^{79}/(3.2*10^{7})=3.2*10^{71}years. The lifetime of the universe is 14 billion or 1.4*10^{10}years, so (3.2*10^{71})/(1.4*10^{10})=2.3*10^{61}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*10^{10}years or 2 lifetimes, much much less time than it would take to count all the atoms.

- How many lifetimes ?
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 10**^{2}years. If the universe is 14 billion or 1.4*10^{10}years old, then there are (1.4*10^{10})/(10^{2})=1.4*10^{8}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 10**^{6}years, then the proportion of the lifetime of the universe that we've existed is: (10^{6})/(1.4*10^{10})=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*10^{7}seconds in a year, so (7.1*10^{-5})*(3.2*10^{7})=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.

- How many galaxies ?
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*10**^{5}light-years, and that the distance between the Milky Way and Andromeda is 2.2*10^{6}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*10^{5}light-years. Dividing 2.2*10^{6}by 2*10^{5}, 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*10**^{5}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*10^{5}= 1.4*10^{6}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*10^{15}meters in a light-year, which means there are 9.5*10^{12}kilometers in a light-year. So if Proxima Centauri is 4.2 light-years away, that's about 4*10^{13}km. That distance divided by the diameter of the Sun, gives us our answer: (4*10^{13})/(1.4*10^{6})=2.9*10^{7}. 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).**