Download Astronomy 202 Solutions to Homework #1 Problem 1: Scientific

Astronomy 202
Solutions to Homework #1
Problem 1: Scientific Notation Review and Estimation
b. The mass of the Sun is 2 * 1033 grams. There are about 100 billion (1011) stars in the
Milky Way Galaxy, and about 100 billion galaxies in the Universe. Assuming the Sun is
a typical star, and the Milky Way is a typical galaxy, approximately what is the total
mass of all the stars in the universe?
Total mass = 2*1033 * 1011 * 1011 ≈ 2 * 1055 grams
Most of you got this one.
c. The mass of one hydrogen atom is 1.67 * 10-24 g. Approximately how many atoms are
there in all the stars in the universe?
Total atoms = 2 * 1055 g / (1.67 * 10-24 g/atom)≈ 1079 atoms
Some of you multiplied rather than divided here. You cand do a quick “sanity check” to
make sure you did this the right way around: you know from 1b how many grams there
are in the universe, and from this problem that an atom has a mass of much, much less
than 1 gram; so you'd better have more atoms than grams in the universe!
d. The closest star to the Sun is Alpha Centauri, about 4.12 * 1013 km away. The Apollo
11 moon rocket travelled at about 11 km/sec. Approximately how long (in seconds)
would it take the rocket to reach Alpha Centauri?
Time = 4.12 1013 km / (11 km/sec) ≈ 4 * 1012 seconds (about 10 5 years)
4
Some of you had problems with this one. Don't let the big numbers get in the way of the
setup; if you don't understand how to set it up, put in numbers you can understand.
For instance: Phoenix is about 200 km away. Your car can go about 100 km/hr. About
how long will it take you to get to Phoenix? All of you should be able to do that
problem; the one about the moon rocket and Alpha Centauri isn't any harder, it just uses
bigger numbers!
Problem 2: Inverse Square Law Reasoning
a. Imagine you have a 60-watt reading lamp on your desk, 50 cm from your book. The
convenience store across the street has a 10,000 watt floodlight in the parking lot 100
meters away. Could you turn off your lamp and read by the light coming in the window?
Why or why not?
Fluxlamp = 60 watts/(4*(50 cm)2) = 1.9 * 10-3 watts/cm2
Fluxfloodlight = 104 watts/(4104 cm)2) = 7.9 * 10-6 watts/cm2
You cannot turn off your lamp and still be able to read; the light from the floodlight is
over 1000 times fainter on your desk.
Note that you could do this problem simply by noting that (100/0.5)2 is more than
10,000/60.
Some of you failed to convert the distance to the floodlight to centimeters.
b. How many 60-watt light bulbs would you need in your desk lamp for the light from the
lamp to be as bright as sunlight (that is, for the flux on your book to be the same)?
Flux = (Nbulbs * 60 watts)/(4*(50 cm)2) = 3.9 * 1026 watts / (4* (1.5 * 1013 cm)2 )
The factors of 4 cancel, leaving us with
Nbulbs = (1.73 watts/cm2) / (0.024 watts/cm2)
Nbulbs = 72 bulbs
Problem 3: Flux Calculations
3. What is the flux from the Sun at:
a. Venus (0.72 AU from the Sun)
Flux = 3.9 * 1033 ergs/sec / (4 * (0.72 * 1.5 * 1013 cm)2)
Flux = 2.7 * 106 ergs/sec/cm2 = 0.27 watts/cm2 = 6.07 * 1025 W/AU2
b. Mars (1.52 AU from the Sun)
Flux = 3.9 * 1033 ergs/sec / (4 * (1.52 * 1.5 * 1013 cm)2)
Flux = 5.99 * 105 ergs/sec/cm2 = 5.99 * 10-2 watts/cm2 = 1.35 * 1025 W/AU2
c. Europa (a moon of Jupiter, 5.2 AU from the Sun)
Flux = 3.9 * 1033 ergs/sec / (4 * (5.2 * 1.5 * 1013 cm)2)
Flux = 5.2 * 104 ergs/sec/cm2 = 5.2 * 10-3 watts/cm2 = 1.17 * 1024 W/AU2
Lots of people had trouble with the units here; usually they would give an answer in
W/cm or W/AU. That's wrong! If you ever have problems, just write down the units
and carry them through the problem; the cm (or m, or AU) get squared along with the
numbers. This is important; flux is light hitting an area. If that's not what your units are,
it isn't flux!