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Homework 1
Author: Student, Amazing A.
September 8, 2016
Chapter 1
Problem 23
Which of the following correctly lists our ‘cosmic address’ from small to large?
(a) Earth, solar system, Milky Way Galaxy, Local Group, Local Supercluster, universe.
Problem 25
The star Betelgeuse is about 600 light years away. If it explodes tonight, ...
(c) we won’t know about it until about 600 years from now.
With light as our only messenger for the explosion of Betelgeuse, we must
wait the 600 years for the light of the supernova to reach us.
Problem 46
Driving Trips: Imagine that you could drive your car at a constant speed of
100 km/hr, even across oceans and space. How long would it take to drive (a)
around Earth’s Equator? (b) from the Sun to Earth? (c) from the Sun to Pluto?
(d) to Alpha Centauri?
This problem requires the adoption of a distance measure, but these can
vary by method of measurement. NB: Try to use scientific notation whenever
possible in astronomy!
(a) For this exercise, we will set the circumference equal to d = 40, 000 km.
From
T ravel T ime =
40, 000 km
Distance
=
= 400 hrs ≈ 16.67 days
V elocity
100 km/hr
(1)
(b) To keep things simple, we will use a distance from the Sun to the Earth of
roughly 1 AU≈ 1.5 × 108 km. Then
T ravel T ime =
1.5 × 108 km
= 1.5 × 106 hrs ≈ 6.25 × 104 days = 171 years
1.0 × 102 km/hr
(2)
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(c) The distance to Pluto is not as straightforward, as it is on an eccentric orbit.
The semi-major axis is approximately d = 40 AU, which is d ≈ 6.0 × 109 km.
6.0 × 109 km
= 6.0×107 hrs ≈ 2.5×106 days = 6.85×103 years
1.0 × 102 km/hr
(3)
It would take nearly 7000 years to get to Pluto!
(d) To get to Alpha Centauri, which is 4.4 light years away, a traveler would
need to pack a lot of provisions. In kilometers, 4.4 lyrs=4.18 × 1013 km, as one
light year is d = (3.0 × 105 km/s) ∗ (60 sec/min) ∗ (60 min/hr) ∗ (24 hrs/day) ∗
(365 days/yr) ∼ 9.5 × 1012 km (the speed of light ≈ (3.0 × 105 km/s)).
T ravel T ime =
4.18 × 1013 km
= 4.18×1011 hrs ≈ 1.74×1010 days = 4.77×107 years
1.0 × 102 km/hr
(4)
You would arrive at the nearest start system in roughly 48 million years.
T ravel T ime =
Extra Credit: Problem 47
Faster Trip: Suppose you wanted to reach A. Centauri in 100 years. (a) How
fast would you have to go in km/hr? (b) How many times faster is the speed
you found in part (a) from the speeds of our fastest current spacecraft (50,000
km/hr)?
(a) We will use our distance from the previous problem (d = 4.18 × 1013
km). Converting our travel time to hours, t = (100 yrs) ∗ (365 days/yr) ∗
(24 hrs/day) = 8.76 × 105 hrs and so our speed will need to be
Distance
4.18 × 1013 km
=
= 4.77 × 107 km/hr
T ime
8.76 × 105 hrs
This is quite speedy.
(b) How much faster than our current speediest probe?
Speed =
Distance
4.77 × 107 km/hr
=
= 954 ≈ 1 × 103
T ime
5.0 × 104 hrs
Nearly 1000 times faster.
(5)
(6)
Chapter 2
Problem 26
This morning I saw the Full Moon setting at about the same time the Sun was
rising.
True. Full moons are setting at roughly 6 a.m. while the sun is rising. See
the Schematic in Figure 2.22.
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Problem 33
A week after the Full Moon, the moon’s phase is ...
(b) 3rd quarter. Roughly once a month, the moon completes its phases. A
week after full (half of the full cycle), would correspond to 3/4 of the cycle.
Problem 42
Assume you live on the moon near the center of the face that looks towards
Earth. (a) Suppose you see a full earth in your sky. What phase of the Moon
would people on Earth see? Explain. (b) Suppose people on Earth see a Full
Moon. What phase would you see for Earth? (c) Suppose people on Earth see a
waxing gibbous moon. What phase would you see for Earth? (d) Suppose people
on Earth are viewing a total lunar eclipse. What would you see from your home
on the moon?
(a) If the Earth is completely sun-lit, that corresponds to a new moon visible
to Earth-dwellers. It is easiest to understand this by drawing a figure similar to
2.22 in the book, using the direction of sunlight as a point of reference.
(b) If Earthlings witness a full moon, that means that they are facing directly
away from the sun in the sky, which would mean a new Earth would be visible
to the Mooninites.
(c) This can be solved a number of ways, but if we recognize the pattern from
the previous examples, we learn that the phases of the moon and Earth are
opposite for viewers of each (a new moon for Earthlings will be a Full Earth
for Mooninites). Thus, a waxing gibbous moon will correspond to a waning
crescent Earth as seen by the Mooninites.
(d) A total lunar eclipse for Earthlings only occurs during a full moon. A
Mooninite would therefore see a new Earth during an Earthling total lunar
eclipse (or a total solar eclipse for a Mooninite).
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