Astronomy Assignment #5 Solutions
Unit Number
8
10
11
Questions for Review
5, 6, 8
7, 8
1, 2, 3, 4, 6, 7, 8
Problems
13
10, 11, 12, 13
Unit 8 Problems
13. You see a half-full moon visible in the early morning sky. How long will you have to wait for
the next full moon?
This problem depends on your knowing which phase the moon is in and the time between the
principle moon phases (New, 1st Quarter, Full, and 3rd Quarter).
Let’s determine the phase of the Moon first. A “half-full moon” must mean the Moon is at either
1st quarter or 3rd quarter. We need to use the other given piece of information that the Moon is
seen in the early morning sky to determine which of the two quarters the Moon is in. If the
Moon is visible in the early morning sky, then the Moon must have risen before the Sun. This
means that the Moon is west of the Sun. If you can visualize the whole sky map and imagine
that the Sun is somewhere on the ecliptic and the Moon must be west of the Sun – that is to the
right of the Sun – then identifying the phase of the Moon is simple. The quarter moon phase that
applies when the Moon is to the right of the Sun, or west of the Sun, must be the 3rd Quarter
since the Moon is moving faster eastward than the Sun and will soon overtake the Sun implying
that the New Moon is the next principle phase. So the Moon is at 3rd Quarter.
Since each of the principle phases is separated by about 1 week (really 29½ days divided by 4 =
7.4 days between principle phases), the next full moon will be about three weeks away. To be
more specific, it will be 1 week from 3rd Quarter to New Moon, a second week from New Moon
to 1st Quarter and a third week to the next Full Moon. A more precise answer would be 22.1
days, but 3 weeks is precise enough.
Unit 10 Problems
10. On average, Mercury is 0.387 times Earth’s distance from the Sun, and Pluto is 39.5 times
Earth’s distance from the Sun. What is the Sun’s angular diameter as seen from Mercury? from
Pluto?
Before beginning the “hard-core” mathematics of this problem, let’s remind ourselves what the
answers to the question ought to be. Since Mercury is 0.387 times closer to the Sun than the
1
 2.65 times bigger than it does from Earth. You may
Earth, the Sun ought to look
0.378
remember or have read that the angular size of the Sun (and Moon) is about 0.5. This implies
that from Mercury the Sun should have an angular diameter of around 1.32. Similarly, since
1
 0.0253
39.5
times smaller than it does from Earth. This implies that from Pluto the Sun should have an
angular diameter of around 0.0127. Let’s do the math and see if our inferences are correct.
Pluto is 39.5 times farther from the Sun than the Earth, the Sun ought to look
A
L
This is an angular size problem, so we will use the angular size relation
. Solve for


2D
360
A, the angular size.
A
L


2D
360
360   L 360   2  696,000 km 
A 

 1.374 
8
2D
2 0.387  1.5  10 km

A

As seen from Mercury, the Sun has an angular diameter of 1.374. This is equivalent to about 82
arc-minutes.
In like fashion, solving for the Sun’s angular size from Pluto
A
L


2D
360
360   L 360   2  696,000 km 
A 

 0.0135 
2D
2 39.5  1.5  10 8 km 
A
As seen from Pluto, the Sun has an angular diameter of 0.0135. This is equivalent to about 0.8
arc-minutes or 48 arc-seconds.
L
12. A cloud directly above you is about 10 across. From the weather
report you know that the cloud is 21,200 m high. How wide is the cloud?
This is an angular size problem, so we will use the angular size relation
A
L

. Solve for L, the true width.

2D
360
A
L


2D
360

A
10 
L
 2D 
 2 2,200m   384m
360 
360 
The cloud is 384 meters wide.
D
A
13. The great galaxy in Andromeda has an angular diameter along its
long axis of about 5. Its distance is 2.2 million ly. What is its linear
diameter?
This is an angular size problem, so we will use the angular size relation
A
L
. Solve for L, the true width.


2D
360
A
L


2D
360

A
5
L

2

D

 2 2.2  10 6 ly  192,000ly
360 
360 

L
D
A

The galaxy is 192,000 light years wide.
5. Suppose you see a person whom you know is 1.8 m tall, standing at a distance of where he
appears to have an angular height of 2. How far away is he?
This is an angular size problem, so we will use the angular size relation
D, the distance.
A
L


2D
360
360  L 360  1.8m
D  
  
 51.6m
2
2
A
2
The person is 51.6 meters away.
A
L

. Solve for

2D
360