Astronomy Assignment #3 Solutions
Unit Number
10
11
Questions for Review
5, 6
1, 2, 3, 4, 6
Problems
10, 11, 12, 13
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
Pluto is 39.5 times farther from the Sun than the Earth, the Sun ought to look
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.
This is an angular size problem, so we will use the angular size relation
A
L
. 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 
8
2D
2 39.5  1.5  10 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
D
A
L


2D
360

A
10 
L

2

D

 2 2,200m   384m
360 
360 
A
The cloud is 384 meters wide.
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?
A
L

This is an angular size problem, so we will use the angular size relation
. Solve for

2D
360
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.