# Astronomy Assignment #1

```Astronomy Assignment #9: The Sun
Text Problems:
Unit Number
51
52
Questions for Review
4, 6 (refer to the results of
the Two-Layer Model), 9
5, 6
Problems
10, 11
11, 12, 14
Unit 51
1. The Sun’s angular size is 0.5, and its distance is 1.5 x 108 km. Use this information to calculate
the Sun’s diameter.
A
L
This is an angular size problem. Use the angular size relation
and solve for the true


2D
360
diameter of the Sun L.
A
L


2D
360
A
0.5
8
L  2D 

2

1
.
5

10
km

 1.3110 6 km
360 
360 

 
The diameter of the Sun is about 1.31 million kilometers.
2.
Calculate the Sun’s average density in kg per liter. The Sun’s mass is approximately 2x1030 kg.
Its radius is approximately 7x108 m. Show that its average density is approximately 1.4 kg/L.
This is a density problem so we use the formula for density of a sphere  

M
2 10 kg 

4
 7 10 m 
3
30
M
.
4 3
R
3
kg
kg  1 m 3 
kg
 1,392 3  1,392 3  
  1.392
L
m
m 1,000 L 
3
4 3
8
R
3
The average density of the Sun is about 1.4 kg/L. Note: The Sun has the same density as the typical
Jovian planet indicating that they are made of the same “stuff”. The Sun and the Jovian planets are
chemically identical (almost). Jupiter, if it were about 70 times larger in mass, could have been a star.
Unit 52
11. Worldwide, humans use about 4x1020 Joules of energy each year. A typical power station
generates about 109 watts of power while the Sun’s luminosity is about 4x1026 watts. (Recall that 1
watt = 1 Joules/second.)
a. How many power stations are needed to produce power equivalent to the Sun’s?
Answer this question with a ratio.
4 10 26 W
 4 1017 Power Plant
W
110 9
Power Plant
About 4x1017 typical power plants are needed to produce the power the Sun produces. Note:
The current population of humans on the Earth is nearing 10 billion. If every single one of
the 10 billion humans on the Earth owned 10 million typical power plants each , then that
would be the equivalent of the Sun’s energy output. As you can see, it is kind of hard to
make meaningful analogies to the Sun’s power output.
b. How many typical power plants are needed to supply current human energy usage?
There are 3.15x107 seconds in a year. Thus, a single typical power plant will produce 109 J/s
x (3.15x107 s) = 3x1016 joules in a year. Since humans use about 2x1020 Joules of energy
each year, the number of power plants required to produce this energy is
4  1020 J
 1.33  104 Power Plants .
J
3  1016
Power Plant
Thus, about 13,300 power plants are needed to provide yearly energy usage for humans.
c. If we could store all of the energy that the Sun produces in 1 second, how long would it last
humans at the current energy use rates?
The Sun in 1 second produces about 4x1026 J. Since humans use 4x1020 J of energy each
year, the energy produced by the Sun in 1 second would last the human race about 1 million
years. Wow! Note: The human race has walked the Earth for about 4 million years. If we
could have captured all the energy that the Sun produced in just 4 seconds, we could have
more than powered all of human history.
12. Using E=mc2, calculate how much mass is turned into energy during a year by a typical power
station. How much mass is turned into energy each year by worldwide energy usage? (Use the
data given in problem 2.)
There are 3.15x107 seconds in a year. Thus, a single typical power plant will produce 109 J/s x
(3.15x107 s) = 3.15 x 1016 joules in a year
E  mc 2
E
3.15  1016 J
m  2 
 0.35 kg
2
c

8 m
 3  10

s

A typical power plant converts about 1/3 of a kilogram of mass into energy in the course of one year.
According to problem #2, worldwide, humans use about 4x1020 Joules of energy each year.
E  mc 2
m 
E
4  10 20 J

 4,444 kg
2
c2 
8 m
 3  10

s

2
The 2x1020 J of energy used by humans each year is equivalent to the energy released by the
destruction of 4,444 kg of mass.
14. One “kiloton” of explosive energy is 4.2x1012 Joules. Using Einstein’s formula, calculate the
mass that is converted into energy in a 1-megaton H-bomb explosion.
One megaton of energy is 1,000 kilotons. So the energy in one megaton is 4.2x1015 Joules.
E  mc 2
E
4.2 1015 J

 .047 kg  47 grams
2
c2 
m

8
 3 10

s

The mass converted into energy in a 1-megaton H-bomb explosion is equivalent to 47 grams. Note:
This mass of 47 grams is equivalent to the mass of four half-dollar (50 cent) coins. The atomic
bombs that devastated the Japanese cities of Hiroshima and Nagasaki were
equivalent to about 15 kilotons of energy and would have required that only 0.75
grams of mass be converted into energy – that is equivalent to the mass of about 1/3
of a US Penny. When you know that the Sun converts 4.3 million metric tons of
mass into energy each second, then you can better imagine the barely –imaginable
amount of energy the Sun produces.
m 
3
```