Astronomy 100. H. Tahsiri
Chapter 0 . Preview
Answers to Selected Problems
1. RSun = 7x105 km. REarth = 6.4x103 km. We can determine how much bigger the Sun
is than the earth by dividing the Earth's radius into the Sun's. Therefore RSun/ REarth=
7x105 km/ 6.4x103 km = 7/6.4 x(105/103) = 1.09x(105-3) = 1.09x(102) = 109.
2. The average distance between the Earth and the Sun is called the astronomical unit. 1
AU = 1.5x108 km. The distance traveled by a light ray in 1 year is called a light year (ly).
1 ly = 1x1013 km. We can find the number of AU in a light year by dividing the number
of km in a light year (1013) by number of km in an AU (1.5x108). Thus,
Number of AU in a light year = 1013/(1.5x108) = 1013-8/1.5 = 105/1.5 = 6.7x104 =
about 68,000.
6. (4x108)3/(5x10-6)2 = 43x(108)3/(52x(10-6)2)=64x1024/(25x10-12)=2.56x1036.
7. t=D/v. For light, v=c, so t=D/c. Pluto is about 40 AU from the Sun and 1 AU
=1.5x108 km. Thus, Pluto is 40x1.5x108 = 6x109 km from the Sun. The speed of light is
3x105 km/s. Thus, t = 6x109/3x105 = 2x104 seconds. Divide by 60 to get the number of
minutes = about 330 minutes.
9. (3x104)2/(4x10-6)1/2 = 32x104x2/41/2x10-6x1/2 = 9x108/2x10-3 = 4.5x108-(-3) = 4.5x1011.
10. Given an object's speed and distance from another, we can find how long it took
them to move apart by using the relation D=vt, where D is the distance, v is the speed,
and t is the time. Solving for t gives, t=D/v. Now inserting values for v and D, we find t
= 3x108 ly/6,000 km-seconds-1. Notice that are units are mixed here. We have ly on top
and km below. Thus, we need to express ly in km. From the appendix we can find that 1
ly = 9.5x1012 km. Thus, t= 3x108 lyx9.5x1012 km/ly/6,000 km-seconds-1 =
(3x9.5/6)x108+12-3 seconds = 4.8x1017 seconds. To express this in years, we need to divide
by the number of seconds in a year = about 3.2x107. Thus, t = 4.8x1017 seconds/3.2x107
seconds/yr = 1.5x1010 yrs. This is one way that astronomers deduce the age of the
Universe.
Chapter 1
Answers to Selected Problems
3. If P = 64 years, you can use Kepler's 3rd Law to estimate its distance from the Sun.
P2 = a3, where P=period in years and a=average orbital radius in AU. We can find a by
taking the cube root of both sides to get P2/3 = a, or a = P2/3 = (64)2/3 = 16 AU
4. Aliens have a 1Mo star and are 4 AU from it. According to Kepler's 3rd law
P2 = a3 P=square root of (43) = (43)1/2 = 8 years.
5. The Andromeda galaxy has an angular diameter of 5 degrees at a distance of 2.2x106
ly. We can find its true size by using the angular diameter formula
L = 2DA/360, where D = distance, A = angle subtended, and L = linear diameter.
Thus, L = 2 x 2.2x106 ly x 5 degrees / 360 degrees = 1.92x105 ly
6. A shell of gas has an angular diameter A = 0.1 degrees and a linear diameter L = 1 ly.
We can find its distance D by reversing the procedure in the previous problem. We begin
by writing out the angular diameter formula L = 2DA/360. We next solve it for D by
dividing both sides by 2A and multiplying both sides by 360 to get
D = 360L/(2A). Next inserting values for L and D we find
D = 360 x 1 ly/( 2 x 0.1) = 573 ly.
8. This is an application of Kepler's third law, P2 = a3, where a is in AU and P is in years.
If P = 125 yrs, then a3 = 1252. Solving for a, we take the cube root of both sides to get
a = (1252)1/3, where we have used the fact that the cube root of a number is the number to
the 1/3 power. Using your pocket calculator, you will find that a = 25 AU. If the planet's
orbit is circular, then that is also the planet's orbital radius.
9. This problem is another application of Kepler's third law, P2 = a3, where a is in AU and
P is in years. In this case, we are given a, and are asked to find P. Thus P2 = a3 = 163.
Solving for P by taking the square root and recalling that the square root is the number to
the 1/2 power, we find that P = (163)1/2 = 64 yrs. (Note: in solving this problem, you can
simplify the math by reversing the order of the power and the square root. That is take
the square root of 16 (=4) and then cube it to get 64.
Chapter 2
Answers to Selected Problems
1. Since
F(Moon-Earth) =
G MMoon MEarth
,
R2Moon-Earth
the gravitational force exerted by the Moon on the Earth is the same as that exerted by the
Earth on the Moon.
2. The escape velocity for the Earth is given as follows:
Vesc = (2GM/R)1/2, where G=7x10-11 m3/kg/s2, M(Earth)=6x1024 kg and
R(Earth)=6x106m. Inserting these values in the equation, we find that
Vesc = (2GM/R)1/2 = (2 x 7x10-11 x 6x1024 / 6x106)1/2 =1.18x104 m/s
Or Vesc = approx. 12 km/s
3. The escape velocity for the Sun may be found using the formula for the escape
velocity Vesc = (2GM/R)1/2 with the Sun's mass and radius inserted. This gives
Vesc = (2 x 7x10-11 x 2x1030/ 7x108)1/2 = 6.3x105 m/s = approx. 630 km/s
4. To compare the escape velocity of Mars and Saturn, we write out the expression for the
escape velocity for each planet. Thus, Vesc(Mars) = [2GM(Mars)/R(Mars)]1/2,
Vesc(Saturn) = [2GM(Saturn)/R(Saturn)]1/2.Next, we divide the expression for Saturn's
escape velocity by the expression for Mars' escape velocity to get,
Vesc(Saturn)/Vesc(Mars) = [M(Saturn)/R(Saturn)]1/2/[M(Mars)/R(Mars)]1/2,
We then evaluate each expression using the appropriate mass and radius. Thus,
M(Mars) = 0.1 M(Earth), M(Saturn) = 95 M(Earth), R(Mars) = 0.5 R(Earth) and
R(Saturn) = 9.4 R(Earth). Vesc(Saturn)/Vesc(Mars) = (95 x 0.5/9.4 x 0.1)1/2 = 7.1.
5. To compare the escape velocity of the Earth and Moon, we use the same approach as
in the previous problem. Since M(Earth)/M(Moon) = 81 and R(Earth)/R(Moon) = 3.8,
Vesc(Earth)/Vesc(Moon) = [M(Earth)/M(Moon) x R(Moon)/R(Earth)]1/2
= (81/3.8)1/2 = 4.6
6. STUDENTS NEED TO DO THIS.
The weight of a body is just the gravitational force exerted on it. You can therefore find
your weight on the Earth or on the Moon from their surface gravities. The Earth's surface
gravity is 9.8 m/s2 while the Moon's is 1.7 m/s2. The Moon's is thus 9.8/1.7 = about 5.8
times smaller than the Earth's. Thus your weight on the Moon is your weight on the
Earth divided by 5.8.
8. The Milky Way Galaxy has a mass of about 1011 M(Sun), and the Sun is 2.6x1020 m
from the center of the galaxy.
v = (GM/R)1/2 = (6.7Ex10-11 x 2x1030 x 1011/2.6x1020) v = 2.3x105 km/s
P = 2R/v = 2  x 2.6x1020/2.3x105 = 7.1x1015 s. Given that 1 year is about 3.2x107
sec, the period in years is 7.1x1015 s/3.2x107, or the Sun makes one orbit of the galaxy
in 2.2 x108 years.