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Tutorial 1: Universe and Astronomical Measures,
Concept of Energy and Atomic Structure.
About your tutor
 Name: CHING Chee Leong

: S12-03-17, Phys. Depart.

:65162534
: [email protected]
 Research interest: Gravitational physics and cosmology under Loop
Gravity approach. Foundations of quantum mechanics, field theory and
symmetries.
 Consultation:
Tuesday: 4-5pm; Wednesday: 2-3pm; Friday:3-4pm
or email me for other convenient timings.
About tutorial
Tutorial timing: Each tutorial class lasts for roughly 50-55 minutes. After that
students are allowed to leave the tutorial room. Please kindly sign the attendance
list.
During the tutorial:
• Discuss questions posted in tutorial assignment.
• Clear up any doubts. If time is permit, I will conduct some
physical demonstrations in the class to enhance your
understanding.
Question 1: [Astronomical Unit]
Astronomical distances are so large compared to terrestrial ones that much
larger units of length are needed. An Astronomical Unit (AU) is equal to the
average distance from the Earth to the Sun, about 92.9 x 106 miles. A parsec
(pc) is the distance at which 1 AU would subtend an angle of exactly 1 second
of arc (see fig. below). A light year (LY) is the distance that light, traveling
through a vacuum with a speed of 186,000miles/sec (or 300,000km/sec)
would cover in 1.0 year.
An angle of exactly 1 arc sec
Earth
1 pc
1 AU
Sun
1 pc
pc stands for parallax seconds
Recall on some trigonometry
q
r
S
Note: For small θ (in radians),
Sinq 
180   radian 1 
o
o
s
s
 q   s  rq
r
r

180
radian
1o  60 min  60  60 sec
60  60 arc sec 

rad
180

1
1 arc sec 

rad
180 3600
a) Express the distance from the Earth to the Sun in parsecs and in light years.
1 AU
q 1 arc sec
 1 AU  r  1 arc sec  1 par sec  1 arc sec

1
 1AU 

pc  4.85  10 6 pc
180 3600
r
since s = rθ
s

Given that 1AU= 92.9x106 miles and
1 light-years (LY)= speed of light x one year time
=186,000miles/sec x (365 days x24hrs x60mins x60secs)
= 5.87x1012 miles.
Hence, distance from Earth to Sun is
1 AU 92.9 106 miles
5


1
.
58

10
1LY 5.87 1012 miles
 1AU  1.58 10 5 LY
1b) To express 1ly and 1pc in miles,
1 light-years (LY)=186,000miles/sec x (365 days x24hrs x60mins x60seconds)
= 5.87x1012 miles
1 AU
1 par sec 
1 arc sec
92.9 106 miles
 1 par sec 
 1.911013 miles

1

180 3600
2). [Stellar Parallax] There are many ways to measure the distance of a, say,
star from the earth, and one way to do that is through this technique, called
stellar parallax.
a) Explain how this is done?
Parallax is the apparent shift of a foreground object relative to some distant
background as the observer’s point of view changes. Astronomers make use this
trick as a measurement method.
January
July
Stellar Parallax
Measure the angular shift θ θ
of the foreground star with
respect to background
stars, and compute P
P
P=
1/
Once we know the value of parallax
angle P, we can compute D easily
parallax angle
which implies
2θ
D = …?
since sin P ≈ P, if P is very small
Baseline = 1 AU
January
July
Note: 1 AU = 1 Astronomycal Unit, which is the sunearth distance ≈150 million
kilometers
Stellar Parallax (cont’d)
b) What are the advantages and (c) disadvantages of this technique?
It’s arguably the most straightforward and simplest measurement technique…
well, except that it only works for relatively nearby stars >_<
The parallax shift is getting smaller and
smaller as the distance increases (click on
figure beside), making it more and more
difficult to do measurement accurately. Only
few hundreds stars whose distance can be
measured this way
Ancient astronomers believed in geocentric theory for they could not detect the
parallax shift of stars with their naked-eye. Thus, they concluded that either the stars
had to be very farrrrrr away, or earth was not moving…
Well… they chose the wrong conclusion for thousands years apparently >_<
3) [Hubble’s Law and Cepheids]
a) What is Hubble’s Law and how it changes our view on universe? Discuss.
Hubble’s Law: Mathematically expresses the idea that more distant
galaxies move away form us faster; its formula is ν=Ho x d, where ν is the
galaxy’s speed away from us, d is its distance, and Ho is Hubble’s
constant, 72.6 ± 3.1 (km/s)/Mpc.
Our Universe is expanding
and NOT static!
Main Scientific Consequences:
i) There is an origin for our universe (standard
referred as “Big Bang”).
ii) Hubble’s Law deduces that our universe has a
finite age (not equal to finite space!).
3b) The Andromeda (nearest galaxy to our Milky Way) is approaching with a
speed of about 300km/sec. What will be the detected wavelength of the
stationary Hydrogen 656nm line?
Since the source velocity is small compare to velocity of light, vr/c << 1,
vr  zc;
convention:
 '
z

vr  0 source is approaching
vr  0 source is receding
vr  '

c

 300 km / s
 '656 nm


300,000 km / s
656 nm
  '  655.3 nm
blue shifted
3c) . If we have a type-I Cepheid with the period of luminosity being
approximately 140 hours and an apparent luminosity of 5 x 10-12 watt. What is
it's distance from us in light years?
From Figure 1.10 in the text book, we get the absolute luminosity (L) for
period 140 hrs = 5.04 x 105 s,
L ≈1000 Ls =1000 x (3.86 x 1026)W
= 3.86x 1029 W
The apparent luminosity is
L
LA 
4d 2
L
3.86 10 29
19
d


7
.
84

10
m
12
4 LA
4 (5 10 )
 2.54 kilo pc  8287 light  years
Method used to measure distances in universe
Method
Distance
(pc)
Object
Time(yr)
Hubble's Law
109-1010
Quasars
1011
Apparent luminosity - Galaxies
108
Virgo Cluster
108
106-107
Andromeda
Cluster
107
105
Andromeda
Cluster
105
Cepheid Variables - Type I
101-104
Center of Milky
Way
103
Parallax
100
Alpha Centauri
101
Apparent luminosity - Super Giants
Cepheid Variables - Type II
A big question (worth billion dollar)
4) [Mathematic and Group Theory]
Do all the even numbers form a group? Explain your answer. Also, why the
theory of group is so important in nature?
In order to form a group, four conditions must behold: (i) Closure, (ii)
Associativity, (iii) Identity exist and (IV) Inverse.
Examples:
Group action:
Group action: Addition
Multiplication (NOT a group).
(YES). We can check 4 four criteria
(i) 2 x 2=4 (still even number)
are satisfied.
(ii)(2x4)x10=2x(4x10)
(i) 2+2=4 (still even number)
(iii)Identity= 1 such that 2x1=1x2=2 (but
(ii)(2+4)+10=2+(4+10)
the identity is not in the set (even
(iii)Identity= 0 such that
number)
2+0=0+2=2
(iv)Inverse 2 x 1/2=1 (1/2 is not even
(iv)Inverse 2-2=0
number)
Nature prefers to manifest herself with a lot of symmetries. Group theory is the
natural language to study the symmetries of the systems, i.e. rotation, translation
inertial frames idea (in classical physics), or gauge theory, supersymmetries (in
quantum theory), etc. Symmetries naturally lead to the Conservation laws, i.e
energy, momentum and electric charges conservation.
5) [Mechanical energy] We are on top of a 100m high tower with our friend
“Frednotsoclever” and would like to investigate the laws of nature.
a) What is the potential energy of an iron ball with a mass of 1kg exactly at the
height of the tower?
Potential energy has no absolute meaning, only the difference (between point
considered and reference point) in potential energy carries physical
significant.
U  mgh 1kg  (9.81ms 2 )  (100m)
 981J
b) If we drop the ball. What is the kinetic energy of the iron ball an infinitely
short time before hitting the ground?
Assume initially velocity of ball is zero (i.e. at rest), and due to energy
conservation,
KE  U  981J
c) “Frednosoclever” wants to determine the impact velocity by dividing the
height of the building by the time it takes to fall from the top to the bottom.
Of course this is wrong. How much will his calculation differ from the correct
result?
From energy
conservation,
1 2
KE  mv  981J
2
 v  44.3ms 1 correct result
From kinematics,
44.3ms1
v  u  at  t 
 4.52s
2
9.81ms
Frednosoclever’s
100m
v' 
 22.12ms 1 false result
calculation:
4.52s
The discrepancy in impact velocity is around 50%. Fred result is wrong because
his calculation is the average velocity and not the instantaneous one (in fact,
the velocity is varying along the trajectory).
6) [Conservation of Energy]
a) If the asteroid that hit the Yucatan Peninsula was travelling at 10km/s when
it released an energy of 1023J, estimate its mass.
1 2 1
mv  m (10 103 ) 2  1023 J
2
2
 m  2 1015 kg
KE 
b) One kilogram of the explosive TNT releases about 4.184x106J of energy.
Estimate theTNT equivalent of the energy released in part (a) above.
10 23 J
16

2
.
39

10
kg of TNT
6
4.184 10 J
 2.39 1013Tons of TNT
c) The atomic bombs that destroyed the cities of Hiroshima and Nagasaki
during World War II were about the equivalent of 15 kilotons of TNT each.
Estimate the energy released by the asteroid in terms of the equivalent
number of atomic bombs of this size.
1023 J
9

1
.
59

10
atomic bombs
6
6
15 10  4.184 10 J
7) [Atomics Size] Arrange the following objects according to their size in decreasing
order: Chlorine Nucleus, Chlorine Molecule (Cl2), Chlorine Atom (Cl), Chlorine Ions
(Cl - ). Find out the quantitative values for the above objects.
Some terminologies in chemistry:
Atomic radius: Distance from the
atomic nucleus to the outermost
stable electron orbital in an atom that
is at equilibrium. (commonly
known as size of atom)
Covalent radius: Corresponds to half
of the distance between two identical
atomic nuclei bound by covalent bond.
Cl
Cl
ratom; cl
rcov; cl2
From experiment rcov; cl2≈0.099 nm
Cl
Let us approximate size of Cl2 := 2 x rcov; cl2 = 0.198 nm
rcov; cl2
Negative ions are larger than their
atoms! How about positive ion?
Smaller.
-
Cl
Ionic radius of
Chlorine,
rion;cl= 0.181 nm
+
Cl
From experiment ratom; cl≈0.100
nm
Na
Na
ratom; cl
Cl
From nucleus physics,
Hence, r nucleus; cl = 1.2 x 10-15 x (35.5)1/3
rnucleus= ro A1/3
= 3.944 fm
where A= mass number; Acl= 35.5 and
ro= 1.2 x 10-15 m (empirical constant)
8) [Nuclear Radioactivity] The radioactive isotope 57Co decays by electron
capture with a half-life of 272 days.
a) Find the decay constant and the lifetime.
Half Life 272days  2.35107 s
Life time (or mean life-time) is defined as,
Tmean
T1/ 2 2.35 107
 

 ln 2
0.693
1
 3.39 107 s
decay constant

1
Tmean
 2.95 10 8 s 1
half life is not mean life time!
b) If you have a radiation source containing 57Co, with activity 2.00µCi
(1Ci≡3.70x1010 decays/sec), how many radioactive nuclei does it contain?
dN (t )
A
 2.00 Ci
dt
 7.40 104 decays / s
7.40  10 4 s 1
N (t )  


2.95 10 8 s 1
 2.511012 Nuclei
dN (t ) / dt