Tutorial 1: Universe and Astronomical Tutorial 1: Universe and Astronomical Measures, Concept of Energy and Atomic Measures, Concept of Energy and Atomic Structure.Structure.

About your tutorName: CHING Chee Leong

: S12-03-17, Phys. Depart.

:65162534 : phyccl@nus.edu.sg

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 tutorialTutorial 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.

1 AU

An angle of exactly 1 arc sec

1 pc

1 pcSun

Earth

pc stands for parallax seconds

Recall on some trigonometry

Note: For small θ (in radians),

S

r

rsr

s

r

sSin

radianradian oo

1801180

radarc

radarc

o

3600

1

180sec1

180sec6060

sec6060min601

a) Express the distance from the Earth to the Sun in parsecs and in light years.

since s = rθ

pcpcAU

arcpararcrAU

arc

AUsr

61085.43600

1

1801

sec1sec1sec11

sec1

1

Given that 1AU= 92.9x106 miles and1 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

LYAU

miles

miles

LY

AU

5

512

6

1058.11

1058.11087.5

109.92

1

1

1b) To express 1ly and 1pc in miles,

1 light-years (LY)=186,000miles/sec x (365 days x24hrs x60mins x60seconds) = 5.87x1012 miles

milesmiles

par

arc

AUpar

136

1091.1

36001

180

109.92sec1

sec1

1sec1

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

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?

Stellar Parallax

January July

Baseline = 1 AU

θMeasure the angular shift θ of the foreground star with respect to background stars, and compute P

P = 1/2 θ

P

parallax angle

D = …?

Once we know the value of parallax angle P, we can compute D easily

which implies

since sin P ≈ P, if P is very small

Note: 1 AU = 1 Astronomy-cal Unit, which is the sun-earth distance ≈150 million kilometers

Stellar Parallax (cont’d)b) What are the advantages and (c) disadvantages of this technique?

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 >_<

It’s arguably the most straightforward and simplest measurement technique… well, except that it only works for relatively nearby stars >_<

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,

'

; zzcvrconvention:

recedingissourcev

gapproachinissourcev

r

r

0

0

nm

nm

nm

skm

skmc

vr

3.655'

656

656'

/000,300

/300

'

blue shifted

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

24 d

LLA

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?

yearslightpckilo

mL

Ld

A

828754.2

1084.7)105(4

1086.3

419

12

29

Method used to measure distances in universe

MethodDistance

(pc) Object Time(yr)

Hubble's Law 109-1010 Quasars 1011

Apparent luminosity - Galaxies 108 Virgo Cluster 108

Apparent luminosity - Super Giants 106-107

Andromeda Cluster 107

Cepheid Variables - Type II 105

Andromeda Cluster 105

Cepheid Variables - Type I 101-104

Center of Milky Way 103

Parallax 100

Alpha Centauri 101

Examples:Group action: Addition(YES). We can check 4 four criteria are satisfied. (i)2+2=4 (still even number)(ii)(2+4)+10=2+(4+10)(iii)Identity= 0 such that 2+0=0+2=2(iv)Inverse 2-2=0

Group action: Multiplication (NOT a group). (i)2 x 2=4 (still even number)(ii)(2x4)x10=2x(4x10)(iii)Identity= 1 such that 2x1=1x2=2 (but the identity is not in the set (even number)(iv)Inverse 2 x 1/2=1 (1/2 is not even 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.

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?

A big question (worth billion dollar)

In order to form a group, four conditions must behold: (i) Closure, (ii) Associativity, (iii) Identity exist and (IV) Inverse.

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.

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,

JUKE 981

J

mmskgmghU

981

)100()81.9(1 2

From kinematics, sms

mstatuv 52.4

81.9

3.442

1

1

2

3.44

9812

1

msv

JmvKEFrom energy conservation, correct result

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).

Frednosoclever’s calculation:

112.2252.4

100' ms

s

mv false result

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?

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.

kgm

JmmvKE

15

23232

102

10)1010(2

1

2

1

b) One kilogram of the explosive TNT releases about 4.184x106J of energy. Estimate the TNT equivalent of the energy released in part (a) above.

TNTofTons

TNTofkgJ

J

13

166

23

1039.2

1039.210184.4

10

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.

bombsatomicJ

J 966

23

1059.110184.41015

10

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.

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)

Some terminologies in chemistry:

Cl

ratom; cl

ClCl

rcov; cl2

Covalent radius: Corresponds to half of the distance between two identical atomic nuclei bound by covalent bond.

ClCl

rcov; cl2

From experiment rcov; cl2≈0.099

nmLet us approximate size of Cl2 := 2 x rcov; cl2 = 0.198 nm

Cl

ratom; cl

From experiment ratom;

cl≈0.100 nm

Cl

-

Ionic radius of Chlorine, rion;cl= 0.181 nm

Negative ions are larger than their atoms! How about positive ion? Smaller.

Cl

From nucleus physics,rnucleus= ro A1/3 where A= mass number; Acl= 35.5 andro= 1.2 x 10-15 m (empirical constant)

Hence, rnucleus; cl = 1.2 x 10-15 x (35.5)1/3

= 3.944 fm

Na Na

+

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.

sdaysLifeHalf 71035.2272

Life time (or mean life-time) is defined as,

s

TTmean

7

72/1

1039.3

693.0

1035.2

2ln

1

half life is not mean life time!

decay constant

181095.21 s

Tmean

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?

sdecays

Cidt

tdNA

/1040.7

00.2)(

4

Nuclei

s

sdttdNtN

12

18

14

1051.2

1095.2

1040.7/)()(