Download here

GEM2507
Physical Questions in Everyday
Life
Tutorial 1
Where are We??????
Name: Setiawan
Office: S13-02-09, Phys. Departm.
Phone: 6516-2988
Email: [email protected]
Consultation: Wed. 10.00-12.00
1. What is a distance of 5 kilo parsec
in lightyears?
A minute of arc, arcminute, (MOA) = 1/60 of 1o
 An arcsecond = 1/60 * arcminute
 The parsec ("parallax of one arcsecond", symbol
pc) = 30 trillion km or 3.3 light years.


Defined as the distance from the sun at which two
imaginary lines—one projected from the Earth, and
one projected from the sun at a right angle to a third
line connecting the Earth and the sun—intersect in
space at an angle of 1 arcsecond
Calculating the value of a parsec
A
B
Calculating the value of a parsec



S = Sun
E = Earth
D = point in space 1 parsec from the Sun
1 AU = 149,597,870,700 m
1 parsec ≈ 3.085 678×1016 m ≈ 3.261 564 light-years.
Answer for question 1

5 kiloparsec = 5 * 3261.636
= 16308.181 light years
Distances less than a parsec

Distances measured in fractions of a
parsec usually involve objects within a
single star system. So, for example:

one astronomical unit (AU) - the distance
from the Sun to the Earth - is 4.85×10−6 pc.
Parsecs and kiloparsecs

Distances measured in parsecs include distances
between nearby stars, such as those in the same
spiral arm or globular cluster. For example



Second Nearest star to the Earth (Proxima Centauri)
1.29 parsecs away.
The center of the Milky Way is about 8 kpc from the
Earth, and the Milky Way is about 30 kpc across.
Andromeda Galaxy (M31), the most distant object
visible to the naked eye, is a little under 800 kpc away
from the Earth.
Megaparsecs and gigaparsecs

Astronomers typically measure the
distances between neighboring galaxies
and galaxy clusters in megaparsecs. For
example:

The Andromeda Galaxy is 0.77 Mpc away
from the Earth.
2. Would it be a good idea to measure the distance to the
closest galaxy outside our own by triangulation?
No, as the parallax angle would be extremely small if it was used to
measure the distance to the closest galaxy from our own galaxy.
Indeed, the maximum distance measurable by triangulation is about
50 parsecs, only enough to explore our sun’s immediate
neighborhood and far short of even being sufficient to map our own
galaxy, the Milky Way
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
3. 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 lightyears?

From Figure 1.9 in the lecture note, we get the
absolute luminosity (L) for period 140 hrs = 5.04
x 105 s,
L = Ls = 5 * 4 x 1026 W
= 2 x 1027 W

The apparent luminosity is
L
LA 
4d 2
L
2 x1027
18
d


5.63
x
10
m  182.62 pc  596.4lightyears
12
4 LA
4 (5 x10 )
4. Joan (weighing 70kg) is running at a constant speed down a slope
with an angle of 6 degrees with a speed of 10km/h. What is her
acceleration? What is her momentum? (Use standard units!)
y
N

f
W
x
Constant speed = no acceleration
 Wy
v = 10 km/h = 10/3.6 = 2.78 m/s
p = m v = 70 x 2.78 = 194.6 kg m/s
5. Give the value of the Hubble constant in terms of
seconds only (i.e. no other units)
H0 = 7.2 x 104 m/s / Mpc

= 7.2 x 104 m/s / 3.0857x 1022m

= 2.33 x 10-18 /s


Hubbles law
V = H0 * d
Question 6
6. Galileo was one of the first to systematically try to
measure the speed of light. His attempt involved two
observers positioned in two towers that were about 10km
apart. The idea was that the first observer opens a shutter in
a lantern and then as soon as the second observer sees the
light from the first lantern, opens his shutter. Galileo would
then measure the time it takes from opening the first shutter
to seeing the light from the second lantern arrive at the first
lantern. Unfortunately for Galileo, this experiment turned out
to be inconclusive. Why is that so? Would it have been
sensible to choose towers that are further apart?
Answer for question 6

It is not possible because the velocity of light is
tremendously large. As a matter of fact, light
only takes 0.000005 seconds to travel one mile
In order to measure the speed of light. It is
extremely important to measure the time
accurately which needs a sufficiently long
distance and extremely negligible reaction time
of the observer
 How to measure speed of light?
 http://www.colorado.edu/physics/2000/waves
_particles/lightspeed_evidence.html