name: setiawan office: s13-02-09, phys. departm. phone: 6516-2988 email:...
TRANSCRIPT
Name: SetiawanOffice: S13-02-09, Phys. Departm.Phone: 6516-2988 Email: [email protected]: Wed. 10.00-12.00
Tutorial 1
Where are We??????
GEM2507
Physical Questions in Everyday Life
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 m1 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
MethodDistance
(pc) ObjectTime(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
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
24 d
LLA
2718
12
2 105.63 10 182.62
4 4 (5 10 )A
L xd x m pc
L x lightyears4.596
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!)
N
WWy
x
y
f
Constant speed = no acceleration
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 lawV = H0 * d
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?
Question 6
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/l
ightspeed_evidence.html