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Transcript
BEN-GURION UNIVERSITY
AN INTRODUCTION TO ASTRONOMY
Dr. Uri Griv
Department of Physics, Ben-Gurion University
Tel.: 08-6428226 Email: [email protected]
Astronomy Picture of the Day
Discover the cosmos! Each day a different image or photograph of our fascinating universe is featured,
along with a brief explanation written by a professional astronomer.
2001 October 4
M74: The Perfect Spiral
Credit: Gemini Observatory, GMOS Team
Explanation: If not perfect, then this spiral galaxy is at least one of the most photogenic. An island
universe of about 100 billion stars, 30 million light-years away toward the constellation Pisces, NGC
628 or M74 presents a gorgeous face-on view to earthbound astronomers. Classified as an Sc galaxy, the
grand design of M74’s graceful spiral arms traced by bright blue star clusters and dark cosmic dust
lanes, is similar in many respects to our own home galaxy, the Milky Way. Recorded with a 28 million
pixel detector array, this impressive image celebrates first light for the Gemini Multi-Object
Spectrograph (GMOS), a state-of-the-art instrument now operational at the 8-meter Gemini North
telescope. The Gemini North Observatory gazes into the skies above Mauna Kea, Hawaii, USA, while
its twin observatory, Gemini South, is scheduled to begin operations later this year from Cerro Pachón in
central Chile.
Tomorrow’s picture: A Flock of Stars
The Birth of Astronomy
• To most people astronomy means stars
• The Solar system:
(a) all the planetary orbits are nearly circular
and (b) lie within only several degrees of the
plane of ecliptic
2
The Birth of Astronomy
• The radius of the Earth: Erastothenes’
method
• SUGGESTION: The Sun is far enough
away that the rays from the Sun are virtually
parallel when they strike the Earth
3
The Birth of Astronomy
• Aristarchos’ method: the relative distance of
the Moon and the Sun
• SUGGESTION: The angular sizes of the
Moon and the Sun do not change
appreciably with time → they maintain
nearly constant distances from the Earth (in
other words, the orbits are circular)
4
The Birth of Astronomy
• The reason for the seasons
5
The Birth of Astronomy
• The signs of the Zodiac
6
The Birth of Astronomy
• The tides
7
The Tides
8
The Birth of Astronomy – Tides
GMe Mm
GMe Mm
+
FS − FC = −
2
(R − re )
R2
re
≈ −2GMe Mm 3
R
9
Tidal Instability
10
Accretion Disks
• Roche lobe
11
Kepler’s Three Laws
•
a31
P12
= 3
2
P2
a2
12
Kepler’s Laws – 1st Law
13
Kepler’s Laws – 2nd Law
14
Kepler’s Three Laws – 2nd Law
• Consider a planet Mp moving about the sun
in an elliptical orbit
• The gravitational force F ||r
• Then the torque acting on the planet due to
this central force τ = r×F = 0
• Recall that the torque equals the time rate of
change of angular momentum L, or
τ = dL/dt. Therefore, because τ = 0
L = r×p = mr×v = const
(1)
• Since L = const, the planet’s motion at any
instant is restricted to the plane formed by r
and v
15
Kepler’s 2nd Law
• The radius vector r sweeps out an area dA
in a time dt
• Since dr = vdt, we get
1
1
L
dA = |r×dr| = |r×vdt| =
dt
2
2
2Mp
so
L
dA
=
= const → Kepler′ s second law
dt
2Mp
• The law applies to any central force,
whether inverse-square or not
• PROOV: Mp va ra = Mp vp rp , where the
subscripts a and p represent aphelion and
perihelion, respectively
16
Kepler’s Three Laws – 3rd Law
• Consider a planet Mp in a circular orbit
about the sun of mass Msun
• The gravitational force on the planet is
equal to the centripetal force
GMpMsun
Mp v 2
=
2
r
r
(2)
• But the orbital velocity v = 2πr/T , then
(2πr/T )2
GMsun
=
r2
r
• Finally,
2
T =
where Ksun =
= const
2
4π
GMsun
4π 2
GMsun
r3 = Ksunr3 ,
(3)
= 2.97 × 10−19 s2 /m3
17
Kepler’s Three Laws – 3rd Law
• Kepler’s 3rd law is obtained from the
inverse-square law for circular orbits,
F ∝ 1/r2
• Useful planetary data
• Note that the Pluto is no longer a planet
18
Kepler’s Three Laws – 3rd Law
• EXAMPLE: Calculate the mass of the sun
using Tearth = 3.156 × 107 s and Rearth =
1 AU = 1.496 × 1013 cm.
• Using Eq. (3), we get
Msun
3
4π 2 Rearth
33
=
=
1.99
×
10
g
2
GTearth
• Note that the sun is 333 000 times as
massive as the earth !
19
Possible Planetary Orbits
20
Planetary Orbits
• Possible Earth’s orbits, Vrot = 0 − 50 km/s
21
Eccentricity
• Two elliptical and one circular orbits;
the same major axis
22
Gravitational Potential Energy
• The energy associated with the position of a
particle m
• The change in the gravitational potential
energy
Z rf
Z r2
dr
F (r)dr = GMe m
Uf − Ui = −
2
ri r
r1
• Uf − Ui = −GMe m r1i − r1f
• Taking Ui = 0 at ri = ∞, we obtain
GMe m
U (r) = −
negative!
r
23
Gravitational Potential Energy
• The gravitational potential energy
associated with any pair of separated
particles m1 and m2
Gm1 m2
U =−
r
• Binding energy Gm1 m2 /r
• That is, U becomes less negative as r
increases
• Consider Earth–particle system: Total
energy is const
1 2 GMe m
E = mv −
2
r
• Newton’s 2nd law:
GMe m
GMe m
mv 2
1
2
=
→
mv
=
r2
r
2
2r
24
(4)
Conservation of Energy
• Substituting this into Eq. (4),
GMe m
GMe m
em
E = GM
−
=
−
≡
2r
r
2r
U
2
• E is negative !
• EXAMPLE: Calculate the work required to
move an earth satellite m from a circular
orbit of radius 2Re to one of radius 3Re
•
GMe m
Initial energy Ei = − 4Re
em
Final energy Ef = − GM
6Re
em
Work W = Ef − Ei = GM
12Re
25
Conservation of Energy
• EXAMPLE: Escape velocity
• An object m is projected vertically upward
from the earth’s surface with vi
26
Escape Velocity
• When the object reaches its maximum
altitude, vf and rf = rmax
• Because the total energy of the system is
conserved,
1 2 GMe m
GMe m
mvi −
=−
2
Re
rmax
1
• → vi2 = 2GMe R1e − rmax
• From this expression we can find
h = rmax − Re
• Settingq
rmax = ∞ and vi = vesc , we get
e
vesc = 2GM
Re – independent of m
27
Escape Velocity
• Escape velocities from planets
• Atmosphera
28
First Cosmic Velocity
• First cosmic velocity v =
• If r ≈ re , v ≈ 8 km/s
29
q
GMe
r
Libration Points
30
Conservation of Energy
1
1
2
total energy = E = m1 v1 + m2 v22
2
2
Gm1 m2
+ −
= const
r
• The bound elliptical orbit
• First cosmic velocity: v =
p
GM/r
p
• Second cosmic velocity: v = 2GM/r
31
Conservation of Energy
• The orbits
• The system is bound: E < 0
• Unbounded orbits: E > 0
• The transition case: E = 0
32
Angular Momentum – Conservation
• For a single particle, j = r×p,
where p = mv
33
Angular Momentum – Conservation
• EXAMPLE: j = m(r×v) = const
34
Angular Momentum – Conservation
• EXAMPLE: Solar system j = m(r×v) =
const
35