Astronomy 1 - Fall 1998
Lecture Notes Sept 28
Topics:
Synodic Period
Retrograde Motion
Kepler's 3rd Law
Transits of Mercury and Venus
Newton's Laws
Law of Gravitation
Centre of Mass Motion
READ Text Chapter #2
-- READ especially the "More Precisely" s
2-1 (Some Properties of Planetary Orbits)
2-2 (Newton's Laws of Motion and Gravitation)
2-3 (Weighing the Sun)
ASTRONOMICAL:
Friday Sept 25 - Mercury was at superior conjunction (
with the Sun)
today Sept 28- Moon at First Quarter
Tuesday Sept 29 - Venus at Greatest Helicentric
Longitude North (explain)
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LECTURE
Continue unfinished material
Kepler's 2nd and 2rd Laws- Sept 23 Lecture
Kepler's 3rd Law P2 is proportional to a3
(See Table 2.1) chapter 2 Chaisson
Period Semi-Major P2/a3
(yrs) (AU) (yr2/AU3)
Mercury 0.241 0.387 1.002
Venus 0.615 0.723 1.0008
Jupiter 11.86 5.203 0.9986
Neptune 163.73 30.110 0.982
It works for Satellites also (with different constant)
Saturn's Satellites: (days) (106 km) (10-18d2/km3)
Tethys 1.888 0.295 138.8
Dione 2.737 0.377 139.8
Rhea 4.518 0.527 139.5
Titan 15.945 1.222 139.3
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Transits of Mercury and Venus
Inferior Planets: Mercury and Venus should come between
the Earth and the Sun sometimes. Not every revolution because
inclination of orbits. Mercury Orbit = 7.0o
Venus Orbit = 3.4o
DIAGRAM of Inclination:
Only when Earth is at the line of nodes of the planets orbit can there
be a transit of Venus or Mercury.
Transit of Mercury Nov 15, 1999
Node of Mercury - May and November
Node of Venus - June and December
Transits of Venus(from Fred Espenak's Transit Page)
-Because Venus's orbit is considerably larger than Mercury's orbit,
transits of Venus are much rarer. Indeed, only six such events have
occurred since the invention of the telescope
(1631,1639, 1761,1769, 1874 and 1882).
-Transits of Venus are only possible during early December and
June when Venus's orbital nodes pass across the Sun.
-Transits of Venus show a clear pattern of recurrence at intervals
of 8, 121.5, 8 and 105.5 years.
Transits of Venus: 1901-2200
Date Universal
Time
2004 Jun 08 08:19
2012 Jun 06 01:28
2117 Dec 11 02:48
2125 Dec 08 16:01
Synodic Period of Venus:
Siderial Period = 0.6152 years = 224.7 days
Synodic Period = 584 days = 1.60 years = 8/5 years
ie. 5 Venus synodic periods = 8 Earth years
Every 8 years Venus has the same location in the sky.
1998 Observer's Handbook : Inferior Conjunction Jan 16
1990 Observer's Handbook : " " Jan 18
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Newton's Laws
(1) Inertia (once in motion an object tends to stay in motion)
(2) Acceleration is proportional to applied force (inverse mass constant)
a = F/m
Velocity, v, is the change in distance per time (units = m/s)
Acceleration, a, is change in velocity per time (units = m/s2)
Both velocity and acceleration have direction as well
as magnitude (size) and hence are vectors. Here vectors
are denoted with bold face font.
units: mass: [m] = kg
Force: [F] = kg m/s2 = Newton (N)
(3) for every action there must be an equal reaction
<---O O-->
Force of gravity - acceleration of gravity ; Earth Surface
Weight = W = mg | g | = 9.8 m/s2
EXAMPLES:
-Weight of 70 kg person on Earth's Surface
= 9.8 m/s2 (70) = 686 Newtons
-Acceleration of 2000 kg automobile (no friction)
with push of 500 N (~110 pounds)
= 500 N/2000 kg = 0.25 m/s2
Law of Gravitation:
Newton's Method used astronomy
Acceleration of the Moon
Circular Motion and 'Falling' of the Moon toward the Earth
In the case of the Moon orbiting the Earth, its changing
orbital velocity represents an acceleration toward the
center of the Earth and hence is 'Falling' toward the
Earth just as something dropped on the surface of the
Earth. The Moon must be accelerating or it would be
moving in a straight line as demanded by Newton's first
law. The difference is that the Moon is moving with a
velocity perpendicular to that direction and never
reaches the Earth and always stays in its circular orbit.
Circular Motion
radial acceleration = ar = v2/R
[Explanation of the v2/r Relationship]
v = 2(pi)R/P (speed = distance around circumference / time)
P = period = 27.3 days x 84,400 s/day = 2.36x106 s
Moon, v = (384,000 km) 2(pi) / 2.36x106 s = 0.1023 km/s
= 1023 m/s
ar(Moon) = (1023)2/(384,000,000 m) = 0.00273 m/s2
ar(surface R = 3670 km) = 9.8 m/s2
Ratio ar(Earth surface)/ar(Moon) = 9.8/0.00273 = 3590
Ratio of distances = 384,000/6370 = 60.3
note that 602 = 3600
So Newton deduced that acceleration ~ 1/R2
Must be proportional to Mass
g = GM/R2
Force F = mg = GmM/R2
G = Universal Gravitation Constant = 6.67x10-11 N m2/kg2
Weigh the Earth:
M = gR2/G = (9.8)(6,370,000)2/G = 5.96x1024 kg
Motion of Centre of Mass
m1 r1 = m2 r2 center of mass
m2 O---------------|--------o m1
r2 r1
Earth-Moon System
ME = 6x1024 kg
MM = 0.0735x1024 kg ME/MM = 81.4
rM/rE = 81.4 Since rM + rE = 384,000 km
thus rE = 4660 km = 0.73 RE
DIAGRAM of the EARTH-MOON center of Motion
----------------------
Kepler's 3rd Law from Newton's Law of Gravitation
P2 = 4(pi)2a3/[G(M+m)]
M = mass of central body
m = mass of orbiting body
a = mean distance between the bodies
P = period of the orbits.
Solar System:
Mass of the Sun MSun >> MEarth
so MSun + MEarth ~ MSun
P2 ~ 4(pi)2a3/[GM]
So
MSun = 4(pi)2a3/(G P2)
= 4(pi) (1.5x1011 m)3/[6.67x10-11 Nm2/kg2* (3.16x107)2]
= 1.98x1030 kg
= 1/3 million times M(Earth).
Note: The constant 4(pi)2/GMSun = 2.97x10-19 s2/m3
= 1 yr2/AU3
Earth and Satellites:
The constant 4(pi)2/GMEarth = 9.91x10-14s2/m3
Example of Earth Satellite:
What is the period of a satellite orbiting the Earth 500 km above its surface?
[P(satellite)/P(Moon)]2 = [a(satellite)/a(Moon]3
P(Moon)= 27.3 days
a(Moon)= 384,000 km
a(Satellite) = 6370 km + 500 km = 6870 km
So
P(satellite) = P(Moon) [a(satellite)/a(Moon)]3/2
= 27.3 days (6870 km/384,000 km)3/2
= 0.065 days = 1.57 hours