Astronomy 111 Overview of the Solar system This is a composite image of all the planets in our solar system. From the top is Mercury (taken by the Mariner 10 spacecraft); Venus (taken by the Pioneer Venus Orbiter); a full disk of the Earth (taken by the Apollo 17 astronauts); the Moon and then Mars (captured from ground-based telescopes); and Jupiter, Saturn, Uranus and Neptune (as seen by the Voyager spacecraft). Credit NASA/JPL. ❑ Contents of the solar system ❑ What we can measure and what we can infer about the contents ❑ How do we know? Initial explanations of four important facts about the solar system ❑ Important Units ❑ The Celestial sphere ❑ Working with Angles 1
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Astronomy 111 Overview of the Solar system
This is a composite image of all the planets in our solar system. From the top is Mercury (taken by the Mariner 10 spacecraft); Venus (taken by the Pioneer Venus Orbiter); a full disk of the Earth (taken by the Apollo 17 astronauts); the Moon and then Mars (captured from ground-based telescopes); and Jupiter, Saturn, Uranus and Neptune (as seen by the Voyager spacecraft). Credit NASA/JPL.
❑ Contents of the solar system ❑ What we can measure and what we
can infer about the contents ❑ How do we know? Initial
explanations of four important facts about the solar system
❑ Important Units ❑ The Celestial sphere ❑ Working with Angles
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Inventory of the Solar System
Q: What would an outside observer see if they were looking at our Solar system? A: The Sun. The luminosity of the Sun is 4×108 times as
luminous as the second brightest object, Jupiter. The Sun contains 99.8% of the mass (but NOT most of the angular momentum).
L◉=4×1026 Watts, M◉=2×1033 g
The solar system is: The Sun + some debris.
Imag
e by
Robe
rt G
endl
er
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Aside: observations of other planetary systemsIn most other stellar (extrasolar) systems, all we know about are the stars, because they are overwhelmingly bright.
– But, planets have been detected around other stars, with surprising properties (high eccentricities, small radii from star, sparse and packed systems). – Our detailed understanding of the Solar system has
helped us interpret/understand the formation and evolution of planetary and stellar systems.
– However, what we are finding about extrasolar systems is challenging previous views of solar system formation and evolution.
Astronomer-artist’s conception of HD 209458 and its largest planet (Robert Hurt, SSC/Caltech)
Components of the “debris”
❑The Giant planets (mostly gas and liquid)
❑Terrestrial planets and other planetesimals, (mostly rock and ice)
❑The Heliosphere (plasma)
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Pale Blue Dot" photograph of the Earth taken by the Voyager 1 spacecraft on July 6,1990
Giant Planets❑ Jupiter dominates with a mass of 10-3 times that of the Sun (1/1000 M◉ ) or 300 times that of the Earth
❑ Saturn’s mass is 3×10-4 M◉
❑ Neptune and Uranus are about 1/6 the mass of Saturn. ❑ Giant planets at 5, 9, 20, and 30 times further from the Sun than
Earth is (a = 5, 9, 20, 30AU; a is commonly used to denote semi-major axis, covering the possibility of an elliptical orbit).
❑ Saturn and Jupiter have roughly Solar type abundances (composition) which means mostly hydrogen and helium. Saturn probably has a rocky core; Jupiter might not.
❑ Neptune and Uranus are dominated by water (H2O) and ammonia (NH3) and have a larger percentage of their mass in rocky/icy cores.
❑ Jupiter and Saturn are called gas giants. ❑ Neptune and Uranus are called ice giants.
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Terrestrial Planets and Other Debris
• Earth and Venus, radius R~6000km, a=1.0, 0.7 AU respectively • Mars, R~3500km, a=1.5AU • Mercury R~2500km, a=0.4AU • Asteroid belt, a~2-5 AU • Moons and satellites • Kuiper Belt, including Pluto a>30AU • Oort Cloud, including Sedna
a~104AU • Planetary Rings • Interplanetary dust
Mars D
aily Global Im
age M
ars Global Surveyor M
OC
April
1999
Messenger image of lava flooded craters on Mercury October 6, 2008 6
The HeliosphereAll planets orbit within the heliosphere, which contains magnetic fields and plasma primarily of solar origin.
The solar wind is a plasma traveling at supersonic speeds.
Where it interacts or merges with the interstellar medium is called the heliopause.
Cosmic rays (high energy particles) are affected by the magnetic fields in the heliosphere.
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Planetary properties that can be determined directly from observations
❑ Orbit ❑ Age ❑ Mass, mass
distribution ❑ Size ❑ Rotation rate
and spin axis ❑ Shape ❑ Temperature ❑ Magnetic field ❑ Surface composition ❑ Surface structure ❑ Atmospheric structure
and composition
Saturn, Venus, and Mercury, with the European Southern Observatory in the foreground (Stephane Guisard).
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Other planetary properties that can be inferred or modeled from the observations
❑ Density and temperature as functions of position, of the atmosphere and interior
❑ Formation ❑ Geological history ❑ Dynamical history
These will be discussed in context with the observations and astrophysical processes as part of this course. Cut-away view of the Sun’s convection zone
(Elmo Schreder, Aarhus University, Denmark).
UNITS• 1 Astronomical unit AU = 1.5×1013cm Is the distance between the Sun and Earth. More precisely is the earth’s semi-major axis in heliocentric coordinates. • Time: seconds or years. 1year ~3×107s • Angles: 360o ~ 2π radians 60 arcminutes per degree, 60’ 60 arcseconds per arcminute, 60”
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Angles on the sky
• One arcsecond, 1”
• 1” is the angle that we can resolve at optical wavelengths
without atmospheric blurring (seeing). • d=Dθ d = size of object D = distance to object (same units as d)
θ = angle in radians
6o
1' 1 radians1" 4.8 10 radians60" 60' 180
o π −× × × = ×
D
θ
d
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Q. How far away could we resolve an Earth?Q. Is Neptune resolved?
• Rearth ~ 6500km
• Rneptune ~25,000 km, a~30AU
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How do we know?Note that most of what we will say about the planets this semester is founded upon a lot of very direct evidence, rather than theory or reason-aided handwaving. For example: 1. The Earth is approximately spherical. Direct observation, and known since about 500 BC: ❑ Lunar phases indicate that
it shines by reflected sunlight. ❑ The full moon is
nearly on the opposite side of the sky from the Sun.
❑ Therefore lunar eclipses are the Earth’s shadow cast on the moon.
❑ The edge of this shadow is always circular; thus the Earth must be a sphere.
Lunar eclipse (16 Aug 2008) sequence photographed by Anthony Ayiomamitis
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How do we know? (continued)
2. The planets orbit the Sun, nearly all in the same plane. Also direct observation. ❑ The planets and the Sun always lie in a narrow band across
the sky: the zodiac. This is a plane, seen edge on. ❑ Distant stars that lie in the direction of the zodiac exhibit
narrow features in their spectrum that shift back and forth in wavelength by ±0.01% with periods of one year. No such shift is seen in stars toward the perpendicular directions, or in the Sun. The shift is a Doppler effect, from which one can infer a periodic velocity change of Earth with respect to the stars in the zodiac by ±30 km/sec. Thus the Earth travels in an approximately circular path at 30 km/sec, centered on the Sun.
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How do we know? (continued)
The Doppler effect: Suppose two observers had light emitters and detectors that can measure wavelength very accurately, and move with respect to each other at speed v along the direction between them. Then if one emits light with pre-arranged wavelength the other one will detect the light at wavelength that is, where is the speed of light.
λ0 ,
λ λ
λ λ
λ
" #= +$ %& '−
=
0
0
0
1 ;
,
vc
v c
= × 10 -12.99792458 10 cm secc
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How do we know? (continued)
3. The Sun lies precisely 1.50×1013 cm (1 AU) from Earth. Direct observation: it follows from the previous example, but we can measure the same thing more precisely these days by reflecting radar pulses off the Sun or the inner planets, measuring the time between sending the pulse and receiving the reflection, and multiplying by the speed of light. ❑ Note that knowing the AU enables one to measure the
sizes of everything else one can see in the Solar system. ❑ Using Venus or Mercury gives the most accurate results, as
the Sun itself doesn’t have a very sharp edge at which the radar pulse is reflected. It works with Venus or Mercury as follows:
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How do we know? (continued)
Suppose you were to send a radar pulse at Venus when it appeared to have a first-or third-quarter phase, so that the line of sight is perpendicular to the Venus-Sun line. Then
Sun
Earth
1 AU
Venus
θ dθ
Δ=
=
= × 13
2
1 AUcos1.4959787069(3) 10 cm
c td
d
(corrected for planet size and averaged over the orbit)
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How do we know? (continued)
4. The solar system is precisely 4.6 billion years old. Direct observation, again, on terrestrial and lunar rocks, and on meteorites, nearly all of which originate in the asteroid belt. We will study this in great detail: ❑ When rocks melt, the contents homogenize pretty thoroughly. ❑ When they cool off, they usually recrystallize into a mixture of
several minerals. Each different mineral will incorporate a certain fraction of trace impurities.
❑ Those trace impurities that are radioactive will vanish, in times that are measured accurately in the lab.
❑ Thus the ratio of amounts of radioactive trace impurities tell how much of the tracer the rock had when it cooled off, and how long ago that was.
❑ The oldest rocks in the Solar system are all 4.567 billion years old, independent of where they came from, so that’s how long ago the Solar system solidified.
The Celestial Sphere
Astronomy was the first precision science. For about 200 years, astronomy was considered
mathematics. Positions of objects can be predicted extremely well
(e.g., eclipses). To do this we require a good coordinate system.
This we call the celestial sphere. It is a way for us to specify positions of objects on the sky.
Geocentric reference frame
Rochester is close to 75o W of Greenwich, England or 5h W of Greenwich England. So 24h = 360o, or 1h = 15o on the equator. Mees observatory latitude is 42o 42m. Relation between time and angle for longitude is not a coincidence.
Coordinate system tied to rotation axis of the Earth
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Coordinate system on the sky
❑ North Celestial Pole (NCP) and the South Celestial Pole (SCP) – intersections of the north and south pole rotation axes with the celestial sphere.
❑ Celestial Equator – Intersection of the earth's equatorial plane with the celestial sphere. The celestial equator is not aligned with the ecliptic.
❑ The ecliptic is the plane containing the planets.
❑ The celestial sphere is an imaginary sphere of infinite radius with the earth located at its center.
❑ An astronomer can only a hemisphere of see the sky at a time, that is, only half the sky is above the horizon at any time. The sky moves as the earth rotates. Just as the sun rises and sets every day, so do many stars in the sky each night.
Declination (DEC, δ)❑Declination is the angle
between the celestial equator and the object. It is similar to latitude.
❑The north star has a declination ~+90o , and is approximately stationary on the sky.
❑Objects below -15o declination are difficult to observe from Rochester.
❑The circle which contains the rotation axis of the earth and the zenith is called the meridian Objects reach their highest altitudes (elevations in the sky) when they are on the meridian.
Right ascension is similar to longitude. It is defined with respect to the vernal equinox, (Mar 21). Remember longitude is with respect to Greenwich England.
Sun is in ecliptic! Sun intersects celestial equator twice in the year
Vernal equinox
❑Vernal equinox is a moment of time or a direction. Consider the the intersection of the ecliptic with the earth’s equatorial plane.
❑The vernal equinox is the moment in time when the sun crosses this point.
❑The sun crosses the vernal equinox at 21 March, noon UT (Universal time)
Coordinates on the celestial sphere
Just as latitude and longitude specify a position on the glob, declination and right ascension specify a location on the celestial sphere.
Vernal equinox
❑ The Earth’s equator and the ecliptic cross at two points on the celestial sphere or directions. One of these sets the vernal equinox, the other the autumnal equinox. ❑ Location of stars given with respect to the vernal
equinox direction. ❑ Orbital elements for orbits of bodies in the solar
system are also given with respect to this direction. ❑ The great circle containing both equinoxes and the
poles defines the zero-point of right ascension. ❑ Note: since the earth’s spin axis precesses, the
angular position of the vernal equinox depends on the year or epoch.
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Right Ascension (RA, α)
❑ Right ascension is often given in units of time (hours, minutes seconds). Similar to longitude on Earth, the Right Ascension of an object on the celestial sphere is measured eastward along the celestial equator, from a zero-point (i.e. where α = 0o), defined to be the vernal equinox – the place where the Sun is on the sky on March 21.
❑ If you look directly in the opposite direction this means objects at RA~12h are cresting (rising to the meridian) at midnight on Mar21.
❑ You can’t look at things that have an RA of 0h in the spring because the sun is in the way.
❑ In the fall the Earth is on the opposite side of the Sun. You can observe stars at an RA of 0h in the fall.
❑ 24 hours in RA, and 12 months. This means each month increases the RA of objects which are on the meridian at midnight by 2hours.
Local Sidereal Time (LST)
❑ Sidereal is with respect to the locations of stars. ❑ Sidereal time is the RA that is cresting or on the
meridian at a particular time. ❑ To estimate LST you need to know the time and
date as well as your location. ❑ Sidereal Time ST is = the right ascension of a star
on the meridian. Thus ST = 0h at noon during the vernal equinox (Mar 21). At this time the Sun is on the meridian.
❑ You cannot observe things with RA=0 during the vernal equinox. Why not?
❑ At midnight on the vernal equinox, LST=12h. Objects with RA=12h reach the meridian at midnight.
Elevation and Hour Angle
❑ The Hour angle, HA is the angular measure in h,m,s from the meridian westward along the celestial equator to the foot of the hour circle (like longitude circle) through an object.
ST = HA + α Sidereal time = hour angle + right ascension ❑ HA = 0 when an object is on the meridian. ❑ HA<0 before it reaches the meridian. ❑ HA>0 afterwards.
❑ Altitude or Elevation describes how high an object is above the horizon.
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What RA is up at midnight?
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24 hours in RA Each month of movement in Earth’s orbit gives 2 hours in RA
RA=0 hours is up Sept 21 at midnight RA=2 hours is up Oct 21 at midnight RA=12 hours is up at midnight Mar 21
The sidereal and solar day
Careful observation of the sky will show that any specific star will cross directly overhead (on the meridian) about four minutes earlier every day. In other words, the day according to the stars (the sidereal day) is about four minutes shorter than the day according to the sun (the solar day). If we measure a day from noon to noon - from when the sun crosses the meridian (directly overhead) to when the sun crosses the meridian again - we will find the average solar day is about 24 hours.
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Sidereal day vs solar day
Ps = Sidereal period: Orbital period of a rotating planet as seen from an external fixed observer
Pd = Solar day: Time between noons as seen by a person on the surface
Po = orbital period
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Elevation
Q. The elevation of an object at meridian depends on what?
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Elevation
Q. The elevation of an object at meridian depends only on what?
A. The object’s declination and your latitude.
If you are on the equator the north and south poles are always on the horizon. Objects with DEC=0 pass straight overhead.
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Elevation
Q. What is the maximum elevation of an object that has a declination of +80o at Mees Observatory (with latitude +42o 42’)?
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Elevation
Q. What is the maximum elevation of an object that has a declination of +80o at Mees Observatory (with latitude +42o 42’)?
A. Polaris has elevation ~43o
Object has maximum elevation 10o higher or 53o
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EPOCH
oRA is defined with respect to the vernal equinox. However the earth’s rotation axis precesses with a period of about 26,000 years. The rotation axis moves in a circle of ~23.5 degrees.
oWhen you type in an RA and DEC into a telescope computer system you must specify an epoch for the coordinate system. Standard epochs include J2000 (Julian 2000). 1950 is also a popular epoch. Whenever you write down astronomical coordinates, you need to note the epoch of the coordinates.
oA TCS (telescope control system) computes the position of the object after precessing the coordinates to the current epoch.
Converting between RA and DEC and angular distance in radians
Often you want to know the distance between two points on the sky which are specified in RA and DEC. RA is typically given in hours, minutes and seconds. On the equator (DEC=0), 24h=2π=360o. 60m=1h, 60s=1m. However, for DEC≠0, 24h≠2π radians. You must take into account the cosine of the DEC.
21hour RA cos(DEC) radians24π
× =
Angular distances
Q. 1 second in RA is how many arcseconds of angular distance at a DEC of 45?
Q. 1 second in DEC is equivalent to many arcseconds?
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Angular distances
Q. 1 second in RA is how many arcseconds of angular distance at a DEC of 45?
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Answer: 24 hours in 2π radians unit of time!!!!! 360 degrees in 2π radians
60 minutes per hour, 60 seconds per minute 60 arcminutes per degree, 60 arc seconds per arc minute
To convert seconds to arcseconds 360/24 = 15 On the celestial equator there are 15 arc seconds per second of time
Angular distances
Q. 1 second in RA is how many arcseconds of angular distance at a DEC of 45o?
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Answer: On the celestial equator there are 15 arc seconds per second of time
Multiply by cos(45o) =1/√2 Answer: 15/√2 =10.6 There are 10.6 arc seconds per second of time in RA at a DEC of 45 degrees
Angular distances
Q. 1 second in DEC is equivalent to many arcseconds?
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Angular distances
Q. 1 second in DEC is equivalent to many arcseconds?
Answer: 1 DEC is in degrees, arc seconds in DEC are true arc seconds
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Solid Angles, Spherical coordinates
[ ] where 0,2 on equatorcos off the equatord
dφ φ πθ φ
∈
,2 2
d π πθ θ # $∈ −' () *
𝜙 azimuthal angle (like longitude or RA) 𝜃 like latitude or DEC
Integrating Solid Angles
cosdA d dθ θ φ=
]/ 2
/ 2
/ 20 / 2
Integrating over the sphere Area of solid angle
cos sin 2
4 radians
A d d
A
π ππ
ππ
φ θ θ π θ π
π
2
−−
= = 2 × = 2 ×
=
∫ ∫
just as we needed the cosine of the dec to compute an angular distance from the RA we need it here to compute solid angles
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Props: Celestial sphere Dopper ball or whirly hose Astroscan for looking at sunspots
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