Transcript PHY216_lect3_2014_sub

PHYS216
Practical Astrophysics
Lecture 3 – Coordinate Systems 2
Module Leader:
Dr Matt Darnley
Course Lecturer:
Dr Chris Davis
Calendars & Julian Date
• Gregorian calendar - used universally for civil
purposes
• Julian calendar - its predecessor in the western
world
The two differ only in the rule for leap years: the
Julian calendar has a leap year every fourth year,
while the Gregorian calendar has a leap year
every fourth year except century years not exactly
divisible by 400.
The Julian Date (JD) - a continuous count of days
and fractions of days since Greenwich noon on 1st
January, 4713 BCE.
Modified Julian Date (MJD) – a more convenient
form of JD.
MJD = JD – 2400000.5.
Swedish version of the
Gregorian Calendar, c. 1712.
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Calculating Julian Date
The Julian Day Number is calculated as follows:
1) Express the date as y m d, where y is the year, m is the month number (Jan = 1,
Feb = 2, etc.), and d is the day in the month.
2) If the month is January or February, subtract 1 from the year to get a new y, and
add 12 to the month to get a new m. (This is because we consider January and
February as being the 13th and 14th month of the previous year).
3) Dropping the fractional part of all results of all calculations (except JD), let
A = rounddown(y/100)
B = 2 – A + rounddown(A/4)
C = rounddown(365.25 x Y)
D = rounddown(30.6001 x (m + 1))
JD= B + C + D + d + 1720994.5
e.g. JD on 2014 July 31 –
A = rounddown(2014/100) = 20
B = 2 – 20 + roundown(20/4) = -13
C = rounddown(365.25 x 2014) = 735613
D = rounddown(30.6001 x (7+1)) = 244
JD = -13 + 735613 + 244 + 31 + 1720994.5
= 2456869.5000 (at 0:00 hrs)
This is the Julian Day Number for the beginning of the date in question at 0:00
hours (Greenwich time). Note half day extra because the Julian Day begins at noon.
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Calculating Julian Date
1) Express the date as y m d, where y is the year, m is the month number (Jan = 1,
Feb = 2, etc.), and d is the day in the month.
2) If the month is January or February, subtract 1 from the year to get a new y, and
add 12 to the month to get a new m.
3) Dropping the fractional part of all results of all calculations (except JD), let
A = integ(y/100)
B = 2 – A + integ(A/4)
C = integ(365.25 . y)
D = integ(30.6001 . (m + 1))
JD= B + C + D + d + 1720994.5
e.g. JD on 2014 July 31 at 00:00 hrs JD = 2456869.5000
at 06:30 hrs JD = 2456869.7708
Correspond to seconds!
JD is the Julian Day Number for the beginning of the date in question at 0 hours
(Greenwich time).
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Universal Time
Universal Time is the name by which Greenwich Mean Time (GMT) became known
for scientific purposes in 1928.
UT is based on the daily rotation of the Earth. However, the Earth’s rotation is
somewhat irregular and can therefore no longer be used as a precise system of
time.
Versions of UT:
UT1:
• The mean solar time at 0° longitude (Greenwich).
• The Sun transits at noon, UT1.
• Derived from observations of distant quasars as they transit the Greenwich
meridian.
UTC – Coordinated Universal Time:
• Time given by broadcast time signals since 1972
• Derived from atomic clocks.
• UTC is kept to within 1 second of UT1 by adding or deleting a leap second
Remember: UT (or UTC) = GMT
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Solar vs Sidereal Time
Solar Day: Time between successive transits of the sun (noon to noon)
Sidereal Day: Time between successive transits of distance stars and galaxies
… a sidereal day isn’t quite as long as a normal day!.
A sidereal day is about 23 hours, 56 minutes, 4 seconds in length.
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4 mins/day
=
2 hrs/month
=
24 hrs/year
Earth must rotate almost 1o more ( ≈ 360/365) to get the Sun to transit.
Takes approx 4 mins to rotate through 1o
Hence:
• a Sidereal Day is 4 mins shorter than the (mean) Solar Day
• the Local Sidereal Time (LST) gets 4 mins later at a given clock time every day.
Things to remember:
1. LST is the Hour Angle of the Vernal Equinox, g, (shown later)
2. The RA of a star = its Hour Angle relative to g.
3. At meridian transit of any star, LST = RA
4. LST tells us which RA is currently going through transit
5. LST - RA of an object = Hour Angle of the object
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When is an object observable?
• On March 21st, the Sun is at the Vernal Equinox, i.e. on March 21st
the RA of the Sun = 00h
• On March 21st at noon, LST is exactly 12 hrs ahead of the local time
(synchronise watches!)
At transit, RA = LST, and the Sun transits at midday, so….
At midday on March 21st, LST = 0 hrs
At midnight on March 21st, LST = 12 hrs
– this means that targets at RA = 12hrs are transiting
Each month sidereal time moves 2 hours ahead of clock (solar) time
At midday on April 21st, LST = 2 hrs
At midnight on April 21st, LST = 14 hrs
… and so on, in an annual cycle.
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When is an object observable?
• At 00:00 hrs LST on March 21st, a person in Greenwich facing due south would be
staring right at the meridian-transiting Sun… (would probably still need a spray tan)
• On March 21st, as the earth rotates, the HA of g and the LST both increase together.
• No matter what day it is, LST = always equals the Hour angle of the Vernal Equinox
LST = 15 hrs
Local time = 1 am
g
LST = 12 hrs
Local time = 10 pm
g
To Vernal
Equinox, g
LST = 0 hrs
Local time
= noon
LST = 12 hrs
Local time = midnite
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When is an object observable?
LST = 18 hrs
Local time = 2 am
LST = 15 hrs
Local time = 11 pm
LST = 12 hrs
Local time = 8 pm
g
g
To Vernal
Equinox, g
LST = 12 hrs
Local time = midnite
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When is an object observable?
Example:
The Hyades (open cluster) has RA ≈ 04h 30m, Dec ≈ +15o
Ideally want to observe it on a night when LST = 04h 30m at midnight (why
midnight?)
• LST = 12 hrs at midnight on March 21st (Sources with RA = 12 hr transiting…)
• 04h 30m is 16.5 hours later than 12 hrs
• LST moves on by 2 hours/month w.r.t solar time
• 16.5 hours difference = 8.25 months
• 8.25 months after March 21st is …
 Late November is
 the best time to
 observe the Hyades.
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What is the approximate Local Sidereal Time
Example:
Its 22:05 PDT on 1st June 2014 at the Mount Laguna Observatory, near San
Diego, California. What is the LST and thus the RA of transiting sources?
KEY: on March 21st, RA~12hrs transits at ~midnight (local time)
1st June is 2-and-a-bit months later, so add 2 hours per month:
RA ~ 16 hrs transits at midnight
But we’re observing about 2 hours earlier, so RAs that are 2 hrs less transit
RA ~ 14 hrs transits at about 10pm at the end of May
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How does this relate to GMT (UT)?
Earth viewed from above…
June 21st
(toward
RA ~ 18 hrs)
UT (time in
Greenwich)
Is 05.05am
LST ≈ 22 hrs
June 1st 2014
Time in California
PDT = 10.05 pm
LST ≈ 14 hrs
g
March 21st
(toward RA ~ 12 hrs)
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Calculating Local Sidereal Time
To the nearest hour (good enough for a small telescope)
Convert local time at the Observatory to UT/GMT
•
UT = tloc + Dt
tloc is the local time in decimal hours; Dt is the time difference between local and GMT/UT.
Calculate the LST at the Observatory
•
•
LST ~ (UT - 12) + Dd . (4/60) – l . (4/60)
LST = GMST – l . (4/60)
Where ’12’ corrects for the 12 hr time difference between LST and UT on 21 March.
• Dd is the number of days AFTER the Vernal Equinox (noon on 21 March, when LST = 0 hrs)
• l is the longitude WEST, in decimal degrees.
The factors 4/60 convert both Dd and l to decimal hours. Your answer will therefore be in
decimal hours.
Try this example:
• What is the LST at the Armagh Observatory, l = 6.6500o W, at 19.00
BST on 28 March?
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Calculating Local Sidereal Time
more precisely
The precise formula for calculating LST must take into account the Earth’s
Nutation and Precession (see e.g. the Astronomical Almanac published by the US
and UK Nautical Almanac Offices: aa.usno.navy.mil/faq/docs/GAST.php).
1. Convert UT Date and Time to a precise Julian date
•
JD - see slides at start of this lecture …
2. Calculate the number of days, D, since 1 January, 2000 at 12 hrs UT.
• D = (JD – 2451545)
3. Calculate GMST
• GMST = 18.697374558 + (24.06570982 . D)
(this number will probably be large; reduce it to within 24hrs by subtracting some multiple of 24)
4. Correct for Longitude of observatory (add if E of Greenwich, subtract if W)
•
LST = GMST +/- l . (4/60)
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Calculating Local Sidereal Time
An example: LST at 20.00 UT on 31st July, 2014 in Armagh (6.65o W)
1. Convert UT Date and time to a precise Julian date
•
JD = 2456870.3333
2. Calculate the number of days, D, since 1 January, 2000 at 12 hrs UT.
• D = (2456870.3333 – 2451545) = 5325.3333
3. Calculate GMST
• GMST = 18.697374558 + (24.06570982 . 5325.3333)
= 128176.6241 hrs (- 5340 . 24) = 16.6241 hrs
4. Correct for Longitude of observatory (add if E of Greenwich, subtract if W)
• LST = 16.6241 – 0.4433 = 16.1807 hrs, or 16:10:51
(Calculation done with Excel. You may get slightly different numbers depending
on how your calculator handles these very big numbers; but you should get
to within a minute of time.)
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Messier objects at 14 hrs RA?
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M 101
Messier objects at 14 hrs RA?
M3
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Calculating Alt-Az from RA, Dec, and Sidereal Time
So how do I point my telescope at M3 ?
Need to know:
• RA (a) and Dec (d)
• Latitude of the observatory, f
• Local Sidereal Time, LST
Remember: to calculate Alt and Az, you ONLY need HA, d , and f.
Need to remember:
• HA is the time since the target transited
• LST is equivalent to the RA that is transiting
• Therefore: HA = LST - RA
Example:
1. Target is M3, RA: 13h 42m 11.6s Dec: +28° 22’ 38.2″ (current epoch)
2. LST is 14 hrs 03 min on Mount Laguna (latitude, f = 32.8400o)
3. Now, calculate (i) HA from the RA and LST and (ii) the Altitude and Azimuth
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Other Things Which Affect Sky Positions – 1
1. Nutation
A 9 arcsec wobble of the polar axis along
the precession path - caused by the Moon’s
gravitational pull on the oblate Earth.
Main period = 18.66 years.
R = Rotation of earth
P = Precession
N = Nutation
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Other Things Which Affect Sky Positions – 2
2. Refraction
Displaces a star's apparent
position towards the
zenith.
R ≈ tan z
where R is in arcminutes
and z, the zenith distance,
is in degrees.
(only accurate for
z << 90o,
because
tan90 = ∞ )
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How does refraction affect the sun’s
appearance at sunrise/sunset?
Due to refraction, the Sun appears to set 2 minutes AFTER it actually does set!
Need a more precise empirical formula:
R = cot ( 90-z + 7.31/[90-z+4.4] )
At z = 90o: R = cot (7.31/4.4) = 34.4 arcmin. R ≈ 0.5 deg.
If it takes 6 hrs for the sun to move from zenith to the horizon, i.e. through 90 deg,
it takes 6 hrs x 0.5/90 = 0.033 hrs = 2 minutes to move 0.5 deg.
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Other Things Which Affect Sky Positions – 3
3. Height above sea level
Observer's height above sea level means that the observed horizon is lower
on the celestial sphere, so the star's apparent elevation increases.
Measured angle of elevation, q , above the observed horizon = q + a
where displacement, a, in arcmins is given by:
a = 1.78 √h
(h = height above sea level, in metres)
Q. Which is perpendicular
to the radius of the Earth,
the Celestial or the
Observed Horizon?
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Mauna Kea Observatory
Big island, Hawaii
The summit of Mauna Kea in Hawaii is 4200 m above sea-level,
• h = 4200 m; therefore, a = 115 arcmin – that’s almost 2 degrees!
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Other Things Which Affect Sky Positions – 4
4. Stellar Aberration
Caused by velocity of the Earth around the Sun ( ≈ 30 km/s).
Need to point the telescope slightly ahead in the direction of motion.
The amount depends on the time of year and the direction of the star.
Maximum effect ≈ 20 arcsec
LEFT: The angle at which the rain appears to be falling depends on the speed of the rain and the speed at
which the person is running: sin q = vman / vrain.
RIGHT: For a star near the ecliptic pole, or for a star in the plane of the ecliptic and at right angles to the
direction of motion of the Earth around the sun: sin q = vearth / c
vearth = 30 km/s and the speed of light, c = 300,000 km/s. Therefore, q = 0.0057 deg = 20 arcsec
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Angular Separations
and converging lines of RA
Stars 1 & 2:
RA: 10h and 12h
Dec: 0o
Stars 3 & 4:
RA: 10h and 12h
Dec: +60o
★3 ★4
Stars 1 & 2 are 2 hrs apart in RA
Stars 3 & 4 are 2 hrs apart in RA
But
the angular separation of stars 1
& 2 is NOT the same as for stars
3 & 4 because lines of right
ascension converge towards the
poles!
★1
★2
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Angular Separations
and converging lines of RA
Stars 1 & 2:
RA: 10h and 12h
Dec: 0o
Stars 3 & 4:
RA: 10h and 12h
Dec: +60o
★3 ★4
Angular sep of Stars 1 & 2 in
degrees:
2 hrs = 360o x 2/24 hrs x cos d
= 360o x 2/24 hrs x cos 0
= 30o
Stars 3 & 4:
2 hrs = 360o x 2/24 hrs x cos d
= 360o x 2/24 hrs x cos 60
= 15o
★1
★2
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Small Angular Separations
How to calculate the angular separation, q, of 2 objects on the sky
For two objects, A and B, with coordinates (RA
and Dec) aA , dA and aB , dB
Dd = dA  dB
Da = (aA  aB) cos dmean
Where dmean is mean declination of both objects,
in degrees.
Angular separation, q :
q = √ (Da2 + Dd2)
*** This is only valid if q < 1o ***
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Small Angular Separations
Star A:
18h 29m 49.6s +20o17’05”
Star B:
18h 29m 46.0s +20o16’25”
Dd = 0
dmean = +20o16’45”
20:17:00
(an example)
A
q
Dd
+20o16.67’
=
= +20.28o
Da = 3.6 seconds of time
B
Key: 1 sec of time = 15”.cosdmean
Angular separation, q , is given by:
20:16:00
Therefore:
3.6 sec of time = 3.6 × 15” × cos20.28o
Da = 51
Da
18:29:50
48
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q = √ (Da2 + Dd2) =  (0×0 + 51×51) = 65 arcsec
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Large Angular Separations
To calculate the angular separation, q , of 2 objects with a large separation (q > 1o) or
in the general case, the following formula can be used:
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Tangential or “Proper” motions
Stars move with respect to “stationary” background galaxies.
The brightest star in the sky, Sirius, has the following position and proper motion, m:
Correct the RA and
Dec coordinates
separately
NOTE: These
are SPEEDS !!
The get the precise, current epoch coordinates of the star you need to:
(a) precess the coordinates AND
(b) correct for the star’s proper motion.
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Proper motions
• Typical proper motions of nearby
stars ≈ 0.1 arcsec/year
• Star with highest proper motion is
Barnard’s star; PM = 10.25
arcsec/year
Barnard’s star
Evolution of the Great Bear:
The changing appearance of the
Big Dipper (Ursa Major) between
100,000 BC and 100,000 AD.
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Asteroids and Comets
Asteroids and comets can have very high
proper motions (arcseconds per
second!)
Ephemeris – a table of coordinates over a
range of dates
2 July
EXAMPLE
Near-Earth Asteroid NEO 2012-DA
On 1 July at 12.00 UT its coordinates are:
18h 29m 49.6s +20o17’05”
Dd
Its Proper motions is
ma cos d = 1.0 arcsec/minute
md = 2.0 arcsec/minute
1 July
Da
Q. What are its coordinates on 2 July at
12.00 UT?
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And finally…. A bit of astrology!
RA~3hr
RA~1hr
RA~5hr
(Sorry,
couldn’t
resist)
At mid-day on March 21st, the sun
(when viewed from the Earth) is at RA
= 0 hrs, midway between the
constellations of Aquarius and Pisces
(according to the IAU)…
What’s the star sign of someone born
on March 21st? And why is it “wrong”?
E
RA~13hr
RA~17hr
RA~15hr
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