Practical Astronomy

Ast 401/Phy 580

Fall 2015

QuizPhil has observing time on the mighty 6.5-meter MMT next Tuesday (Sept 15) to observe M31 (00:40 +41:00). If he wants to wait until M31 is at least 30o above the horizon, what is the earliest he can observe? (Specify both in local time and UT.)

Reminder:

sin(alt)=sin(Φ)sin(δ)+cos(Φ)cos(δ)cos(HA)

where Φ = 31.7o

Is your answer reasonable?

sin(alt)=sin(Φ)sin(δ)+cos(Φ)cos(δ)cos(HA)

Φ = 31.7o

δ= +41o

alt=30o

cos(HA)=[sin(30o)-sin(31.7o)sin(41o)]/cos(31.7o)cos(41o)

=0.155/0.642=0.241

HA=arcos(0.241)=±76o = ±5.1 hours

LST=00:40 ± 5.1 hours = -4.4 to 5.8 = 19.6 to 5.8 hr LST

When is this?

LST midnight Sept 15 will be 23:30hr (1 week before Sept 22)

19.6 LST will be 3.9 hours before midnight = 8pm MST

5.8 LST will be 5.3 hours after midnight =5:15am MST

Do we expect this to be exact? If not, why not?

Time Zones

We assume that Sept 20 midnight = 00:00 LST is exact. But that’s only true for one longitude within our time zone. Mt Hopkins is at 110.9deg west of Greenwich, or about 7.4 hours...not 7.0 hours west of Greenwich. Result is that it is only 23:30 at midnight on Sept 20. So, probably we need to add about 20 minutes to these values.

How far is Flagstaff of the standard time zone?Longitude 111.6o

111.6/15 = 7.44 hours. But our time zone is 7 hours behind Greenwich, not 7.44! That’s 26 minutes, or nearly half an hour

Sept 20, midnight isn’t 00:00:00 LST!!!!!!

Instead it will be:

A. 23:30

B. 00:30

skycalc

Written by John Thorstensen....

secz is the “airmass” (more about this later, and an altitude of 30o (zenith distance z of 60o) corresponds to an airmass of 2.0

So, roughly 19:00 (7pm) to 5:10am.

Stellarium

Free planetarium software for Windows/Linux/Mac: http://www.stellarium.org

http://www.stellarium.org

Catching up on some nomenclature

• Right ascension = RA = α (usually XX:YY:ZZ or XXhYYmZZs but very occasionally decimal degrees by multiplying by 15o/hr).

• Declination = dec = δ (usually XXoYY'ZZ")

• Hour angle = HA (usually XX:YY:ZZ or XXhYYmZZs but can convert to degrees by multiplying by 15o/hr).

• Altitude = Alt (usually in decimal degrees)

• Azimuth = Az (usually in decimal degrees)

• Zenith distance = z (usually in decimal degrees) =90o - altitude

• Galactic longitude = l or lII (usually in decimal degrees)

• Galactic latitude = b or bII(usually in decimal degrees)

• Local Sidereal Time=LST (usually XX:YY:ZZ or XXhYYmZZs)

Precession

It's not enough to know the right ascension and declination: you must also specify the equinox. (Some astronomers incorrectly refer to this as the "epoch".)

Why?

The Earth is a spinning top where gravity (primarily the sun's) acts as the torque. The direction that the pole is pointing CHANGES over time, ipso facto changing the coordinates of stars over time.

Precession

Precession

Takes 26,000 year; the half-angle of the cone is 23.5o

Changing location of NCP over 26,000 year

Precession

How large an effect is this?

Field of view of most telescopes are a few arcminutes; in general, you'd like to aim a telescope to a few arcseconds. Vernal equinox moves by 50" per year!

Precession

Thus, all modern telescope control systems allow you to enter RA, DEC, and EQUINOX, and they then precess the coordinates to the current day, hour, minute, and second in pointing the telescope. "Standard" positions of stars are usually tabulated for 2000.0 ("J2000") or (more rarely) 1950.0 ("B1950"). Occasionally other equinoxes crop up. (1975.0 for many Magellanic Cloud objects.

If you were pointing a telescope by hand (no computer system) you would have to do the precession yourself.

PrecessionVery good (but approximate) formulae are:

Example: Consider a star at 00:42:44.3 and +41o16’ 09” (J2000). What are its coordinates in mid 2013 equinox?

Precession

α= 00:42:44.3 = 10.6846o

δ= +41o16’ 09”= 41.2692o

Δα=(2013.5-2000.0) x (3.075s + 1.336s x sin 10.6846 x tan 41.2692)=44.4s

Δδ(2013.5-2000.0) x (20.043” x cos (10.6846)=265.9” (=4’ 25.9”)

Precession

Using IRAF:

Input (J2000)Output

Note that the difference is 44.5 s of RA and 4' 25.8" of dec.

but wait, kids, that's not all!

Nutation

There's an extra wobble that needs to be taken into account because of the moon. The variation in the Earth-Sun and Earth-Moon distances introduce a 9" wobble on top of precession, which is called nutation. Again, modern telescope control systems take this into account.

Precession

Given that modern telescope control systems take precession into account, why do you have to think about this?

• Someone has to WRITE the telescope control system.

• Fiber placement in spectrographs.

• Might be pointing a telescope manually

someday!

Proper motions

So far, we've only discussed the issue with the earth. But, the stars themselves move. (Not galaxies..) They're zipping around the Milky Way with velocities on the order of 10-100 km/sec.

Consider a very nearby star, at a distance of 10 parsecs, and a star moving tangentially at 30 km/sec. How much does angular change in 1 year?

Proper motions

There are roughly π x 107 sec/year. So, the star will move by 30 km/sec x π x 107 sec/year = 9 x 108 km. Call it a billion.

Distance to star in km: 10 pc = 10 pc x 206265 AU/pc x 150 x 106 km/AU = 3 x 1014 km.

So angular change (in radians) is just 1x109/3x1014= 3.3x10-6 radians. Multiply by 206265"/radians ---> 0.7". Oooh! That matters, partically over 10 or 20 years...A few arcseconds!

s=rθ

θ=s/r

Proper motions

Some stars are closer or moving even faster, resulting in even larger proper motions. Barnard’s star is the record holder, changing by 10” per year. (It would still take 180 years for it to change its position on the sky by 1/2 degree, the size of the full moon).

Proper motions

Canopus

Canopus is the brightest star in the southern hemisphere, outshining even Sirius. Its coordinates are 06hr24min -52o42’.

When can see this star from Flagstaff? (Latitude 35o.)

A. June.

B. Dec.

C. Never.

D. Always.

Canopus

Canopus is the brightest star in the southern hemisphere, outshining even Sirius. Its coordinates are 06hr24min -52o42’.

When can see this star from Flagstaff? (Latitude 35o.)

A. June.

B. Dec.

C. Never.

D. Always.

Canopus

Canopus is the brightest star in the southern hemisphere, outshining even Sirius. Its coordinates are 06hr24min -52o42’.

When can we see this star from Las Campanas Observatory (latitude -29 deg) in Chile? DISCUSS!

Catalogs usually list the two components of proper motions as:

μα cosδ and μδ

with units of mas/yr (milli-arcsecs per year). For instance, Barnard’s star is listed as having

μα cosδ= -798.7 mas/yr and μδ =10337.8 mas/yr.

Proper motions

Coordinates of such stars are given not only with a particular equinox but also with a particular epoch; usually these are the same. (I.e., J2000 coordinates and the epoch is also 2000.) Many spectrophotometric standard stars are nearby white dwarfs, and you have to manually correct for proper motion before you tell the telescope where to point. Don’t forget the cosδterm in correcting back to ΔRA!

Proper motions

Aberration of starlight

Another effect that affects the location of an object when you point the telescope is the aberration of starlight. Same effect as rain/snow falling vertically viewed from a car moving horizontally---watch the streaks on the window. It’s just vectors. The earth is moving around the sun at 30 km/sec, and if you’re looking at a star that is nearly perpendicular to that motion, it will be shifted by a tiny angle: about v/c (in radians), or 30/3x105 or 10-4 x 206265 = 20”! Telescope control systems take this into account.

Refraction

A final issue that will affect the location of a star is due to the earth’s atmosphere: refraction.

RefractionThe shift is always perfectly along elevation, and will make a star appear to be higher in the sky than it actually is. How large an effect?

Zenith: 0

45o elevation: 1’ (yikes!)

10o elevation: 5.3’ (double yikes!)

The sun appears flattened at sunrise/sunset because refraction is 34’ on the horizon but only 29’ half a degree above it. (Sun’s disk is half a degree in size). So, flattens it by 5/30, about 17%.

Refraction

Displacement in Az is approximately 60” tan z,

where z is the distance to the zenith (90-alt).

Aiming a telescope

Which of the following is NOT a function of position on the sky:

a) The amount of precession.

b) The amount of nutation.

c) The amount of refraction.

d) The amount of proper motion.

e) The amount of aberration.

(1) A star is listed in the UCAC3 as having J2000.0 coordinates of

12:13:13.141 +38:12:15.12 J2000 with

μα cosδ= 112.2 mas/yr and μδ =-62.5 mas/yr.

You go to observe it at the DCT and need the coordinates of where the star will be tonight in J2000 coordinates. What coordinates do you send over to the telescope control system?

Problem 2 on next page…

Homework

Due Sept 22

(2) Estimate the approximate size of the following on the position of a relatively nearby star. Be clear about your assumptions. Keep the math simple.

a) Precession

b) Nutation

c) Proper motion

d) Aberration of starlight

e) Refraction

Homework

Due Sept 22