Mock Revsion Help Sheet

Question 1

Moon parts

Answer 1

. Sea of crisis
. Apennine Mountains 
. Sea of tranquility 
. Tyco crater
. Kepler crater
. Copernicus crater
. Ocean of storms

Question 2

Constellations + pointers

Answer 2

. Cassiopeia
. Cygnus
. Plough- Arcturus and Polaris
. Orion - Sirius, Aldebaran and Pleiades from Orion’s Belt
. Great Square of Pegasus - Fomalhaut and Andromeda
. Southern Cross
. Summer Triangle

Question 3

Celestial equator

Answer 3

Line on sky above the equator, 0 declination

Question 4

Ecliptic

Answer 4

Plane on which the Earth orbits the sun, and so the line across the sky upon which the sun appears to move

Question 5

Zodiacal band

Answer 5

Region of sky a few degrees either side of the ecliptic where the sun and all the planets are always found

Question 6

Observers meridian

Answer 6

Imaginary line across the sky from the observers southern horizon to the observers northern horizon, also point where objects culminate

Question 7

Observers zenith

Answer 7

The location directly above the person in the sky, 90 degrees from ever horizon, 90 degrees altitude

Question 8

Autumnal equinox

Answer 8

Where the ecliptic and the celestial equator meet and the Sun moves from the northern sky to the southern sky (declination 0°)

Question 9

Winter solstice

Answer 9

Sun has lowest possible declination of -23.5° so the northern hemisphere is tilted directly away from the Sun and it is furthest south from the celestial equator it will get

Question 10

Vernal equinox

Answer 10

Where the ecliptic and the celestial equator meet and the Sun moves from the southern sky to the northern sky (declination 0°),

Question 11

Summer solstice

Answer 11

When the Sun has highest possible declination of +23.5° so the northern hemisphere is tilted directly towards the Sun and it is furthest north from the celestial equator it will get.

Question 12

Tropics

Answer 12

The Tropic of Cancer is L = +23.5° and the Tropic of Capricorn is L = -23.5°. This angle is because that is the tilt of the Earth from 90° away from the ecliptic. Therefore the tropics are the most northerly and southerly latitudes respectively where the Sun can appear on the observer’s zenith (which it does on 21st June for Cancer and 21st December for Capricorn).

Question 13

What is the Right Ascension of the First Point of Aries, and on which date is the Sun in front of it

Answer 13

FPoA RA = 0 h. The Sun is in front of it at midday on the vernal equinox (about 20th March).

Question 14

Dave observes that the Sun is due south at his location on a particular date when his watch (set to GMT) says 11:48. The value of the Equation of time for that date is –4 minutes.
What is his AST

Answer 14

1200 – Sun due south means it is on the meridian, culminating, shadows pointing due north, shadows shortest.

Question 15

MST equation

Answer 15

MST = AST – EoT

Question 16

Dave observes that the Sun is due south at his location on a particular date when his watch (set to GMT) says 11:48. The value of the Equation of time for that date is –4 minutes.
Calculate MST

Answer 16

MST = AST – EoT = 1200 – (-4) = 1204

Question 17

Longitude equation

Answer 17

L = ¼(MST – GMT)

Question 18

Dave observes that the Sun is due south at his location on a particular date when his watch (set to GMT) says 11:48. The value of the Equation of time for that date is –4 minutes.
Calculate Dave’s longitude

Answer 18

L = ¼(MST – GMT) = ¼(1204 – 1148) = ¼(16) = +4°. Positive means E, so 4°E.

Question 19

What is the radiant of a meteor shower?

Answer 19

The radiant is the point on the sky from which all the meteors seem to come. It is there because as the Earth moves into the region of debris, the first point on the Earth’s atmosphere will be where the rocks fall in from – perspective.

Question 20

How to name a meteor shower

Answer 20

The names of the meteor showers are from the constellation that is behind the radiant. Eg Perseids has the radiant in front of Perseus.

Question 21

How to find out where a meteor’s radiant is

Answer 21

• Go out with a star chart or graph paper on a dark, clear night.
• Wait 20 minutes+ to get dark-adapted eyes.
• Watch for a meteor, and put the streak on the star chart (if using graph paper, you’d have to plot a good few stars first!).
• Repeat this process for many meteors.
• Feed the line of each streak backwards until you find a point where all the lines cross.
• Mark on the radiant at this point, and identify the constellation it is in front of to name the meteor shower.

Question 22

How to prepare for naked eye astronomy

Answer 22

. You must dress warmly (clear sky = cold night)
. Spend at least 20 minutes outside without looking at light sources, so your eyes become dark-adapted (the pupils expand so that more light can enter and you can pick out fainter objects).
. Any light to help should have a red filter on it as the pupil does not respond much to red. The cells at the edges of our retina are more light-sensitive than those in the middle, so looking past/to the side of an object (averted vision) rather than directly at it means it will appear brighter.

Question 23

Seeing conditions vs weather

Answer 23

Seeing conditions is about the stillness of an object on the sky (whether it is twinkling or not) caused by the amount of movement of air. Weather affects how transparent the sky is (clouds etc).

Question 24

What is magnitude

Answer 24

Absolute magnitude is the brightness of an object as viewed from a distance 10 pc. Apparent magnitude is the brightness of an object as viewed from Earth. With the naked eye, we can make out apparent magnitudes down to +6 (dimmest). The brightest apparent magnitude in the sky is about 0.

Question 25

How to determine effect of light pollution

Answer 25

* Research and record the accepted apparent magnitude of some reference stars. Do this for stars with apparent magnitude ranging from 0 to +6. * Go out and observe these stars and record the apparent magnitude according to your judgement. * Determine which stars you researched cannot be seen, and write down the apparent magnitude that is not visible.

Question 26

Why do the phases of the Moon occur?

Answer 26

Phases occur due to the arrangement of the Earth, Sun and Moon. Eg when the Moon is between Earth and Sun, it’s NEW (not visible). When it is on the far side of the Earth from the Sun, it is FULL.

Question 27

Why do we only see one side of the Moon?

Answer 27

The Moon is TIDALLY LOCKED to the Earth – the heavier side rotates to face us, so the rotational period (spin) of the Moon is exactly the same as the orbital period of the Moon – both 27.3 days.

Question 28

Lunar libration

Answer 28

Effect of the Moon’s wobble - a slight change in the side facing us so we can see slightly more than 50% of the whole surface – due to the elliptical shape of its orbit and the slight inclination of its orbit (not quite on the ecliptic). The elliptical shape causes it to wobble left and right; the inclination causes it to wobble up and down.

Question 29

Precession

Answer 29

Wobble of the Earth’s axis over thousands of years. At the moment, Polaris is almost exactly above the north pole (so very close to the celestial north pole) but it will move away before returning in 26000 years’ time.

Question 30

The altitude of a star of declination +5° for Sheila at latitude 20°N

Answer 30

``` L = 90 – A + D A = 90 – L + D = 90 – 20 + 5 = 75° ```

Question 31

The declination of a star if it is seen with altitude 15° for a Trixie at latitude 0°,

Answer 31

``` L = 90 – A + D D = L – 90 + A = 0 – 90 + 15 = -75° ```

Question 32

The altitude of Polaris for an observer at any latitude

Answer 32

Polaris’ declination is 90°, so the equation becomes: L = 90 – A + 90 so L = 180 – A and A = 180 – L. That means that whatever your latitude is, IF YOU FACE NORTH INSTEAD OF SOUTH, then Polaris will be that altitude above your northern horizon. Eg: L = 50°N. A = 180 – 50 = 130° above southern horizon. That’s the same as 50° above the northern horizon.

Question 33

Observer latitude

Answer 33

Observer Latitude = 90 – object Altitude at culmination + object Declination L = 90 – A + D

Question 34

Umbral contacts

Answer 34

U1 is the 1st umbral contact, U2 is the 2nd umbral contact, U3 is the 3rd umbral contact, U4 is the 4th umbral contact. • Measure the time from 1st to 2nd umbral contact (U1 to U2). • Measure the time from 1st to 3rd umbral contact (U1 to U3). • Do the second time (time to cross the umbra) divided by the first time (time to enter the umbra). The answer is the number of times the Earth’s diameter is bigger than the Moon’s.

Question 35

Local Sidereal Time and Local Hour Angle

Answer 35

The local sidereal time (LST) is the value of right ascension that is due south for the observer (the RA on the local meridian). The local hour angle is the angle away from the meridian an object appears (H). The right ascension of an object is the angle it is away from the First point of Aries (vernal equinox point), α. So the right ascension of a star can be calculated by doing: RA (α) = LST – LHA (H).