| Year | Previous Year Winter Solstice1 | 2nd New Moon after Winter Solstice2 | Paschal Full Moon3 | Chinese New Year | Easter Sunday | Ash Wednesday |
|---|---|---|---|---|---|---|
1996 Feb 19 07:30 BST |
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2004 Jan 22 05:05 BST |
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2005 Feb 9 06:28 BST |
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2007 Feb 18 00:14 BST |
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2014 Jan 31 05:40 BST |
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2014 Dec 22 06:59 BST |
2015 Feb 19 07:48 BST |
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2018 Feb 16 05:06 BST |
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2019 Feb 5 05:04 BST |
2http://aa.usno.navy.mil/data/docs/MoonPhase.html (1996-2013), http://users.hartwick.edu/hartleyc/moon/newmoon6.html (2014-2019)
3http://users.sa.chariot.net.au/~gmarts/easter.htm Back to Main Menu
Celestial Sphere
PROBLEM
Determine the optimal day of the year for observing the following cosmic X-ray sources, with the indicated RA and Dec coordinates:
(a) Crab Nebula (5h 34m 31s, +22d 01m) (Green 2004)
(b) SN1006 (15h 02m 50s, -41d 56m) (Green 2004)
(c) Ophiuchus Cluster (17h 12m 26.9s, -23d 22m 08.4s) (http://www.astro.isas.jaxa.jp/astroe/proposal/ao1/ao1list.html)
(d) Cygnus Loop (20h 51m 00s, +30d 40m) (Green 2004)
SOLUTION
(a) Convert RA into hours. Then divide by 24 hours and multiply by 365.25 days to determine how many days after the autumnal equinox the observation should be made. Assume the autumnal equinox occurs on Sept. 22, which would be ideal for observing a source with RA = 0.
RA = 5h 34m 31s = 5.575 h => 85 days => Dec. 16
(b) RA = 15h 02m 50s = 15.047 h => 229 days => May 9
(c) RA = 17h 12m 26.9s = 17.207 h => 262 days => June 11
(d) RA = 20h 51m 00s = 20.85 h => 317 days => Aug. 5
Green, D. A. 2004, "A Catalogue of Galactic Supernova Remnants," Bulletin of the Astronomical Society of India, 32, 335-370. (http://www.mrao.cam.ac.uk/surveys/snrs/tables-l.pdf)
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Telescopes
PROBLEM
The Meade 114EQ-AST telescope comes with an objective mirror with diameter D = 4.5" = 114.3 mm and focal length fo = 1000 mm and eyepieces with focal lengths fe = 25 mm and 9 mm.
(a) What magnifications are possible?
(b) What is the focal ratio of the telescope?
(c) Estimate the angular size of the Moon without a telescope and as viewed using this telescope with the lower magnification.
(d) Estimate the angular sizes of Mars, Jupiter, and Saturn at opposition, without a telescope and as viewed using this telescope with the higher magnification.
SOLUTION
(a) The magnification is given by
M = fo / fe
So, using the 25 mm eyepiece, the magnification is
M = (1000 mm) / (25 mm) = 40
and, using the 9 mm eyepiece, the magnification is
M = (1000 mm) / (9 mm) = 111
(b) The focal ratio is the ratio between the focal length fo and the diameter D.
FR = fo / D = (1000 mm) / (114.3 mm) = f/8.749
(c) The Moon's orbit has semimajor axis a = 384400 km and eccentricity e = 0.055. The Moon's radius is Rmoon = 1738 km. Its perigee and apogee distances are
rp = a(1 - e) = 363258 km
and
ra = a(1 + e) = 405542 km
The minimum and maximum angular sizes of the Moon as seen from Earth without a telescope are
qmin = tan-1(2Rmoon/ra) = tan-1[(2)(1738 km) / (405542 km)] = 8.571 x 10-3 rad = 0.4911°
and
qmax = tan-1(2Rmoon/rp) = tan-1[(2)(1738 km) / (363258 km)] = 9.569 x 10-3 rad = 0.5482°
With a magnification of M = 40, the minimum and maximum angular sizes of the Moon as seen from Earth are
qmin' = qminM = (8.571 x 10-3 rad)(40) = 0.9523 rad = 54.56°
and
qmax' = qmaxM = (9.569 x 10-3 rad)(40) = 1.063 rad = 60.92°
(d) Mars' orbit has semimajor axis a = 227.9 x 106 km and eccentricity e = 0.093. Mars' radius is Rmars = 3397 km. Its perihelion and aphelion distances are
rp = a(1 - e)
and
ra = a(1 + e)
The radius of Earth's orbit, assumed to be circular, is re = 149.6 x 106 km. The minimum and maximum center-to-center distances from Earth to Mars are
rmin = rp - re = a(1 - e) - re = (2.279 x 108 km)(1 - 0.093) - 1.496 x 108 km = 5.711 x 107 km
and
rmax = ra - re = a(1 + e) - re = (2.279 x 108 km)(1 + 0.093) - 1.496 x 108 km = 9.949 x 107 km
The minimum and maximum angular sizes of Mars as seen from Earth without a telescope are
qmin = tan-1(2Rmars/rmax) = tan-1[(2)(3397 km) / (9.949 x 107 km)] = 6.829 x 10-5 rad = 0.2347' = 14.09"
and
qmax = tan-1(2Rmars/rmin) = tan-1[(2)(3397 km) / (5.711 x 107 km)] = 1.190 x 10-4 rad = 0.4090' = 24.54"
With a magnification of M = 111, the minimum and maximum angular sizes of Mars as seen from Earth are
qmin' = qminM = (6.829 x 10-5 rad)(111) = 7.587 x 10-3 rad = 0.4347°
and
qmax' = qmaxM = (1.190 x 10-4 rad)(111) = 1.322 x 10-2 rad = 0.7574°
Jupiter's orbit has semimajor axis a = 778.4 x 106 km and eccentricity e = 0.048. Jupiter's radius is Rjup = 71398 km. For Jupiter, we get
rmin = 5.914 x 108 km
rmax = 6.662 x 108 km
qmin = 2.144 x 10-4 rad = 0.7369' = 44.21"
qmax = 2.414 x 10-4 rad = 0.8300' = 49.80"
qmin' = 2.382 x 10-2 rad = 1.365°
qmax' = 2.683 x 10-2 rad = 1.537°
Saturn's orbit has semimajor axis a = 1425.6 x 106 km and eccentricity e = 0.054. Saturn's radius is Rsat = 60000 km. For Saturn, we get
rmin = 1.199 x 109 km
rmax = 1.353 x 109 km
qmin = 8.869 x 10-5 rad = 0.3049' = 18.29"
qmax = 1.001 x 10-4 rad = 0.3441' = 20.64"
qmin' = 9.855 x 10-3 rad = 0.5646°
qmax' = 1.112 x 10-2 rad = 0.6371°
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Moon
PROBLEM
The Moon rotates about its axis in 27.322 days. Its orbit is approximately elliptic with semimajor axis a = 384,400 km and eccentricity e = 0.055. Because its orbital velocity varies according to Kepler's second law ("The radius vector of a planet sweeps equal areas in equal amounts of time"), the side of the Moon facing the Earth varies slightly during its orbit. Calculate the total fraction of the Moon's surface which is eventually visible from the Earth as a result of this.
SOLUTION
The radial distance r from the center of the Earth to the center of the Moon can be written as
r = 1 / (B + A cos q)
where q is its angular displacement from its perigee position,
A = e / [a(1 - e2)] = 1.43514 x 10-7 km-1 = 1.43514 x 10-10 m-1
and
B = 1 / [a(1 - e2)] = 2.60935 x 10-6 km-1 = 2.60935 x 10-9 m-1
B is also equal to
B = - mK/L2
where
K = - GMem
G = 6.67 x 10-11 N m2/kg2 is the gravitational constant, Me = 5.974 x 1024 kg is the mass of the Earth (Karttunen et al. 1987), m is the mass of the Moon, and L is the orbital angular momentum of the Moon. Thus,
B = mGMem/L2 = GMe(m/L)2 => L/m = sqrt(GMe/B) = sqrt[(6.67 x 10-11 N m2/kg2)(5.974 x 1024 kg)/(2.60935 x 10-9 m-1)] = 3.908 x 1011 m2/s
Now the orbital angular momentum of the Moon is constant, and since the mass of the Moon is also assumed constant, L/m is also a constant. The orbital angular momentum is
L = r x p => L = rmvq = rm r dq/dt = mr2 dq/dt => L/m = r2 dq/dt
where
vq = r dq/dt
is the velocity component perpendicular to the radius vector. It follows that r2 dq/dt is also a constant
r2 dq/dt = L/m = 3.908 x 1011 m2/s
So
dq/dt = C/r2 = C(B + A cos q)2 (I)
where C = 3.908 x 1011 m2/s.
We will now calculate the amount by which the visible side of the Moon varies during its orbit. From the above equation (I) for dq/dt, we know the orbital angular velocity of the Moon as a function of its angular position q. Define t = 0 as the time when the Moon is at perigee. The above expression for dq/dt can be integrated to get the time at which the Moon is at some orbital angular position q0.
t = ∫0q0 dq / [C(B + A cos q)2]
The integral may be done numerically using a step size of 2p/360, for example. Set up a spreadsheet in Excel with the following columns:
Column A: an index number (i = 1, 2, 3,..., 361) representing positions in the orbit. There will be a total of 361 rows in the spreadsheet. The (first, last) row corresponds to q = (0, 2p).
Column B: the constant A = 1.43514 x 10-10 in units of m-1
Column C: the constant B = 2.60935 x 10-9 in units of m-1
Column D: the constant C = 3.908 x 1011
Column E: the angular position qi = (i - 1)(2p/360)
Column F: the angular velocity (dq/dt)i = C(B + A cos qi)2
Column G: the time interval (dt)i elapsed between the (i-1)th row and the ith row, (dt)i = (qi - qi-1) / {(0.5)[(dq/dt)i-1 + (dq/dt)i]} (i.e., the change in q divided by the average value of dq/dt during the time interval), which is set equal to zero for the first row
Column H: the time ti corresponding to the ith row, which is the sum of the values of dt for all rows up to and including the ith row
Column I: the angular orientation ai of the Moon about its rotation axis relative to its orientation at perigee, ai = (i-1)(2p/360)
Column J: the difference qi - ai
Lines 1, 2, 3, and 361 should have an appearance equivalent to the following for Columns A, E, F, G, H, I, and J (i, qi, [dq/dt]i, [dt]i, ti, ai, and qi - ai):
A E F G H I J
i qi (dq/dt)i (dt)i ti ai qi-ai
1 0.000000 2.96141E-06 0.000000 0.000000 0.000000000 0.000000
2 0.017453 2.96137E-06 5893.611095 5893.611095 0.015609964 0.001843
3 0.034907 2.96122E-06 5893.798267 11787.40936 0.031220424 0.003686
361 6.283177 2.96141E-06 5893.611123 2372241.406 6.283177198 2.8E-06
The last entry in Column H is the number of seconds elapsed by the time the Moon gets back to perigee, which is 2372241.406 s = 27.4565 days. This is slightly different from the Moon's measured orbital period of 27.322 days because the real orbit is slightly different from an ellipse. The formula for period t as a function of the semimajor axis is
t = sqrt(4p2a3/GMe) = sqrt[4p2(3.844 x 108 m)3/(6.67 x 10-11 N m2/kg2)(5.974 x 1024 kg)] = 2372242.291 s = 27.4565 days
which is consistent with the above.
The minimum and maximum values of qi - ai in Column J are ± 0.11003 rad, which indicates that (2)(0.11003 rad) = 0.22 rad of the Moon's surface eventually comes into view due to the variation in its orbital velocity. Thus, the total fraction of the Moon's surface that eventually comes into view is 0.5 + (0.22 rad)/(2p rad) = 0.535 or 53.5%.
Note: The process discussed here, where the variation in the Moon's orbital velocity allows more of its surface to be seen, is an example of a libration. There are other librations or processes that allow us to see more of the Moon's surface. As a result of these other librations (http://www-spof.gsfc.nasa.gov/stargaze/Smoon4.htm), the total fraction of the Moon's surface that can be observed from the Earth at one time or another is about 59%.
At any instant of time, slightly less than 50% of the Moon's surface can be seen from the Earth. Imagine a cone whose vertex is at the observer and which extends towards the Moon so that its side is tangent to the Moon's surface. Assume that the center-to-center distance between the Earth and the Moon is equal to the semimajor axis a = 3.844 x 105 km. The distance from the observer to a point of contact between the cone and the Moon's surface is
r = sqrt[(a - Re)2 - Rm2]
where Rm = 1738 km is the radius of the Moon. Thus,
r = sqrt[(3.844 x 105 km - 6.37 x 103 km)2 - (1738 km)2] = 3.780 x 105 km
The angle g between the center-to-center line between the Moon and the Earth and the radius from the center of the Moon to a point of tangency is
g = tan-1(r/Rm) = tan-1[(3.780 x 105 km) / (1738 km)] = 1.566 rad = 89.74°
The total solid angle of the Moon's surface which is visible from the Earth is
W = ∫ dW = ∫01.566 sin q dq ∫02p df = 2p(cos 0 - cos 1.566) = 2p(0.995402) ster
which is [2p(0.995402) ster] / (4p) = 0.497701 = 49.7701% of the total solid angle. This is the fraction of the Moon's surface which is visible from the Earth at any given instant, assuming no cloud cover or other obstructions.
Karttunen, H., Kröger, P., Oja, H., Poutanen, M. and Donner, K. J. 1987, Fundamental Astronomy (Berlin, Germany: Springer-Verlag).
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Mars
PROBLEM
(a) What is the difference between an opposition of Mars and a closest approach?
(b) From a table of oppositions, determine when Mars made its closest approach to Earth in the past 100 years and when it will make its closest approach to Earth in the next 100 years.
For the closest approach in the next 100 years:
(c) What will be the magnitude and size of Mars as viewed from Earth?
(d) What telescope magnification will be needed so that Mars will appear as large as the Moon viewed with the naked eye?
SOLUTION
(a) An opposition of Mars occurs when the ecliptic longitudes of Mars and the Sun are 180° apart, which happens when Earth is between Mars and the Sun. A closest approach between Mars and Earth occurs when the distance between the two planets is minimized. Because of the shapes and orientations of their orbits, oppositions and closest approaches occur at approximately but not exactly the same time. They could occur on different days.
(b) Tables of Mars oppositions and closest approaches are available at http://www.seds.org/~spider/spider/Mars/marsopps.html. From the table of closest approaches, those occurring between 1906 and 2106 are listed in the table below.
Year
Opposition
Date
Closest Approach
Date
Distance (km)
Distance (AU)
1924 Aug 23 Aug 22 55,776,939 0.37284581
2003 Aug 28 Aug 27 55,758,006 0.37271925
2050 Aug 14 Aug 15 55,957,000 0.374051
2082 Sep 1 Aug 30 55,883,780 0.373564
Between 1906 and 2006, Mars made its closest approach to Earth on 27 August 2003. Between 2006 and 2106, Mars will make its closest approach to Earth on 30 August 2082.
(c) The diameter of Mars is d = 4212 miles = 6.779 x 106 m. At a distance of r = 5.588 x 1010 m, the diameter of Mars subtends an angle of
q = tan-1(d/r) = tan-1[(6.779 x 106 m) / (5.588 x 1010 m)] = 1.213 x 10-4 rad = 25.02"
= 6.951 x 10-3 ° = q2082
According to Norton's Star Atlas, Twentieth Edition, during the 2005 opposition, Mars subtended an angle of q2005 = 19.9" and had a magnitude of M2005 = - 2.3. A difference of five magnitudes corresponds to a factor of 100 difference in brightness. Thus, one magnitude corresponds to a factor of 1001/5 = 2.511886432 difference in brightness. The brightness of an object is inversely proportional to the square of its distance. The angular size of a distant object is inversely proportional to its distance. Thus, the brightness of a distant object is proportional to the square of its angular size. It follows that the ratio between the brightness of Mars during its 2082 opposition and during its 2005 opposition is
B2082/B2005 = (q2082/q2005)2 = (25.02"/19.9")2 = 1.581
and Mars will therefore appear 1.581 times brighter during its 2082 opposition than it did during its 2005 opposition. The difference in magnitude is proportional to the difference in the logarithm of the brightness. The proportionality constant is - 5/2. Thus,
M2082 - M2005 = - (5/2)(log B2082 - log B2005) = - (5/2) log(B2082/B2005) = - (5/2) log 1.581
= - 0.4972 => M2082 = M2005 - 0.4972 = - 2.3 - 0.4972 = - 2.797
During the 2082 opposition, Mars will have a magnitude of M2082 = - 2.797.
(d) The diameter of the Moon is d = 2160 miles = 3.476 x 106 m. Its mean distance from Earth is r = 2.389 x 105 miles = 3.844 x 108 m. Its mean angular size is
qMoon = tan-1(d/r) = tan-1[(3.476 x 106 m) / (3.844 x 108 m)] = 0.5181°
The ratio between the angular size of the Moon and the angular size of Mars during the 2082 opposition is (0.5181°) / (6.951 x 10-3 °) = 74.54. Thus, a telescope whose magnification is 75 would be needed to make Mars appear as large as the Moon appears to the naked eye.
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| Year | Opposition Date |
Closest Approach Date |
Distance (km) | Distance (AU) |
|---|---|---|---|---|
Between 1906 and 2006, Mars made its closest approach to Earth on 27 August 2003. Between 2006 and 2106, Mars will make its closest approach to Earth on 30 August 2082.
(c) The diameter of Mars is d = 4212 miles = 6.779 x 106 m. At a distance of r = 5.588 x 1010 m, the diameter of Mars subtends an angle of
q = tan-1(d/r) = tan-1[(6.779 x 106 m) / (5.588 x 1010 m)] = 1.213 x 10-4 rad = 25.02"
= 6.951 x 10-3 ° = q2082
According to Norton's Star Atlas, Twentieth Edition, during the 2005 opposition, Mars subtended an angle of q2005 = 19.9" and had a magnitude of M2005 = - 2.3. A difference of five magnitudes corresponds to a factor of 100 difference in brightness. Thus, one magnitude corresponds to a factor of 1001/5 = 2.511886432 difference in brightness. The brightness of an object is inversely proportional to the square of its distance. The angular size of a distant object is inversely proportional to its distance. Thus, the brightness of a distant object is proportional to the square of its angular size. It follows that the ratio between the brightness of Mars during its 2082 opposition and during its 2005 opposition is
B2082/B2005 = (q2082/q2005)2 = (25.02"/19.9")2 = 1.581
and Mars will therefore appear 1.581 times brighter during its 2082 opposition than it did during its 2005 opposition. The difference in magnitude is proportional to the difference in the logarithm of the brightness. The proportionality constant is - 5/2. Thus,
M2082 - M2005 = - (5/2)(log B2082 - log B2005) = - (5/2) log(B2082/B2005) = - (5/2) log 1.581
= - 0.4972 => M2082 = M2005 - 0.4972 = - 2.3 - 0.4972 = - 2.797
During the 2082 opposition, Mars will have a magnitude of M2082 = - 2.797.
(d) The diameter of the Moon is d = 2160 miles = 3.476 x 106 m. Its mean distance from Earth is r = 2.389 x 105 miles = 3.844 x 108 m. Its mean angular size is
qMoon = tan-1(d/r) = tan-1[(3.476 x 106 m) / (3.844 x 108 m)] = 0.5181°
The ratio between the angular size of the Moon and the angular size of Mars during the 2082 opposition is (0.5181°) / (6.951 x 10-3 °) = 74.54. Thus, a telescope whose magnification is 75 would be needed to make Mars appear as large as the Moon appears to the naked eye.