1 MOVEMENTS OF THE EARTH (Apparent movement of the sun, time determination and geographic...
-
Upload
archibald-cook -
Category
Documents
-
view
219 -
download
0
Transcript of 1 MOVEMENTS OF THE EARTH (Apparent movement of the sun, time determination and geographic...
1
MOVEMENTS OF THE EARTH(Apparent movement of the sun,
time determination and geographic coordinates)
PROBLEM SOLUTIONS
Teaching Team:Prof. Alfonso Calera Belmonte (Dpt. Applied Physics, UCLM)Prof. Antonio J. Barbero (Dpt. Applied Physics, UCLM)Consultant:Prof. Kathy Walsh (Dpt. Modern Languages, UCLM)
Physics
Environmental
LESSON 1
2
Physics
Environmental
Eratóstenes of Cirene (284-192 B.C), a greek astronomer, geographer, mathematician and philosopher, was the first wise man who got an accurate measure of the circumference of the Earth. His procedure to do that based on the following: he knew that at noon of the summer solstice day the rays of the sun dropped vertically on the bottom of a well in the ancient city of Siena (near the actual Assuan, Egipt), located almost exactly on the Cancer tropic. Using a gnomon* he could measure the inclination angle of the solar rays in the city of Alexandria, the same day at the same time. Alexandría was located about 800 km to the North from Siena (Eratostenes was director of the famous Library of Alexandría). The angle measured by Eratostenes was 7º14’.From those data, determine the Earth circumference (or its radious).
(At the Eratostenes time, the most difficult data to obtain was an accurate value for the distance between the two cities).
PROB. 0101 / MEASURING THE EARTH RADIUS
http://www.astromia.com/biografias/eratostenes.htm
* Gnomon: a vertical stick over a flat platform, by measuring the length of the shadow it is possible calculating the incident angle of the rays of the sun.
3
Physics
Environmental
Siena
Alexandria
TROPIC OF CANCER
0
0
d
R
36º
24º
28º
32º
http://www.lib.utexas.edu/maps/africa/egypt_pol97.jpg
dR
= 7º14’ = 7.23º d 800 km
d
R km6340
18023.7
800
km398342 R
PROB. 0101 / MEASURING THE EARTH RADIUS (CONTINUED)
4
The geographic coordinates of Palma de Mallorca are 39º34’ N, 2º39’ E and those of Edinburgh are 55º57’, N 3º10’ WCompare the following quantities for the days December 4th and June 4th in both cities:A. The official times of sunrise and sunset.B. The lenght of the day.C. Solar elevation and azimut at noon (official time). D. What time (official) is the sun crossing the local meridian? You have to have in mind the fact that in Spain the official winter time and the official summer time are forwarded respect GMT by 1 hour and 2 hours, respectively. In Britain, the official summer time is forwarded by 1 hour, but the official winter time has no forward respect GMT.
PROB. 0102 / TIME AND DAY LENGTH
Physics
Environmental
5
ΦΦcoscos
Φcos tantan
sinsins
Sunrise hour angle:
33611.0)º57.39()(-22.14ºtΦcos tanantantans º36.70s
Declinations: º14.22
June 4th (J=155) º34.22
December 4th (J=338)
Palma Mallorca º57.39'34º39
Edinburgh
º95.55'57º55
(Dec 4th)
33972.0)º57.39()(22.34ºtΦcos tanantantans º86.109s (Jun 4th)
60209.0)º95.55()(22.34ºtΦcos tanantantans º98.52s (Dec 4th)
60821.0)º95.55()(22.34ºtΦcos tanantantans º46.127s (Jun 4th)
PROB. 0102 / TIME AND DAY LENGTH (CONTINUED) PART A)
Physics
Environmental
6
horas
/º15)(º
00:00:12 (LAT) time Sunrise
Local apparent time at sunrise:
PROB. 0102 / TIME AND DAY LENGTH (CONTINUED) PART A)
Physics
Environmental
7
Sunrise time (LST) LST = LAT - 4(Ls-Le) - Et
Palma M.4-dic, Calculus. Exemple
LST = 7:18:35 - 4(0-(-2.65º)) - 9.59
Converting longitude degrees to minutes
Time equation of the day
Local longitude (decimal fraction of a degree)
LAT in minutes
Standard meridian longitude (in this case, Greenwich)
= 438.56 min -10.60 min -9.59 min = 418.39 min
= 6.973 h = 6:58:24
PROB. 0102 / TIME AND DAY LENGTH (CONTINUED) PART A)
Physics
Environmental
8
Sunrise time (LST)
LST = LAT - 4(Ls-Le) - Et
Winter (Dec 4th): Official time = LST + 1 = 7h 58m 23s
Summer (Jun 4th): Official time = LST + 2 = 6h 27m 53s
Palma de Mallorca:
Official time:
Edinburgh:
Winter (Dec 4th): Official time = LST = 8h 31m 09s
Summer (Jun 4th): Official time = LST + 1 = 4h 40m 43s
PROB. 0102 / TIME AND DAY LENGTH (CONTINUED) PART A)
Physics
Environmental
9
horas
/º15
)(º2day theofLength
52.98
127.46
4-dic
4-jun
º sEdinburgh
4-dic
4-jun
º sPalma M.
70.36
109.86
Length of the day.
PROB. 0102 / TIME AND DAY LENGTH (CONTINUED) PART B)
Physics
Environmental
10
zsinsinsin coscosΦcoscosΦ
Φcoscos
ΦΨcos
sinsinsin
Solar elevation
Azimut
Official time LST
Palma M., 4-dic 12:00:00 11:00:00
Palma M., 4-jun 12:00:00 10:00:00
Edimburgo, 4-jun 12:00:00 11:00:00
Edimburgo, 4-dic 12:00:00 12:00:00
LAT
11 56 56
10 49 26
11 20 12
10 12 42
h m s
PROB. 0102 / TIME AND DAY LENGTH (CONTINUED) PART C)
LAT = LST + 4·(Ls-Le) + Et
Physics
Environmental
11
)12(15 HSL
Hour angle determination (degrees) from LAT
Palma M., Dec 4th
Palma M., Jun 4th
Edinburgh, Jun 4th
Edinburgh, Dec 4th 0.8
17.6
)(º
10.0
26.8
11.91 0.73
53.95 28.44
27.60 10.41
61.46 60.86
)(º )(º
zsinsinsin coscosΦcoscosΦ
Φcoscos
ΦΨcos
sinsinsin
Solar elevation
Azimut
PROB. 0102 / TIME AND DAY LENGTH (CONTINUED) PART C)
Physics
Environmental
12
Determination of the official time when the sun passes across the local meridian. We add 1 or 2 hours to LST corresponding to LAT = 12:00
Palma M., Dec 4th
Palma M., Dec 4th
Edinburgh, Jun 4th
Edinburgh, Dec 4th 11 50 24
11 57 54
11 50 24
11 57 54
h m s
11:50:24
12:57:54
12:50:24
13:57:54
LST Official time
The sun is passing through the local meridian
PROB. 0102 / TIME AND DAY LENGTH (CONTINUED) PART D)
Physics
Environmental
13
Physics
Environmental
On the evening of the 15th April Her Majesty’ agent 007, James Bond, is kidnapped in London by enemy agents from a secret organization attempting to swap him with one of their own main ringleaders, who is getting a holiday paid by the government as a convict in a britain prison.Bond is inmediately carried by plane out of the country and that night he is locked up in an secret hiding-place abroad. However before the sunrise, a few hours later, our clever agent 007 escapes and finds a safe refuge in the tower of a church from which he can watch over an extensive area around him.Once he’s convinced to have eluded his kidnappers, Bond waits for the sunrise and when the sun is just rising in the horizon he uses the Polar star as the north point reference and, using two straight sticks and his splendid analogical watch (the stupid kidnappers did not take it away, they didn’t notice that these watch was also a pocket computer), he gets a measure of the angle between the rising sun and the north direction. The result of that measure is 70º towards the East. After that, he catches a carrier pigeon from the pigeon loft of the tower, and uses some old wooden boards he finds there to built a cage to keeping the animal. Furthermore, he uses a flat board and another stick to build a simple ‘gnomon’ (see next page picture).
PROB. 0103 / DETERMINATION OF LATITUDE AND LONGITUDE
70º
Polar star
14
Physics
Environmental
When the sun is about to reach its maximum elevation, Bond watches carefully the shadow of the stick and writes down the time of the tower clock at the moment when the shadow reaches its minimum length. This clock indicates 11:44 h, whereas on his own watch he can see 9:44 h. Using these measurements along with a table he gets from the memory of his wonderful watch, and after using also the calculator that of course it has, Bond pulls out a sheet from his notebook, writes on it the geographical coordinates of the place where he actually stands and a brief note addressed to the local government of the country where he actually is, asking for a rescue helicopter. Straight away, he ropes the piece of paper to the leg of the carrier pigeon and releases it. Then, he sits calmly down and waits for the rescue.
Which are the geographic coordinates? The government of which country has he asked for rescue?
Gnomon
Minimum lenght
PROB. 0103 / DETERMINATION OF LATITUDE AND LONGITUDE (CONTINUED)
At this moment, the tower clock indicates 11:44 and Bond’s watch indicates
9:44
15
Physics
Environmental
W
E
S N
= 110º
70º
Sun crossing over the local meridian
at 11:44 LST
Besides we know from the tower clock that when the sun reaches the meridian it is 11:44 (LST) and of course it is 12:00 (LAT)
PROB. 0103 / DETERMINATION OF LATITUDE AND LONGITUDE (CONTINUED)
If the angle to the north is 70º at sunrise, then the solar azimut is
= 180º-70º = 110º
Moreover, from Bond’s watch we also know that in London it is 9:44
16
Physics
Environmental
Φcoscos
ΦΨcos
sinsinsinRelationship between azimuth
and latitude, declination and solar altitude angles:
At sunrise = 0 Ψcos
Φcossin
The kidnapping occurred on April 15th, and Bond escapes the following day, which is the 106th day on the year (assuming no leap year).
The latitude is obviously 60º N for the observer can see the Polar star.
Latitude determination: 500.0º011cos
º84.9Φcos sin
º60500.0cosΦ 1
April 1st
PROB. 0103 / DETERMINATION OF LATITUDE AND LONGITUDE (CONTINUED)
April 16th
For day J = 106 we find = 9.84º, and Et = 0.02 min, so we’ll consider negligible the Et contribution.
17
Physics
Environmental
Longitude determination
LAT = LST + 4 (Ls-Le) + Et
Local standard time Time equation
Local apparent time
12:00 = 11:44 + 4 (Ls-Le) + Et
0:00
Longitude correction
Longitude correction 4 (Ls-Le) = 12:00 - 11:44 = +0:16
+16 minutes towards E from standard meridian = +4º from standard meridian of this place
What is the standard meridian of this place?
PROB. 0103 / DETERMINATION OF LATITUDE AND LONGITUDE (CONTINUED)
Because the sun takes 4 minutes to going over ONE grade
Difference of longitude between the standard meridian
Ls and the local meridian Le
(Ls-Le) = 4º
18
Physics
Environmental
Standard meridian of this place
Bond’s watch indicates London time: there it is 09:44 LST (the same in Greenwich).The tower clock indicates local standard time: it is 11:44 LST.So, we conclude from these difference (2 hours) that the standard meridian lies 30º E from Greenwich: Ls = -30º.
(Ls-Le) = 4º Le Local meridian longitudeLs Standard meridian longitude
Greenwich
0º
Ls Le
-30º
-4º
E
-34º
(-30º-Le) = 4º -Le = 30º+4º=34º
(towards W, longitude > 0; towards E, longitude < 0)
Le = -34º (34º E)
PROB. 0103 / DETERMINATION OF LATITUDE AND LONGITUDE (CONTINUED)
19
Physics
Environmental
34º E
60º N N
Geographic coordinates: 60º N, 34º E
Country: Rusia
PROB. 0103 / DETERMINATION OF LATITUDE AND LONGITUDE (CONTINUED)
20
On 31st January 2005 a loner navigator happens on some point of the North Atlantic ocean. He has a sextant aboard and also an accurate clock keeping GMT time. By using his sextant the navigator finds that when the sun is passing through the local meridian its altitude upon the horizon is 30º23’24’’, and at this moment the clock indicates that GMT is 14:00:00.
a) What is the geographic position of this navigator (latitude and longitude).
b) The navigator goes always to the bed at sunset. What time will he go to the bed this day? Give the result both in LAT and GMT.
Use table of declination and time equation for January.
PROB. 0104 / DETERMINATION OF LATITUDE AND LONGITUDE ON THE SEA
Mes de ENERODía Declinación (º) Ec. tiempo (min)1 -23.06 -2.902 -22.98 -3.353 -22.89 -3.794 -22.80 -4.235 -22.70 -4.676 -22.59 -5.097 -22.47 -5.528 -22.35 -5.939 -22.21 -6.34
10 -22.07 -6.7411 -21.93 -7.1412 -21.77 -7.5213 -21.61 -7.9014 -21.45 -8.2715 -21.27 -8.6316 -21.09 -8.9917 -20.90 -9.3318 -20.71 -9.6619 -20.51 -9.9920 -20.30 -10.3021 -20.09 -10.6022 -19.87 -10.8923 -19.64 -11.1824 -19.41 -11.4525 -19.17 -11.7026 -18.92 -11.9527 -18.67 -12.1928 -18.42 -12.4129 -18.15 -12.6230 -17.89 -12.8231 -17.61 -13.00
January
Day Declination (º) Time Equation (min)
21
Celestial equator
S
E
W
N
30º23’24’’
PROB. 0104 / DETERMINATION OF LATITUDE AND LONGITUDE ON THE SEA(CONTINUED)
Maximum solar altitude
Daily solar path on 31st Jan
22
PROB. 0104 / DETERMINATION OF LATITUDE AND LONGITUDE ON THE SEA(CONTINUED)
=30º23’24’’
= 90º - (-)
= -17º36’36’’
S
W
Ecuador celeste
N
E
31st January = -17.61º = -17º36’36’’
Et = -13.00 minutes =30º23’24’’= 30.39º
Latitude:
= 90º - (-) =90º - (30º23’24’’-(-17º36’36’’)) =
= 90º - (30º23’24’’-(-17º36’36’’)) = 90º-48º = 42º N
= 42º N
23
Calculation of longitude
LAT = GMT + 4(Ls-Le) + Et
LAT = 12:00 h (noon)
GMT = 14:00 h
Et = -13 min
4(Ls-Le) = 12:00 – 14:00 – (-0:13)
Ls= standard meridian longitude
Le= local meridian longitud
Ls= 0º (Greenwich)
4(Ls-Le) = -120 min – (-13 min) = -107 min
-Le = -107 min/4 (min/grado) = -26.75º
Le = +26.75º = 26º45’ W
= 42º N
Le = 26º45’ W
Le = 26º45’ W
Ls, Le
>0 towards W
<0 towards E
PROB. 0104 / DETERMINATION OF LATITUDE AND LONGITUDE ON THE SEA(CONTINUED)
This is LST in Geenwich
24
ΦtantanΦcoscosΦsinsin
cos
s
Sunrise time on 31st January at 42º N
28580.024tan)61.17tan(cos s
º39.73)28580.0(cos 1 s
min 53 hours 4hours 89.4hour/º15
º39.73
Sunrise hour angle:
At sunset the hour angle is the same than it was at sunrise, but towards the west.The number of hours passed from noon until sunset is:
Sunset time (LAT): 12:00 + 4:53 = 16:53 hours
Sunset time (GMT)
LAT = GMT + 4(Ls-Le) + Et GMT = LAT - 4(Ls-Le) - Et
GMT = 16h 53 min - 4(0-26.75) – (-13) min = 16 h 53 min + 107 min + 13 min = 18 h 53 min
PROB. 0104 / DETERMINATION OF LATITUDE AND LONGITUDE ON THE SEA(CONTINUED)
25
Determine the local apparent time on each of the following cities for each of the below listed days when it is 12:00:00 UTC. Use the Spencer formula for the time equation. It is recomended to use an Excel sheet to perform all calculations.
PROB. 0105 / LOCAL APPARENT TIME
September 21stOctober 22ndNovember 23rdDecember 24thJanuary 25thFebruary 26thMarch 27thApril 28thMay 29thJune 30th
City Latitude Longitude Day
26
LAT = LST + 4 (Ls-Le) + Et
So Ls = 0º0’0’’ and LST = UTC = 12:00:00. We have to calculate a the time equation for each day from Spencer formula.
)18.229)(2sen04089.02cos014615.0
sen032077.0cos001868.0000075.0(
tE
beeing the daily angle 365
12
J (J is the number of the day of the year)
Ciudad Latitud Longitud Día J (rad) Et (min) LAT (HSL)Atlanta 33º45’ N 84º23’ W 21-sep 264 4.5273 6.90 06:29:22Damasco 33º31’ N 36º18’ E 22-oct 295 5.0610 15.66 14:40:51Katmandú 27º49’ N 85º21’ E 23-nov 327 5.6118 13.30 17:54:42Lisboa 38º40’ N 09º10’ W 24-dic 358 6.1455 0.78 11:24:07Madrid 40º24’ N 03º41’ W 25-ene 25 0.4131 -11.70 11:33:18Montevideo 34º50’ S 56º10’ W 26-feb 57 0.9640 -13.39 08:01:57Moscú 55º45’ N 37º37’ E 27-mar 86 1.4632 -5.97 14:24:30Nairobi 01º18’ S 36º47’ E 28-abr 118 2.0141 2.57 14:29:42Pekín 39º55’ N 116º23’ E 29-may 149 2.5477 2.99 19:48:31Tokio 35º41’ N 139º44’ E 30-jun 181 3.0986 -3.26 21:15:41
Since we have to calculate LAT at 12:00:00 UTC, our reference meridian will be that of Greenwich.
PROB. 0105 / LOCAL APPARENT TIME (CONTINUED)
27
An archeological team aboard an scientific ship looking for sunken vessels of historical interest finds on February 13th a wreckage in the Mediterranean sea, near of the Spanish coast. As the GPS aboard is out of order, the captain takes manually some measures to record the position of the shipwreck:
1º) At the sunrise time, the solar azimut is 72.71º.2º) When the sun is crossing the local meridian, the
spanish official time, indicated by the clock of the ship, is 13:17:23.
A) Determine the geographical position of the wreckage. Point out on the map these coordinates.
B) What is the official time for the sunrise this day on that point? What is the solar elevation upon the horizon at 12:00:00 LAT?
36º
38º
40º
0º2º
PROB. 0106 / LOOKING FOR A WRECKAGE
28
Ecuador celeste
S
E
W
N
=72.71º(Sunrise)
13:17:23Official time when the sun crosses the local meridian:
February 13th data from tables:
= -13.63º; Et = -14.26 min
Relationship between azimut, declination, latitude and solar elevation:
ΦcoscossinΦsinsin
Ψcos
PROB. 0106 / LOOKING FOR A WRECKAGE (CONTINUED)
29
At the sunrise time the solar elevation is = 0 79288.0)º71.72cos(
)º63.13sin(Ψcos
sinΦcos
Latitude = cos-1(0.79288) = 37.54º = 37º 32’ 40’’
From this latitude we can obtain inmediately the solar altitude at noon (12:00:00 LAT):
= 90º - + = 90º - 37.54 + (-13.63) = 38.83º = 38º 49’ 48’’
Longitude determination: LST = Official time –1 = 12:17:23 (winter schedule)
LAT = LST + 4 (Ls-Le) + Et 4 (Ls-Le) = LAT – LST – Et
4 (Ls-Le) = 12:00:00 – 12:17:23 – (-00:14:16) = -00:17:23 + 00:14:16 = -00:03:07
4 (Ls-Le) = -3.117 min Ls-Le = -0.779º
Ls = 0º (Greenwich) Le = +0.779º = 0º 46’ 45’’ W
Longitude of the shipwreck
PROB. 0106 / LOOKING FOR A WRECKAGE (CONTINUED)
3036º
38º
40º
0º2º
0º 46’ 45’’ W
37º 32’ 40’’ N
Coordinates of the wreckage:
37º 32’ 40’’ N, 0º 46’ 45’’ W
Sunrise hour angle (February 13th):
ΦΦcoscosΦ
cos tantansinsin
s
s = 79.26º = 79.26/15 = 5.284 hours
Sunrise LAT (February 13th):
12-5.284 = 06:42:58
Sunrise LST (February 13th):
LST = LAT - 4 (Ls-Le) - Et = 07:00:21
Sunrise official time at the wreckage position 08:00:21
PROB. 0106 / LOOKING FOR A WRECKAGE (CONTINUED)