Astronomy 1 - Homework #3 Solution
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Chaisson Prob#1 - Astronomical Measurements by Tycho Brahe were accurate
to 1' arc. What size does this angle correspond to for the following distances:
1' arc is a small angle and the size, s, is given by
s = d x (angle in radians)
Since 60' = 1o and 57.3o = 1 radian
1' = 1/(60 x 57.3) = 0.000291 radians
(a) the distance to the Moon (384,000 km)
s = 0.000291 x 384,000 km = 112 km
(b) the distance to the Sun (1 AU = 150,000,000 km)
s = 0.000291 x 150,000,000 km = 43,600 km
(c) the distance to the Saturn (at opposition)
At opposition the Earth is between Saturn and the Sun
Saturn is 9.54 AU and Earth is 1.0 Au from the Sun
Distance to Saturn is 8.54 AU
s = 0.000291 x 8.54 AU = 0.00248 AU
= 0.00248 AU x 150,000,000 km/AU = 373,000 km
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Chaisson Prob#4 - An asteroid has a perihelion distance of
2.0 AU and an aphelion distance of 4.0 AU. Calculate its orbital
semi-major axis, eccentricity and period.
rp =Perihelion = a(1-e) ra = Aphelion = a(1+e)
So rp + ra = a -ea + a + ea = 2a
and
a = (2.0 AU + 4.0 AU)/2 = 3.0 AU
rp = a(1-e)
e = 1 - rp/a
e = 1 - 2.0/3.0 = 1/3 = 0.333
Kepler's 3rd Law: P(yrs)2 = a(AU)3
P = a3/2
= (3.0)3/2 = 5.20 years
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Chaisson Prob#5
Halley's comet has a perihelion of 0.6 AU and an orbital period of
76 years. What is its aphelion distance from the Sun?
The Period gives the semi-major axis of the orbit
a3 = P2
a = P2/3 = (76 AU)2/3 = 17.9 AU
ra + rp = 2a
so ra = 2a - rp = 2(17.9) - 0.6 = 35.2 AU
[Halley's comet reaches maximum distance from the Sun between
the orbits of Neptune and Pluto]
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Chaisson Prob#6
How long will a radar signal take to complete a round trip
between Earth and Mars when the two planets are 0.7 AU apart?
Radar travels at the speed of light (300,000 km/s)
Since
distance = speed x time
time = distance/speed
distance = 2 (0.7 AU) 150,000,000 km/AU = 2.1x108 km
travel time = 2.1x108/300,000 km/s = 700 s
= 11.7 minutes
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Chaisson Chapter 2 Project #1
Use ECU (or look it up in the Observer's Handbook) to
determine the next opposition dates in 1998 (or 1999)
of Mars, Jupiter and Saturn. Determine the length
of time of retrograde motion for each.
Use the fact that the azimuth of a planet will be 180o (on your
local celestial meridian) at midnight when at opposition with the Sun.
A procedure to determine opposition with ECU:
1) Lock on the Planet (animation)
2) Set time to midnight (12:00 am or 0 hrs)
3) Center the Planet on the screen
4) Advance the date muliples of 1 day until
the azimuth reading is 180o
Turn on the trails to follow the motion - note the retrograde motion
(east-to-west) when near opposition. The beginning and end of the
retrograde motion occurs when the planet is stationary.
Opposition Period of retrograde motion
Mars April 27, 1999 Mar 18 to Jun 4
Jupiter Sept 20, 1998 July 18 to Nov 14
or Oct 25, 1999 Aug 14 to Dec 21
Saturn Oct 23, 1998 Aug 16 to Dec 30
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Additional Problems to solve:
#1 -Use ECU to determine the dates and times (UT) of the
next two solar eclipse in the Saros series that contains the
February 26, 1998 total solar eclipse. At what longitude with the
eclipses be on the local celestial meridian at maximum eclipse?
The 1998 Eclipse occured at 2:30 pm on Feb 26
The Saros series of Solar eclipse are separtated by
18 years and 11 days (see class notes).
Setup ECU for Feb 26, 1998 + 18 years 11 days
and center on the Sun. Find the time of the eclipse
by advancing the time until the Moon covers the Sun.
Next eclipse of the series is on:
Mar 9, 2016
Maximum Eclipse (Kentville, NS) is 10:50 pm AST
Second eclipse after 1998 in the series is on:
Mar 20, 2034
Maximum Eclipse (Kentville, NS) at 5:35 am AST
(No eclipse in Nova Scotia - the centre
of the Moon is 35' arc North of the
centre of the Sun at closest approach
This means that one must be at least the
diameter of the Moon [3500 km] farther
south on the Earth to see the Eclipse)
Eclipse is total on the Equator south of NS.
Summary: 1998 Feb 26 2:30 pm AST 18:30 UT
2016 Mar 9 10:50 pm AST 2:50 UT Mar 10
2034 Mar 20 5:35 am AST 9:35 UT
Note that each successive eclipse take place about 7 hours later
and that means the maximum eclipse takes place ~ 15x7 = 105 degrees
longitude farther each each time.
#2 - What is the approximate hour angle of the Moon at Midnight
when the Moon in its waxing cresent phase (3 days past New).
The phase determines the geometry of the Sun, Earth, Moon system
and the time determines where we are on the Earth relative to the Sun. See
the diagram for the phases of the Moon in Chapter #1 of Chaisson.
Below is the solution. Hour Angle is measured westward from the local celestial
meridian to the object of interest. In this case 135 degrees or 9 hours.
