Astronomy 101 HW #2
Solutions
1.Sketch for the following situations, name the lunar phase and give the approx. time the moon sets
(Note: Direction of rotation of earth and revolution of moon both is counter clockwise and the dotted line shows that
visible half of moon from earth)
a) Moon rises at midnight:
Since it rises at midnight, it sets 12 hours later
at noon, and the phase would be last quarter
E
b) On this day a lunar eclipse occurred in China:
Since it was lunar eclipse, the earth is between moon
and sun, so it sets at 6 am and the phase is full moon. Each
day the phase of the moon is the same for all locations on
Earth, since the moon moves slowly in its orbit.
c) Approximately 2 days have passed after a new moon:
The moon rises 50 minutes later daily. The new moon
rose at 6 am so after 2 days the moon will rise 2X50 minutes
(100 minutes = 1 hrs 40 minutes) later then 6 am, which is 7:40 am
so it sets at 7:40pm, the phase after new moon is waxing crescent
d) You see the moon rising in the eastern sky on your
way to 3 pm class:
Since the moon is rising at 3 pm, it will set 12 hours later
at 3 am, the phase would be waxing gibbous
e) You see the moon with earthshine in the eastern sky:
Earth shine is visible when dark portion of the moon is visible
due to the reflected light from Earth, which is possible only
when Moon is on the side of its orbit closer to Sun (the phase is crescent).
When moon can be seen with earth shine in eastern sky,
the phase must be between last quarter phase and the new
phase. The phase would be waning crescent, the approximate
moon rise will be 3 am – 4 am, so it sets 3pm -4pm
f) From college park the moon looks like this
at 8 pm:
Since the moon is half lit on the right side, is a first quarter moon,
It rises at noon time and sets at midnight
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aa
Semi-major
axis= a
AA
Orbit
P = period in years
2Xa
Problem#1: Ceres the asteroid has an orbital period of 4.6 years. Calculate the semi major axis of its orbit.
Ans: Using the Kepler’s law,
P 2=a3 ; P in years and a in AUs
P =4.6 years, so P 2 = 4.6 X 4.6 = 21.16
a3 = 21.16 so a =(21.16)1/3 = 2.765 AU = using significant figures 2.8 AU
Problem#2: The semimajor axis of Comet Hale-Bopp’s orbit is 260 AU. Calculate its orbital period.
Ans: Using the Kepler’s law,
P 2=a3 ; P in years and a in AUs
Since a = 260 AU, so a3 = 260 X 260 X 260 = 17576000
Which is P 2 = 17576000 so P =(17576000)1/2
P = 4192.37 years, using significant figures 4200 yrs
P. 62 RQ#10. What is the difference between a hypothesis, a theory and a law?
Ans:
Hypothesis: Possible answer to a question
Theory: A hypothesis, which has been tested and supported many times
Law: A statement that describes a pattern in nature, the statement must hold under specified conditions
without exceptions
P.124: RQ#6: Why do we think that the solar system formed about 4.6 billion years ago?
Ans: The age of a rock is determined by analyzing the radioactive elements it is composed of.
If a sample is available, a good estimation of the age can be made using such a test in a laboratory.
Radioactive elements have a tendency to decay in a predictable fashion. Uranium is one example of
such materials. Tests indicate that the solar system was formed about 4.6 billion years old based on:
i) Oldest rocks on earth have been dated up to 4.3 to 4.4 billion years
ii) Rocks from moon are dated up to 4.48 billion years
iii) Oldest meteorite so far has been dated to 4.6 billion years