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1
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- Go over HW#1
- Finish Cycles of the Sky
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2
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- 1. If sunlight takes 8 minutes to travel from the Sun to the Earth, how
long does moonlight take to travel from the Moon to the Earth? To find this, use information about
the relative distance of the moon and the sun which you can find in your
text.
- I’ve asked you to set this up in terms of light travel time. We know that the distance from the
Earth to the sun is 1 AU, or about 150,000,000 km. The distance from the Earth to the
moon is in appendix A table A-14, and is 384,000 km. 150 million divided by 384 thousand is
391. So, light will take 1/391
time to travel between the moon and Earth as it does from sun to
Earth. So the travel time is 8min
x 60 sec/min / 391 = 1.2 seconds.
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3
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- 2. How many suns would it take, laid edge to edge, to reach the nearest
star?
- Basically, all you have to do here is divide the distance by the
diameter of the sun. The distance
is 4.28 light-years (table A-7).
Convert to meters by multiplying by 9.46x1015m per
light-year (table A-6). The
diameter of the sun is 2 x radius = 14x108m (table A-6). So, the number of suns is 4 x 1016
m/14 x 108 m = 2.9 x 107 m.
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4
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- 3. What do you get when you multiply together 3 x 1032, 2 x
1056, 0.5 x 1018, and 3 x 10100 ?
- Many calculators won’t do this!
Here is the power of scientific notation. Just multiply the leading numbers: 3 x
2 x 0.5 x 4 = 9. Add the
exponets: 32 + 56 + 18 + 100 = 206.
The product is therefore 9 x 10206.
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5
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- 4. If light takes about eight minutes to travel from the Sun
to the Earth, and about forty minutes to travel from the Sun to Jupiter,
how many astronomical units (AU) is Jupiter from the Sun?
- Easiest solution – look it up in table A-13 (5.2 AU)! Otherwise this is really the same as
the moonlight problem. Jupiter is
(40/8 = 5) times farther from the sun than the Earth is. The Earth is 1 AU from the sun, so
Jupiter is about 5 AU from the sun.
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6
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- 5. What is the ratio of the mass of Earth to the mass of the
moon? (Hint: look in appendix A of the text for these values.)
- Just take the ratio of masses.
From table A-6, we have 6x1024 kg/7x1022
kg. 1024/1022
is 102 = 100. 6/7 is
close to 1. So the answer is
about 100 without needing a calculator.
Scientific notation makes it easy, and often in astronomy we’re
not concerned with super precise answers. Remember that Earth is about 100 times
more massive than the moon, not 81 times.
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7
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- 6. If we have two stars in fictional constellations, and the
two stars are named Alpha Buffalo and Beta Cowboy, which is brighter and
by what factor? The apparent visual magnitude of Alpha Buffalo is 3.2
and the the apparent visual magnitude of Beta Cowboy is 2.7.
- Magnitudes are “backwards” so the star with the smaller magnitude (Beta
Cowboy) must be brighter. By what
factor? Use the equation IA/IB
= 2.512 (mB-mA) =
2.512-0.5 = 0.63
- So IB/IA = 1/0.63 or about 1.6.
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8
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9
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10
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11
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- Longest possible total eclipse is only 7.5 minutes. Average is only 2-3 minutes.
- Shadow sweeps across Earth @ 1000 mph!
- (Compare with scene in The Mummy Returns!)
- Birds will go to roost in a total eclipse. The temperature noticeably drops.
- Totally predictable (even in ancient times, e.g., the Saros Cycle,
eclipse pattern repeats every 6585.3 days or 18 years, 11 1/3 days).
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12
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13
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14
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15
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- How big is the moon? Have you
ever seen the moon near the horizon?
Has it looked huge, much larger than when it is high in the sky?
- Last full moon was Wed. night, Sep. 10.
Next one is Oct. 10.
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16
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Notes
