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1
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- Solutions given here on these pages.
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2
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- 1. 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
exponents: 32 + 56 + 18 + 100 = 206.
The product is therefore 9 x 10206.
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3
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- 2. 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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4
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- 3. 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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5
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- 4. 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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6
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- 5. Everyone looked up the constellations Andromeda and Pegasus in the
text. This question encourages
you to look at the figures, not just read the text.
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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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- 7. If you were kidnapped and transported to a distant location on Earth,
but noticed that Polaris was 20 degrees above the horizon, what could
you figure out about your whereabouts?
c. The latitude of my
location is 20 degrees north.
Think about the celestial sphere and where Polaris would be in
the sky if you were on the equator (0 degrees latitude) or on the North
Pole (90 degrees latitude).
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9
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- 8. Which of the following sets of stars have ALL been the "north
star" (or pretty nearly) at one time or another?
c. Polaris, Vega, and
Thuban. The precession figure in
the textbook allows you to answer this one.
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10
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- 9. The sun is on the celestial equator at the times of
c. the autumnal equinox and the vernal equinox. These points define the transition of
the sun between the northern and southern skies, when its path crosses
the equator.
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11
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- 10. Relatively large "spring" tides occur
a. at new moon and full moons – the textbook provides this answer
in words as well as in a figure.
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12
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- 11. What must be the phase of the moon if a solar eclipse is to take
place?
- New! We’ll see that today in a
demonstration.
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13
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- 12. One of the nearest and brightest stars in the (southern) sky is
Alpha Centauri. Assume that its
radius is the same as that of the sun.
The distance to Alpha Centauri can be found in Appendix A of the
text. Assuming a perfect
telescope and no atmospheric turbulence, what is the angular diameter
(not radius!) of the star as seen from Earth?
- 7 x 10-3 arcseconds.
- Angular diameter = 206265 arcsec x 2Rsun
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4.3 light years
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Notes