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Transcript
Astronomy 10: Introduction to General Astronomy
Instructor: Tony Piro, [email protected]
Homework #3: Solutions
Chapter 8
(1) page 208, question 6
Wien’s law tells us that λpeak T = 0.3 cm K. Plugging in T = 60 K, we find
λpeak =
0.3 cm K
= 0.005 cm = 50 µm
60 K
Pluto is brightest in the infrared.
(2) page 208, question 9
The Kuiper belt is a group of thousands of objects, orbiting slightly exterior to Neptune. They are
thought to be left over from our Solar System’s formation. Pluto is a member of the Kuiper belt (although
Pluto was discovered earlier because of it’s large size and high albedo, which makes it easier to observe).
(3) page 208, question 10
There is nothing chemically burning in a comet’s tail. The tail is formed out of gas and dust that is
being blown off the comet from the Sun’s solar winds. For this reason, the comet always points away from
the Sun.
(4) page 208, question 18
A shooting star is a meteoroid: a rock falling through the Earth’s atmosphere, which is being heated due
to friction.
(5) page 209, question 25
Certain groups of asteroids have orbits than the cross the orbit of the Earth around the Sun. Over time,
all of these asteroids will eventually be “swept up” by Earth. Some of these collisions will undoubtedly have
dire consequences for life on Earth. It is important to monitor the orbits of these asteroids so that we will
have a chance to prevent the collision.
Chapter 9
(6) page 224, question 5
The most common method for discovering an extra-solar planet is via the Doppler shift of the parent
star. As the star wobbles back and forth from the pull from its planet, the light from the star shift to be
bluer, then redder, and so forth.
Other methods for discovering planets that we discussed were microlensing and transits. Although they
have not been as fruitful as the Doppler effect so far, these will be used more and more in the future.
(7) page 224, question 7
The question tells us that the speed of the reflex motion (or wobbling, as we have referred to it in class)
is inversely proportional to mass. If the speed of the Earth is VEarth = 30 km/s, then the corresponding
speed of the Sun (from being pulled on by the Earth) is
VSun =
6.0 × 1024 kg
MEarth
× 30 km/s =
× 30 km/s = 9 cm/s
MSun
2.0 × 1030 kg
1
(8) page 224, question 9
When we measure velocities of stars using Doppler shifts, it can only measure the radial velocity (that is,
the velocity directly toward or away form us). In general, the orbit can be at an angle from us, so that the
true orbital velocity is equal or greater than what we measure. In turn the inferred mass is equal or greater
than what we get from the measured velocity.
(9) page 224, question 13
The fractional change in brightness of a star from a transiting planet is proportional to the relative
diameter squared. For example, if you have a planet with diameter d and star with diameter D, then
∆L
=
L
d
D
2
This was derived in Lecture 10. Thus if the transiting planet has 10% the diameter, then the brightness
decreases by 0.12 = 0.01 = 1%
Chapter 10
(10) page 242, question 5
Things that vary with the solar-activity cycle: sunspots, solar flares, coronal mass ejections, prominences,
and the total energy output of the Sun.
(11) page 242, question 10
Although the corona is very hot, it is also very faint. This makes it difficult to see against the everyday
blue sky. Nevertheless, if we go to especially high elevations on Earth, or wait for a solar eclipse, we can
study the inner parts of the corona with telescopes on Earth.
Using satellites, we can also study the corona by making an artificial eclipse covering the Sun’s photosphere. Since they are in space, we don’t have to worry about the Earth’s atmosphere scattering light and
making the corona difficult to observe. Unfortunately, this artificial eclipse blocks the inner regions of the
corona, so the satellites mainly study the outer part.
So in general, we study the outer parts with space satellites, and the inner parts with ground telescopes.
(12) page 243, question 27
Sunspots are regions in which a locally higher magnetic field prevents hot material from bubbling up
from the Sun, so it looks darker according to the Stefan-Boltzmann law. So the answer is c
(13) page 243, question 30
Remember from the Stefan-Boltzmann law, that the b ∝ T 4 , thus the answer is
4000 K
6000 K
4
=
so the answer is e
2
4
2
3