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PHYS 175, FALL 2014
HW #3 S OLUTION
1
6.1 Describe refraction and reflection. Explain how these processes enable astronomers
to build telescopes.
Refraction is the bending of light due to the differences in the speed of light as light
passes from one medium into another. Reflection is the bouncing of light from a
surface. Telescopes improve on the human eye by gathering more light and magnifying images. Lenses (refraction) and mirrors (reflection) each can accomplish these
tasks. Both lenses and mirrors are used to focus images on detectors or films.
6.3 Explain why light rays that enter a telescope from an astronomical object are essentially parallel.
Since telescopes are used to image distant objects, the rays of light spread out very
little over the length of the telescope’s optical system.
6.6 Do most professional astronomers actually look through their telescopes? Why
or why not?
Working astronomers haven’t really looked through their telescopes for serious observations since photography became available to the field in the mid-1800s. Film,
while inefficient, has a much longer integration time than the human retina, so it can
image much fainter objects by being exposed for a long time. The advent of the
charge coupled device (CCD) in the 1970s saw astronomers quickly abandon film
for the high efficiency of the new imagers, and the ability to digitize the images for
analysis.
6.25 What are the optical window and the radio window? Why isn’t there an X-ray
window or an ultraviolet window?
These are two bands within the electromagnetic spectrum in which the Earth’s atmosphere is nearly transparent – visible light and a fair portion of radio waves travel
through the atmosphere with very little loss of intensity. While some UV light does
get through the atmosphere, it is still heavily attenuated (else we’d get sunburned
much more easily!). The same goes for X-rays, as they are almost completely absorbed in the upper atmosphere. UV and X-ray telescopes must be placed above the
atmosphere, in space, in order to ”see” anything.
PHYS 175, FALL 2014
HW #3 S OLUTION
2
6.36 The four largest moons of Jupiter are roughly the same size as our Moon and are
about 628 million (6×108 ) km from Earth at opposition. What is the size in km of
the smallest surface features that the Hubble Space Telescope (resolution of 0.1
arcsec) can detect? How does this compare with the smallest features that can be
seen on the Moon with the unaided human eye (resolution of 1 arcmin)?
For this we’ll make use of the formula in the text in Box 1-1, p. 9
D=
αd
.
206, 265
In the first part of our problem, D is the size of the smallest feature we can resolve
with the HST at the distance d and having an angular size of only α (expressed in
arcsec).
D=
0.1 · 6 × 108
αd
=
≈ 290 km .
206, 265
206, 265
We repeat the calculation for the Moon using d ≈ 3.85×105 km (the average distance
from Earth to the Moon) and need to convert 1 arcmin to arcsec in order to use the
formula. There are 60 arcsec/arcmin, so α = 60 arcsec in the second part of the
problem. Using the formula, we find that the naked eye can resolve a feature of
about 110 km. This is not surprising, when we consider that the HST has 600 times
the resolving power of the human eye, but Jupiter’s moons are about 1,500 times
more distant. So a quick estimate of the change in feature resolution may be made
by multiplying by a factor of 600/1,500 = 10/25. HST can resolve roughly 300 km
on a Jovian moon, so the eye should be able to resolve a feature roughly 10/25th the
size, or about 120 km.
7.8 The absorption lines in the spectrum of a planet or satellite do not necessarily
indicate the composition of the planet or satellite’s atmosphere. Why not?
The reflected sunlight from the body’s atmosphere also has absorption lines from
the Sun’s gaseous nature, as well as any gasses in space that the sunlight travelled
through. These can be confused for absorption lines actually caused by the body’s
own atmosphere.
7.9 Why are hydrogen and helium abundant in the atmospheres of the Jovian planets
but present in only small amounts in the Earth’s atmosphere?
For gasses to escape a planet’s atmosphere, they must have a speed of about 1/6th
of the planet’s escape velocity. The Jovian planets are relatively massive and cool.
This keeps the gasses moving slowly enough (average kinetic energy is related to
temperature) and makes the escape velocities high enough to trap the lighter gasses
like H and He.
PHYS 175, FALL 2014
HW #3 S OLUTION
3
7.14 What is one piece of evidence that impact craters are actually caused by impacts?
There are two mentioned in the text: circular shapes are expected at large distances
due to shockwaves caused by impacts (like artillery craters on Earth), or that there
are often central peaks in the middle of the circular craters (again, as seen on Earth).
7.19 How is the magnetic field of a planet different from that of a bar magnet? Why is
a large planet more likely to have a magnetic field than a small planet?
While planetary magnetic fields are dipolar, like a bar magnet, their origin is due
to currents, not aligned crystals of magnetized metal. A planetary magnetic field is
induced by a dynamo of circulating, molten metal, which carries a current. Large
planets can gravitationally generate enough pressure at their cores to maintain a
liquid metallic state required for such a dynamo.
7.23 Mars has two small satellites, Phobos and Deimos. Phobos circles Mars once
every 0.31891 days at an average altitude of 5,980 km above the planet’s surface.
The diameter of Mars is 6794 km. Using this information, calculate the mass and
average density of Mars.
This is an excellent example of using physics and astronomy to ”weigh” a planet!
Box 4-4, p. 93, of the text takes you through this calculation using Newton’s form of
Kepler’s 3rd law
4π2
2
a3 .
P =
G (m1 + m2 )
Assuming that m1 m2 , as is the case between a small moon and a planet, we can
solve for the combined mass and take it to be effectively the mass of the planet. This
is done in the text as well, giving
4π2 a3
.
GP2
For this problem we need to take a as half the diameter of Mars plus the average
altitude of Phobos, and express the result in m. The period must be expressed in
seconds, not days, given Newton’s form and the universal gravitational constant G.
Using all this we find
m1 + m2 =
m1 + m2 =
4π2 ([6794/2 + 5980] · 1000)3
6.67 ×
10−11
· ([0.31891 · 86400])
2
≈ 6.43 × 1023 kg .
This value is very reasonable, and for a density ρ, we just need to divide this by the
volume of Mars as
m
6.43 × 1023
ρ= 4 3 = 4
≈ 3, 920 kg/m3 .
3
πr
π (6794/2 · 1000)
3
3