Download ASTR 1120H – Spring Semester 2010 Exam 1 – Answers The AU is

ASTR 1120H – Spring Semester 2010
Exam 1 – Answers
PART I (70 points total) – 14 short answer questions (5 points each):
1. How is the Astronomical Unit (AU) defined? Give an example of a situation in which
this unit of measure would be convenient to use.
The AU is the average distance between the Earth and the Sun. The AU
is useful for measuring distances in the solar system.
2. Consider the angular diameter of the planet Saturn as viewed from Earth. Does this
value change with time? If so, how? If not, why not?
As the Earth orbits the Sun it gets closer, then farther, then closer, etc.,
to Saturn. As a result, the angular size of Saturn gets larger, then
smaller, then larger, etc.
3. Using his heliocentric model, Copernicus was able (a) to calculate the orbital periods
of the planets and (b) to estimate the relative distances of the planets from the Sun. How
did he do these two things?
Copernicus (a) used the synodic periods to determine the sidereal
periods and (b) used the sidereal periods and simple geometry and
trigonometry to determine the planets' relative distances.
4. What was Tycho Brahe's objection to the heliocentric model?
Brahe thought that if the Earth moved (as it must in the heliocentric
model), then as a result of its motion parallax of the nearest stars should
be observed. He didn't see parallax, so he concluded that the
heliocentric model could not be correct.
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5. What observations did Galileo make that reinforced the heliocentric model? Why did
these observations contradict the older model of Ptolemy? Why could these observations
not have been made before Galileo's time?
The most important ones were his observation of the complete cycle of
phases of Venus (which cannot occur in a geocentric model) and the
discovery of the moons of Jupiter (demonstrating that not everything
orbited the Earth, as the geocentric model contends). Neither of these
phenomena can be observed with the naked eye.
6. What is the difference between Kepler's 3rd law and Newton's version of Kepler's 3rd
law? Explain why the difference is important.
Kepler's 3rd law applies only to objects orbiting the Sun; Newton's
version is universal. The important difference between them is that
since Newton was able to derive Kepler's 3rd law from his (Newton's)
laws of motion and his law of universal gravitation he demonstrated
that the 3rd law is valid because of the two objects' mutual gravitational
attraction.
7. How did 17th century Danish astronomer Olaus Rømer measure the speed of light?
Rømer measured the time interval between successive eclipses of the
moons of Jupiter and noticed that that interval increased as the Earth
moved away from Jupiter and decreased as the Earth moved toward
Jupiter. He attributed the difference in time to the finite speed of light.
Knowing the distance of the Earth to the Sun, he was able to estimate
the speed of light.
8. How did Newton show that a prism breaks white light into its component colors, but
does not add any color to the light?
Sending white light through a prism breaks it into its component colors.
Sending only one of those colors through a prism results in that same
color, and only that color, emerging from the prism.
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9. Suppose we have two heated spheres, A and B, both the same size, but sphere A has a
Kelvin temperature twice that of sphere B. How do their energy fluxes compare? Which
sphere’s maximum emission will peak at a shorter wavelength?
Using the Stefan-Boltzman law (F = σ T4) we can see that doubling the
temperature will increase the flux by a factor of 24 = 16, so A's flux is 16
times greater than B's. Using Wien's law (λ max = (0.0029 K m)/T) we
can see that doubling the temperature will halve the wavelength of
maximum emission, so A's λ max is half that of B's.
10. Consider a hypothetical atom in which electron transitions can involve three energy
levels: highest energy (15 eV), middle (10 eV), and lowest energy (0 eV). What
transition will yield the shortest wavelength emission line? What transition will yield the
longest wavelength absorption line?
From E = (hc)/λ we know that short wavelength means large energy
and long wavelength means small energy. We also know that emission
lines are produced when electrons go from high energy levels to low
energy levels and absorption lines are produced when electrons go from
low energy levels to high energy levels. So, the shortest wavelength
emission line will occur in the transition from highest to lowest energy
level and the longest wavelength absorption line will occur in the
transition from middle to highest energy level.
11. Explain some of the disadvantages of refracting telescopes compared to reflecting
telescopes.
The main disadvantages of refracting telescopes are that their objective
lenses must be totally free of defects, they suffer chromatic aberration,
some of the incident light is absorbed in the lens, and the objective lens
can only be supported from its edges. None of these drawbacks occur in
reflecting telescopes.
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12. What is active optics? What is adaptive optics?
Active optics is a method of trying to correct for any mirror
deformation. Adaptive optics is used to try to overcome the seeing
effects caused by turbulence in the atmosphere.
13. Why have CCDs replaced photographic film for recording astronomical images?
CCDs have much higher quantum efficiency, better angular resolution,
larger dynamic range, a more uniform color response, and their images
can be digitized.
14. What parts of the electromagnetic spectrum emitted by an astronomical object can be
observed with ground-based telescopes? What parts require space-based observatories?
Why?
The optical (or visible) and radio parts of the spectrum can be observed
with ground-based telescopes. Infrared, ultraviolet, x-rays, and gamma
rays must be observed from space because the Earth's atmosphere is
opaque to those parts of the spectrum.
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PART II (30 points) – 6 mathematical problems (5 points each):
1. The angular size of the Moon as seen from the Earth is α = 1,867″ (about ½°). What
was the angular size of the Earth as seen by the Apollo astronauts who walked on the
Moon? (The Earth's diameter is 12,756 km. The Moon's diameter is 3,476 km.)
The distance from the Earth to the Moon is the same whether you're
observing from the Earth or from the Moon. So, if the Earth's diameter
is bigger than the Moon's diameter by a factor of 12,756/3,476 = 3.67,
then the angular size of the Earth as seen from the Moon will be 3.67
times larger than the angular size of the Moon as seen from the Earth,
i.e.,
3.67 × 1,867″ = 6,852″ ≈ 1.5°
2. Imagine a hypothetical planet like the Earth orbiting a hypothetical star with 9 times
the mass of the Sun. If the semi-major axis of the hypothetical planet's orbit is 1 AU,
what would be the hypothetical planet's orbital period (in days)?
For the real Earth, PEarth = 365 days. This can be obtained from
Newton's version of Kepler's 3rd law:
PEarth2 = (4π 2 a3)/(G(m1 + m2)) = (4π 2 a3)/(GmSun).
For this hypothetical planet orbiting this hypothetical star, we have
P2 = (4π 2 a3)/(G(9mSun)) = (1/9) (4π 2 a3)/(GmSun) = (1/9) PEarth2 , so
P = (1/3) PEarth= (1/3) (365 days) ≈ 122 days.
3. The human eye has evolved over time so that its peak wavelength sensitivity is about
0.5 μm (1 μm = 10−6 m). Use Wien's law to calculate the temperature of blackbody
radiation to which the eye is most sensitive.
λ max = (0.0029 K m)/T
⇒
T = (0.0029 K m)/λ max = (0.0029 K m)/(0.5 × 10-6 m) = 5800 K
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4. Hydrogen gas emits a strong spectral line of red light with a wavelength of λ =
656.281 nm (the Hα or Balmer-α line). This emission line is seen in the spectrum of a
distant galaxy but at a wavelength of λ = 662.844 nm. How fast is this galaxy moving
with respect to Earth, in terms of the velocity of light, c? Is it moving towards us or away
from us?
Δλ/λ lab = (λ obs – λ lab)/λ lab = (662.844 – 656.281)/(656.281) = 0.010 = ν/c
⇒ν = 0.100 c, away from us, since ν is positive resulting in a redshift.
The Arecibo radio telescope in Puerto Rico, the largest single dish radio telescope in the
world, has a diameter D = 305 m. The VLA (Very Large Array) radio interferometer in
Socorro, NM, has 27 dishes, each of which has a diameter D = 25 m. In its largest
configuration, the VLA's effective diameter is 36 km.
5. What is the ratio of the light-gathering power of the Arecibo telescope compared to
that of the VLA? Which of these two facilities is better for detecting faint objects?
Light-gathering power ∝ area of radio dish(es) ∝ diameter(s) squared.
For Arecibo, it's simply (305 m)2 = 93,025 m2. For the VLA, it's 27 × (25
m)2 = 16,875 m2. So, Arecibo can detect fainter objects.
6. If observing a celestial object at a radio wavelength λ = 6 cm, the Arecibo telescope
will yield an angular resolution of θ = 49". What is the VLA’s angular resolution when
observing at 6 cm in its largest configuration? Which of these two facilities is better for
seeing fine details in celestial objects?
In its largest configuration, the VLA has an effective diameter of 36 km
or 36,000 m. Its angular resolution at 6 cm (or 0.06 m) is
θ = (2.5 × 105)λ/D
= (2.5 × 105)(0.06 m)/(36,000 m) = 0.417 arcseconds = 0.417″
So, the VLA has better angular resolution.
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