Download Astronomy 1400: Homework 5

Astronomy 1400: Homework 5
Due in class, Friday, October 10
Name:
Instructions: To receive partial/full credit you must show your work or explain your answer thoroughly. Please circle your final answer to each problem
if it is a number.
1. This problem explores observing the granulation on the surface of the Sun.
(a) (10 points) Calculate the typical angular size of the granulation cells. The typical
size of the cells is 1000 km.
θ=
1000 km
s
=
=
d
1.49 × 108 km
...
»
»»
6.71 × 10−6 »
radians
360¢◦
60′¤ 60′′
×
× ′ =
»
»»
2π »
1¢◦
radians
1¤
× ...
1.38′′
(b) (10 points) Use the answer from the previous part to calculate the minimum size
of telescope needed to see these cells. You are trying to view them with your eyes
through a telescope either by projection or with a proper solar filter (λ = 550 nm).
Express your answer in inches or mm, which are the most commonly used units for
amateur telescopes. Warning: watch your angular units.
We need to use the diffraction limit . . .
θ = 1.22
»
λ
λ
550 »
1 mm
nm
=⇒ d = 1.22 =
× 6 »=
−6
d
θ
6.71 × 10 radians 10 »
nm
81 mm = 3.2 in
You could have calculated your angles in whatever units you want. Radians and
arcseconds should be most common. You also could have used the formula for
the diffraction limit from the book in Math Insight 6.2 that already did the unit
conversions for you to give the answer in arcseconds (the constant is 2.5 × 105′′
instead of 1.22 in the formula above), but then you would have had to calculate the
answer to part a) in arcseconds. The important part is to use consistent units.
(c) (5 points) Would you be able to see sunspots with the size telescope you found in
the previous part (are they larger or smaller than the granulation cells)?
Sunspots are a range of sizes, but are generally much larger (the magnetic field
lines cross many granulation cells), so, therefore, you would also be able to easily
see them if you can see the granulation. The picture in Figure 14.15 demonstrates
this nicely.
Page 1 of 4
(d) (10 points (bonus)) Do a little online research to find the typical sizes (remember
that this means diameter) of amateur telescopes. Write down the range of typical
sizes you find for under about a few hundred dollars. For a given size, are reflectors
or refractors generally more expensive? Why do you suppose that is? Finally, for
the range of typical sizes you find for under a few hundred dollars, would you be
able to resolve the solar granulation cells? Ignore possible atmospheric effects.
In this price range, you should find telescopes about ∼ 2 – 10 in (∼ 60 – 300 mm).
Refractors are definitely more expensive that reflectors of the same size because
making a lens (that light has to travel through) is more difficult that making a
mirror (only the surface has to be good). You should be able to find both reflectors
and refractors larger than 3.2 in or 81 mm for a few hundred dollars.
2. (10 points) The Sun has a brightness of 1360 W/m2 as seen from Earth. How bright
does the Sun appear to be at Pluto? [Hint: Use Pluto’s distance from the Sun listed in
Appendix E in the back of your book.]
We can compare Pluto and Earth’s brightnesses and distances to find out how bright
the Sun is at Pluto because the Sun is not changing brightness. We are using the Sun
as a “standard candle” in this way.
©
BP luto
1
LSun /4πd2P luto
d2Earth
1©
AU
)2 =
=
=
=
(
2
2
©
BEarth
LSun /4πdEarth
dP luto
40 ©
1600
AU
BP luto =
1360 W/m2
BEarth
=
=
1600
1600
0.85 W/m2
3. (10 points) Why are sunspots ∼ 4 – 5× darker than their surroundings? Why is there
such a big difference in the amount of light (flux) from them as compared to their
surroundings (hint: what law tells us this)?
Sunspots are darker because they are cooler (∼ 4000 K as opposed to the 5800 K plasma
that surrounds them; see page 476 of the book). The strong magnetic fields prevent
hot material from rising up through them. The Stephan-Boltzmann law, F = σT 4 , (the
Sun is essentially a blackbody) tells us that a small change in temperature means a big
change in flux.
4. (5 points) True or False: Fusion occurs only in the core of the Sun. Briefly explain why.
True. This is the only place where it is hot and dense enough for there to be enough
particles traveling at high enough speeds to stick together.
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5. Calculate the hydrogen-burning lifetime of the Sun following the steps below.
(a) (5 points) When the Sun was first formed its total mass was about the same as it
is today, 2.0 × 1030 kg, and but its chemical make up was slightly different. About
75% of Sun’s mass was hydrogen (compared to 70% today). Compute the mass of
hydrogen in the newborn Sun.
MH = 0.75 × 2.0 × 1030 kg =
1.5 × 1030 kg
(b) (5 points) Only about 13% of the Sun’s hydrogen ever becomes available for fusion.
Calculate the mass of hydrogen available for fusion in the newborn Sun.
MHf = MH × 0.13 = 1.5 × 1030 kg × 0.13 =
2.0 × 1029 kg
(c) (5 points) During the hydrogen fusion process, 0.7% of the original hydrogen mass
is converted into energy. Calculate the amount mass from part b) that would
disappear to become energy.
Mconv = 0.007 × MHf = 0.007 × 2.0 × 1029 kg =
1.4 × 1027 kg
(d) (5 points) Now use Einstein’s famous formula, E = mc2 , to calculate the energy
released by the mass from part c). Express your answer in Joules.
E = mc2 = (1.4 × 1027 kg)(3 × 108 m/s)2 =
1.3 × 1044 J
(e) (5 points) The number you calculated above is the total energy available from hydrogen fusion in the Sun. Assuming that the Sun’s luminosity stays constant at
3.8 × 1026 J/s, how long could the Sun shine? Express your answer in years.
L=
©
©
½
E
1©
min
E
1.3 × 1044 J¢
1½
hr
17
s
×
=
3.4
×
10
=⇒ t =
=
×
¢
© × ...
©
t
L
3.8 × 1026 J¢ /s
60 ¢s
60 ©
min
...
©
1 yr
1©
day
×
©=
©
½
24 ½
days
hr 365 ©
1.1 × 1010 yrs
(f) (5 points) Our best age estimates of the solar system place it at 4.6 billion years
old. What fraction of the Sun’s total lifetime has been used up?
f=
4.6 × 109
=
1.1 × 1010
0.42
So, roughly 42% of the Sun’s hydrogen burning lifetime has been used up.
Page 3 of 4
Multiple Choice: Choose the best answer.
6. (5 points) The James Webb Space Telescope (JWST) is an infrared telescope to be
launched in 2018. JWST has a 6-meter-wide mirror, bigger than the Hubble Space Telescope’s (HST’s) 2-meter-wide mirror. However, while HST is most sensitive at around
500 nanometers, JWST is most sensitive at around 4.5 micrometers. How will JWST’s
resolution compare to HST’s at their respective wavelengths?
A.
B.
C.
D.
E.
JWST will have a resolution about 9 times better than HST.
JWST will have a resolution about 3 times better than HST.
JWST will have a resolution about the same as HST.
HST will have a resolution about 3 times better than JWST.
HST will have a resolution about 9 times better than JWST.
θHST
6m
1
500 nm
λ
=⇒
×
=
=
d
θJW ST
2m
4500 nm
3
Remember, smaller angle means better resolution. Because the change is wavelength is
3× larger than the change in diameter, θHST is still 3× smaller/better than θJW ST !
θ = 1.22
7. (5 points) Hydrogen fusion in the Sun requires a temperature (in Kelvin) of . . .
A.
B.
C.
D.
E.
...
...
...
...
...
thousands of degrees.
millions of degrees.
billions of degrees.
trillions of degrees.
any temperature, as long as gravity is strong enough.
8. (5 points) Molten lava can be as “cool” as 1000 K. Humans, at ∼ 300K, emit the most
brightly at a wavelength of 10µm. At what wavelength does molten lava peak in its
emission?
A.
B.
C.
D.
E.
33µm
30µm
3µm
330nm
300nm
9. (5 points) We can learn a lot about a star by observing its spectrum. Which of the
following statements about spectra is NOT true?
A. The peak wavelength of the star’s spectrum tells us its temperature.
B. The total amount of light in the star’s spectrum tells us its radius.
C. We can determine the star’s composition by recognizing the pattern of atomic and
molecular transition lines in the spectrum.
D. We can look at Doppler shifts of atomic and molecular lines in the spectrum to
determine the star’s velocity either toward us or away from us.
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