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hwk01ans
Ast 4001, 2015 Sept 8--17
Homework set 1 answers -- Distances and absolute magnitudes
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1. At what distance would the Sun have an apparent angular diameter of 1 arcsec?
Answer: As seen from Earth at distance 1 AU, the Sun's apparent diameter is about 0.53
degree  1900 arcsec. (This is a famous fact, but confirm it by calculating the linear
diameter / distance.) Therefore the Sun would subtend 1 arcsec at a distance of about
1900 AU  2.8 × 10 14 m, about 60 times as far as Neptune. Since 1 parsec = 206265 AU
(same as the number of arcsec in a radian), that distance corresponds to about 0.0092 pc,
1/140 the distance of alpha Centauri. We could also say "11 light-days."
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2. A particular star cluster has a diameter of 7 pc. If its distance is 2300 pc, calculate the
cluster’s apparent angular size expressed in degrees, arcmin, and arcsec.
Answer: 7 pc / 2300 pc  0.00304 radian  0.17 deg  105 arcmin  630 arcsec.
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3. At maximum brightness, a typical supernova might have visual absolute magnitude
M = about 18.
(a) Compare its visual-wavelength luminosity to the sun, L (SN) / L (sun).
Answer: The Sun has absolute visual magnitude = +4.8 (look it up), so the magnitude
difference is 18  4.8 = 22.3.
This corresponds to a ratio of 10 8.92  8 × 10 8 .
(b) At what distance would the supernova appear as bright as Sirius, m =  1.4 ?
Before looking up a formula, try to figure this out for yourself based only on the
definition of absolute magnitude M .
Answer: According to the definition of absolute magnitude, the SN would have apparent
magnitude m = 18 at a standard distance of 10 pc. This is 16.6 magnitudes brighter
than Sirius, a brightness ratio of 4.4 × 10 6 . According to the inverse-square law, the
SN's apparent brightness will be reduced by a factor of 4.4 × 10 6 if we move it farther
away by a factor of about 2100, compared to the standard-reference distance 10 pc.
So our answer is about 21,000 pc. Of course the official formula gives the same result.
(c) At what distance would it appear as bright as the full Moon?
Answer: If the full Moon has m =  12,7, the SN would appear that bright at a
distance of about 115 pc. This can be done in the same ways as part (b).
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4001hwk01ans - p2
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4. Look up information about Proxima Centauri, a.k.a. alpha Centauri C, the most famous
red dwarf star. Find out its absolute visual magnitude M and its approximate distance
from the brighter central star, alpha Centauri A. Estimate Proxima Centauri’s apparent
magnitude as seen from alpha Cen A, and alpha Cen's magnitude seen from Proxima.
Answers: According to one reference, alpha Cen C is located about 15,000 AU
away from the double star alpha Cen AB; or about 0.073 parsec. (Stars A and B
are never more than 37 AU apart.)
Proxima (or C) has absolute visual magnitude M = +15.49, which gives it
apparent magnitude m = +4.8 as seen from alpha Cen A or B.
On the other hand Alpha Cen A, with absolute magnitude M = 4.4, would have
m = 6.3 as seen from Proxima.
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5. Estimate the bolometric absolute magnitude M of a 100-watt light bulb. Clearly explain
your reasoning.
Answer: The Sun has luminosity L = about 3.9 × 10 26 W, which is more powerful
than the light bulb by a factor of 3.9 × 10 24 . This factor corresponds to a
magnitude difference of about 61.5. Since the Sun has absolute bolometric
magnitude + 4.8, the light bulb must have roughly
M  61.5 mag fainter than the Sun  4.8 + 61.5  +66.3.
We can see whether this is reasonable by calculating the distance where the light
bulb would appear as bright as the Sun, apparent magnitude m  27. The
standard formula gives D  roughly 2 × 10  parsec  6 cm, about twice
the bulb's radius. This seems credible if you imagine how much it would warm your
hand or hurt your eye at that location. If this calculation gave 6 m or 6 mm instead
of 6 cm, then we’d suspect an error somewhere.
(Pertinent fact worth remembering: At noon on a clear day, radiation from the Sun
amounts to roughly 1000 Watts per square meter.)
--- Questions like this are harder to concoct with modern LED's and fluorescent
lights! Believe it or not, an old-fashioned incandescent light has about the
same temperature and color as a cool red M-type star; most of its radiation
is infrared, making it inefficient for visible light. But LEDs and fluorescents
have weird non-thermal colors that don't match any star. --_________________________________________________________________________________________________________________
----- one more problem is on the next page -----
4001hwk01ans - p3
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6. Turn in your solution to this problem on Thursday Sept. 17. Explain your reasoning and
calculations clearly ... and please write legibly!
The figure shows observations of a visual binary star with period = 60 years. If we take random
observational errors into account, the data indicate a circle shown on the right. But the true orbit
in space cannot be a circle, because the primary star is far off-center. The orbit must be a highly
eccentric ellipse seen from an oblique angle. The apparent diameter of the circle is 1.8 arcsec.
(a) Use the right-hand figure to estimate the orbit’s eccentricity . Explain.
Solution: Step one, the major axis is a line that passes through the center of the circle
and also the primary star; it's obviously horizontal in the figure. Step two, if we
carefully measure the figure, the primary star is 0.72 arcsec from the center of the
circle. (The center of the circle is not exactly at the center of the square figure.)
The projected semi-major axis is 0.9 arcsec, the radius of the circle. (Diameter
was stated to be 1.8 arcsec.) Step three, the ratio of these two projected lengths
is the orbit eccentricity:  = 0.72 / 0.9 = 0.8.
Before continuing, note that the axis of the tilt must be vertical in the figure,
i.e., parallel to the minor axis. That is the only way to get vertical symmetry
while foreshortening the major axis -- and the only way it can look like a circle.
(b) Estimate the orbit’s true semi-major axis a expressed in arcsec, i.e., the value
that we would see if the orbit plane were perpendicular to our line of sight.
Solution: The axis ratio of the ellipse is b / a = ( 1 –  2 ) = 0.6 , since we
found  = 0.8 in part (a). We can measure b directly in the figure, because
we view it from a perpendicular direction: b = vertical radius = 0.9 arcsec.
(In other words, vertical distances in the figure are not foreshortened by our
perspective view.)
Since b / a for this elliptical orbit is 0.6, and b = 0.9 arcsec, we conclude that
a = 1.5 arcsec.
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4001hwk01ans - p4
----- problem 6 continued: -----
(c) Estimate the orbit inclination angle i that makes this orbit appear circular.
Inclination i is defined to be zero for an edge-on orbit.
Answer: Oops, "zero for an edge-on orbit" is not standard!! It should have said
"zero for an orbit that is perpendicular to our line of sight", which is 90° different
from the statement. Let's start by accepting what the problem said, even though it
was an unusual definition. Then { projected a } / { true a } = sin i , which is zero
for an edge-on orbit. In this problem, sin i = 0.9 / 1.5 = 0.6 based on parts
(a) and (b), so we get i  arcsin 0.6 = approximately 37° . This is correct for
"the problem as stated."
If we switch to the normal definition of i , then i  arccos 0.6 = 53° = 90°  37°.
Anyone who noticed this glitch and mentioned it will get a little extra credit.
(d) This binary star system is at a distance of 15 pc from us. Use your result in part (b)
to calculate the true linear value of a expressed in AU.
Answer: At D = 15 pc, 1 arcsec corresponds to 15 AU. Therefore a = 1.5 arcsec
implies a true linear value a = 22.5 AU.
(e) Deduce the total mass of the system, expressed in units of M sun .
Answer: The orbital period was stated to be P = 60 years and we have found that
a = approximately 22.5 AU. Kepler’s third law is P 2 = a 3 / M in Solar units.
Therefore M tot = a 3 / P 2 = about 3.2 M sun .
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