Download HW #02 Solutions

Astronomy Assignment #10 Solutions
Answer the following questions from the first half of Chapter 11 Determining Star Properties in
Astronomy Notes. Note: There are additional problems with solutions listed below
than those that appear in HW #2. However, ALL the HW #2 problems are
solved below – you just have to look for them.
Review Questions
1. Describe the procedure used to find distances to the nearby stars.
The distance to the nearby stars is determined using stellar
(trigonometric) parallax. The procedure is to take two
pictures of the star 6 months apart, so the Earth is on
opposite sides of its orbit around the Sun. On the pictures of
the star, if the star is close, it will have slightly changed
position relative to the more distant stars. Measure the
angular change in position of the star on the two
photographs using the known field of view of the image. The
angular change in position is twice the parallax angle, p.
1
The distance of the star is given as d pc 
, where d is
p"
expressed in parsecs and p is expressed in arcseconds.
2. What do you need to know in order to get the scale of interstellar space in terms of kilometers or meters?
This is quite the question. I believe that Nick Strobel is asking about the theoretical basis of the stellar
(trigonometric) parallax. In the mathematical development of stellar parallax a vital ingredient is
knowing the baseline of the right triangle that is used with the parallax angle. The short side of that
right triangle is the Observer (Earth)-Sun distance, the AU. If the magnitude of the AU is not known in
meters or kilometers, then the distance to the star in parsecs cannot be related to meters or kilometers
or any other length unit of known length. The practical utility of the stellar parallax lies in converting a
parsec (a distance defined by and angular shift) into known length units like kilometers or light years.
One needs to know the magnitude of the AU in kilometers to really measure the scale of interstellar
space.
3. If the star-Sun distance = 30 parsecs, how far is the star from the Earth in light years…in kilometers?
30 parsecs  30 parsecs 
3.26 ly
3.09 ×1013 m
 97.8 ly and 30 parsecs  30 parsecs 
 9.27 ×1014 m
1 pc
1 pc
30 parsecs is equivalent to 97.8 light years and 9.27 × 1014 m .
4. If you measure the parallax of a star to be 0.1 arc second on Earth, how big would the parallax of the
same star be for an observer on Mars (Mars-Sun distance = 1.5 A.U.)?
The parallax of the same star be for an observer on Mars (Mars-Sun distance = 1.5 A.U.) would be 50%
larger that the parallax measured from Earth because the baseline of the parallax measurement is 50%
larger from Mars and from Earth. So if you measure the parallax of a star to be 0.1 arc second on Earth,
from Mars the parallax will be 0.15 arcseconds.
5. If you measure the parallax of a star to be 0.5 arc second on Earth, and an observer in a space station in
orbit around the Sun measures a parallax for the same star of 1.0 arc second, how far is the space station
from the Sun?
The distance to the star, of course, does not depend on the baseline of the parallax measurement. SO the
way to proceed is to write the stellar parallax distance for formula as measured from the Earth and then
write it again using the unknown baseline of the space station. (Remember, the baseline is the distance
of the measurer from the Sun or, equivalently, the radius of the space stations orbit around the Sun,
expressed in AU). Thus the two distance equations are:
From Earth at 1 AU from the Sun d pc 
1
0.5
From Space Station at x AU from the Sun d pc 
x
1.0
Since the distances are the same, the right hand side of the two equations above must equal each other.
1
x

0.5 1.0
Solve for x, the baseline of the space station measurement. x 
1.0
2
0.5
So the space station is 2 AU from the Sun (just past the orbit of Mars).
6. If you can measure angles as small as 1/50 arc second, how far out can you measure star distances from
the Earth using the trigonometric parallax method? How long do you have to wait between
observations?
d pc 
1
1

 50 pc
p" .02"
If you can measure angles as small as 1/50 arc second, the farthest out can you measure star distances
from the Earth using the trigonometric parallax method is 50 pc or 163 light years. You have to wait for
Earth to move halfway around its orbit to take the two necessary pictures of the star to measure parallax,
so you have to wait six months between observations.
7. If you can measure angles as small as 1/50 arc second, how far out can you measure star distances from
Jupiter (Jupiter-Sun distance = 5.2 A.U.) using the trigonometric parallax method? However, how long
do you have to wait between observations? (Use Kepler's third law to find Jupiter's orbital period and
divide by two.)
5.2 5.2

 260 pc
p" .02"
If you can measure angles as small as 1/50 arc second, the farthest out can you measure star distances
from the Jupiter using the trigonometric parallax method is 260 pc or 847 light years. You would have
d pc 
to wait for Jupiter to move halfway around its orbit to take the two necessary pictures of the star to
measure parallax, so you have to wait about six years between observations.
Review Questions
1. Two identical stars have different apparent brightnesses (fluxes). One star is 10 parsecs away from us
and the other is 30 parsecs away from you. Which star is brighter and by how many times?
The brightness of a star depends on two factors; its luminosity and its distance. This is summarized in
L
the relation B 
. Notice that this brightness relation is an example of an inverse square law.
4  d 2
Since the stars are identical, they have the same luminosities. Thus, since the stars have the same
luminosities and one star is 3 times farther away compared to the other, then the brightness of the more
distance star is 9 times less than the brightness of the closer star.
2. Two identical stars have different fluxes. One star is 5 parsecs away from you and appears 81 times
brighter than the other star. How far away is the dimmer star?
The brightness of a star depends on two factors; its luminosity and its distance. This is summarized in
L
the relation B 
. Notice that this brightness relation is an example of an inverse square law.
4  d 2
Since the stars are identical, they have the same luminosities. Thus, since the stars have the same
luminosities and one star is 81 times brighter compared to the other, then the distance to the dimmer
star is 9 times greater than the distance of the brighter star. Since the brighter star is 5 pc away, then
the dimmer star is 45 pc away.
3. The Earth receives about 1380 Watts/meter2 of energy from the Sun. How much energy does Saturn
receive from the Sun (Saturn-Sun distance = 9.5 A.U.)? (A Watt is a unit for the amount of energy
generated or received every second.)
The brightness of a star depends on two factors; its luminosity and its distance. This is summarized in
L
the relation B 
. Notice that this brightness relation is an example of an inverse square law.
4  d 2
Since Saturn is 9.5 times farther from the Sun than the Earth, the Sun will appear 9.52 = 90.25 times
dimmer as seen from Satturn compared to the Sun as seen from the Earth. Since the Earth receives
1,380 W/m2 from the Sun, Saturn will receive only 15.3 W/m2 from the Sun.
Review Questions
1. What does a magnitude interval of 5 correspond to in brightness? How about an interval of 1? How
about an interval of 3?
A magnitude interval of 5 corresponds to a factor of 100 in brightness. A magnitude interval of 1
corresponds to a factor of about 2.5 in brightness. A magnitude interval of 3corresponds to a factor of
about 2.53 = 15.6 in brightness.
2. Do bright things have larger or smaller magnitudes than fainter things?
Brighter astronomical objects have lower (not smaller) magnitudes than fainter objects. This is because
the magnitude scale is a “backwards” scale that assigns numerically lower values of apparent (or
absolute )magnitudes for brighter (or more luminous) objects.
3. How is apparent magnitude different from absolute magnitude?
Apparent magnitude expresses a stars brightness (that depends on its luminosity and distance), while the
absolute magnitude expresses a star’s luminosity. In fact, the absolute magnitude is the apparent
magnitude a star would have if it were at a standard distance from Earth of 10 parsecs.
4. Put the following objects (given with their apparent magnitudes) in order of brightness as seen from
Earth (faintest first): Sun (-26.7), Venus (-4.4), Barnard's Star (9.5), Sirius (-1.4), Proxima Centauri
(11.0).
(faintest first): Proxima Centauri (11.0), Barnard's Star (9.5), Sirius (-1.4), Venus (-4.4), Sun (-26.7)
5. You receive 8× 10-9 Watts/meter2 of energy from a star 2 parsecs away with an apparent magnitude =
1.3. What is the energy you receive from a star with an apparent magnitude = 5.3?
A difference of 1 in apparent magnitude represents a difference of about 2.5 in brightness (W/m2). So a
star that has an apparent magnitude of 5.3 will be 4 magnitudes dimmer than a star with an apparent
magnitude of 1.3. This 4 magnitude difference in apparent magnitudes means that the dimmer star will
be about 2.54 = 40 times dimmer than the brighter star. Thus the energy you would receive from a star
W
8  10 9 2
m  2  10 10 W .
with an apparent magnitude = 5.3 is
40
m2
6. Two identical stars but star B is 10X farther away than star A. What is the difference in magnitudes
between the two stars?
Star B that is 10 times farther away, but otherwise identical to star A, would appear 100 times dimmer
(since brightness follows and inverse square law). A star that is 100 times dimmer will be 5 magnitudes
dimmer in apparent magnitude. So the difference in apparent magnitudes between the two stars is 5
with star A having the brighter magnitude.
7. What two things does luminosity depend on?
The luminosity of a star depends on two factors; its temperature and its surface area. This is
summarized in the Stefan-Boltzmann Law L  4R 2  T 4 .
8. If our Sun has luminosity = 1 solar luminosity, what is the luminosity of the following stars if they have
the same diameter as the Sun (fill in the table):
star temperature (K) luminosity
Sun
6,000
1
A
12,000
16
B
2,000
.0123
C
36,000
1,296
If the radius remains constant then the
luminosity varies as T4.
So star A is twice the temperature of
the Sun, so its luminosity is 24= 16
times the Sun’s Luminosity.
9. Two stars have proper motions of 0.5 arc seconds/year. Star (A) is 20 parsecs away and star (B) is 30
parsecs away. Which one is moving faster in space?
The proper motion is a measure of the angular shift of the star per year. If two stars have the same
proper motion then they have the same angular shift in their position per year. A star that is more
distant must move faster in space to have the same proper motion as a closer star. See figure below.

, proper motion
Tangential velocity
of the close star
Tangential velocity
of the far star
10. If our Sun has a surface temperature of 5840 K, how many times hotter than the Sun is the hottest Otype star? How many times cooler than the Sun is the coolest M-type star?
The hottest O-type star has a temperature of about 50,000 K and this is approximately 10 times hotter
than the Sun (5,800 K). The coolest M-type star has a temperature of about 2,000 K and this is
approximately 3 times cooler than the Sun (5,800 K).
11. Some stars have temperatures of only 3000 K but have over 100X more luminosity than the Sun. How is
this possible?
The luminosity of a star depends on two factors; its temperature and its surface area. This is
summarized in the Stefan-Boltzmann Law L  4R 2  T 4 . Thus a cool star can have a very large
luminosity if it has a very large radius.
12. Would a red giant have a smaller or larger magnitude in a ``V'' filter than in a ``B'' filter? (Remember the
first rule of magnitudes!)
Red stars are cool stars and emit most of their radiation in the red end of the visual spectrum. They emit
less energy in the blue end of the visible spectrum. So, photographed through a B (blue) filter they
would appear relatively dim compared to a photograph through a visual (yellow) filter. So the V
magnitude of a red stars will be a smaller number than the B magnitude.