Download Solutions Set #2 1. What is the B

Solutions Set #2
1. What is the B-band brightness ratio of Betelgeuse to Rigel? This
requires a bit of on-line sleuthing, particularly since I forgot that Betelgeuse
is a variable star. When I consult SIMBAD, I find the following numbers:
B = 2.35 for Betelgeuse, and B = 0.09 for Rigel. Now it may seem weird
that Rigel has a lower number (and hence, is brighter) than Betelgeuse, when
Betelgeuse is the brightest star in Orion! This is because Rigel is a blue star
and Betelgeuse is a red star. In the B band, Rigel is putting out a lot more
light than Betelgeuse is. If we had picked R band, the answer would be quite
different.
To continue: In order to get the brightness ratio, we use the defintion of
the magnitude scale:
m1 − m2 = −2.5 log (F1 /F2 ),
(1)
where F1 /F2 is the brightness ratio were looking for. So we first solve this
equation for the ratio, and then we can simply plug the numbers in and find
the answer. The brightness ratio is given by
m2 −m1
0.09−2.35
F1
= 10 2.5 = 10 2.5 = 0.125.
F2
(2)
So Rigel is about eight times brighter than Betelgeuse, in the B band.
2. If one star has m1 = 8.4 and the other star has m2 = 5.2, then we
know right away that star two is brighter than star one. We know that
five magnitudes difference is about 100 times brighter, and one magnitude
difference is 2.5 times brighter, but this is 3.2 magnitudes difference, so it
will be between 2.5 and 100, but closer to 2.5. Using the magnitude equation
(Equation 1), we can put 3.2 on the left and solve for F1 /F2 to get that
star two is about twenty times brighter than star one. The ratio of F1 /F2 =
0.0525, to be exact. So if F2 is 5% of F1 , then both of them together must be
105% of F1 . Which is to say, adding star two to star one makes the brightness
increase by 5%.
So what would the magnitude change corresponding to a 5% increase be?
Well, we use Equation 1 again, but now the left side is unknown, and the
right side is −2.5 log (1.0525) = −.05555. This means adding star one will
make the magnitude go down by 0.0556, compared to star two. This means
the magnitude of both will be 5.1, which is indeed less than 5.2 (because
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both stars are indeed brighter than one star), but not much less, because a
5% increase is not a huge change.
You could get the same answer by using star one as the comparison. Then
both stars together would be 20 times brighter than star one. (Star two is
19 times brighter than star one, so the sum is 20 times brighter than star
one by itself.) Again using Equation 1, −2.5 log (20) = −3.255, which is the
change from star one by itself. Since star one is 8.4, we subtract 3.255 to get
5.1, which is the same thing we got the other way!
3. This problem is simply an exercise in getting used to scale and to
thinking about non-visible light, and how all the different parts of the electromagnetic spectrum compare. To compute the answer, you get the frequency
of the light (use c = λν if you dont know it), and then use E = hν to get the
energy. c = 3.00 × 108 m/s, and h = 6.63 × 10−34 Js. So if I get frequency in
hertz (Hz), I will get energy in joules (J).
a) WQFS: 6.03 × 10−26 J
b) Microwave oven: 1.60 × 10−24 J
c) Cell phone: 7 × 10−24 J
d) Human body: 2 × 10−20 J
e) Red photon: 3.1 × 10−19 J
f) Green photon: 3.6 × 10−19 J
g) Blue photon: 4.7 × 10−19 J
h) Dental x-ray: 10−13 J
i) Gamma ray: 10−10 J
A couple patterns I hope you notice: first of all, the visible light spans a
tiny range of energy, compared to all the other sources of light. The difference
between red and blue is less than a factor of two!!! Note that radio waves carry
10 million times less energy than visible light, which in turn carries a billion
times less energy than gamma rays! Overall, the total span of these relatively
common sources of radiation spans seventeen orders of magnitude!!!! Just to
put it in perspective, the ratio of the sun-earth distance to the thickness of
a sheet of aluminum foil is ten times smaller than that!
Another interesting point: note that the energy of a cell phone photon
is actually more than three times higher than that of a microwave oven
photon. And yet a microwave oven will cook your food, while a cell phone
causes no harm. (The people who think cell phones cause brain cancer seem
to be wrong.) I can think of two reasons for this: first, ovens use a LOT
more power than a cell phone, which means although each photon carries
less energy, there are a heck of a lot more of them. Second, the energy of a
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microwave oven photon has been carefully chosen because water “likes” to
absorb that particular energy, so that energy gets absorbed more easily (by
water) than a cell phone photon.
4. You are monitoring a fading star with your telescope. You measure
that the magnitude is increasing at a constant rate of 0.2 magnitudes every
24 hours. How would you describe the change in brightness of the star as a
function of time?
Usually, the concept of magnitudes is introduced in comparing two different stars. However, it works just as well for comparing the same star at
different times. If ∆m = 0.2 in one day, then we can use the definition of
magnitudes to get the brightness ratio between today and yesterday:
!
mtoday − myesterday
Ftoday
.
= 0.2 = −2.5 log
Fyesterday
(3)
Solve for the brightness ratio:
0.2
Ftoday
= 10− 2.5 = 0.832,
Fyesterday
(4)
which is another way of saying that every time 24 hours pass, the star has
dimmed to 83% of its brightness the previous day. Alternately, you could
say that the brightness dims by 17% each day.
5. You have two 100-watt light bulbs, but one of them is eight times as
far from you as the other. What is the difference between their magnitudes?
If one light bulb is 8 times as far away as the other, it must look 82 = 64
times dimmer. Which means the ratio of brightnesses is 64. So I can plug
that into the magnitude equation:
m1 − m2 = −2.5 log (64) = 4.5.
(5)
A brightness ratio of 64 corresponds to a magnitude difference of 4.5, with
the more distant (dimmer) light bulb having the larger magnitude number.
Note that a change in brightness of 100 times corresponds to a magnitude
change of 5, and you found in problem 2 that a change in brightness of 20
times corresponds to a magnitude change of 3.2, so it would make sense that
a magnitude change of 4.5 would be from a brightness change between 20
and 100 times, which 64 is.
6. The sun puts out a total power of about 1026 joules of energy every
second in the form of light.
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a) If you assume it is all emitted as visible light, roughly how many
photons are emitted by the sun every second? In problem 3, you found that
red, blue and green light all have energies of about 10−19 J, so I am going to
just assume that the energy of a single photon is 10−19 J. If the sun is putting
out 1026 J every second, I divide those two numbers to get 1045 photons every
second.
b) Roughly how many photons per square meter are hitting the Earth
from the sun?
The Earth is 1.5 × 1011 m from the Sun. A sphere of radius r has a
surface area of 4πr2 . That means the 1045 photons have to be spread out
among 3 × 1023 m2 . That means for every square meter at the Earth, there
are about 4 × 1021 photons passing through that square meter every second.
c) How many photons enter your eye every second if you look directly at
the Sun? How many photons go in your eye each second if you look at the
Sun through an 8” telescope?
The area of your pupil is roughly 2 × 10−5 m2 . If we multiply this by
the answer from the previous problem, we get 8 × 1016 photons per second,
or 1017 if we round off. So, if you look at the sun for a second, almost ten
thousand billion photons hit your retina. That’s a lot. Of course, if you were
to look at the sun, your iris would contract, shrinking the size of your pupil
to block more photons, but not by a factor of 10! It still hurts!!
The area of the telescope aperture is roughly 103 times bigger than the
area of your eye. So a thousand times as many photons enter the telescope
than enter your eye every second. That’s about 1019 photons entering the
telescope every second. That’s why you never want to look through the
telescope at the sun; its like your pupil just got a thousand times bigger.
(Although the secondary blocks some of that.)
d) The star Sirius is roughly 25 magnitudes dimmer than the Sun. When
you look through the 8” telescope at Sirius, roughly how many photons enter
your eye each second?
By the magnitude formula in Equation 1, if the difference in magnitudes
is 25, that means the brightness ratio is 10−10 . If 1020 photons enter the
telescope every second from the sun, then 1010 photons will enter the sun
every second from Sirius. Remarkable, when you think about it. Sirius is
so unimaginably far away, and the telescope is only 8” across, and yet ten
billion photons from Sirius would go through its opening every second.
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