Astronomy 102 Problem Set #1 Solutions
Problem #1: The Cassini spacecraft is
currently on its way to Saturn and will arrive in 2004. Right now (January
23, 2000) it is 1.5 million km from the asteroid Masurski, and 380
million km from Earth. How long do radio signals from this spacecraft take
to reach controllers here on Earth?
Solution: This one is pretty straightforward
if
you remember that radio waves are light waves. Even though you hear
sound from your radio, the radio waves which propagate from the radio station
to the receiver inside your radio are a form of electromagnetic radiation;
they are light waves with wavelength much longer than that of visible light
waves. As light waves, they propagate at the speed of light, which
is 3 x 108 m/s.
The distance between the spacecraft and Earth is 380 million km, or
380 x 106 km. We can (and should) convert this value to meters
as follows:
-
380 x 106 km = 380 x 106 km x (1000 m/1 km)
-
= 380 x 109 m
-
= 3.8 x 1011 m
Now, distance = speed x time, so time = distance/speed (divide both sides
of the equation by the speed). We know the distance and the speed so,
-
time = distance/speed
-
= 3.8 x 1011 m / 3 x 108 m/s
-
= 1266 seconds
-
= also known as 1266 sec/ 60 (sec/min) = 21.1 minutes
Problem #2: a) How many green photons
(wavelength = 500 nm) does it take to equal the energy of one gamma ray
photon (wavelength = 10-2 nm)?
b) How many of these gamma ray photons would it take to equal
the the energy required to light a 100-watt light bulb for one second (100
J)?
Solution -- Part a: Let's start with the
energy of a photon:
-
energy = Planck's constant x speed of light / wavelength
-
E = h c / l
-
= 6.626 x 10-34 J s x 3.00 x 108 m/s / l
-
= 1.99 x 10-25 J m / l
where the constant listed above will yield an energy in Joules for an input
wavelength measured in meters.
So, the energy of a photon with wavelength 500 nm, or 500 x 10-9
m, is
-
E = 1.99 x 10-25 J m / l
-
= 1.99 x 10-25 J m / 500 x 10-9 m
-
= 3.98 x 10-19 Joules.
Likewise the energy of a photon with wavelength 10-2 nm, or
1.00 x 10-11 m, is
-
E = 1.99 x 10-25 J m / lambda
-
= 1.99 x 10-25 J m / 1.00 x 10-11 m
-
= 1.99 x 10-14 Joules.
OK, so now we have the energy in photons of both wavelengths, and we want
to know how many 500nm photons we would need to equal the energy in one
gamma ray photon. Think about this for a minute, and I bet you'll see that
the number of photons we need is equal to the ratio of the energies of
the gamma ray and 500 nm photons. For example, if a gamma ray photon has
ten times the energy of a 500 nm photon, then you'll need ten 500 nm photons
to equal the energy of one gamma ray photon. If the energy ratio is 20,
then you'll need 20 500nm photons, and so on.
We can express this idea mathematically by saying that the energy in
N
500 nm photons is equal to the energy in one 10-2 nm photon.
We don't yet know what N is but, since the energy in N 500
nm photons is equal to the energy in one 10-2 nm photon, we
can set up the following equation:
-
energy in N 500 nm photons = energy in one 10-2 photon
-
N x energy in one 500 nm photon = energy in one 10-2
nm photon
and, if we divide both sides of the equation by the energy in one 500 nm
photon,
-
N = energy in one 10-2 nm photon / energy in one 500
nm photon
which says that N is equal to the ratio of the energies of the two
photons, which is precisely what we came up with based on our intuition
above. So now we're in a position to answer the question:
-
N = energy in one 10-2 nm photon / energy in one 500
nm photon
-
= 1.99 x 10-14 J / 3.99 x 10-19 J
-
= 50,000.
You might note that this ratio of energies is equal to the ratio of the
wavelengths of the two photons. That's not a coincidence at all. In fact,
there's an even easier way to do this problem, if you set up the ratio
at the very beginning. Consider the following definitions for the energies
of the two photons:
-
E500 nm = h x c / l500 nm
-
E10-2 nm = h x c / l10-2
nm
where the subscripts denote that the quantity (energy and wavelength) pertains
to the 500 nm or 10-2 nm photon. We can then construct the ratio
of the energies directly,
-
E10-2 nm / E500 nm = (h x c /
l 10-2 nm) / (h x c /
l500 nm)
and since the h and c appear on the top and bottom of the ratio, they divide
out, leaving
-
E10-2 nm / E500 nm = ( 1 /
l10-2 nm) / (1 /
l500 nm).
Somewhere deep in your high school math background is the memory that 1/(1/x)
= x (you can't fool me; I know it's there somewhere). Using this perhaps
newly-remembered information, we can simply the above expression to yield:
-
E10-2 nm / E500 nm =
l500 nm / l10-2
nm
-
= 500 nm/ 10-2 nm
-
= 50,000
and with a lot less calculator work, we get the answer that you'll need
50,000 green photons to equal the energy of just one gamma ray photon.
Solution -- Part b: Now we're asked how
many gamma ray photons we'll need in order to supply 100 Joules of energy
to a light bulb. From above, we know that one gamma ray photon contains
1.99 x 10-14 Joules; therefore N gamma ray photons will
contain
-
energy in N gamma ray photons = N x energy in one gamma ray
photon
-
= N x 1.99 x 10-14 Joules.
We want to know what N is if the total energy is 100 Joules, so
-
100 Joules = N x 1.99 x 10-14 Joules.
and therefore
-
N = 100 Joules / 1.99 x 10-14 Joules.
-
= 5.02 x 1015 photons.
Problem #3: a)The public radio station in
Lewisburg broadcasts at a frequency of 100.1 MHz ("mega-hertz"). What is
the wavelength of these radio waves? b)I'm driving in my car at
55 mph (24.6 m/s) directly toward the radio station's broadcast tower.
What is the frequency shift (due to the Doppler Effect) in the signal my
car radio receives? Is the received frequency higher or lower than the
broadcast frequency?
Solution -- Part a: This is just an application
of the speed-wavelength-frequency relation for any type of wave:
-
speed = wavelength x frequency
We are given the frequency of the radio waves in MHz. 1 MHz = 106,
so the frequency is
-
100.1 MHz = 100.1 MHz x (106 Hz/ 1 MHz)
-
= 100.1 x 106 Hz, or
-
= 1.001 x 108 Hz
In addition, since radio waves are just another form of light waves,
we know that the speed of these waves is the speed of light, 3.00 x 108
m/s. So,
-
3.00 x 108 m/s = wavelength x 1.001 x 108 Hz
or, rearranging the equation,
-
wavelength = 3.00 x 108 m/s / 1.001 x 108 Hz
-
= 3.00 m/s / Hz
but what kind of a unit is m/s/Hz? Well, recall that the unit of frequency,
Hz, is just "per second" or 1/s. So the 1/s in m/s and the 1/s from Hz
cancel, leaving units of m. Does this unit make sense for our answer? Sure!
We're calculating a wavelength, which is a length and should have units
of length. Meter are units of length, so these units are appropriate, and
our answer is:
Solution -- Part b: This part of the problem
is a straightforward application of the Doppler Effect, much like the problem
we did at the end of the first lab. The Doppler formula is:
-
shift in frequency / rest frequency = speed of emitter / speed of wave
We already know some of these quantities. The rest frequency is the frequency
emitted by the radio station, 100.1 MHz, or 1.001 x 108 Hz.
The speed of the wave is the speed of light, since
radio waves are light
waves (have I mentioned that before?). In the problem, you're told
that I'm driving in my car toward the radio station at a speed of 24.6
m/s, so that's the difference between the speed of the emitter (the radio
station isn't moving) and the speed of the receiver in my car. Therefore,
we have three out of the four variables in the Doppler relation, so we
can solve for the fourth, which is the shift in the frequency received.
So,
-
shift in frequency / rest frequency = speed of emitter / speed of wave
-
shift in frequency / 1.001 x 108 Hz = 24.6 m/s / 3.00 x 108
m/s
Multiplying both sides by 1.001 x 108 Hz, we get
-
shift in frequency = 1.001 x 108 Hz x 24.6 m/s / 3.00 x 108
m/s
-
= 8.2 Hz.
Note that the units of the answer are Hz because the m/s on the top and
bottom of the right hand side of the equation cancel out. These units make
sense since we're looking for a shift in frequency, which should be measured
in units of frequency, or Hz.
Lastly, we need to decide whether the received frequency is shifted
to a frequency higher or lower than that emitted by the radio station.
Since I'm toward from the radio station, the separation between successive
waves is a little smaller than if I weren't moving. That means that the
received wavelength is a little shorter, and therefore that the received
frequency is little higher.
Incidentally, 8.2 Hz is a tiny change in frequency -- when compared
with 100.1 million Hz -- much smaller that the ability of most radios
to tune. The frequency shift introduced by a car's motion toward or away
from a radio station is so small that you don't have to retune your radio
every time you speed up or slow down.