Physics in Everyday Life
Prof George Lafferty, 13 Oct 2014
Notes for Lectures 7 and 8
1
1.1
The Atmosphere
The composition of the atmosphere
The Earth’s atmosphere is composed mainly of the two gases nitrogen, N2 and
oxygen O2 , with the former making up 78% of the volume and the latter 21%. The
rest is mostly argon (0.9%), water vapour and carbon dioxide. Everyday life relies
of course on the oxygen. When we breathe in, some of the oxygen is absorbed into
our bloodstream, but all of the nitrogen simply goes back out again when we exhale.
However, the nitrogen does serve the vital function of diluting the oxygen to a level
that our bodies can cope with.
1.2
The specific gas constant
Like many common gases at STP, air behaves as an ideal gas to a very high degree
and so the ideal gas equation pV = nRT may be applied. Here p is the pressure
of n moles of gas occupying a volume V at (absolute) temperature T . R is the gas
constant whose value is R = 8.314 K K−1 mol−1 . (R is often called the universal,
molar or ideal gas constant).
The ideal gas equation in the above form is not particularly useful when applied
to the atmosphere, which is effectively an open system. Generally there are no walls
containing a certain volume of air. However, atmospheric physicists often use the
concent of a parcel of air, which is yet another idealisation like the ray of light.
We will express the ideal gas equation in a form that is more useful when
dealing with the atmosphere. As a mixture of nitrogen and oxygen, the air has
an effective relative molecular mass M = 29.0. (In the lectures we called this the
molecular weight, but that is perhaps a confusing term since it is a mass rather
than a weight, and also a weight (or mass) would normally have units.) The relative
molecular mass is the mass of a molecule in atomic mass units i.e. its mass divided
by one-twelfth of the mass of a carbon-12 atom. One mole of any gas has a mass
of M grams, and so an amount of gas that has a mass of m kg contains 1000m/M
moles. For a mass m we can therefore rewrite the ideal gas equation as
pV =
1000m
RT.
M
i
The density ρ of the gas is the mass per unit volume and so ρ = m/V , giving
p=
1000ρ
RT
M
which we write as
p = ρRs T
with Rs = 1000R/M called the specific gas constant. For dry air at STP (but there
is not a lot of dry air around Manchester), Rs = 287 J kg−1 K−1 .
1.3
Variation of pressure and density with altitude
Close to the Earth’s surface the atmospheric pressure has an average value of
105 N m−2 . Variations above and below this average charcterise the weather, with
lower pressure generally accompanying rain and winds and higher pressure associated with settled, sunny weather. In the UK, typical slow variations in atmospheric
pressure are about 2% about the mean value. The largest recorded variations correspond to about 10% changes relative to the average. In Hurricane Katrina, the
lowest pressure recorded was 0.92 × 105 N m−2 . At standard atmospheric pressure
and a temperature of 20◦ C, the density of the air is 1.2 kg m−3 . Since pressure
and temperature don’t change by much (fractionally) near sea level, the density is
always close to this value. Note that 20◦ C, is not double 10◦ C. It has been known
for weather pundits to say that for former is ”twice as hot”. Temperature must of
course be measured on the absolute scale, where it is clear that 20◦ C is 3.5% higher
than 10◦ C.
Since the pressure at sea level is 105 Pa, then each square metre must be
overlain by a mass m of atmosphere corresponding to mg/(1m2 ) = 105 Pa, giving
m = 10.2 tonnes. Let’s look at how atmospheric pressure varies with altitude. In
essence, the pressure at any point in the atmosphere is due to the weight of atmosphere above that point. Consider a cylinder of atmosphere, as shown in figure 1.
We have a vertical axis to represent height. The cylinder has a vertical axis, a crosssectional area A and a height δh. The pressure at the bottom face of the cylinder,
at height h, is p(h) and the pressure at the top face is p(h + δh). Since the bottom
face has more atmosphere above it, then p(h) will be larger than p(h + δh) by an
amount given by the weight of atmosphere in the cylinder divided by the area A.
Therefore we can write
ρ Aδh g
p(h) − p(h + δh) =
A
where ρ is the density. If we now let δh → 0 we have
p(h + δh) − p(h)
dp
→
= −ρg.
δh
dh
ii
Figure 1: An imaginary cylinder of atmosphere, as discussed in the text.
We have from earlier the result that p = ρRs T and so we can write the density as
ρ = p/Rs T . Then
p
dp
= − where h0 = Rs T /g.
dh
h0
Let p(0) be pressure at sea level. Then separating the variables and integrating gives
Z
p(h)
p0
dp
1
=−
p
h0
Z
h
dh
0
which leads to
p(h) = p(0) exp(−h/h0 ).
Therfore, if T is constant, h0 is constant and the pressure falls exponentially with
altitude. The quantity h0 is the scale height, which is the change in altitude over
which the pressure falls by a factor of 1/e. In fact the temperature of the lower
atmosphere is not constant, but the rate of change with altitude is slow enough that
it doesn’t significantly affect the approximation of an exponential drop in pressure
with altitude. (Later we will estimate how temperature varies with height in the
atmosphere.) Note that although the above derivation was done for a cylinder
of atmosphere, the result is quite general. Any vertical column of air could be
subdivided into a large number of regular cylindrical shapes (actually hexagonal
prisms would do the trick nicely, since they fit together).
For T = 273 K the scale height of the Earth’s atmosphere is 8 km. The radius
of the Earth itself is 6400 km, some one thousand times the scale height, which
nicely illustrates the fact that the atmosphere is effectively a relatively thin film of
iii
gas on the surface of the Earth. The Earth’s highest mountain, Mount Everest, has
an altidude of about 10 km, where the pressure (and hence the density) of the air
is only 30% of the value at sea level.
Let’s look at the effect of pressure variation over small differences in altitude,
typically much smaller than the distance of the scale height. We are aware of pressure
differences over such small altitude changes when, for example, our ears “pop” if we
go quickly down a hill. Clearly these pressure changes are large enough, and happen
quickly enough, for our ears to record. We will start at some altitude H + δh and
consider the pressure change as we move to altitude H. For small δh h0 , we have
already shown that
δh
δp
≈− .
p(H)
h0
We see that the fractional change in atmospheric pressure over a small change
in altitude is just equal to altitude change as a fraction of the scale height. (Such a
linear relation is a feature of any exponential curve over a short enough part of the
curve; essentially an exponential function exp(−x/x0 ) looks like a straight line for
small ranges in x x0 .)
Let’s look at an example: a car descends a hill, dropping a vertical distance
of 10 m in 5 s. The fractional change in atmospheric pressure would be −δh/h0 =
−(−10)/8000 = 1.2 × 10−3 , so that the pressure changes by about 1 part in 1000
i.e. 0.1%. It takes 5 s for this to happen, so the rate of change is 1.2 × 10−3 /5 =
2.4 × 10−4 s−1 . Note that this is a rate of a fractional change and so the units are
simple those of inverse time. So the pressure changes by 0.024% per second. Anyone
who has been in a car descending even a moderately steep hill, or who has skied
down a hill rather faster than they ought to have done, or has been in a fast elevator,
will have noticed the effect of this apparently small rate of change of pressure on
their ears. Clearly the ears are sensitive instruments, as we will discuss more when
we come to consider sound waves.
1.4
The Coriolis force and circulation of weather systems
With a view to understanding the motion of air masses near the Earth’s surface,
and in particular the circulation of weather systems, we need to consider frames of
reference. A frame of reference may be considered in terms of a set of co-ordinate
axes. In order to write down equations that describe the motion of an object in
three dimensional space, we need a set of co-ordinates to describe the position as
a function of time. We will work with cartesian co-ordinates (x, y, z), although
sometimes polar co-ordinates are more convenient.
According to Newton’s first law, a particle at rest or moving with constant
iv
speed in a straight line (i.e. with constant velocity) will maintain that status unless
affected by an external force. (Newton’s second law says what happens to the
particle’s momentum if such an external force is applied.) The first law presupposes
the existence of what are called inertial frames of reference. While this definition is
a bit circular, it helps us to consider an inertial frame of reference as one in which
Newton’s first law holds true. For practical purposes, physicists often refer to the
so-called fixed stars as defining such a frame of reference. Over reasonably long
time scales the pattern of stars in the sky does not change, while the planets (the
word coming from the Greek, meaning “wanderer”) wander among this pattern. In
fact the distant stars are moving much faster relative to Earth than are the nearby
planets, and the reason for the apparent fixedness of the stars is simply that they
are so very far away. For the same reason a distant aircraft seems to move quite
slowly as the rate at which its path sweeps out an angular range relative to our eyes
is low because of the large distance.
Every 24 hours the pattern of stars appears to rotate around the Earth, but of
course it is the Earth itself that is rotating. The rate of this rotation is low enough
that we have the impression that we are not moving. At the same time, the Earth is
orbiting the Sun, completing a full orbit once a year, while the Milky Way galaxy is
also rotating and moving within the local cluster of galaxies. All of these rotational
and orbital motions correspond to accelerations (changes of speed or direction), but
we are content to say that when standing still on the Earth’s surface we are unaware
of all this, and that we are not moving. This, indeed, was the view that generally
held right up until the times of Copernicus and Gallileo. The surface of the Earth
is an approximate inertial frame of reference, with respect to which the fixed stars
move very slowly. (A famous experiment known as Foucault’s pendulum provides a
good demonstration that the Earth’s surface is not a true inertial frame.)
The fact that the Earth is not really an intertial frame helps us understand
some aspects of the motion of weather systems. Consider a “parcel” of air, moving
along the x axis in a straight line with constant speed v in an inertial frame of
reference. For simplicity we will consider the motion only in two dimensions. Figure 2 shows the situation, where the red blob represents the air parcel. According
to Newton’s first law, the air will continue to move in a straight line in the absence
of forces. If the parcel starts at the origin at time t = 0 then its (x,y) co-ordinates
will be (vt, 0). Now let’s look at this from the point of view of an observer at the
origin who is rotating relative to the interial frame (but does not necessarily know
this). This observer will have co-ordinate axes x0 , y 0 as shown, rotating at constant
angular speed ω. To this observer the air parcel will appear to spiral out away from
the origin. The co-ordinates of the air parcel as measured by the rotating observer
can be express in terms of those in the inertial frame:
x0 = vt cos(ωt),
v
Figure 2: Motion of a particle moving with constant velocity along the x-axis,
according to a rotating observer, as discussed in the text.
y 0 = −vt sin(ωt).
We can differentiate twice with respect to time to get the velocity and acceleration
in the primed frame:
ẋ0 = v cos(ωt) − ωvt sin(ωt),
ẏ 0 = −v sin(ωt) − ωvt cos(ωt)
and
ẍ0 = −ωv sin(ωt) − ωv sin(ωt) − ω 2 vt cos(ωt) = −2ωv sin(ωt) − ω 2 vt cos(ωt),
ÿ 0 = −ωv cos(ωt) − ωv cos(ωt) + ω 2 vt sin(ωt) = −ωv cos(ωt) + ω 2 vt sin(ωt).
Thus the position r0 , velocity v0 and acceleration a0 of the air parcel as seen from
the rotating frame can be written in terms of unit vectors as follows:
r0 = vt (cos (ωt) , −sin (ωt)) ,
v0 = (ẋ0 , ẏ 0 ) = v (cos (ωt) , − sin (ωt)) − ωvt (sin (ωt) , cos (ωt)) ,
a0 = (ẍ0 , ÿ 0 ) = 2ωv (− sin (ωt) , − cos (ωt))) + ω 2 r (− cos (ωt)) , sin (ωt)).
As seen from the frame rotating with angular speed ω, a particle with constant
velocity v in the xy plane in an inertial frame has an acceleration, which has two components. The first part, 2ωv (− sin (ωt) , − cos (ωt)) points in a direction to the right
of the direction of motion (which is the direction of the unit vector involved) and is
vi
known at the Coriolis acceleration, while the second term ω 2 r (− cos (ωt)) , sin (ωt))
points towards the origin and is called the centripetal acceleration.
The rotating observer interprets these accelerations as being due to the Coriolis
and centripetal forces. Of course, these are not real forces, but inertial or fictitious,
since the body is moving in a straight line in an inertial frame of reference. The
appearance of the fictitious forces has come about because of the acceleration (rotation) of the observer, not of the parcel of air. We have derived this result for a flat
surface in two dimensions. On the Earth’s surface, the magnitude of the Coriolis
acceleration is given by 2ωv cos λ where λ is the latitude, which is 90◦ at the north
pole and 0◦ on the equator.
vii