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PH4 – Oscillations & Fields
PH4.1 – Vibrations
This section includes circular motion, simple harmonic motion, damping and resonance. It is
essentially unchanged from the equivalent sections in the legacy specification. All its
contents are adequately covered in many A-level Physics textbooks and there is no current
intention to publish guidance notes.
For examples of examination questions, see previous PH4 papers.
PH4.2 – Momentum concepts
This short section has been augmented by the introduction of the concept of photon
momentum and hence of radiation pressure. This draws upon the photon ideas in PH2 and
presents the opportunity for synoptic questions. The obvious application is the “light sail”
which is proposed as a method of interplanetary propulsion. No equations in addition to
p
h

hc
will be required. Questions could probe the difference between cases in which

f
photons are absorbed and those in which they are reflected [giving twice the momentum
transfer]. The concept of elastic and inelastic collisions draws upon energy from PH1.
For examples of examination questions, see previous PH4 papers.
PH4.3 – Thermodynamics
Statements PH4.3(a)–(g) deal with the behaviour of ideal gases. They include a simple
treatment of the kinetic theory of gases, including the concept of the mole. It too is essentially
unchanged from the previous specification.
Statements PH4.3(h)–(p) cover the concepts of thermodynamics: heat, work and internal
energy. The 1st Law of Thermodynamics is also included. In spite of its presence in the
current specification, it is a section which many students find obscure and accordingly a set
of notes is provided: go to the WJEC website, www.wjec.co.uk , select Physics and GCE AS/A
under “Find resources” and “view the full list of documents” under Related Information.
For examples of examination questions, see previous PH4 papers.
PH4.4 – Electrostatic and Gravitational Fields of Force
These two fields of force are treated together, in view of their mathematical similarity. The
field line is introduced as indicating the direction of the force upon a test object [charge or
mass, respectively] and leads on to its mathematical expression in the concept of the vector
quantity of field intensity. The scalar potential in a field is defined in terms of the work
required to be done [by an external agent] in bringing a unit test object from a point of zero
potential – infinity for mathematical convenience.
There are many similar equations and candidates will be helped to avoid their misapplication
by the equation sheet included in the question paper.
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The only introduced concept in this section is that the gravitational field outside a sphericallysymmetric body is identical to that of an identical point mass situated at the centre of the
body. The point of introducing this statement [essentially Gauss’s Law] is to allow for the
application of Newton’s Law of Gravitation to approximately spherical planets, moons and
stars and, in the next section, to the hypothetical dark matter in which galaxies are supposed
to be embedded. It is not explicitly stated, but the other aspect of Gauss’s Law will be
assumed, i.e. that the net contribution to gravitation field by those parts of a spherically
symmetric mass distribution lying outside the radius of the point in question is zero.
For examples of examination questions, see previous PH5 papers.
PH4.5 – Application to Orbits in the Solar System and the Wider Universe.
This section of the specification contains traditional kinematics and application of Newton’s
Laws of Motion. Much the theory is covered in A level text books. The section on mutual
orbits is an exception. The applications to missing matter in galaxies and the detection of
extra-solar planets (ESOs) require very little additional theoretical input.
Statements PH4.5 (a) – (d) deal with the application of Kepler’s Laws of Planetary Motion
and Newton’s Law of Gravitation to the orbit of objects around a massive central object.
With the exception of the statement of Kepler’s Laws, this work could have been examined
under the legacy specification. Suitable statements of Kepler’s Laws are:
K1: The planets orbit in ellipses with the Sun at one focus.
K2: The radius vector sweeps out equal areas in equal intervals of time.
K3: The square of the period of orbit is directly proportional to the cube of the semimajor axis.
The whole of PH4.5 will concentrate on circular orbits. Very little work will be set on the
elliptical aspects. Candidates should be qualitatively aware of the ellipse. [eccentricity will
not be explored quantitatively] and the meaning of “semi-major axis.” The implication of K3,
that the period of orbit of an object in circular orbit is the same as that of an object in an
elliptical orbit with the same semi-major axis, should be understood. An example of where
this is important is the Transfer Orbit.
Transfer Orbits
Consider a satellite being carried on the upper stage of its launch rocket. It is currently in a
low circular orbit – say an altitude of 500 km [radius of orbit ~ 7000 km]. It needs to be
transferred to a geosynchronous orbit [radius ~42000 km].
low earth
(parking) orbit
A
geosynchronous
orbit
B
transfer orbit
The major axis of the transfer orbit is 7 000 + 42 000 = 49 000 km, so the semi-major axis is
24 500 km. The time taken to transfer can then be worked out because the time taken to
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complete half an orbit [the dotted line] is the same as the time for half a circular orbit of
radius 24 500 km. Interestingly, though this is not required knowledge, the energy of the
satellite in the transfer orbit is also the same as if it were in a circular orbit of the same radius,
so we can calculate the additional energy [and therefore the impulse] needed to be given at A
and at injection at B.
“Derivation” of Kepler’s 3rd Law
This follows from Newton’s Law of Gravitation, F  G
m1m2
r2
, and the ideas of centripetal
force developed in section PH4.1.
Consider an object of mass m in a circular orbit of radius r about a much more massive object
of mass M. This could be satellite – natural or artificial - in orbit about a planet, a planet
about a star or a star about the supermassive black hole in the centre of our galaxy.
The centripetal force necessary for the [accelerated] circular motion is given by: F  mr 2 ,
4 2
or equivalently by F  mr 2 , where T is the orbital period.
T
So we can write:
Dividing by m and rearranging, we have:
GMm
4 2

mr
r2
T2
4 2 r 3
.
T2 
GM
i.e the orbital period squared is proportional to the radius cubed, which is K3 for a circular
orbit.
Note that, we have assumed that the central body is a point mass, which it will certainly not
be, but it is also correct if the central object is spherically-symmetric [see above]. Note also
that, historically, the derivation was done in the opposite direction, with Kepler’s 3rd Law
being the evidence for the inverse square relationship.
Weighing the Earth
Experiments to determine G, the universal constant of gravitation, used to be described as
weighing the Earth. This is because a knowledge of G and the orbital radius and period of the
Moon enables us to calculate the mass of the Earth.
Data: Radius of Moon’s orbit = 3844  108 m [380 000 km].
Period of Moon’s orbit = 2732 days = 236  106 s.
G = 6673  1011 N m2 kg2.
T2 
4 2 r 3
4 2 r 3
, so M E 
= 604  1024 kg.
G
GM E
Note that, in this analysis, we have assumed that the mass of the moon is negligible and that
the moon orbits about the centre of the Earth. In fact ME ~ 81 MM so the assumption leads to
some inaccuracy albeit small [~1%].
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This type of analysis is very useful in obtaining information about remote objects in the
universe. For example, we can “weigh” other planets and determine their mean densities,
furnishing data which is useful for developing models of their composition. We can also
weigh stars, black holes and whole galaxies using the same technique, see e.g. “how to
measure the mass of a black hole” on the Physics page of the WJEC website.
The relationship between T and r furnishes data which is useful for developing candidates’
graphical skills. Data on Jupiter’s satellites for example can be used in a log-log plot to
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establish the power law relationship. Students could also plot, say, T 3 against a and use the
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gradient to determine MJ. Alternatively, T against a 2 is a possibility. Note that it is
unproductive to plot T2 against a3 as, whereas most of the points are almost at the origin a
couple are a long way out.
Often the speed of an orbiting object is measured directly, e.g. using Doppler shift [see
below] in which case we could use the following analysis to determine the central mass M.
GMm mv 2

r2
r
GM
v2 
Dividing by m and simplifying:
.
r
Dark Matter and he motion of objects in galaxies.
Spiral galaxies are flattened assemblages of stars which all rotate in the plane of the galaxy
around the centre in its gravitational field. In additional to stars, spiral galaxies contain large
quantities of gas and dust from which new stars form. Details of the structure of spiral
galaxies will not be examined. Consider the following observed [simplified] rotational speed
curve for a typical spiral galaxy: [The low radius part of the curve is obtained from
observations of stars and gas clouds in the visible part of the disc. The observed speeds
beyond the visible galactic disc are from orbiting clouds of neutral hydrogen which emit a
characteristic 21 cm line in the microwave region of the spectrum.]
Rotational speed (km s-1)
observed
200
100
central
galactic bulge
extent of visible disc
calculate
d
50 000
100 000
Distance from centre (light years)
[N.B. the inner ellipse is my crude attempt using “Draw” to represent the central galactic bulge]
The “calculated” curve is that predicted by taking into account the observed normal or
“baryonic” matter in the galaxy. N.B. “Observed” doesn’t only mean “light-emitting” – it
also includes dark gas clouds, whose speeds we can detect by their absorption lines in the
light of more distant object which we view through them.
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GM
, where r is the orbital radius, v is
r
the orbital speed and M is the total mass within the orbit [assuming a spherically-symmetric
distribution]
A useful equation in investigating these curves is v 2 
It is worth highlighting two regions of the curves:
(a) The low-radius part of the curves.
Here the curves coincide and the speed is roughly proportional to the orbital radius. In
other words, the wholes central region of the galaxy rotates with roughly the same
angular velocity. This implies that the density of matter is constant within this region:
For a constant density, , the mass within an orbit of radius r = 43  r 3  .
So, using the equation above: v 2  G
4
3
 r3
 43 G r 2 , so v  r .
r
So we can see that, for the central regions of a galaxy, coinciding with the galactic
bulge, the observed rotational velocity is consistent with the observed constant density
of matter and the value of the matter density is consistent value of the rotational speeds.
(b) The high-radius part of the curves.
The approximately constant orbital velocity is explicable if the density of the material
falls off roughly as r 2 :
If   kr 2 , the total mass within the orbit, M   4 r 2  kr 2  dr  4 kr
G  4 kr 
 4 Gk , i.e. v is a constant
r
If we imagined a gas cloud orbiting at, say, 75 000 light years from the centre of the
galaxy, it is doing so in the combined gravitational field of all the matter closer to the
centre. In the case of a spherically-symmetric object we can for such purposes consider
it as a point mass with its whole mass concentrated at its centre. Clearly the visible
galaxy is not spherically symmetric, but it is not a bad approximation to consider it so
for great distances. Beyond the visible disc, where the observed matter density is very
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low, we’d expect the orbital speed, v, to fall off approximately as r  2 , the same
relationship as we observe for the planets in the Solar System. The observation that,
beyond ~ 50 k l-y, the rotational speed is ~ constant implies that the material of the
galaxy extends well beyond the observed galaxy, i.e. the visible galaxy is embedded in
an unobserved cloud of material and also than the whole galaxy has a much greater
mass [~ 10 times] than that of the observable matter.
So v 2 
N.B. It is worth emphasising that using “Dark Matter” to explain the discrepancy
between the observed and calculated orbital speeds is a hypothesis, albeit one which is
widely supported in the theoretical cosmological community. Some theoretical
cosmologists have proposed modifications to the law of gravitation to account for the
observations. The modifications take into account the fact that, at the scale of the Solar
System, the inverse square law works very well. In one such model, by Milgrom, the
modification takes effect at gravitational accelerations of less than 10-9 m s-2 [i.e. ~10-10
g] and for these accelerations the gravitational force falls of as inverse r rather than
inverse r2]. This is a classic example of “watch this space” or “How Science Works.”
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Objects in mutual orbit, leading up to the discovery of Extra-solar Planets.
Centre of Mass
We normally think that planets orbit stars and, to a good approximation, this is true because
the planet is so much less massive than the star. For example, the MEarth = 6  1024 kg and
MSun = 2  1030 kg. For precise work or for situations where the two orbiting bodies are of
similar mass, such as a binary star or the Pluto-Charon system, we need to refer to the Centre
of Mass.
Both bodies orbit around a point which, in the absence of externally-applied forces, is
stationary [or, more strictly, moves with constant velocity]. This point is called the centre of
mass.
Let us consider two spherically-symmetric objects, of comparable mass, orbiting about their
centre of mass. Before we do any algebra, we can infer three things about the system:
1. Symmetry considerations tell us that that the centre of mass must be on the line
joining the centres of the two objects.
2. The centre of mass must be between the objects as the direction of the centripetal
acceleration must be towards it.
3. The angular velocities of the objects must be identical – if this were not the case, the
objects would sometimes be on the same side of the Centre of Mass, which clearly
contradicts point 2.
Time for some algebra: Consider two bodies, of mass m1 and m2 orbiting around their Centre
of Mass, C.
m1
r2
r1
C
m2
d
Each body exerts an attractive force upon the other and, by Newton’s 3rd Law, these are equal
and oppositely directed.
m1r1 2  m2 r2 2
So, we can write
So, dividing by  ,
Substituting for r2

Similarly
m1r1  m2 r2
m1r1  m2 (d  r1 )
m2
r1 
d
m1  m2
m1
r2 
d
m1  m2
The orbits of two massive objects, e.g. a binary star system:
Now that the position of the centre of mass is sorted out, we can use Newton’s Law of
Gravitation to work out the orbital characteristics of the binary system as follows:
Consider the orbit of body 1 about the centre of mass. The centripetal force is provided by the
gravitational attraction of body 2 upon body 1.
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So we can write
m1r1 2 
Substituting for r1:
m1
Dividing by m1m2 and rearranging:
2 
as T 
2

Gm1m2
d2
m2
Gm1m2
d 2 
m1  m2
d2
G  m1  m2 
d3
4 2 d 3
d3
T2 
or T  2
G  m1  m2 
G  m1  m2 
Consider the case of the Earth-Sun system. The Earth-Sun distance is 1496 million km. With
the masses given above, the distance of the centre of mass of the two bodies from the centre
of the Sun is given by:
6 1024
r
1 496 108 km  450 km
30
24
2 10  6 10
In this calculation, clearly the mass of the Earth in the denominator is quite insignificant. The
figure of 450 km compares to a radius for the Sun of 700 000 km – so not large!
Aside: In fact, we could come up with this figure without the above analysis, just by using the
idea of Conservation of Momentum. The argument could go as follows.
Let the speed of the Earth in its orbit be vEarth, so its [linear] momentum is given by:
pEarth = 6  1024 kg  vEarth
Assuming the momentum of the Earth-Sun system is zero, it follows that the momentum of
the Sun is the same, in the opposite direction. So the orbital speed of the Sun is given by:
vSun 
6 1024  vEarth
 3 106 vEarth .
30
2  0 10
As the two bodies take the same time, T, to orbit the Centre of Mass:
T
2 rEarth 2 rSun

vEarth
vSun
So
2 rSun
2 1 496 1011 m

vEarth
3 106 vEarth
So
rSun = 440 km
which agrees to within the accuracy of the data. Can you spot the approximation?
We’ll return to the idea of using momentum conservation when we analyse extra-solar
planetary systems.
Question 1: If the mass of the Earth were 100 times as great [6  1026 kg], what would be the
effect on:
1. the length of the year;
2. the position of the centre of mass of the Earth-Sun system;
3. the orbital speed of the Earth;
4. the orbital speed of the Sun.
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Question 2: The dwarf planet, Pluto, has a
mass of 127  1022 kg. Its moon, Charon has a
mass of 19  1021 kg. The mean separation of
their centres is 19 640 km. Use these data to
determine:
1. the position of their centre of mass;
2. the orbital period of the two bodies;
3. the orbital speeds of the two bodies.
Nice pic of Pluto,
Charon and the
recently discovered
Nix and Hydra
Measuring speeds using the Doppler Effect.
Many objects that we study, including stars and gas clouds, have emission and/or absorption
spectra with identifiable lines. If such an object is moving towards or away from us, the
wavelength of the radiation which we receive is shifted. This shift is towards longer
wavelengths [red shift] if the object is moving away from us and towards shorter wavelengths
if it is moving in our direction.
We shall only use the low-velocity approximation for the Doppler shift,
 v
 .
i.e.
 c
The velocity v in this equation is the component of the objects velocity relative to the
observer along the line joining the observer to the object. This is known as the radial velocity,
which can be slightly confusing, if we are considering an object in orbital motion about
another]. In this low-velocity approximation [“low velocity” is relative to c, the speed of
light, so speeds up to (say) 107 m s-1 would be considered “low”], any motion at right angles
to the line of sight produces a negligible Doppler shift.
Sign convention: In the equation above,  will be positive if v is positive, so we measure v
away from the observer.
Alternative forms of the equation: Because the frequency of radiation, f, is inversely
proportional to , the same equation holds, with the slight complication that there is now a
minus sign:
f
v
 .
f
c
Of course, if you are happy to remember that a positive v produces a smaller f you can forget
about a sign convention.
Information about stars from the Doppler Effect.
Suppose we observe a star which has a massive star in orbit or, more correctly, a star and
massive planet in mutual orbit. Normally, we would not be able to see the planet, but we
would infer its presence from data about the speed of the star.
In questions we would always assume that we see such a system edge-on. In reality, the
situation is more complicated. All candidates might be asked to consider is what effect it
would have on our observations if the system were tilted.
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Such a system might look as follows:
star
light from the star to observers
orbit
on the Earth
massive
planet
For such a system the data could be presented graphically, e.g. variation in the received
wavelength of the sodium D2 line which has a laboratory wavelength of 58900 nm.
 / nm
5895
0
58900
58850
0
05
10
15
20 Time /days
From this graph, you can determine:
(a) the mean radial speed of the star system, from the mean wavelength [~5891 nm];
(b) the star’s orbital speed, from the amplitude of the wavelength variation [~02 nm];
(c) the period of the orbit.
Notice here the use of the word “radial”. Question (a) asks you to find the component of
the binary system’s velocity in the direction directly away from the Earth.
N.B. It is not only the wavelength [and frequency] of the radiation itself which
undergoes Doppler shift. Recently astronomers noticed that the period of the pulsations
from a pulsar [a neutron star] vary in a periodic way. This is attributable to the effect of
an orbiting planet or companion star. They used the Doppler equation in the form:
T v
 ,
T
c
where T = period of the pulsations, in the same way as the wavelength in the example
above to work out the orbital parameters and the masses involved.
Where it all comes together:
Most of the information about the masses of stars and the evidence for the existence of extrasolar planets has come from Doppler measurements in orbiting systems.
For the case of Extra-solar planets we can assume that mP mS [where mp is the mass of the
planet and mS is the mass of the star]. With this approximation, the equation for the period of
the mutual orbit reduces to:
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d3
(1)
T  2
GmS
Algebraic manipulation gives the following approximations for the orbital speeds:
G
(2) and
vS  mP
mSd
vP 
GmS
(3)
d
For a given star system, normally we would know the mass of the star, mS, from our
knowledge of stellar models and the observations would be the received wavelength of a
spectral line against time. For an edge-on system this variation would be sinusoidal and we
can undertake the following steps:
1.
2.
3.
4.
5.
From the graph, determine the orbital period, T.
From the amplitude of the observed Δ, determine vS .
Use equation (1) to determine the separation of the star and planet, d.
Use equation (2) to determine the planetary mass mP and (3) to determine its speed.
If the planet occults the star [passes in front of it – if it’s a true edge-on system it
should do but many will just miss occulting the star], we can further estimate the
planet’s diameter from the period of occultation and its speed and the ratio of the
stellar to planetary diameters from the fractional decrease in the observed light.
The data needn’t be presented graphically, e.g.:
Example
The wavelength of the H line which, in the laboratory has a value of 4861 nm, in the
radiation emitted from a star is observed to fluctuate with an amplitude of ± 105  10-3 nm
with a period of 125  106 s. The mass of the star is 30  1030 kg. Assuming that this
behaviour is caused by an orbiting planet and that we observe the system edge-on:
1. calculate the distance of the planet from the star;
2. calculate the mass of the planet and its orbital speed.
STOP PRESS: Astronomers studying this star have noticed that its brightness drops by ~1%
once in every orbit of the planet. This dimming lasts for 454 hours. They suggest that this
dimming is caused by the planet blocking of the light from the star as it passes in front as
seen from the Earth.
3. Use this information to estimate the diameter of the star and that of the planet. Calculate
also the planet’s density.
Binary Star Systems
Most information about stellar masses comes from a study of binary star systems, i.e. a pair
of stars in mutual orbit. Because both objects in such a system emit light, the orbital
velocities of the two stars can be found directly and so the masses can be calculated. Consider
the following graphs of the radial speeds to two stars in close mutual orbit:
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What information can we glean from this without any algebra?
 The mean radial velocity is + 40 km s-1, i.e. the binary system is receding from
us at this speed.
 Assuming we see the system edge on, the speeds of the two components are
75 km s-1 and 25 km s-1 [these are the amplitudes of the speed variations].
 The period of the orbit is 176 days [152  106 s]
We can do some sums before we need to apply complicated theory:
1. The ratio of the masses is 3:1, i.e. one component [the faster one] has ¼ of the total
mass of the system and the other has ¾ of the total mass. This comes from
momentum considerations: in the frame of reference in which the centre of mass is
at rest, the momentum of each component must be equal and opposite. The speeds
are in the ratio 1:3 so the masses must be in the ratio 3:1.
Another way of looking at this idea is as follows:
The gravitational forces on the components are equal: i.e. m1r112  m2 r22 2 .
Dividing by : m1r11  m2 r22 , i.e. m1v1  m2v2
2. We can work out the circumference [and then the radius] of each of the orbits:
e.g. The slower [more massive] component:
circumference  orbital speed  orbital time
 25kms-1  1 52 106 s
 38 million km
So the radius of the orbit is calculated at 604 million km [circumference = 2r]
Likewise the orbital radius for the faster component is 1814 million km.
3. From the two orbital radii, we can infer that the separation of the stars, d, is
242 million km [the sum of their orbital radii].
Now the earlier formulae click in:
4. We can apply the formula T  2
d3
to find the total mass, m1 + m2, of
G  m1  m2 
the system – which comes out at 36  1030 kg, so the combined mass of the stars is
roughly twice that of the Sun.
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5. From this, we can work out that the masses of the individual stars are 09  1030 kg
and 27  1030 kg [remember the 3:1 ratio].
More or less useful references for gravity, spectra and mutual orbits
Check out the applets on:
http://www.ioncmaste.ca/homepage/resources/web_resources/CSA_Astro9/files/html/applets.
html - more GCSE than GCE for Stars, Spectra and Kepler’s Laws
On http://jersey.uoregon.edu/vlab/elements/Elements.html you can find the wavelengths of
spectral lines [put the mouse cursor on the line and click]
Doppler spectroscopy: http://en.wikipedia.org/wiki/Doppler_spectroscopy
An example of a radial velocity curve: http://www.howstuffworks.com/planet-hunting2.htm
Mutual orbit simulation: http://www.howstuffworks.com/framed.htm?parent=planethunting.htm&url=http://exoplanets.org/doppler.html
This site also has mutual orbit simulation and a plethora of other applets:
http://phet.colorado.edu/new/simulations/sims.php?sim=My_Solar_System
Data for 51Peg: http://zebu.uoregon.edu/51peg.html
Overview of detecting ESOs: http://astro.unl.edu/naap/esp/detection.html
Another overview : http://www.esa.int/esaSC/SEMYZF9YFDD_index_0.html
GCE Physics – Teacher Guidance
4 December 2007