Download Advancing Physics A2

Name ………………………………………………………
Advancing Physics A2
John Mascall
The King’s School, Ely
Chapter 11 Out into space
Student Notes
August 2009
Assessable learning outcomes for all of Chapter 11
Candidates should demonstrate evidence of:
1. knowledge and understanding of phenomena, concepts and relationships by describing,
and explaining cases involving:
(i) force as rate of change of momentum;
(ii) work done, including cases where the force is not along the line of motion;
(iii) conservation of momentum; Newton’s 3rd Law as a consequence;
(iv) changes of gravitational and kinetic energy;
(v) motion in a uniform gravitational field;
(vi) the gravitational field and potential of a point mass;
(vii) motion in a horizontal circle and in a circular gravitational orbit;
2. scientific communication and comprehension of the language and representations of
physics,
by making appropriate use of the terms:
(i) force, kinetic and potential energy, momentum, gravitational field, gravitational potential,
equipotential surface;
by sketching and interpreting:
(ii) graphs showing gravitational potential as area under a gravitational field versus
distance graph;
(iii) graphs showing force as related to the tangent of a graph of gravitational potential
energy versus distance;
(iv) diagrams of gravitational fields and the corresponding equipotential surfaces;
3. quantitative and mathematical skills, knowledge and understanding by making calculations
and estimates involving:
(i) gravitational potential energy change mgh;
(ii) energy exchange, work done, ΔE = F Δs ; no work done when the force is
perpendicular to the velocity;
(iii) momentum, p = mv, F = Δmv/Δt;
(iv) a = v2/r, F = mv2/r;
(v) for the radial components: Fgrav = - GmM/r2, g = Fgrav/m = - GM/r2 ;
(vi) gravitational potential energy Egrav = - GmM/r ;
(vii) Vgrav = Egrav/m = - GM/r .
A Revision Checklist for Chapter 11 can be found on the Advancing Physics
CD-ROM.
Page | 2
Ch 11.1: Rhythms of the heavens
Learning outcomes

To make an object move in a circular path, a force must act perpendicular to its
velocity.

A centripetal acceleration can be found by a = v2 / r and the force required to cause
it must be F =mv2 / r or F = mω2r.

No work is done when a force on a body acts perpendicular to its motion.
A remarkable way of thinking about the heavens began with the Greeks. Effectively they
‘geometrised’ the Universe, so that celestial motions took place within an empty threedimensional space.
Question for discussion
To what extent are we aware of basic unaided eye observations (movements of Sun,
Moon, planets, stars)? Do we make everyday connection between these observations and
the accepted model of the solar system?
Notes:
A historical overview of ideas leading up to our accepted model of the solar system can be
found on pages 31- 37 of the Advancing Physics A2 text. This is supported by the
PowerPoint ‘Rhythms of the heavens’.
This work can be supported by Activity 10S Software Based 'Watching the planets go
round' using File 10L Launchable File 'An electronic orrery'
This shows the relative positions (not to scale) and periodic times of the planets.
Elliptical motions
●
It makes sense to place the Sun at the centre of the solar system, since it
represents 99.8% of the total mass.
●
The retrograde motion of planets, as seen from Earth, provides another inducement
for accepting the Copernican model: the apparent motion is physically (rather than
merely mathematically) explained by the relative motions of Earth and other
planets.
The student's book describes how Kepler ‘cracked’ the orbit of Mars by discovering that it
follows an elliptical path. Kepler’s model consolidated and simplified the Copernican idea
of the Sun at the centre but, most important, Kepler introduced the notion that the Sun
physically causes the motion of the planets.
Page | 3
Kepler’s laws are described in the student's book (page 36) but are not in the specification.
They are summarised below:
1 a planet moves in an ellipse with the Sun at one focus
2 the line from the Sun to a planet sweeps out equal areas in equal times
3 square of orbital time is proportional to cube of orbital radius
From a forces point of view you should see that if the Sun attracts a planet then if there is
a component of force in the direction of travel (or against it) so the planet’s kinetic energy
will increase as potential energy falls (and vice versa). A comet is the extreme case: in
close to the Sun it moves very fast, but as it recedes it slows down.
◦
Try drawing a Sun at one of the foci shown in the diagram below. Given that the
force on the planet acts towards the Sun you should be able to find positions on the
ellipse where the planet will speed up and positions where it will slow down.
This might be modelled with a marble rolling around inside a bowl which has sloping walls
or with 1/r potential well. Speeds change in elliptical orbits because there is a component
of force along the path. Circular motion is a special case, with the force component being
zero.
Display Material 10O
OHT 'Eccentricity of planet orbits'
Ellipses and orbits
b
focus
focus
major axis
a
ae
You may find it interesting to draw an ellipse with the aid of a board, two drawing pins and
a piece of string held at a fixed length on the board by the drawing pins. When you move a
pencil around keeping the string taut you should be able to produce an ellipse – if there is
enough slack in the string. The drawing pins represent the foci. What would you draw if
you effectively brought the two drawing pins together so that the foci coincided? You will
only need one drawing pin of course.
Page | 4
Display Material 50O
OHT 'Geometry rules the Universe'
Kepler: Geometry rules the Universe
Law 1: a planet moves in an ellipse with the Sun at one focus
Astronomy
Geometry
planet
Mars
a
b
focus
Sun
focus
Ellipse: curve such that sum of a and b is constant
Orbit of Mars an ellipse with Sun at a focus
Kepler: Geometry rules the Universe
Law 2: the line from the Sun to a planet sweeps out equal areas in equal times
Astronomy
Geometry
planet
Mars
fast
slow
focus
Sun
Speed of planet large near Sun, smaller away from Sun
Areas swept out in same time are equal
Page | 5
Kepler: Geometry rules the Universe
Law 3: square of orbital time is proportional to cube of orbital radius
2
4
Mars
Orbital period against
orbital radius
O rbital period squared
against orbital radius
cubed
Mars
3
1
2
Earth
Venus
1
Earth
Venus
Mercury
Mercury
0
0
0
50
100
150
200
0
250
1
2
4
radius3 /AU3
radius/million km
Display Material 30O
3
OHT 'Kepler’s third law'
For simplicity we can assume that the Sun is at the centre of the solar system. Taking from a spreadsheet
data on mean orbit radius of the planets and their orbit period, how might you derive Kepler’s third law?
Size of orbit - time taken to orbit
time of orbit
size of orbit
20
80
15
60
10
40
5
20
0
0
Mercury Venus Earth Mars Jupiter Saturn Uranus
Mercury Venus Earth Mars Jupiter Saturn Uranus
Orbital time increases more rapidly than orbital radius.
The scales are all linear. Comparing the two bar charts, you will see that the orbit period rises faster than
mean orbit radius. To make them go up in similar steps, you guess that it might be possible by raising them
to different powers. The one that rises more slowly (radius) will need to be raised to a higher power. Out of
intelligent guesswork like this came the third law.
Page | 6
Retrograde motion can be shown using Activity 20S Software Based 'Retrograde motion'
Using File 20L Launchable File 'Retrograde motion of Mars'
You may be able to find other simulations of retrograde motion on the Internet.
Display Material 20O
OHT 'Retrograde motion'
Mars’ retrograde motion
apparent
path
Sun
Earth
Mars
Seen from m oving Earth, Mars seem s to make a backward loop as Earth overtakes on the inside
Page | 7
Centripetal acceleration and force
We begin with Galileo’s insight: every body continues to move with constant speed unless
acted on by a resultant force. From this, together with Descartes’ idea that a force is
needed to change direction, Newton established his first law of motion. Galileo’s thinking,
leading up to his 'pin and pendulum’ experiment, provides a useful introduction to circular
motion.
Display Material 40O
OHT 'Galileo's ramps'
Galileo imagines motion continuing for ever
accelerating
downhill
decelerating
uphill
equal slopes: loss of
speed uphill equal to
gain of speed downhill
accelerating
downhill
decelerating
uphill
shallow uphill slope:
loss of speed uphill
equal to gain of speed
downhill, but takes
longer
accelerating
downhill
constant
speed
no uphill slope: motion
continues without loss
of speed
Galileo’s experim ents with sloping ram ps showed that
distance travelled is proportional to (time)2 .
He reasoned that speed change depends on height, not
on slope. So a ball rolled from a given height should return
to the same height, in the absence of friction.
If there is no slope, the ball should roll on for ever.
Page | 8
Activity 30D
Demonstration 'Galileo’s frictionless experiment'
Notes:
In orbit terms, circular motion is a special case.
● The body neither slows down nor speeds up, but moves at a constant speed.
● The effect of a constant force perpendicular to velocity is simply to make the direction
change at a constant rate, thus circular motion.
● A centripetal force does no work on that body. This is consistent with ‘work done is force
times distance moved in the direction of the force’, which we met in chapter 9.
A number of examples of circular motion are given on page 39 of the student text. You
should be able to think of many more examples.
Notes:
Page | 9
Display Material 60O
OHT 'Centripetal acceleration'
Acceleration towards centre of circular orbit
A
circular path
radius r
speed v
v1
N.B.
This theory requires
the use of radians.
B
r
r
v2

velocity turns through
angle  as planet
goes along circular
path in short time t
radius turns through 
velocity turns through 
speed v
A

B

v1
v
v2
r
arc AB
r
change of velocity v
towards centre of circle
arc AB = distance in time
t at speed v
arc AB = v t
v t

r

v t
r

multiply by v: v
divide by t:
v
v
v
v
v t
v
r
v
v2

= acceleration
t
r
2
Acceleration towards centre = v
r
A full discussion of the ideas involved is given on pages 37-39 of the student text.
Page | 10
OHT 'Centripetal acceleration is proportional to v2'
Display Material 70O
Centripetal acce leration proportional to v squared
Speed v
speed v carries object
round angle  in time t
v
v


v
v
v
v = v 
Speed v
2
speed v / 2 carries object
round angle / 2 in time t
v
2

2
v
2
v
2
v
2
v =
4

2
v
4
( v2 ) ( 
2 )
This display demonstrates that an object travelling in a circle at half speed has only a
quarter the acceleration of an object with full speed.
Page | 11
The following activity illustrates many relevant points:
Activity 60S Software Based 'Driving round in a circle' using File 40L Launchable File
'Modelling circular acceleration'
It is easy to mistake the adjective ‘centripetal’ as describing a new, fundamental force of
nature. What we refer to as a ‘centripetal force’ describes its net effect on the body’s
motion resulting from one or more forces acting. In the case of the planets, the centripetal
force of course is gravity.
We have established the equation for centripetal acceleration
a = v2/r
This acceleration results from an unbalanced force acting towards the centre of the circle
(centripetal force). As with all unbalanced forces, we can apply Newton’s second law:
F = ma, in which case the expression for centripetal force becomes
F = mv2/r
An experiment can be used to show the relationship works within the limit of accuracy of
measurements.
Activity 40E Experiment 'Testing F = mv2/ r ‘
It is easy to show that angular velocity ω = v/r and that the centripetal force F = mω2r.
You may find these forms of the equation useful.
Notes:
You may wish to read about the training of astronauts in Reading 50T. This article refers to
parabolic flight used to induce a sense of weightlessness, and the ‘high-g’ wheel used to
enable the occupant to experience very high accelerations. Horizontal circular motion such
as that used in the ‘high-g’ wheel is similar to planetary orbits in that a single force acts.
The astronaut's weight is perpendicular to the centripetal force. You may also wish to think
about the physics involved in a centrifuge. There are many uses for these devices.
Page | 12
Section 11.2 Newton’s gravitational law
Learning outcomes

F = - G m1m2/ r2

The concept of gravitational field, and diagrams to represent it

The gravitational field strength g = - GM / r2 and its graphical representation

Calculation of the orbit radius and time of a satellite or planet.
What is essential in this section is the gravitational force, introduced quantitatively and
developing its field description. You will need to become confident using equations,
graphs, diagrams and computer models to represent the effects of gravity. Some of the
historical background is covered in the student text as is the way in which we use
Newton’s ideas more than 300 years after Newton established them.
The gravitational law
It is essential that you understand that a body in a circular gravitational orbit is in free fall.
Newton’s thought experiment on page 43 of the student text gives a taste of the thinking
behind this.
Notes:
You should already know that the gravitational field strength on the surface of the Earth is
9.8 Nkg-1 and an object on the surface experiences a free-fall acceleration of 9.8 ms-2. The
similarity of the two numbers is not a coincidence!
The force of gravity provides the unbalanced force to cause the free-fall acceleration.
Page | 13
Display Material 90P Poster ‘Acceleration of the Moon’
Newton thought about the force keeping the Moon in its orbit. The calculation shows that
the acceleration of the Moon is just caused by a diluted gravitational force from the Earth.
The acceleration of the Moon and the inverse square law
Acceleration of the Moon
Diluting Earth’s gravity
Time
1 Moon month = 27.3 days
= 27.3  24  3600 s
= 2.35  106 s
Distance
circumference = 2 3.84  108 m
= 24.1  108 m
Travelling around
the Earth once a
month
g = 9.8 m s–2 at surface
Moon’s orbit
radius = 384 000 km
Earth’s radius
radius = 6400 km
Moon
orbit radius
384000 km
Ratio of Earth’s radius
to Moon’s orbit
6400 km
ratio =
384000 km
1
=
60
Moon
orbit radius
384000 km
diluted
gravitational
pull
Earth
Speed
24.1  108 m
v=
2.35  106 s
v = 1020 m s–1
Acceleration
v2 m
a= r
a=
Gravity diluted by
inverse square law
Earth’s surface:
g = 9.8 m s–2
At Moon’s orbit:
9.8 m s–2
(1020 m s–1)2
acceleration =
3.84  10 m
8
a = 0.0027 m s–2
Earth
acceleration found
from motion
602
a = 0.0027 m s–2
same acceleration
found from inverse
square law
The acceleration of the Moon is simply diluted Earth gravity. The acceleration measures the gravitational field
The geometrical argument for the inverse square law is shown in the Advancing Physics
A2 student's book page 42. The diagram has been reproduced below. In Newton’s time it
was well known that light intensity from a point source falls off this way because the
spherical surface through which it passes grows as 4r2.
It is not difficult to illustrate this principle with a very bright small light source, a screen and
a square hole at the centre of a thick card. Start with the screen just behind the ‘hole’. With
the ‘hole’ fixed in position, the same light is spread over four times the area when the
distance between lamp and screen is doubled. It should be possible to show that the light
intensity falls by a factor of 4, but that assumes that you have a correctly calibrated light
sensor and no background light!
Newton established that F  m1m2 / r2
which becomes F = - G m1m2
r2
The minus sign indicates that the force is always attractive. All distances are measured
from the centre of mass of the masses involved.
Page | 14
Note that the very small value of G means that the forces involved only become significant
when the masses are very large such as those on an astronomical scale.
Question
Estimate the gravitational force between two students sitting in class.
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Page | 15
The CD-ROM provides several resources which look at interesting phenomena explained
by the law: tidal forces, discovering that other stars have planets, supernovae and black
holes.
You can use Question 80W Warm-up Exercise 'Newton’s gravitational law' to build
confidence in using this equation.
The Modellus model below will help you to get a feel for how gravity works.
Activity 70S Software Based ''Variations in gravitational force' using File 50L Launchable
File 'Newton’s gravitational law'
We can use this model to show the radial field idea and how we can represent the
variation of field with distance graphically. With a radial gravitational field the magnitude of
the field is the same at any given distance from the centre of the mass causing the field.
Activity 80S Software Based 'Gravitational universes' using File 60L Launchable File
'Gravitation between two bodies'
Use this to model a one-planet solar system with masses that are very different. If the
masses are comparable, a binary star can be modelled. In all cases it is very important to
have the correct initial velocity!
The calculation below shows how to calculate the orbital radius for a geostationary satellite
and also how to prove one of Kepler’s laws. Geostationary satellites are important for
communications purposes.
[Notice that we conveniently removed the minus sign from Newton’s equation. That is
because, strictly, centripetal force mv2/r should be preceded by a minus sign as this
equation also describes the force towards the centre of the circular orbit i.e. away from the
direction in which r is measured.]
Page | 16
Display Material 110O
OHT 'A geostationary satellite'
Geostationary satellite
m = mass of satellite
R = radius of satellite orbit
v = speed in orbit
G = gravitational constant
= 6.67  10 –11 N kg –2 m2
M = mass of Earth
= 5.98  10 24 kg
T = time of orbit
= 24 hours = 86400 s
N
gravitational
force
orbit radius
R = 42 000 km
satellite orbit turns at same
rate as Earth turns
S
Calculating the radius of orbit
Gravitational force on
satellite
F orce producing
acceleration to centre
mv
R
equal
2
GMm
R2
Forces are equal:
mv2
R
divide by m:
v2
R
multiply by R :
=
=
v2 =
speed in orbit depends
on time of orbit and
radius
GM m
R2
GM
v =
R2
GM
R
equal
equate ex press ions for v2:
rearrange to calc ulate R:
v2 =
2R
T
42 R2
T2
4 2R 2
GM
=
R
T2
GMT 2
=
42
R
3
Kepler’s third
law deduced
insert values of G, M and T : R = 4.2  10 4 km
R = 6.6  radius of Earth (6400 km)
Note that Kepler did not know Newton’s laws so he had to work entirely from geometrical
arguments.
The Classroom Video ‘Satellites, orbits and gravity’ examines how satellites are launched,
the orbits they follow and their uses. This is well worth looking at if time is available.
Page | 17
Gravitational field strength
In this episode, we make the step from forces acting on real bodies to the abstract world of
‘what if?’ It is important to know what a ‘field’ looks like.
We use both graphical and diagrammatic representations and need to be aware that g is a
vector field which points always in the opposite direction from r (in the inverse square field
of a point mass).
We know that F = - G Mm / r2 where M represent the mass producing the gravitational field
and m is a test mass. We know from Chapter 9 that gravitational field strength is the force
per unit mass at a point in the field and is given by g = F/m. It follows that
g = - GM / r2.
You also learned in chapter 9 that acceleration due to gravity and field strength at a point
are the same, though they differ in their units (ms-2 and Nkg-1). This follows since a = F/m
and g = F/m so a = g as suggested earlier.
The field strength at a point depends solely on position and masses; however, because of
its rotation, the Earth is oblate and so distance to the Earth’s centre depends on latitude.
Variations in altitude and local topography also affect field strength. [Because of the
Earth’s rotation, measured acceleration of free fall and so weight can be less than the field
strength, depending on latitude.]
Page | 18
This section of the student's book ends with the Apollo mission to the Moon. The two sets
of related questions here show how 1/r2 can be obtained empirically.
Activity 110S Software Based 'Probing a gravitational field'
using File 80T Spreadsheet Data Table 'Acceleration data for Apollo'
Display Material 100O
OHT 'Apollo goes to the Moon'
The path of Apollo 11
Moon (shown at 24 h intervals)
homeward
outward
time
09:57:60
speed
2619 m s–1
Earth
Typical Apollo flight path to the Moon, showing outward journey on
the right and homeward journey four days later on the left
Page | 19
Apollo 11 goes to the Moon
Pairs of observations of speed and distance
Pairs of observations made 600 s apart in time. Distances r taken from centre of Earth. Apollo 11 is coasting with rockets
turned off.
speed/m s–1
distance/106 m
mean distance r/10 6 m
mean acceleration/m s –2
pair A
5374
5102
26.3
29.0
27.65
–0.453
pair B
3633
3560
54.4
56.4
55.40
–0.122
pair C
2619
2594
95.7
97.2
96.45
–0.042
pair D
1796
1788
169.9
170.9
170.40
–0.013
Graph of gravitational field against r
0.0
pair D
pair C
–0.1
pair B
–0.2
–0.3
Velocities are only
approximately directed away
from Earth. Acceleration is thus
only approximately equal to g
–0.4
pair A
–0.5
0
50
100
150
200
distance r from centre of Earth/106 m
Graph of gravitational field against 1/r 2
0.0
–0.1
pair D
pair C
pair B
–0.2
–0.3
Graph approximates to a
straight line through the
origin.
Gravitational field
proportional to 1/r 2
–0.4
–0.5
0
pair A
0.0005
–2
0.0010
6
r /(10 m)
0.0015
–2
Can you see how to find the mass of the Earth from the graph above?
Page | 20
Ch 11. 3 Arrivals and departures
Learning outcomes

Momentum = mass × velocity.

Force is a rate of change of momentum.

The thrust of a jet is the momentum carried away per second by the jet.

The law of conservation of momentum: the vector sum of momentum is unchanged
in an interaction between objects.

In consequence of the law of conservation of momentum, interaction forces come in
equal and opposite pairs.
This section encompasses all three of Newton’s laws of motion.
In the Advancing Physics A2 student's book Ch 11.3, we develop the concept of
momentum by considering collisions from several inertial frames of reference (observers
moving at constant velocity relative to each other), together with an argument from
symmetry. This gives a deep explanation of the concept of momentum and its
conservation. Inertial mass is reintroduced from this viewpoint, and Newton’s third law
arises naturally.
In the middle of the chapter, we rediscover F = ma together with the more general form
‘force equals rate of change of momentum’ written as F = p/t. A wide range of examples
applying these ideas is given and we use these ideas to explain why time of contact
affects the impact force.
Momentum and its conservation
Momentum, p, is defined as the product of mass and velocity.
p = mv
Since velocity is a vector and mass is a scalar, momentum must also be a vector.
The units of mass  velocity are kg m s-1 but you should also be able to argue that
momentum can have units of N s. The reason for this will soon be apparent.
Arguments for momentum and its conservation are given in the student’s book pages 4850.
Experiments will be carried out to reinforce these ideas which are also very well presented
in the ‘Collisions’ video from Classroom Video. The video shows collisions between cars,
lorries, aircraft, meteorites and people – some of them spectacular.
Page | 21
In the example below, we assume that Mir and Progress have the same mass. This is not
actually the case, but it makes the example easier to understand.
Display Material 130O
OHT 'Two craft collide'
Two crafts approach one another and dock together
View 1 Observation craft hovers where the craft will
meet
+v
–v
P rogress
M ir
observation craft
video of collision seen from observation craft
–v
+v
P rogress
Mir
View 2 O bservation craft travels alo ngside M ir
–v
+v
Progress
Mir
–v
observation craft
video of collision seen from observation craft
+v
+2v
Progress Mir
The same event looks different
from two different points of view
Page | 22
One event seen from two points of view
Before collision
–v
+v
Progress
Mir
momentum before
= +mv – mv = 0
After collision
velocity = 0
Progress
Mir
momentum after
=0
velocity of this frame relative to frame above
Before collision
velocity = 0
+2v
Progress
Mir
momentum before
= +m(2v) = +2mv
After collision
2m
+v
Progress Mir
momentum after
= (2m)v = 2mv
Momentum is different in the two views of the same event, but in
each case: momentum after = momentum before
Page | 23
Display Material 150O
OHT 'Conservation of momentum'
Conservation of mom entum p = mv
Before collision:
p1
p2
[total momentum p] before = [m1 v 1 + m2 v2 ]be for e
m2
m1
After collision:
p1
p2
[total momentum p] after = [m1 v1 + m 2 v2 ]after
During collision: momentum p goes from one mass to the other
before:
M o men tu m co nse rved
p1
los es  p
p2
p
gains  p
p
after:
[p 2] a fter = [p 2] b ef or e + p
therefore:
p1
[p ]t ota l = 0
[p 1] a fter = [p 1] b ef or e – p
p2
[ p 1 + p 2] a fte r = [ p1 + p 2] b efo r e
Ch an ges o f velo city:
m 1 v 1 = –  p
m 2 v 2 = + p
therefore:
–
 v2
 v1
=
m1
changes of momentum are equal and oppos ite
changes of v elocity are in invers e proportion to mas s
m2
M omentum just goes from o ne ob ject to the other. Th e to tal momen tum is co nstant
Page | 24
Here are six more collisions. Notice that in every case the total momentum before equals
the total momentum after.
Display Material 160O
equal masses, inelastic collision
OHT 'More collisions'
total
momentum
before
unequal masses, inelastic collision
total
momentum
before
velocity v
velocity –v
velocity v
before
0
during
velocity zero
before
during
after
after
after
0
both velocities
zero
after
velocity a little
less than v
equal masses, elastic collision
total
momentum
before
unequal masses, elastic collision
total
momentum
before
velocity v
velocity zero
velocity v
before
during
velocity zero
before
during
after
after
after
velocity zero
velocity a little
less than v
velocity v
equal masses, elastic collision
total
momentum
before
velocity v
after
velocity much
less than v
unequal masses, elastic collision
before
velocity –v
velocity v
before
0
during
velocity zero
before
during
after
after
after
0
velocity –v
total
momentum
velocity v
after
velocity much
less than v
velocity up to
2v
Page | 25
Activity 120E Experiment 'Low friction collisions and explosions'
In this experiment we look at elastic, inelastic and explosive interactions between two air
track vehicles.
Activity 130E Experiment 'Newton’s cradle'
This is a short demonstration of elastic collisions using a well-known executive toy!
Activity 160S Software Based 'Modelling collisions'
using File 80L Launchable File 'Simulation of a collisions'
We can use this model to see what happens when masses and initial velocities are
changed. We can also change the coefficient of restitution, but this is not a concept that
appears in the specification.
Use the table below to summarise the types of interaction you have studied.
Type of interaction
Elastic
Momentum
Kinetic energy
Inelastic
Explosions
Note that momentum is conserved in all interactions.
Strictly we should say that momentum is conserved in a closed system where no
external forces act.
Question
A ball falling towards the Earth gains momentum because there is an unbalanced force
applied to it due to the earth. Can you explain how momentum can be conserved in this
case?
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
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Page | 26
Softening the blow
This section in the student’s book concentrates on reducing the peak or average forces in
collisions.
From F = ma and a = v/t we have F = mv/t
Since p = mv, p = m v for a fixed mass, and so F = p/t
This says that force is the rate of change of momentum. The force is determined by
how quickly the momentum changes.
There are many cases where we aim to reduce the peak or average forces in collisions by
increasing the time over which the momentum changes i.e. we reduce the rate of change
of momentum. You should see that this is the physics behind the use of egg boxes to
protect eggs, seat belts and crumple zones in cars to protect the occupants, parachute
landing etc.
When hammering in a nail, we use a steel hammer head so that the hammer head loses
its momentum over a short time interval and the force exerted is very large. However, we
use rubber headed hammers for laying concrete paving slabs!
It is useful to consider the force-time graph during a collision. Remember that, from
Newton’s third law, the magnitude of the force acting is the same on each of the colliding
objects.
Page | 27
Questions
1
A parachutist of mass 80 kg takes 0.2 s to reduce her speed from 6 ms -1 to zero
when she hits the ground.
(a) Calculate the average force exerted on her legs during impact with the ground.
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
(b) Explain why this force would be larger if she landed without bending her knees.
…………………………………………………………………………………………………
…………………………………………………………………………………………………
2
A hammer is used to drive a nail into a piece of wood.
(a) Explain why the hammer exerts a force on the nail.
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
(b) Describe the best design of hammer that should be used for chiselling masonry.
Explain the physics involved in your design.
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
3
A fast travelling cricket ball hurts the hands if caught badly, but can be caught
without discomfort if the hands are drawn back with the ball.
Explain how this works using sketch graphs of force against time to illustrate your
answer.
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
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Page | 28
The equation F = mv/t takes on a new meaning if rewritten as Ft = mv.
Ft is called the impulse of the force and this is equal to the change of momentum.
You should see why momentum is sometimes given the units of Ns.
The change of momentum is determined by the impulse and hence by the size of the force
acting and the time for which it acts. In impact sports such as golf, more momentum is
transferred to the ball if the club is in contact with the ball for a longer time. This is why
sportsmen are trained to follow through with their shots so that the impact time is made as
large as possible.
The ‘Golf’ example given on Multimedia Motion from Cambridge Science Multimedia is an
excellent way of bringing some of these ideas together. We can analyse the motion of the
club head and ball separately using x-t, v-t and a-t graphs and checking for consistency
between the methods. Look at the shapes of the v-t, p-t and F-t graphs. Calculate the
change of momentum p for the ball and use it to deduce the average force exerted on it.
This can be checked against F=ma. Remember to distinguish between average force
(measured form the area under the p-t graph) and peak force. We can use momentum
conservation to establish the effective mass of the club and discuss changes in kinetic
energy if time permits.
Activity 180S Software Based 'Crunch – gently!'
using File 100L Launchable Files 'Crunch'
Newton’s third law follows from the principle of conservation of momentum as
follows:
We have shown that p1 + p2 = 0 as there is no overall change of momentum in an
interaction. It follows that p1 = - p2 so that F1t = - F2t. Notice that the force acts
for the same time on each of the objects! From this we can say that F1 = - F2 which
is Newton’s third law.
The following experiments may be used to consolidate some of the ideas met so far.
Activity 140D Demonstration 'Kicking a football'
Using the equation F t = m ∆v, it is possible to calculate the force used to kick a soccer
ball.
Activity 150E Experiment 'Shove ha’penny’
When one moving object collides elastically with a stationary object of the same mass,
kinetic energy as well as momentum and total energy are conserved. In this simple
experiment, one 2 pence piece is projected at a row of similar coins.
Page | 29
Display Material 140P
Poster 'Momentum, invariance, symmetry'
Thinking about momentum and forces
Principle 1 symmetry
+v
Principle 2 invariance
–v
+v
–v
seen differently is the same as:
identical objects
+2v
result predictable from symmetry
Conservation of momentum
–p
+p
crunch
Split ‘crunch’ into
forces F on each
mass M
mass m
–p
force –F
acts for
time t
change of momentum = –F t
force F =
p
t
force +F
acts for
time t
same time t.
forces equal
and opposite
if define force F =
+p
change of momentum = F t
p
t
mv
then F = t = ma
thus F = ma
From symmetry and invariance (looking differently can’t change events):
1.
2.
3.
4.
momentum is conserved
define mass from change of velocity in collision
define force as rate of change of momentum, giving F = ma
forces on interacting objects act in equal and opposite pairs
Page | 30
Jets and rockets
The rationale for introducing jets and rockets is to use ‘force equals rate of change of
momentum’ in the form F = -vm/t.
Display Material 170O
OHT 'Jets and rockets'
Jets and rockets
rocket velocity V
increases by V
in time t
momentum carried by gas plus
momentum change of rocket = 0
rocket mass M
p
–p
change of momentum of
rocket:
p = MV in time t
momentum carried
away by jet:
p = v m in time t
for jet:
v m = –p
mass m ejected
in time t
equal and
opposite
for rocket:
p = M V
MV = –v m
V =
gas velocity v
–v m
M
M V
–v m
p
thrust =
=
=
t
t
t
Rocket thrust = –v m
t
The following question is useful but has some tricky sections. It is best done in class.
Question 180S Short Answer 'Jets and rockets'
Page | 31
Section 11.4 Mapping gravity
Learning outcomes

Gravitational potential energy difference in a uniform field = mg∆h.

Gravitational potential is gravitational potential energy per unit mass.

Gravitational field strength = – gravitational potential gradient.
GM

Gravitational potential around a point mass V  
r

The total energy of a body orbiting in a gravitational field is gravitational potential
energy plus kinetic energy.
Assuming you had a rocket engine, how would you get a spacecraft out of the grip of
Earth’s gravity? How are geostationary orbits placed at the right orbit radius? These
questions demand an understanding of the concept of potential in a radial field. In the
student's book, visualisation in terms of potential wells is stressed. These lead easily to
making the link between gravitational field strength and potential gradient.
Gravitational potential in a uniform field
You are familiar with the analysis of forces in play as a body moves through a gravitational
field, so we now develop a second way of analysing motions in gravitational fields: by
returning to a consideration of energy changes. Just as the force acting on a particular
body leads to the more general idea of field strength, so gravitational potential energy
leads to gravitational potential, a scalar field description.
Over short distance ranges, particularly at the surface of the Earth, the value of g does not
change appreciably and so analysis of energy changes is relatively simple.
The following question revises the relevant energy ideas met so far.
Question 200W
Warm-up Exercise 'Pole vaulting'
Notice that the height of the pole vaulter will be independent of the vaulter’s mass. This
follows because the kinetic energy and gravitational potential energy are both directly
proportional to mass so the mass cancels from each of these energy terms when they are
equated. Put another way, mgh = ½ m v2 so h = v2/2g which is independent of mass.
The next experiment reinforces this point.
Activity 190D Demonstration 'Exploring potential with a tennis ball'
The area under the g against h graph has a physical meaning. It shows the work done per
unit mass in moving vertically through a uniform gravitational field, in other words
gravitational potential difference between the two points in the field.
Compare this with the area under a force-distance graph, which represents the work done
for a particular mass moved against a force. If this process was reversible, it would be
possible to store energy by doing work against a force, increasing ‘potential energy’.
Notes:
Page | 32
Display Material 180O
OHT 'Graph showing g against h'
Gravitational field and gravitational potential energy
Uniform gravitational field
Field p icture
Po tential en erg y p icture
potential energy
change  E
m ass m
h
mas s m
force mg
lift by height h
1
40 J kg –
30 J kg –
1
20 J kg – 1
field s trength = g
10 J kg – 1
forc e on m ass in grav itationa l field = m g
1 m
potential energy change
= forc e  dis tance = m gh
field = forc e/m ass = g
gravitational potential difference
= potential change/m ass
= mgh/m
= gh
Field is slope of potential hill
Poten tial difference is area under graph
field g = slope
g
gravitational
potential
difference
h
displacement upwards
area g h
=
gravitational
potential
difference
h
displacement upwards
The field is the rate of change of potential with displacement
Note that a change in gravitational potential energy Ep = mgh
so a change in gravitational potential difference Vgrav = Ep / m = gh
The unit of gravitational potential energy is the joule (J) whereas the unit of
gravitational potential difference is the joule per kilogram (Jkg-1).
We have now developed the concept of equipotential surfaces (which in 2-dimensional
diagrams appear as lines). These must be perpendicular to field lines as no work is done
unless there is a component of displacement in the direction of the force (of gravity).
Page | 33
Gravitational potential in a radial field
We now move on from the uniform field to the radial field. The way that field strength
varies with distance in a radial field has already been established (g  1/r2). We now go on
to show that potential in a radial field varies as 1/r.
The neatest way of showing this is with mathematics that some will not understand. This
involves finding the area under the g against r graph, which really means integration. This
can be done by numerical methods (either counting squares or using modelling software)
for those not familiar with integration of 1/r. We will adopt this method first and then offer
an alternative treatment for those who prefer it. The alternative method is shown in the
student's book section 11.4 and uses data for the return journey of Apollo 11.
Before we tackle the integration we need to establish the idea of field strength as a
potential gradient. You must know that g = - V/r. Consider the analogy with hills and
contour lines: at each point the size and direction of the force depends on the steepness of
the slope. See the part of Display Material 190O OHT 'Field and potential' shown below.
Display Material 190O
OHT 'Field and potential'
Gravitational potential well
field down potential slope
level in well is
potential energy
per kg
Gravitational field and radius
0
0
r
g
r
V grav = area gr
g
radius r
Gravitational potential and radius
0
0
r
Vgrav
field g
= –slope V grav /r
r
V grav
radius r
The field is the slope of the graph of potential against radius
The difference in po tential is the area u nder the g raph of field again st radius
Page | 34
Here you see the vital starting point that Vg = area gr which follows from the definition
of Vg given earlier. The relationship g = - V/r follows since a positive value of r
causes an increase in potential making V positive and making - V/r negative. This is
right as the field strength g acts in the opposite direction to r.
The integration to find the area under the g-r graph is shown below.
Reading 110W: Well actually/but also
Gravitational potential due to a spherical mass
We can find the change in gravitational potential by using calculus.
M
x
x
P
r
A
B
The diagram shows a spherical mass M. To move a unit mass from A to B, a small
distance  x, a force must be applied to the right.
Force applied to the unit mass = G M / x2 to the right.
The work done to move the unit mass from A to B will be:
W 
GM
x2
 x.
The total work done in moving the unit mass from some point P to infinity will be:

 GM
r
x2

 GM 
 GM   GM  GM
dx  
 

  
 r .
 x r
    r 
Gravitational potential at infinity is therefore higher than at P. Since the definition implies
potential at infinity is zero, the potential at any point P will be given by:
Vg  
GM
.
r
If the sphere has a radius R, the gravitational potential at its surface will be:
Page | 35

GM
.
R
You could adopt a more rigorous approach if you wish (using g = - V/r and
g = - G M / r2 ) but only if you like the maths. In this case you would integrate between
points r1 and r2 and then say that the potential is zero when r2 is infinity. All that really
matters is that you know the result!
The interconnection between the relationships can be seen in the next part of Display
Material 190O OHT 'Field and potential' shown below.
Gravitationa l field and gravitational potential
If the potential varies as 1/r then the field varies
as 1/r 2
radius r
0
r
r + r
0
Assume 1/r variation of potential and calculate difference
in potential
at radius r :
Vgrav at r + r
V grav at r
at radius r + r :
V = – GM
r + r
Difference in potential V between r and r + r
is:
field g = – slope V/r
r
V
Field = – dV/dr
If V = GM/r, then
field = – dV/dr = – GM/r 2
V = – GM
r
field g = – V
r
GM
GM
V = –
r – ( – r + r )
– GM (r + r) + GMr
V =
r (r + r)
GMr
i f r is small: V =
r2
Thus:
field g = – G M
r2
The less mathematical approach to establishing the equation for gravitational potential in a
radial field involves analysis of Apollo 11 data as shown on pages 58-59 of the student’s
book.
When Apollo 11 made the homeward journey from the Moon, its speed increased as its
kinetic energy increased as a result of a decrease in potential energy. The total energy
remained constant.
We have Etotal = Epotential + Ekinetic or Et = Ep + Ek
This means that Ep = Et - Ek
Since Vgrav = Ep / m we have Vgrav = Et / m - Ek / m = constant – ½ v2
That means we can use – ½ v2 to represent gravitational potential and it will be accurate
except for a constant that is linked to the total energy. A graph of – ½ v2 against 1 / r
Page | 36
should produce a straight line, but that line would only go through the origin if the total
energy were zero.
The data and graphs from the Apollo mission are shown below.
Display Material 200O OHT 'Apollo returns from the Moon'
Apollo 11 com es back from the Moon
Pairs of observations of speed and distance
Apollo 11 is coasting home ‘downhill’ with rockets turned off. Distances r taken from centre of Earth
distance / 106 m
speed / m s –
1
–
2
6
1
v / 10 J kg–
8
10 m / r
241.6
209.7
1521
–1.16
0.414
1676
–1.40
0.477
170.9
1915
96.8
56.4
2690
3626
–1.83
–3.62
0.585
1.033
1.774
28.4
5201
–6.57
–13.53
13.3
7673
–29.44
0
0
–10
–10
1
2
1
2
3.518
7.513
Variation of gravitational potential with 1 / r
Variation of gravitational potential with distance
–20
1
2
mv 2 is the kinetic energy
v2 is the kinetic energy per kilogram
1
2
2
– v is the change in potential energy
per kilogram
–30
–20
–30
0
50
100
150
distance r / 1000 km
200
250
0
2
4
108 m / r
6
8
Use Activity 260S Software-based ‘Probing gravitational potential’ together with File 160T
Spreadsheet data table ‘Data from Apollo 11 Mission’ to do your own analysis.
Page | 37
Gravitational potential energy in a radial field
If a satellite of mass m was placed in the gravitational field of the Earth the two masses
would then have potential energy measured in joules. Work would have to be done to
The equation for the potential energy of two masses is
Egrav = - GMm
r
where r is the distance between the centres of the two masses.
A more general equation for two masses would be Egrav = - Gm1 m2
r
Warning: this equation does not appear on the formula sheet but it can be useful.
separate the two masses to infinity and that work would represent the magnitude of the
potential energy.
Questions
[G = 6.67  10–11 N m2 kg–2, Earth’s mass = 5.97  1024 kg,
Radius of the Earth = 6.37  106 m , Radius of the Progress orbit = 6.69  106 m]
1
Calculate the gravitational potential at the surface of the Earth.
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
The full analysis shows not just that potential a distance r from a spherical mass
varies as 1/r but that potential is given by the equation:
Vg = - GM / r
Vg is measured in J kg-1 of course.
Notice that at very large distances the potential approaches zero (from below) and at
infinity it is zero.
2
Calculate the gravitational potential energy of the Earth-Progress system if
spacecraft Progress has a mass of 7 tonnes.
Page | 38
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
We can now look at the relationships in more detail.
Activity 250S Software Based 'Variations in field and potential'
using File 120L Launchable File 'Modelling field strength and potential'
You should be aware of the shape of equipotentials around the earth as shown below.
Display Material 210O
OHT 'Relationship between g and Vg'
Gravitational gradients around Earth
Equipotentials are spheres
contour of constant gravitational potential
gravitational field
Page | 39
Display Material 210O
OHT 'Relationship between g and Vg'
These four panels summarise two important ways of representing gravitational fields, and
show the links between them.
Gravitational equations and graphs
Descriptions using forces
gradient of curve
Gravitational field strength
GM
g=– 2
r
Descriptions using energies
Gravitational potential
GM
V=–
r
per
kilogram
r/m
Gravitational forces
F=–
r/m
Gravitational potential energy
GMm
r2
E=–
GMm
r
for a
mass
r/m
area under curve
r/m
Question [Hard]
A Meteosat weather satellite with a mass of 320kg was placed in a geostationary orbit
around the Earth in 1997.
G = 6.67  10–11 N m2 kg–2, Earth’s mass = 5.97  1024 kg,
Radius of the Earth = 6.37  106 m
1
Use Newton’s law of gravitation, and the fact that it is in a circular orbit with a period
of 24 h, to calculate the radius of the orbit.
Page | 40
…………………………………………………………………………………………………
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…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
2
Calculate the velocity of the satellite and hence its kinetic energy whilst in orbit.
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
3
Calculate the gravitational potential energy of the satellite relative to the surface of
the earth. Hint: you cannot use Ep = mgh for this!
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
4
Calculate the minimum energy needed to put the satellite into orbit. Suggest why
your answer is a minimum value.
…………………………………………………………………………………………………
…………………………………………………………………………………………………
Page | 41
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
Total energy and escape
A satellite needs to have the correct velocity as well as being placed at the right height.
Here you can try different combinations of height and velocity, noting their effect on the
stability of the orbit, and on the energy of the satellite in the orbit.
Activity 290S Software Based 'Setting up energetic orbits'
using File 150L Launchable File 'Simulation of satellite orbiting Earth'
Escape velocity ideas are optional, but are worth considering.
The total energy of a space craft Et = Ep + Ek must be less than zero if it to stay within the
potential well, more than zero if it is to escape from the potential well, and equal to zero if it
is reach infinity with a speed of zero.
To escape from the potential well, Ep + Ek  0
If Ek is the kinetic energy on launch, when the space craft has a velocity v, Ep will be the
gravitational potential energy of the Earth-space craft system on launch,
i.e. Eg = - GMm / r.
We have - GMm / r + ½ mv2  0
or - GM / r + ½ v2  0 if the masses are cancelled.
This leads to v2  2 GM / r
Question
Calculate the escape velocity for the earth.
G = 6.67  10–11 N m2 kg–2 , Earth’s mass = 5.97  1024 kg,
Radius of the Earth = 6.37  106 m
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
Page | 42
The diagram of gravitational wells in space
suggests why it might be necessary for
certain space vehicles to reach the escape
velocity.
The arguments here have been based on the interplay between kinetic and gravitational
potential energy in radial fields.
Understanding how to maximize the kinetic energy of a spacecraft draws on ideas of the
Earth’s circular motion. The text (page 60) suggests why it might be worthwhile to launch a
space vehicle near the Equator. Reference is also made to ‘gravity assist’
The questions deal with getting satellites up into higher orbits through a transitional
elliptical orbit (Hohmann orbit) and revisit ideas from section 11.1. Escape velocity ideas
link with work on black holes.
The following activities may also be used:
Activity 200D Demonstration ‘Loop-the-loop
Activity 220S Software-based ‘Storing energy with’
Activity 230S Software-based ‘Inferring fields’
Page | 43
Questions and activities additional to those listed in the Student Notes
Section
Essential
Optional
11.1
Read A2 text pp 31-39
A2 text p 40 Qu 1-6
Question 70W Warm-up Exercise 'Radians and
angular speed'
Question 20W Warm-up Exercise
'Orbital velocities and acceleration'
Question 50C
Question 30S Short Answer
'Centripetal force'
Question 40S Short Answer
‘Circular motion – more challenging’
Question 10D Data handling 'Using
Kepler’s third law'
Comprehension 'Centrifuges'
Display Material 10O
orbits'
OHT 'Eccentricity of planet
Display Material 80P
Poster 'Speeds and
accelerations in the solar system'
Display Material 90P
Poster 'Acceleration of the
Moon'
Reading 10T
Text to Read 'Brahe and Hamlet'
Reading 20T
Text to Read 'Hubble'
Reading 30T
Text to Read 'The problem of
longitude'
Reading 40T
Text to Read 'Kepler's second law and
angular momentum'
Reading 50T
Text to Read 'Training for movement
in space'
Activity 50H
Home Experiment 'Going around
in the kitchen'
The following spreadsheet exercise could be used to:
(i)
verify Kepler’s Laws
(ii)
calculate the density of planets
(iii)
establish the relationship between orbital
velocity and radius (v2  1/r)
File 30T
Spreadsheet Data Table 'Planetary
orbit data'
11.2
Read A2 text pp 41-46
A2 text p 47 Qu 1-6
Question 10W Warm-up Exercise
Question 60C Comprehension 'How Cavendish
didn’t determine G and Boys did'
Question 90C Comprehension 'Are there planets
around other stars?'
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'Testing for an inverse-square law
Question 80W Warm-up Exercise
'Newton’s gravitational law'
11.3
Question 130C Comprehension 'Variations in g'
Question 110S Short Answer
'Finding the mass of a planet with a
satellite'
Reading 60T
Text to Read 'Forces on real objects'
Reading 70T
Text to Read 'Gravity can pull things
apart'
Reading 100T Text to Read 'Supernovae and black
holes'
Question 120D Data handling ‘The
gravitational field between the Earth
and the Moon’
Activity 90S Software Based 'Gravitation with three
bodies' using File 70L Launchable File 'Gravitation
between three bodies'
Read A2 text pp 48-54
A2 text p 55 Qu 1-6
Question 170C Comprehension 'Collision with
spaceship Earth'
Question 190C Comprehension 'Getting a satellite up
to speed'
Question 140W Warm-up Exercise
'Change in momentum as a vector'
Question 150S Short Answer
'Impulse and momentum in collisions'
Question 160S Short Answer
'Collisions of spheres'
Question 180S Short Answer 'Jets
and rockets'
File 100L
File 90M
of view'
Launchable file 'Crunch'
Movies 'Collisions from different points
Activity 170D
engine'
Demonstration 'Testing a rocket
Activity 100E
Experiment 'Crater making'
Reading 80T
Text to Read 'Starting with
momentum'
Reading 90T
Text to Read ‘Sling-shotting’
spacecraft or ‘gravity assist’
11.4
Read A2 text pp 56-60
A2 text p 60 Qu 1-6
Question 200W Warm-up exercise
‘Pole vaulting’
Question 210S Short Answer
'Gravitational potential energy and
gravitational potential'
Question 250S Short Answer
'Summary questions for chapter 11'
Question 220D Data Handling
'Gravitational potential difference,
field strength and potential'
Summary
Question 230D Data Handling 'Changing orbits’
Question 240D Data Handling 'Why is a black hole
black?’
Activity 210D Demonstration 'Gravitational slides'
Activity 270D Demonstration 'Measurements with
gravitational slides'
Activity 240S Software Based 'Analysing data from
the Apollo 11 mission to the Moon'
Activity 280S Software Based 'Relating field and
potential'
Reading 130T Text to Read 'The Cassini-Huygens
mission to Saturn'
A2 text p 63 Qu 1-9
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These notes draw almost exclusively on the resources to be found in Advancing Physics A2 Student’s Book and CD-ROM published by
Institute of Physics Publishing in 2000 and 2008. They are intended to be used in conjunction with these resources and others not
specified.
John Mascall
The King’s School, Ely, Cambs
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