Physics
HSC Course
Stage 6
Space
Part 4: Once around the block then home, James
Contents
Introduction ............................................................................... 2
In orbit ..................................................................................... 4
Uniform circular motion ........................................................................4
Universal gravitation and orbits ...........................................................8
The period of an orbit .........................................................................10
Types of orbit......................................................................................13
Investigating the orbital velocity of a satellite .......................... 16
The slingshot effect ................................................................. 22
Orbital decay ........................................................................... 27
Re-entry .................................................................................. 29
The re-entry angle..............................................................................30
Heat build-up ......................................................................................31
g forces in re-entry .............................................................................33
Ionisation blackout .............................................................................34
Reaching the surface .........................................................................34
Summary................................................................................. 36
Suggested answers................................................................. 39
Exercises – Part 4 ................................................................... 43
Appendix ................................................................................. 49
Part 4: Once around the block then home, James
1
Introduction
In the part you examined the history and nature of rockets, how they
work and what they do, including using them to escape the Earth’s
surface. In this part you will further analyse the circular motion of a
simple orbit, as well as learning about different types of orbits. You will
learn about something called the slingshot effect, a manoeuvre used to
speed up spacecraft in space. Finally, you will examine some of the
difficulties space craft encounter returning to the surface of the Earth
from space.
Before beginning this topic you must have already studied certain
concepts. In particular you must be able to:
Â
r
r
F = ma
•
describe and apply Newton’s second law of motion:
•
•
describe and apply Newton’s law of universal gravitation:
r
mm
F = G 12 2
r
r
r
state the definition of momentum: p = m v
•
state the definition of kinetic energy: E k = 12 mv2
•
state the definition of gravitational potential energy: E p = - G
•
use the conservation of momentum and kinetic energy to analyse
one-dimensional elastic collisions.
m1m 2
r
In this part you will be given opportunities to learn to:
2
•
analyse the forces involved in uniform circular motion for a range of
objects including satellites orbiting the Earth
•
compare qualitatively and quantitatively low Earth and geostationary
orbits
•
discuss the importance of Newton’s Law of Universal Gravitation in
understanding and calculating the motion of satellites
Space
•
describe how a slingshot effect is provided by planets for space
probes by applying:
r3
GM
=
2
T
4p 2
•
account for the orbital decay of satellites in low Earth orbit
•
discuss issues associated with safe re-entry into the Earth’s
atmosphere and landing on the Earth’s surface
•
identify that there is an optimum angle for re-entry into the Earth’s
atmosphere and the consequences for not achieving this angle.
In this part you will be given opportunities to:
•
perform an investigation that demonstrates that the closer a satellite
is to its parent body, the faster it moves to maintain a stable orbit
•
solve problems and analyse information to calculate centripetal force
acting on a satellite undergoing uniform circular motion about the
Earth
•
solve problems and analyse information using:
r3
GM
=
2
T
4p 2
•
plan, choose equipment or resources for, and perform an
investigation to model the effect that removal of the Earth’s
gravitational force would have on the direction of satellite motion
•
plan, choose equipment or resources for, and perform a first-hand
investigation to model the effect of friction and heat on a range of
materials, including metals and ceramics.
Extract from Physics Stage 6 Syllabus © Board of Studies NSW, 1999. The
original and most up-to-date version of this document can be found on the
Board’s website at http://www.boardofstudies.nsw.edu.au.
Part 4: Once around the block then home, James
3
In orbit
When Apollo 11 began its historic mission to the Moon in 1969, the first
step was to get the rocket up into orbit around the Earth. The enormous
Saturn rocket that was used to do this had three stages. As it lifted off it
banked over slightly, and then more and more as it gained altitude until,
by the time it had reached an altitude of 190 km, it was travelling nearly
parallel with the ground.
Two stages of the Saturn rocket had been used by this time and the rocket
was high enough to orbit the Earth. If they had no more available rockets
to fire, however, their mission would have been doomed – they would
have soon fallen back to Earth in a long spiral. The reason is that they
were only travelling about 25 000 kmh-1, not fast enough yet for a stable
orbit at 190 km above the ground.
The third rocket stage was fired and the spacecraft was accelerated
horizontally to the required speed of 28 000 kmh-1, before the rocket
engines were shut down (leaving sufficient fuel to accelerate away from
the Earth later on).
In order to properly understand why it was important that this speed be
reached, you must be able to analyse the simplest form of orbital motion.
This motion is known as uniform circular motion.
Uniform circular motion
Uniform circular motion is circular motion with a uniform orbital velocity.
Try a little thought experiment.
Can you imagine tying a ball to a length of rope and then grabbing the other
end of the rope and whirling the ball evenly around in a large, horizontal
circle? (If you can’t imagine it then you can try it if you like!)
What is necessary to keep the ball on its circular path? The answer might be
obvious to you, but first consider why it is an important question to ask.
4
Space
Newton’s first law of motion states that an object will continue in
uniform velocity in a straight line (or a state of rest) unless acted upon by
an external force. In other words, if something is moving and has no
external force applied, then it will travel in a straight line, not a circle.
The object has to be forced to do something else.
Returning to the ball on the string, what is the ball that is forcing it to not
travel in a circular path. Where is the external force that Newton’s first
law says must exist? The answer is the tension (or pulling force) applied
through the rope. If, when whirling the ball around, you were to let go of
the rope then the ball would fly off at a tangent to the circle (travelling in
a straight line). Without that tension, the circular motion could not exist.
Examine the tension a little closer – what is its direction? It is directed
back along the string towards your hand, is at the centre of the circular
motion. As the ball makes its way around the circle, the tension / force is
always towards the centre of the circle. This idea can be generalised for
all types of circular motion.
For uniform circular motion to exist, there must always be a force
applied towards the centre of the circle. This force is called centripetal
force. The equation for centripetal force is:
Centripetal force,
where
r
mv 2
FC =
r
FC = centripetal force, in newtons (N)
m
= mass of object in motion, in kg
v
= orbital velocity of the mass, in ms-1
r
= radius of circular motion, in m
Centripetal force exists in every case of circular motion, including a
spacecraft or satellite orbiting the Earth. The following diagram shows a
few examples.
If you are travelling in a car as it corners, your own body will try to
continue travelling on in a straight line. The force you feel is the vehicle
/ seat / restraints forcing your body over, onto the circular path. In other
words, the force you feel is a centripetal force.
Part 4: Once around the block then home, James
5
ball
Fc
(tension
in rope)
Fc
(friction)
ball on a rope
cornering car
satellite
Earth
Fc
(gravity)
Fc
(gravity)
Sun
Earth
Earth orbiting Sun
satellite orbiting Earth
Sample problem 1
If, in the thought experiment used earlier, the ball has a mass of 180 g,
the rope is 2.0 m long and its orbital velocity is 8.5 ms-1, what is the
value of the tension in the rope?
Solution
Remember to convert all values to kg, m or s before substituting into the
formula.
In this case, the tension will be the centripetal force, so:
Centripetal force,
6
r
r
mv 2
FC =
r
0.180 ¥ 8.52
=
2.0
= 6.5 N
Space
1
Now it is your turn to try a similar calculation. A truck of mass 25 000
kg rounds a bend of radius 35 m. If the truck is travelling at 30 kmh-1,
what centripetal force is required (in the form of friction between its
tyres and the road)?
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
2
Here is another problem for you. In this problem a satellite of mass
150 kg is following a circular path as it orbits the Earth. It has an
altitude of 250 km and a velocity of 27 900 kmh-1.
a) Calculate the radius of its orbit, in metres from the centre of the
Earth. Use a value of 6 380 km for the radius of the Earth.
_________________________________________________
_________________________________________________
_________________________________________________
b) Convert the velocity of the satellite to ms-1.
_________________________________________________
_________________________________________________
c) Calculate the centripetal force operating to keep this satellite in
its orbit.
_________________________________________________
_________________________________________________
_________________________________________________
_________________________________________________
_________________________________________________
Check your answers.
You should now attempt Exercise 4.1.
Part 4: Once around the block then home, James
7
Universal gravitation and orbits
Focus now on the orbital motion of a satellite around the Earth. The
centripetal force enabling this motion is the gravitational attraction
between the satellite and the Earth. You should recall from Part one of
this module that the strength of this force is described by Newton’s law
of universal gravitation, as shown here.
r
m m
FG = G E 2 S
r
where
mE = mass of the Earth
mS = mass of the satellite, in kg
r
= radius of the orbit, in m
G
= universal gravitation constant
If this expression for the gravitational force is made equal to the
expression for centripetal force, then it is possible to get an equation for
the orbital velocity v of a satellite.
r
m E m S m Sv2
G 2 =
r
r
r
Gm E
\v =
r
You may recall this equation from the module The cosmic engine. It is
important to realise that the radius of the orbit r is the sum of the radius
of the Earth and the altitude of the orbit. r = rE + h
Therefore, the equation for the orbital velocity of a satellite becomes:
r
\v =
Gm E
rE + h
r
where v = orbital velocity, in ms-1
G
= universal gravitation constant = 6.67 ¥ 10-11 Nm2kg-1
mE = mass of the Earth = 5.97 ¥ 1024 kg
rE
= radius of the Earth = 6.38 ¥ 106 m
h
= altitude of orbit, in m
This equation shows you that the orbital velocity required by a satellite in
a specific orbit around the Earth depends upon the mass and radius of the
Earth (which have fixed values) as well as the altitude of the orbit.
8
Space
Further, it is clear that a lower altitude means a higher orbital velocity
required for a stable orbit.
Sample problem 2
Determine the orbital velocity required by a satellite at an altitude of
190 km and 500 km. Use the values of mass and radius of the Earth
given earlier.
Solution
Orbital velocity at 190 km:
r
v=
=
Gm E
r
(6.67 ¥ 10 -11 )(5.97 ¥ 10 24 )
(6.38 ¥ 10 6 + 190 ¥ 10 3 )
= 7 790 ms-1 = 28 000 kmh -1
Notice that this is the speed mentioned earlier as the orbital velocity of
the Saturn / Apollo spacecraft at this altitude.
Orbital velocity at 500 km:
r
v=
=
Gm E
r
(6.67 ¥ 10 -11 )(5.97 ¥ 10 24 )
(6.38 ¥ 10 6 + 500 ¥ 10 3 )
= 7 610 ms-1 = 27 400 kmh -1
You should note that in this case with a greater altitude, the required
orbital velocity of a satellite is less than at 190 km.
Use this method to calculate the orbital velocity of satellites stable in orbits
at altitudes of a)1000 km, and (b) 40 000 km.
_________________________________________________________
_________________________________________________________
_________________________________________________________
_________________________________________________________
Check your answers.
Part 4: Once around the block then home, James
9
You should now attempt Exercises 4.2 and 4.3.
The period of an orbit
In the module The world communicates you were introduced to the idea
of the period of a cyclical motion. At that time you were studying waves,
however the same idea can be used with circular motion. You should
recall that the period T of a cyclical motion is the time taken for one
complete cycle.
Since the period T of an orbit is the time taken for one orbit, it is related
to the velocity of a satellite along that orbit and the length (or
circumference) of the orbit. This relationship is:
r circumference
v=
period
r 2p r
\ v=
T
This is a general relationship that applies to all types of uniform circular
motion. If this expression is equated with the expression derived above
for the orbital velocity v of a satellite, another relationship can be
derived.
2p r
=
T
Gm E
r
Ê 2 p r ˆ = Gm E
Ë T ¯
r
2
4pr 2 Gm E
=
T2
r
multiply both sides by r,
r 3 Gm E
=
T2
4p 2
This equation expresses the relationship between the radius and the
period of the orbit of a satellite orbiting the Earth (mE). It can be
generalised into an equation that applies to any satellites orbiting any
other celestial mass (that is, planet or star).
r 3 GM
=
T 2 4p 2
10
Space
where
r = radius of the orbit, in m
T = period of the orbit, in s
G = universal gravitation constant = 6.67 ¥ 10-11 Nm2kg-1
M = mass of object being orbited, in kg
p = constant
One consequence of this expression is that if comparing different
satellites of the same central body, then the right side of the equation has
a fixed value. This means that the ratio r3/T2 must be the same for each
satellite.
Johannes Kepler first noticed this in the very early 1600s. He had at his
disposal a large volume of meticulously detailed observations of the
motions of the then-observable planets. These observations were made
by his mentor Tycho Brahe and were passed to him on Brahe’s death.
Kepler was trying to find a mathematical relationship that described these
planetary motions when, after a great deal of work, he discovered that the
ratio r3/T2 had the same value for every planet. This has become known
as Kepler’s law of periods:
Êr3 ˆ
Ê r3 ˆ
ÁÁ 2 ˜˜ for planet 1 = Á 2 ˜ for planet 2
ÁT ˜
ËT ¯
Ë ¯
Sample problem 3
Determine the period of the temporary orbit of the Saturn / Apollo rocket
mentioned earlier, at an altitude of 190 km.
Data:
orbital velocity = 7 790 ms-1
radius of the Earth = 6 380 km
Solution
The radius of the orbit
= radius of the Earth + altitude
= 6.38 ¥ 106 + 1.9 ¥ 105
= 6.57 ¥ 106 m
Part 4: Once around the block then home, James
11
Therefore, T =
=
2p r
v
2 p ( 6.57 ¥ 10 6 )
7790
= 5 300 s = 88.6 min
Sample problem 4
Use Kepler’s law of periods to determine the radius of the orbit of Mars
in astronomical units (au), which is the average distance between the Sun
and the Earth.
Data:
radius of Earth’s orbit = 1.0 au
period of Earth’s orbit = 1.0 Earth year
period of the Martian orbit = 1.88 Earth years
Solution
Ê r3
ÁÁ 2
ËT
ˆ
Ê r3 ˆ
˜˜ for Mars = ÁÁ 2 ˜˜ for Earth
¯
ËT ¯
3
rMars
1.0 3
=
1.88 2 1.0 2
3
rMars
= 1.88 2 = 3.5
\ rMars = 3 3.5 = 1.5 au
Since this expression relies on ratio comparison, it is not essential that SI
units be used. In this solution astronomical units and Earth years were
used to simplify the numbers involved.
Earlier you calculated the orbital velocity of a satellite at an altitude of
500 km.
a)
Calculate the period of the orbit of this satellite
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
12
Space
b) Use the comparison technique of Kepler’s law of periods, to
determine the period of a satellite placed at an altitude of 40 000 km.
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
Check your answers.
You should now attempt Exercise 4.4
Types of orbits
Up until this point orbits have been assumed to be circular. In fact, most
orbits are not. Instead they are ellipses (or ovals) of various eccentricity
(the amount of stretch). Most of the planets have nearly circular orbits,
while most comets have extremely eccentric (stretched out) elliptical
orbits. The eccentricity of satellite orbits can vary from satellite to
satellite. Kepler realized the elliptical nature of planetary orbits and
stated the fact as his first law. His third law, the law of periods, is still
applied to elliptical orbits.
When a satellite or other spacecraft is placed into an orbit around the
Earth, it will usually be inserted into one of two general types of orbit,
a ‘low Earth orbit’ or a ‘geostationary’ orbit. The essential difference
between them is altitude.
A low Earth orbit is one that is above the Earth’s atmosphere but below
the lower Van Allen radiation belt. The minimum altitude required to
avoid significant drag from the upper atmosphere is approximately 250
km – the drag of air resistance on anything lower than that will soon have
the satellite slowing down and spiralling back to Earth. The maximum
altitude possible (above the equator) that remains under the Van Allen
belts is approximately 1000 km. The Van Allen belts are regions of high
radiation that pose a threat to the well being of life and equipment. You
have already learned about them in The cosmic engine, and they will be
examined further in Part 5 of this module.
Part 4: Once around the block then home, James
13
The altitude range, therefore, for low Earth orbits is approximately
250 km to 1000 km. This type of orbit can be utilised for a number of
different purposes. For example, the Space shuttles use orbits with
altitude between 250 km and 400 km, and take approximately 90 min to
orbit the Earth. The Hubble space telescope (HST) uses a low Earth
orbit, which places it within reach of the Space shuttle for repair
purposes. An orbital variation used by spy satellites and surveying
satellites, is to orbit the Earth in an almost north-south plane – with the
Earth rotating beneath it. Such satellites are able to observe the entire
globe once every 24 hours.
A geostationary orbit is one in which a satellite does not appear to
move in the sky, as seen from the surface of the Earth. A geostationary
orbit was first suggested by Arthur C Clarke in 1945 who recognised its
potential as a platform for communications. For this to occur the period
of the satellite’s orbit must equal the period of the Earth’s rotation. In
other words, the satellite must take as long to orbit the Earth as the Earth
takes to rotate once on its axis relative to the stars Note that this length
of time is one sidereal day (not one solar day), which is a period of 23
hours, 56 minutes and 4 seconds, or 86 164 seconds. .
The altitude that corresponds to this period can be calculated using the
radius-period relationship derived earlier, as follows.
r 3 Gm E
=
T2
4p 2
(6.67 ¥ 10-11 )(5.97 ¥ 1024 )
r3
=
86 164 2
4p 2
r 3 = 7.50 ¥ 10 22
\ r = 4.2 ¥ 10 7 m
recall that radius of the orbit = radius of the Earth + altitude
so altitude = 4.2 ¥ 10 7 - 6.38 ¥ 10 6
= 3.57 ¥ 10 7 m
= 35 700 km
When allowances are made for other factors, the effective altitude of
such an orbit is 35 880 km above the equator, which is quite high in
comparison to a low Earth orbit. This altitude places a satellite above the
upper Van Allen belt, which reduces the risk of damage caused by it.
Unfortunately, with only the outer envelope of the Earth’s magnetic field
(called the magnetosphere) above there is an increased danger from solar
radiation. This unique position does, however, make it a useful place for
scientific experiments or observations, and so some satellites are placed
there for this purpose.
14
Space
The principal uses of satellites in this orbit are for communications and
weather satellites. If in an orbit over the equator, such a satellite has a
fixed position in the sky relative to the ground. This means that a
receiving dish can be aimed directly at it. No further position
adjustments are required in order to receive optimum signal strength
from the satellite. If the satellite is not positioned over the equator then,
as seen from the ground, it will appear to trace out a figure 8 in the sky
over the course of 24 hours. This is known as a geosynchronous orbit.
The receiving dish will then need to track the satellite (following its
movement in the sky) in order to avoid a natural waxing and waning of
the signal strength (equipment designers may simply choose to tolerate
this and compensate using amplifiers).
geostationary
orbit
eccentric
orbit
polar orbit
Types of orbit around the Earth.
You should now attempt Exercises 4.5 and 4.6.
Part 4: Once around the block then home, James
15
Investigating the orbital
velocity of a satellite
You may do this experiment with your teacher, however you can also try
this at home. The aim of this activity is to compare the velocity - radius
relationship affecting standard uniform circular motion to that of the orbital
motion of a satellite.
You will need:
•
some specially-prepared glass tubing (if attempting this at home try
substituting an empty pen casing)
•
string
•
a mass to swing around safely such as a rubber stopper
•
a 50 g mass carrier
•
several 50 g slotted masses to go with it (you can try substituting
large washers and a bolt and nut for the mass, but you will need to
know their mass)
•
a metre rule or tape
•
stopwatch
•
alligator clip (or paper clip).
Method:
16
1
Arrange this equipment as shown in the diagram, ensuring that the
string is approximately 1.5 m long.
2
Hold the tube and swing the rubber stopper around in a large
horizontal circle. The centripetal force that maintains the circular
motion is provided by the weight of the masses on the hanging end
of the string.
Space
v
radius
~ 1m
glass tube 15 cm X 5 mm
3 mm space
kept constant
alligator
clip
fishing line
steel washers ~ 5 g each
on 50 g mass carrier
3
Begin by examining uniform circular motion with a constant
centripetal force. As you have already learned, the expression for
the centripetal force in uniform circular motion is, in general:
r
r
mv 2
FC =
r
This can be rearranged to show the relationship between v and r:
mv 2
r= r
FC
or
r µ v2
Expressed in words, the radius of the motion is directly proportional
to the square of the orbital velocity, provided the centripetal force is
kept constant.
According to this relationship, what happens to the orbital velocity if the
radius of the orbit is decreased?
_____________________________________________________
Check your answer.
4
Use a mass of 50 g as the hanging mass and measure the length of
the string (the radius of the circle) to 40 cm. Attach the alligator clip
(or paper clip) to the string just below the tubing. This marker will
help you judge when you have the circular motion just right.
5
Hold the tubing upright and begin to swing the rubber stopper
around in a circle. You will need to hold the apparatus above your
head so that you don’t hit yourself, and make sure that you have
plenty of open space around you so that you don’t damage anything
else.
In this position the marker clip will be at about eye level.
Part 4: Once around the block then home, James
17
While watching it, adjust the speed of rotation until the clip remains
just below the tubing, without touching it. When you have reached a
horizontal stable motion, begin timing ten full revolutions. (It is
useful to have a second person to do the timing, though not
essential.)
Record the time in the table below before adjusting the string to the
next indicated length and repeating the process. Continue this until
all the indicated lengths, from 40 cm to 1 m, have been used.
Complete all columns in the table.
Radius r
(m)
Hanging mass
(g)
0.3
50
0.4
50
0.5
50
0.6
50
0.7
50
0.8
50
0.9
50
1.0
50
6
Time for 10
revolutions (s)
Period
(s)
Orbital velocity
pr/T)
(ms-1) (=2p
v2 (m2s-2)
Using one of the grids provided in the Appendix, draw a graph of
radius versus orbital velocity squared. Since radius was the variable
that was uniformly varied, it should be placed on the horizontal axis.
When you have plotted your data, estimate and then draw a line of
best fit. This graph should reflect the relationship shown in the
equation above.
a) What happens to the value of the orbital velocity as the radius is
increased?
_________________________________________________
_________________________________________________
18
Space
b) Describe the relationship shown between radius and orbital
velocity for uniform circular motion with a constant centripetal
force.
_________________________________________________
_________________________________________________
Check your answers.
7
Look now at the velocity – radius relationship for a satellite orbiting
a planet. The centripetal force in this case is provided by the
gravitational force of attraction, and is described by Newton’s law of
universal gravitation:
r
m m
FG = G E 2 S
r
Notice that the force varies with distance. As distance increases, the
value of the force does not remain the same but reduces quickly.
This is an inverse square relationship, that is, the force is inversely
proportional to the square of the distance between the two bodies.
8
In order to simulate the action of gravity the hanging mass will now
be varied, as indicated in the table below. These masses were
calculated by assuming that with a radius of one metre, a mass of 50
grams would be used for the hanging mass (this is the mass of a
mass carrier on its own). For the smaller radii this mass was
multiplied by the inverse of the square of the radius, and then
rounded off.
By adjusting the hanging mass to these values whenever the radius
of the circle is adjusted, the centripetal force will vary in the manner
described by Newton’s law of universal gravitation.
You have already seen that if the above expression is substituted into
r
the equation for centripetal force then the relationship between v
and r for a satellite becomes apparent –
r
v=
Gm E
r
GM
\ r = r2
v
1
or r µ r 2
v
This is quite different to standard circular motion. Here, radius of a
satellite’s orbit is inversely proportional to the square of the orbital
velocity. According to this relationship, what happens to the orbital
velocity if the radius of the orbit is decreased?
_____________________________________________________
Check your answer.
Part 4: Once around the block then home, James
19
9
Repeat the previous procedure, timing ten revolutions for each of the
radii shown, this time adjusting the hanging masses to the values
indicated. (The most difficult radius will be 40 cm because
everything happens quite quickly. It gets easier after that.)
Record your measurements below and then complete the table.
Radius r
(m)
Hanging mass
(g)
0.3
550
0.4
300
0.5
200
0.6
140
0.7
100
0.8
80
0.9
60
1.0
50
Time for 10
revolutions
(s)
Period
(s)
Orbital
velocity
(ms-1)
pr/T)
(=2p
v2 (m2/52)
1/v2
(s2/m-2)
10 Use another of the grids provided in the Appendix to draw a graph of
radius versus orbital velocity squared, from the data in this table.
Radius should again be plotted across the horizontal axis. This
graph should reflect the relationship derived above for a satellite,
and should look quite different to the last graph drawn.
a) For a satellite, what happens to the value of the orbital velocity as
the radius is increased?
_________________________________________________
b) Describe the relationship shown between radius and orbital
velocity for the orbital motion of a satellite.
__________________________________________________
__________________________________________________
Check your answers.
20
Space
This relationship should have produced a hyperbolic shaped graph.
In order to examine this further, on your third grid (in the Appendix) plot
a graph of radius versus 1/v2, again with radius on the horizontal axis.
Does this graph indicate an approximate straight-line relationship
between r and 1/v2? ________________________________________
If you answered ‘yes’ then this is experimental confirmation of the
inverse square relationship expected.
Part 4: Once around the block then home, James
21
The slingshot effect
In February 1992 a space probe called ULYSSES approached the planet
Jupiter. It passed very close to the giant planet, close enough for the
giant planet’s gravity to change its direction, swinging it around behind
the planet and speeding it up, before the spacecraft’s increased velocity
caused it to be flung out again, away from the planet. This was a classic
example of the slingshot effect, also known as a ‘planetary swing-by’ or
‘gravity assist’ manoeuvre. As a result of this manoeuvre, ULYSSES
was sped up and flung out of the plane of the planets rotating around the
Sun in the solar system, although still orbiting the Sun, becoming the first
spacecraft to do so.
north polar pass
June-September
1995
Jupiter orbit
Jupiter encounter
February
1992
Earth orbit
ecliptic crossing
February
1995
Sun
launch
October
1990
south trajectory
south polar pass
June-November
1994
100 days
Ulysses location
(approx.)
Ulysses mission profile.
The slingshot effect was first employed by the Mariner 10 space probe,
which swung by Venus on its way to Mercury. These days it is a
commonly employed manoeuvre to achieve an increase in speed and a
change of direction. It does this by utilising the kinetic energy of a
planet rather than use its own energy (that is, fuel which needs to be
carried until used).
22
Space
But how does the manoeuvre work? In the module Moving about you
examined one-dimensional collisions and saw how the conservation of
momentum allows you to predict what will happen to the bodies
colliding. The same technique can be employed here for, even though
this interaction is not actually a collision, it is certainly an interaction that
conserves momentum.
Consider the scenario as represented in the diagram below, which shows
a ‘before’ and an ‘after’ situation. An important point to appreciate is
that the mass of the planet is very much greater than the mass of the
spacecraft. This is indicated here by using Km as the mass of the planet,
where m is the mass of the spacecraft and K is a very large number.
Vi
before:
planet
spacecraft
Vf
after:
vi
vf
spacecraft
planet
The slingshot effect – before and after.
Conservation of momentum in this situation allows us to say:
initial momentum = final momentum
r
r
r
r
p i of planet + p i of spacecraft = p f of planet + p f of spacecraft
KmVi + m(- v i ) = KmVf + mv f
KVi - v i = KVf + v f
where
(1)
vi = initial velocity of spacecraft, in ms-1
vf = final velocity of spacecraft, in ms-1
Vi = initial velocity of planet, ms-1
Vf = final velocity of planet, in ms-1
Note that all these velocities are relative to the Sun, not each other.
Since this interaction is elastic, with no losses in energy due to any actual
collision, it is also possible to apply the conservation of kinetic energy.
Part 4: Once around the block then home, James
23
When this is done another expression arises:
initial kinetic energy = final kinetic energy
KE i of planet + KEi of spacecraft = KE f of planet + KE f of spacecraft
1
2
2
2
KmVi + 1 2 mvi =
2
2
1
2
2
KmVf + 1 2 mvf
2
KVi - vi = KVf + vf
2
2
(2)
Equations 1 and 2 can be solved simultaneously, although not easily –
several assumptions must be made along the way. This leads to a simple
expression for the final velocity of the spacecraft – relative to the sun.
r
r r
v f = v i + 2Vi
Expressed in words, the exit speed of the spacecraft is equal to its
original speed plus twice the speed of the planet. It turns out that the
scenario considered here represents the maximum speed increase that can
be produced by the slingshot effect – when other angles are introduced
the speed increase is less.
An important question to consider is the source of the energy increase of
the spacecraft. If it has increased its speed then it has also gained kinetic
energy. Where has this energy come from? It may seem that it has
magically appeared, but remember that the kinetic energy of the planetspacecraft system has been conserved. What has happened is that the
planet has slowed very slightly, losing some kinetic energy. This energy
has been transferred to the spacecraft, allowing it an increased speed.
Because of the space crafts relatively small mass the increase in velocity
of the space craft is significant.
Try this exercise to help you visualise the speed increase offered by the
slingshot effect. A collision between a tennis ball and a large, heavy crate
held by a forklift will be considered. It will serve as an analogy to the
interaction between the spacecraft and the planet.
A collision between a ball and a crate – an analogy.
24
Space
1
The mass of the crate and forklift is very much greater than the
tennis ball. If the ball strikes the crate with a speed of 10 ms-1, with
what speed will it rebound assuming there are no energy losses
during the collision?
_____________________________________________________
2
The forklift now drives forward at 5 ms-1. To the driver of the
forklift any further collisions of the ball will appear as they did when
the forklift was stationary. If the ball is now thrown at the crate,
what will be the approach speed of the ball as seen by the driver of
the forklift?
_____________________________________________________
3
What will be the rebound speed of the ball as seen by the forklift
driver?
_____________________________________________________
4
Given your answer to the last question, and the forward speed of the
forklift, what is the rebound speed of the ball as seen by the thrower
of the ball?
_____________________________________________________
5
How does this answer compare to the equation given above for the
speed increase offered from the slingshot effect?
_____________________________________________________
_____________________________________________________
_____________________________________________________
Check your answers.
Consider also a tennis ball on top of basketball. Bounce the two and
watch what happens to the tennis ball after collision with the ground.
The tennis ball has been slingshot back with a much greater speed but
this time the collision between tennis ball and basketball bouncing back
from the ground is a real one.
Sample problem 5
A 110 kg space probe approaches Saturn ‘head-on’ with a velocity of
8 000 ms-1. Saturn’s orbital velocity around the Sun is approximately
9 600 ms-1.
a)
What is the expected maximum possible velocity of the space probe
after the manoeuvre?
Part 4: Once around the block then home, James
25
b) Calculate the change in kinetic energy of the space probe if this
velocity is achieved. Indicate the source of this additional energy.
Solution
a)
r
r r
v f = v i + 2Vi
= 8 000 + 2 ¥ 9 600
= 27 200 ms-1
b)
DKE = KE f - KE i
=
1
=
1
2
mv 2f - 1 2 mv i2
2
¥ 110 (27 200 2 - 8 000 2 )
= 3.4 ¥ 1010 J
This extra 3.4 ¥ 1010 J of energy has been transferred to the space
probe from the planet. At the orbital velocity indicated, relative to
the Sun, Saturn has a kinetic energy of 2.6 ¥ 1034 J, so that the
amount removed from it is an extremely small portion – only about
one million billion billionth of its kinetic energy.
Try this similar problem. If the same space probe proceeds on to Uranus,
and manages another head-on swing-by manoeuvre, calculate:
a)
the maximum exit velocity of the probe
_____________________________________________________
_____________________________________________________
_____________________________________________________
b) its increase in kinetic energy. Neptune’s orbital velocity, relative to
the Sun, is approximately 10 600 ms-1.
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
Check your answers.
You should now attempt Exercises 4.7.
26
Space
Orbital decay
A stable orbit represents a particular amount of mechanical energy for a
satellite, and the higher the orbit the greater is that energy. To appreciate
why this is so you will need to recall expressions for gravitational
potential energy Ep and kinetic energy Ek, for the mechanical energy of
an object is the sum of these energies.
You will recall from the first part of this module that the equation for the
Ep of a satellite of mass m, a distance r from a planet of mass M, is:
Ep = - G
mM
r
You should also recall from the module Moving about that the equation
for Ek is:
E k = 12 mv2
In addition, you now know an equation for the orbital velocity of a
satellite in a stable orbit:
r
v=
GM
r
r
Substituting this expression for v into the equation for Ek gives an
expression for the kinetic energy of a satellite of mass m, in a stable orbit
around a planet of mass M:
E k = 12 G
mM
r
Part 4: Once around the block then home, James
27
It is now possible to construct an expression for the mechanical energy E
of a satellite in the stable orbit:
E = Ek + Ep
mM
mM
- G
r
r
mM
\ E = - 12 G
r
= 12 G
Look now at this expression. Note also that this energy can be released
as the satellite approaches the centre of the Earth. This energy is
potential and would be gained as kinetic if the object were free to hit the
Earth’s centre. Of course it is not as it would hit the surface. But the
gain in KE of a satellite as it falls back to Earth is significant and
represents a conversion of energy into KE.
For a given satellite orbiting the Earth, the values of m and M are fixed,
while G is a constant. This means that the total mechanical energy of the
satellite is inversely proportional to the radius of its stable orbit. The
higher the orbit of a satellite or spacecraft, the greater is its energy (recall
that this energy came from the rocket engines during its launch).
Conversely, the lower the orbit of a satellite or spacecraft, the lower is its
energy. This fact becomes important when considering what happens to
a satellite over time.
Satellites in low Earth orbits are near the extremities of the atmosphere.
However, the atmosphere has no altitude at which it suddenly stops, so
that it simply becomes less and less dense with increasing altitude – even
as far out as 1000 km. This means that all satellites in a low Earth orbit
are subjected to some degree of friction with the atmosphere around it, as
sparse as that may be. This friction causes a loss of kinetic energy due to
heating, and a loss of energy means that the satellite’s orbit will lose
altitude, since lower orbits have lower energy. The term used to describe
this process is orbital decay.
Unfortunately, when a satellite’s orbit decays to a lower altitude this
places it into a denser atmosphere, which will cause further decay.
The trend will continue, with the satellite continuing to lose energy and
plunging to ever increasing atmospheric densities. When the satellite has
reached an altitude of 200 km its fate is sealed, with only hours left in the
sky. As its plunges down past 100 km the friction and heat build-up is so
great that the satellite will probably be vaporised. It is seldom that any
satellite remains reach the ground.
28
Space
Re-entry
As mentioned previously, when an orbiting spacecraft (or satellite) loses
kinetic energy (and velocity) then it begins to fall in towards the Earth
(or other central body). This happens because the gravitational force of
attraction is greater than the centripetal force required for circular motion
at the new, lower orbital velocity. The result is that the spacecraft is
forced closer to the Earth, and the process of orbital decay begins.
The initial loss of velocity can be accidental or deliberate. Accidental
orbital decay occurs when the Earth’s atmosphere expands unexpectedly,
such as during a solar maximum (see Part 5 of this module for more
information). Solar maximums were discussed in the module The cosmic
engine. Deliberate orbital decay occurs when a manned spacecraft must
be returned to the Earth.
In either eventuality, the spacecraft experiences an ordeal of extreme heat
and forces as it re-enters the Earth’s atmosphere. A graphic example of
this was the re-entry of the world’s second ever satellite, Sputnik 2.
The former Soviet Union had decided to send a living creature up with
their second satellite, which was a follow up of their technical triumph,
Sputnik 1. The creature selected was a small dog, called Laika. She had
been chosen for her ability to remain calm and quiet under chaotic
circumstances, and November 3, 1957, she became the first living
creature from Earth to be launched into space. Unfortunately, Sputnik 2
had no re-entry facility and it wasn’t long before its orbit decayed. Laika
died as the cabin overheated, and she burned up along with the rest of
Sputnik 2 during re-entry.
If the occupants of a spacecraft are to return to the ground safely they
must survive this ordeal of re-entry as well as addressing problems of
re-entry angle, a communication blackout and a survivable landing.
To get some idea of re-entry and it’s problems you may want to watch the
movie Apollo 13 which graphically describes all the problems outlined
above as a drama.
Part 4: Once around the block then home, James
29
The re-entry angle
The angle at which a re-entering spacecraft strikes the atmosphere will
determine much of what follows. NASA regards the ‘entry interface’
(the point at which the atmospheric density becomes sufficient to cause
significant re-entry effects) to be at an altitude of 120 km. The re-entry
angle is measured between this interface and the trajectory of the
spacecraft, as shown in the following diagram.
re-entry angle
e
rfac
y inte
entr
120 km
The re-entry angle is critical.
The angle must be chosen very carefully and is specific to the velocity of
the spacecraft involved. There is a small range of acceptable angles that
will result in a satisfactory re-entry, with destructive consequences
should the actual angle stray either side of this range.
∑
If the angle is too shallow then the spacecraft will bounce off the
atmosphere, being propelled back outward. Even a sparse
atmosphere can present an apparently solid barrier when struck a
glancing blow at 30 000 kmh-1.
∑
If the angle is too steep then the spacecraft will descend too quickly,
causing non-survivable heat and g forces.
As an example of the narrow re-entry ‘corridor’ available to spacecraft,
returning Apollo capsules needed to ensure that they had a re-entry angle
between 5.2∞ and 7.2∞.
30
Space
Heat build-up
You have already read how the decay of an orbit means that a satellite or
spacecraft is losing energy and falling to orbits of lower altitude and
energy. During re-entry this process happens rapidly, so that a great deal
of energy is converted to heat very quickly. When this build-up of heat
happens faster than heat is shed to the surroundings, the spacecraft’s
temperature increases – sometimes to frightening levels – even if the
optimum angle for re-entry is used.
Surviving these extreme temperatures presents a design challenge for the
builders of manned spacecraft. The original research into this question
was done in the USA by the designers of intercontinental ballistic
missiles (ICBM) in the early 1950s. These are rocket missiles carrying
nuclear war heads that can be launched from one continent and then land
in another.
On its journey, a typical ICBM can reach altitudes of 1400 km (that is,
higher than low Earth orbit altitude) and, when re-entering the
atmosphere, would experience temperatures of up to 7500°C. At this
temperature the original design of long, slender, needle-shaped missiles
began to vaporise, which naturally limited their effectiveness.
Shape
After some work it was realised that a blunt shape had distinct
advantages over a streamlined one during re-entry. A blunt shape creates
a shockwave ahead of itself, which absorbs most of the heat generated.
This led to the design of a nose cone for ICBMs – one that would detach
from the rocket while still above the atmosphere and re-enter on its own,
facing upward and presenting its blunt rear to the atmosphere as it fell.
This same design was used for the Mercury, Gemini and Apollo
spacecraft – in fact, the first Mercury rockets from the 1960s were only
slightly modified ICBMs. The space shuttle uses this idea by keeping its
nose high as it re-enters, thereby presenting its flat underbelly as the
blunt surface.
While the blunt shape design is able to stop a spacecraft from vaporising,
the spacecraft is still subjected to temperatures high enough to kill living
occupants and damage valuable cargo.
Part 4: Once around the block then home, James
31
Protective layers
The challenge was to develop a protective layer for the spacecraft, and its
blunt face in particular. There have been two successful solutions to this
problem, with a third approach planned.
∑
Ablation is the name given to the technique of using a sacrificial skin
on the blunt face. This skin was usually made of fibreglass, or other
ceramic material that will vaporise, or ‘ablate’, before the metal
casing of the spacecraft. The process of vaporising this skin used and
dissipated the heat, as well as creating an impressive fireball around
the spacecraft. This technique was used successfully by all of the
Mercury, Gemini and Apollo capsules.
∑
Insulation is the technique successfully employed on the space
shuttle. Its surface is covered with spongiform glass fibre tiles that
are 90% air. Air is a particularly good insulator, and the technique
works well at preventing the heat from reaching the occupants of the
spacecraft. However, the tiles are also absorbent. Their structure
leads them to absorb moisture from the atmosphere during a mission,
thereby adding mass to the spacecraft. To prevent this, the tiles are
waterproofed with a silicone sealant between every flight (the heat of
re-entry burns away the sealant).
∑
The experimental half-scale model X-33, and planned full-scale
shuttle replacement VentureStar, will utilise a low maintenance
special metal alloy skin with a particularly high melting point.
To ensure that the skin and occupants survive, these craft have
greater flying capability than the space shuttle and will descend much
slower, thereby preventing excessive heat build-up.
Photo: © NASA
32
Space
g forces in re-entry
It is normal to think of something accelerating towards the Earth as it
falls, but with a re-entering spacecraft the opposite is true. The
spacecraft is slowing down, that is, decelerating. Consider that the
spacecraft begins from an orbit in which it has a velocity of
approximately 30 000 kmh-1, and that as it penetrates the atmosphere
friction causes it to slow down.
The atmosphere is exerting a force that opposes the spacecraft’s motion,
accelerating it back in the direction from which it came. So, rather than
feeling the force of gravity pulling them downward, re-entering
astronauts experience an upward g force that will usually reach higher
peaks than those experienced during lift-off.
The situation for the astronauts is quite similar during lift-off and
re-entry, especially for capsule designs. In both cases the astronauts lay
reclined back, experiencing an eyeballs-in application of upward
g forces.
The g forces of re-entry are greater than lift-off because a re-entering
spacecraft is decelerated by an atmosphere of increasing density, whereas
a launched rocket is accelerating into atmosphere of decreasing density.
Much of the deceleration of a re-entering spacecraft occurs quite
suddenly at about 100 km altitude when it encounters atmosphere of
sharply increasing density, which in turn means greater heat build-up as
well as greater g forces. By contrast, as a launched rocket accelerates the
atmosphere becomes thinner. The maximum acceleration achieved by a
launched rocket is in the lowest atmospheric densities.
As an example of the difference between the peak g forces of lift-off and
re-entry, recall the flight of Alan Shepard in the rocket called MR3.
During that flight the astronaut experienced a peak lift-off force of 6.3 g,
while during re-entry he had to tolerate peak forces of 11.6 g.
The space shuttle has a unique approach to limiting the g forces of reentry. Having small wings allows the pilots to steer the shuttle’s descent.
The astronauts fly the shuttle down in a series of sharp turns, zigzagging
its way down. Each turn scrubs off speed and extends the flight of the
shuttle, with the result that the space shuttle re-enters in a much slower
and more controlled fashion than a capsule ever could. This idea is being
taken further with the X33/VentureStar program mentioned earlier.
Part 4: Once around the block then home, James
33
Ionisation blackout
An additional problem associated with the peak period of heat build-up
around the spacecraft, is the ‘ionisation black-out’. As the heat
accumulates, atoms in the air around the spacecraft become ionised.
These ions form a layer around the spacecraft than prevents the passage
of radio signals to and from the spacecraft. The signals simply cannot
penetrate the ionisation layer.
The length of the ionisation blackout depends upon the re-entry profile.
The Apollo capsules experienced it for a period of three to four minutes
while the Space shuttle, with its ability to extend its descent, experiences
the black-out for up to sixteen minutes.
If you were able to watch the movie Apollo 13 you might recall the radio
blackout sequence near the end of the film.
Reaching the surface
Having survived the ordeal of heat and g force during re-entry, astronauts
still face the problem of reaching and landing on the ground safely. Even
though the re-entering spacecraft has been slowed continually during its
descent, if allowed to strike the ground it would still impact too severely
for its occupants to survive. As a result, designers have adopted a
number of different solutions to this problem:
34
•
The first solution to this problem came when the USA sent a
chimpanzee called Ham up in a rocket following a sub-orbital
trajectory. This means that he simply went up and down but did not
orbit the Earth (this was a trial run for Alan Shepard’s flight, which
followed soon after). The re-entering capsule containing Ham was
carried down through the lower atmosphere by parachute and the
capsule ‘splashed down’ into the ocean. A naval vessel then
retrieved the capsule. This technique worked so successfully that it
was employed for all manned missions that followed, up to the
development of the space shuttle.
•
The first person in space was Yuri Gagarin on April 12, 1961.
He had been told that during his flight he was a passenger and was
not to tamper with the controls unless there was an emergency
(mission planners were worried that his brain may not function
correctly in a weightless environment). The plan was that his
spacecraft would descend over mainland Russia, rather than over
water. When he reached a specified altitude he was to bail out of the
spacecraft, completing the descent with his own parachute. This
avoided the problem of a rough landing.
Space
During the descent, however, the capsule became much more
unstable than expected, shaking and spinning wildly. Gagarin made
the decision to bail out early, and subjected himself to a hair-raising
skydive from an altitude of 7 km, finally landing in a farm in Siberia.
•
The space shuttle has the capability to fly down to an airfield and
land upon it as an airplane would. This is the ideal solution,
allowing the craft to be serviced and used again, and is the technique
being used for future projects, such as the VentureStar.
You should now attempt Exercises 4.8, 4.9 and 4.10.
In this part you have learned about the nature of circular motion,
centripetal force and the orbital motion of a satellite. You learned about
different types of orbits as well as the slingshot effect. Finally you
learned of the various difficulties that must be overcome in order for a
spacecraft to safely re-enter the Earth’s atmosphere and return to the
ground. In the next part, you will learn of some of the difficulties of
space travel and communication, before moving on to consider the
beginnings of Einstein’s theory of relativity.
Part 4: Once around the block then home, James
35
Summary
∑
Uniform circular motion is circular motion with a uniform orbital
velocity
∑
Centripetal force is the force required to maintain circular motion, an
example of which is the orbital motion of a satellite (in its simplest
case).
r
r
mv 2
Centripetal force, FC =
r
∑
It is gravity that is the centripetal force for a satellite’s orbital motion.
If Newton’s law of universal gravitation is combined with the
centripetal equation, then a formula for orbital velocity results.
r
v=
36
GM
r
∑
The period T of a satellite’s orbit is the time taken for one complete
orbit to be performed. This is related to its orbital velocity.
r 2p r
v=
T
∑
The radius of a satellite’s orbit is related to its period, as shown in the
following equation. This was the basis of Kepler’s law of periods.
r
3
T
2
=
Gm E
4p 2
∑
Low Earth orbit is any orbit above the atmosphere but below the Van
Allen belts. This places it in an altitude range of approximately 250
km to 1000 km high.
∑
A geostationary orbit is an orbit that places a satellite at a fixed
position in the sky. To do this, the period of its orbit must equal the
sidereal period of the Earth’s rotation on its axis. This requires an
altitude of approximately 35 800 km.
∑
A slingshot effect can be provided by planets for space probes,
through the use of the planetary swing-by, or gravity assist
manoeuvre. It is used to gain an increase the speed and a possible
change of direction.
Space
∑
Satellites in low Earth orbit are subjected to friction with the sparse
atmosphere that extends out to these altitudes. This friction slows the
satellite and causes a loss of energy. The satellite is pulled to a lower
orbit, which represents a lower energy level. This process is called
orbital decay.
∑
Orbital decay will eventually cause a satellite or spacecraft to re-enter
the atmosphere. Several issues must be addressed if this process to
be survivable.
∑
The re-entry and must be within an acceptable corridor. Too small
and angle will cause a failure to re-enter. Too great and angle will
cause a too rapid re-entry.
∑
Friction with the atmosphere during re-entry causes extreme heat
build-up. The occupants of re-entering spacecraft have been
protected by giving the spacecraft blunt shapes and specially
developed surfaces.
∑
The g forces generated by re-entry can be much greater than those
experienced during lift-off.
∑
At its maximum, the heat build-up around a re-entering spacecraft
produces a surrounding layer of ionised particles. This layer prevents
radio communication with the spacecraft.
∑
Most re-entering spacecraft are still travelling too fast when they
reach the ground, to ensure the survival of its occupants. Various
strategies have been employed to overcome this problem.
Part 4: Once around the block then home, James
37
38
Space
Suggested answers
Uniform circular motion
30 kmh-1 = (30 ∏ 3.6) ms-1 = 8.3 ms-1
1
Centripetal force,
r
mv 2
FC =
r
25 000 ¥ 8.32
=
35
= 50 000 N
a) r = 6.38 ¥ 106 + 2.5 ¥ 105 = 6.63 ¥ 106 m
2
b) 27 900 kmh-1 = (27 900 ∏ 3.6) ms-1 = 7 750 ms-1
c)
Centripetal force,
r
mv 2
FC =
r
150 ¥ 7 750 2
=
6.63 ¥ 10 6
= 1 360 N
Universal gravitation and orbits
At an altitude of 1000 km:
r
v=
=
Gm E
r
(6.67 ¥ 10 -11 )(5.97 ¥ 10 24 )
(6.38 ¥ 10 6 + 1000 ¥ 10 3 )
= 7 350 m / s = 26 500 km / h
At an altitude of 40 000 km:
Part 4: Once around the block then home, James
39
r
v=
=
Gm E
r
(6.67 ¥ 10 -11 )(5.97 ¥ 10 24 )
(6.38 ¥ 10 6 + 40000 ¥ 10 3 )
= 2 930 ms-1 = 10 550 kmh -1
The period of an orbit
a)
The radius of the orbit
= radius of the Earth + altitude
= 6.38 ¥ 106 + 1 ¥ 106
= 7.38 ¥ 106 m
Therefore
T=
=
2p r
v
2 p (7.38 ¥ 10 6 )
7350
= 6308 s = 105 min
b)
Ê r3 ˆ
Ê r3 ˆ
Á 2 ˜ for low satellite = Á 2 ˜ for high satellite
ËT ¯
ËT ¯
(6380 + 1 000)3 (6380 + 40 000)3
=
1052
T2
2
6
Thigh
satellite = 2.73 ¥ 10
\ Thigh satellite = 2.73 ¥ 10 6 = 1 650 min = 27.6 h
Since this expression relies on ratio comparison, it is not essential that SI
units be used. In this solution, kilometres and minutes were used to
simplify the numbers involved.
Investigating the orbital velocity of a satellite
3
The velocity is reduced.
6
a) The orbital velocity is increased.
b)
8
r µ v2
Orbital velocity is increased.
10 a) Orbital velocity decreases.
b)
40
1
µ v2
r
Space
The slingshot effect
1
10 ms-1 (back towards the thrower)
2
10 + 5 = 15 ms-1
3
15 ms-1
4
15 + 5 = 20 ms-1
5
The ball is thrown at 10 ms-1 but rebounds at 20 ms-1, which is equal
to its initial speed plus twice the speed of the forklift. This is
consistent with the equation for the slingshot effect.
a)
r
r
r
v f = v i + 2Vi
= 27 200 + 2 ¥ 10 600
= 48 400 ms-1
b)
DKE = KE f - KE i
= 1 2 mv 2f - 1 2 mv 2i
= 1 2 ¥ 110 (48 400 2 - 27 200 2 )
= 8.8 ¥ 1010 J
Part 4: Once around the block then home, James
41
42
Space
Exercises – Part 4
Exercises 1.1 to 1.10
Name: _________________________________
Exercise 4.1
Calculate the centripetal force acting on each of the following objects
undergoing uniform circular motion.
Object
Mass
(kg)
Velocity
(ms-1)
ball on a
string
0.150
12.5
car
rounding
bend in
road
1 250
17
low Earth
orbit
satellite
180
7 700
the planet
Uranus
8.7 ¥ 1025
1.06 ¥ 104
Part 4: Once around the block then home, James
Centre of
motion
hand
nothing
Radius
(m)
Centripetal
force (N)
0.65
15
Earth
6 700 000
Sun
4.5 ¥ 1012
43
Exercise 4.2
Calculate the orbital velocity of each of the bodies in the following table.
Data:
Universal gravitation constant, G = 6.67 ¥ 10-11 Nm2 kg-1
mass of the Earth = 5.97 ¥ 1024 kg
mass of Jupiter = 1.90 ¥ 1027 kg
mass of Sun = 1.99 ¥ 1030 kg
Body
Centre of
orbit
Radius of orbit
(km)
low Earth orbit
satellite (minimum
altitude)
Earth
6 630
low Earth orbit
satellite (maximum
altitude)
Earth
7 380
Geostationary
satellite
Earth
42 180
Jupiter
Sun
Io
Jupiter
Orbital velocity
(ms-1)
7.78 ¥ 108
422 000
Exercise 4.3
The highest altitude at which the Space shuttle would normally orbit is
400 km, however it occasionally needs to service the Hubble telescope,
at an altitude of 600 km. Compare the orbital velocity and centripetal
force in each of these two orbits.
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Space
Exercise 4.4
The following information is available about the orbits of Mercury,
Venus and Mars:
Planet
Radius of orbit (AU)
Period of orbit (Earth days)
Mercury
0.39
88
Venus
0.73
225
Mars
1.5
-
Data:
1 astronomical unit (AU) = 1.5 ¥ 109 m
a)
Use the data to show that Kepler’s third law is obeyed for the planets
Mercury and Venus.
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b) Use the data to find the speed of Mars.
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Exercise 4.5
a)
Define low Earth and geostationary orbits.
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b) Compare the types of satellites that use each of these types of orbits.
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Part 4: Once around the block then home, James
45
c)
Complete the following comparison table.
low Earth orbit
lower limit
upper limit
geostationary
orbit
altitude (km)
period (h)
orbital velocity (ms-1)
Exercise 4.6
Consider that a spacecraft launched from the Earth is still in orbital
motion around the Sun, along with the Earth.
a)
Explain why a spacecraft intended to travel toward the Sun must first
accelerate back along the Earth’s orbital path.
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b) What would happen to the spacecraft if it simply accelerated directly
at the Sun?
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Exercise 4.7
a)
Describe how a planet can provide a slingshot effect for a space
probe. In your answer, refer to the momentum and energy of the
probe and the planet.
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Space
b) What benefits are to be gained from this type of manoeuvre?
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c)
What is the maximum increase in velocity that can be achieved by a
100 kg space probe travelling at 12 000 ms-1, in a gravity assist
manoeuvre around the planet Mars, which has an orbital velocity of
24 200 ms-1?
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Exercise 4.8
Describe the process of orbital decay. In your answer refer to the energy
of the satellite or spacecraft involved, as well as the forces causing the
change of orbit.
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Exercise 4.9
Two of the most harmful difficulties of re-entry through the Earth’s
atmosphere is the enormous heat build-up that occurs as well as the large
g forces produced. Both of these problems reach their peak magnitudes
at the same time.
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47
a)
Explain the close connection between these two problems.
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b) A rocket launch involves accelerating the rocket from a standstill to
orbital velocity at high altitude. A re-entry involves decelerating
from the same orbital velocity to a standstill through the same
atmosphere. Why should the problems of heat and g force be so
much greater for the re-entry than they are for the launch?
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Exercise 4.10
Several of the factors that must be overcome by a spacecraft attempting
to re-enter the Earth’s atmosphere are:
∑
angle of re-entry
∑
heat build-up
∑
high g forces
∑
ionisation black-out
∑
landing on the ground.
For each factor, discuss the consequences or possible effects upon the
spacecraft and occupants, as well as the means employed to minimise
these effects.
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Space
Appendix
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Space
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