PHYSICS
TUTORIAL NOTES
Space
SYLLABUS
TOPIC 9.2
These notes are meant as a
guide only and are designed to focus your thoughts on the dot points mentioned
in the syllabus. They give a very brief overview of the topic and should be
used in conjunction with your class notes, your textbooks and your research
from other areas such as the library and the Internet.
Notes compiled by:
CARESA EDUCATION SERVICES
SPACE
1. “The Earth has a gravitational field that exerts a force on objects both on and around it. ” (syllabus)
Students learn to “define weight as the force on an object due to a gravitational field. ” (syllabus)
Mass is the amount of matter in a body whereas weight is the
force due to gravity acting on a mass. Mass will not change where the
acceleration due to gravity is different but the weight will change.
W = mg where W = weight, m = mass, g = acceleration due to gravity.
Astronauts in space appear weightless even though gravity is acting on them. At the height of a typical space station gravity is about a third of its value on Earth. The astronauts appear weightless because they are falling around the Earth with the same acceleration as their space station. You could get the same situation on Earth if you were in a lift when the cable broke.
True weight is equal to mg for whatever value g has at that location. Apparent weight is equal to the reaction force exerted on the object and is equal to mg + ma where a is the upward acceleration of the object. ( For an object accelerating down, a is negative)
Students learn to “explain that a change in gravitational potential energy is related to work done. ” (syllabus)
When an object is lifted in a gravitational field the work done is equal to the increase in gravitational potential energy. If an object falls then the gravitational field does work on the object. The amount of work done by the gravitational field is equal to the amount of work required to restore the object to its original position.
Students learn to “define gravitational potential energy
as the work done to move an object from a very large distance away to a point
in a gravitational field. E_{p} = G m_{1}m_{2}/r ”
(syllabus)
Gravitational potential energy is the work required to bring an object from infinity to that point. Since gravity does work on the object the potential energy is negative.
E_{p} = Gm_{1}m_{2} / r where E_{p} = potential energy, G = universal gravitational constant, m_{1} and m_{2} are the masses of the two objects and r = distance between the centres of mass of the two objects.
Students: “Perform an investigation and gather
information to determine a value for acceleration due to gravity using pendulum
motion or computerassisted technology and identify reason for possible
variations from the value 9.8 ms^{2}. ” (syllabus)
A Java applet demonstrating the pendulum experiment:
http://www.phy.ntnu.edu.tw/java/pendulum30/pendulum.html
http://www.phy.ntnu.edu.tw/java/Pendulum/Pendulum.html
http://www.walterfendt.de/ph14e/pendulum.htm
Suspend a pendulum bob from a fixed support by means of a light thread. Adjust the length of the thread so that the distance from the support to the centre of mass of the bob is 50 cm. Draw the bob a little (less than 5^{o}) to the side and release it. Use a stopwatch to time 10 complete to and fro oscillations of the bob. Repeat the procedure twice more to obtain a total of three readings. Determine the average period by dividing the average time for 10 swings by 10.Increase the length of the pendulum to 60 cm and again take three readings of the time taken for ten swings and deduce the average period. Do the same for pendulum lengths of 70 cm, 80 cm and 90 cm. Plot a graph of l (vertical axis) against T^{2}. Determine the gradient of the graph by drawing the line of best fit and multiply by 4p^{2}. This will give the value of “g”. l = T^{2}g / 4p^{2}.
The value could differ from the accepted value of 9.8 ms^{2} due to air resistance, friction between the string and support, limitations in the readings of the ruler and stopwatch and random errors in measuring the time.
A site dealing with the pendulum experiment is:
http://theory.uwinnipeg.ca/physics/shm/node5.html
Students: “Gather secondary information to predict the
value of acceleration due to gravity on other planets. ” (syllabus)
The weight of an object on the surface of a planet is equal to mg where g is the acceleration due to gravity on the planet. It is also equal to the force of attraction between the object and the planet and this is given by F = Gmm_{p}/r_{p}^{2} where m_{p} and r_{p} are the mass and radius of the planet respectively and G is the universal gravitational constant. Equating we get mg = Gmm_{p}/r_{p}^{2}
or g = Gm_{p}/r_{p}^{2}
Example: Given that the Universal gravitational constant , G = 6.67 x 10^{11} Nm^{2}kg^{2} and the mass of the Earth is 6.0 x 10^{24} kg and its radius is 6.38 x 10^{6} m, calculate the acceleration due to gravity on the Earth.
Solution: g = Gm_{E}/r_{E}^{2} = 6.67 x 10^{11} x 6.0 x 10^{24}/(6.38 x 10^{6})^{2} = 9.83 ms^{2}
Exercise: The mass and radius of some planets are given in the table below. Calculate the acceleration due to gravity on these planets and complete the table.
Universal Gravitational Constant, G = 6.67 x 10^{11} Nm^{2}kg^{2}
Planet 
Mass (kg) 
Radius (m) 
“g” (ms^{2}) 
Mercury 
3.35 x 10^{23} 
2.4 x 10^{6} 

Venus 
4.9 x 10^{24} 
5.6 x 10^{6} 

Mars 
6.44 x 10^{23} 
3.4 x 10^{6} 

Jupiter 
1.9 x 10^{27} 
7.2 x 10^{7} 

Saturn 
5.7 x 10^{26} 
6.0 x 10^{7} 

Uranus 
8.7 x 10^{25} 
2.6 x 10^{7} 


1.0 x 10^{26} 
2.5 x 10^{7} 

The Moon 
7.4 x 10^{22} 
1.75 x 10^{6} 

Values of “g” (ms^{2}): Mercury 3.88; Venus 10.42;
Mars 3.72; Jupiter 24.45; Saturn 10.56; Uranus 8.58;
Students: “Analyse information using the expression: F =
mg to determine the weight force for a body on Earth and for the same body on
other planets. ” (syllabus)
Determine the weight of 1 kg of potatoes, a 60 kg person and a 2 tonne elephant on each of the planets in the table below. Remember that the force is measures in newtons.
Planet 
“g” (ms^{2}) 
1 kg potatoes 
60 kg person 
2 tonne elephant 
Earth 




Mercury 




Venus 




Mars 




Jupiter 




Saturn 




Uranus 









The Moon 




Check your answers.
Planet 
“g” (ms^{2}) 
1 kg potatoes (N) 
60 kg person (N) 
2 tonne elephant (N) 
Earth 
9.8 
9.8 
588 
19600 
Mercury 
3.88 
3.88 
232.8 
7760 
Venus 
10.42 
10.42 
625.2 
20840 
Mars 
3.72 
3.72 
223.2 
7440 
Jupiter 
24.45 
24.45 
1467 
48900 
Saturn 
10.56 
10.56 
633.6 
21120 
Uranus 
8.58 
8.58 
514.8 
17160 

10.67 
10.67 
640.2 
21340 
The Moon 
1.61 
1.61 
96.6 
3220 
2. “Many
factors have to be taken into account to achieve a successful rocket launch,
maintain a stable orbit and return to Earth. ” (syllabus)
Students learn to “describe the trajectory of an object
undergoing projectile motion within the Earth’s gravitational field in terms of
horizontal and vertical components. ” (syllabus)
PROJECTILE MOTION IN TWO DIMENSIONS
The only
force acting on an object once it has been thrown or fired is that due to
gravity. There are no horizontal forces acting (assuming we ignore air
resistance) and hence there is no horizontal acceleration. Gravity acts
vertically downwards and the vertical motion of the object is governed by the
equations of motion.
v_{y} = final vertical velocity
a_{y} = vertical acceleration
When approaching
problems on projectile motion the first task is to find the initial vertical
and horizontal components of the motion. If an object is fired at an angle q to the horizontal with velocity v then the
vertical component of the velocity v_{y} = v sin q and the horizontal component v_{x}
= v cos q
Since there is no horizontal acceleration the horizontal distance travelled is given by Dx = v_{x}t
The horizontal distance travelled until it hits the ground is called the “range” so another term for Dx is the range of the projectile.
The time of
flight is calculated by using the equations of motion listed above and applying
them to the vertical motion of the object.
If you want
to find the velocity at which a projectile hits the ground, find the vertical
component of its velocity as it hits the ground by applying the equations of
motion to its initial vertical velocity and add this vectorially to the
horizontal component of its velocity (triangle of vectors). This will give both
the speed and direction of the velocity.
^{ }
Students learn to “describe Galileo’s analysis of
projectile motion. ” (syllabus)
Galileo arranged to have objects dropped from the crows nest of moving boats and noted the path followed by the objects. He found that relative to the mast, the objects fell vertically downwards while relative to a person on the shore the objects would travel in a parabolic path.
Students learn to “explain the concept of escape velocity
in terms of the:

gravitational constant

mass and radius of the planet. ” (syllabus)
An object will escape from the Earth’s gravitational field if it has a positive amount of energy i.e. E_{k} + E_{p} >0 ½ mv^{2} – Gmm_{E}/r_{E} > 0 v^{2} > 2Gm_{E} /r_{E}
As can be seen from the equation, the escape velocity is proportional to the square root of the gravitational constant G. If G had a larger value then the escape velocity would be larger and if G was smaller the escape velocity would be less. The value of G was determined experimentally by Henry Cavendish in 1798 and is believed to be constant throughout the Universe.
As can also be seen from the equation, the escape velocity is also proportional to the square root of the mass and is inversely proportional to the square root of the radius. More massive planets have a higher escape velocity than less massive ones while planets of small radius have a higher escape velocity than large radius planets.
Exercise: Show that the escape velocity of Earth is around 11.2 kms^{1}
v_{escape} = \/(2 x 6.67 x 10^{11} x 6 x 10^{24} / 6.38 x 10^{6})
= 11200.6 ms^{1} = 11.2 kms^{1}
Students learn to “outline
Gravity would supply the centripetal force so that mg = mv^{2}/r i.e. g = v^{2}/r or v^{2} = rg
At sea level v^{2} = 6.38 x 10^{6} x 9.8 v = 7900 ms^{1}
At higher altitudes g would be less so that the orbital velocity would be less.
http://galileo.phys.virginia.edu/classes/109N/more_stuff/Applets/newt/newtmtn.html
Students learn to “identify why the term ‘g forces’ is
used to explain the forces acting on an astronaut during launch. ” (syllabus)
“g” forces are multiples of an astronaut’s apparent weight due to acceleration. If the astronaut is accelerating upwards at 9.8 ms^{2} then his/her apparent weight doubles so they are subjected to a force of 2g. Under normal conditions a force of 1g is acting, while a person in free fall is subjected to zero g forces.
Example:
What “g” forces act on an astronaut
(a) Before the rocket is launched?
(1g)
(b) When the rocket is accelerating upwards at 19.6 ms^{2} just after launching?
(3g)
(c) When the rocket is accelerating upwards at 19.6 ms^{2} at a height where the Earth’s gravity has a value of 4.9 ms^{2}?
(2.5g)
(d) When the rocket is orbiting the Earth in a circular orbit?
(0g)
(e)
Just before the rocket lands when it is slowing
down with retardation 4.9 ms^{2}? (1.5g)
Students learn to “discuss the effect of the Earth’s
orbital motion and its rotational motion on the launch of a rocket. ”
(syllabus)
The Earth rotates towards the east with a period of 24 hours. Any rocket launched from Earth would have a vector velocity consisting of this easterly component added to its vertical velocity according to the triangle of vectors.
For a satellite to orbit the Earth v^{2} = rg where r is the radius of orbit (from the centre of the Earth) and g is the acceleration due to gravity at the height of the satellite (not 9.8 ms^{2}). To orbit the Earth the satellite has to be given a tangential velocity at the required height. Since the satellite already has an easterly component to its velocity from launching then less speed has to be added if the satellite is accelerated in an easterly direction. This means that less fuel is used so that the launching rocket can carry a bigger payload. For this reason most satellites are launched towards the east.
The Earth’s orbital motion around the Sun does not matter for a satellite orbiting the Earth since it remains within the Earth’s gravitational field. However if a rocket is launched to the Moon or to another planet it has to first escape the Earth’s gravitational field. To do this most effectively it has to combine the easterly component of the Earth’s rotation with its orbital velocity around the Sun. Such rockets are launched towards the east and in the direction of the Earth’s orbit around the Sun.
Students learn to “analyse the changing acceleration of a
rocket during launch in terms of the:

Law of Conservation of Momentum

Forces experienced by astronauts. ”
(syllabus)
A rocket that is burning fuel at a constant rate is
expelling gases at a constant rate and is exerting a constant force on the
rocket and the fuel that it contains. However the mass of the rocket and fuel
decreases as the fuel is used up. Since F = ma, as m decreases a increases.
Consequently the acceleration of the rocket increases as the fuel is used up.
The rocket accelerates due to the conservation of momentum. Exhaust gases of
mass m are expelled with a velocity v. The change of momentum is mv. Since
momentum is conserved the rocket and remaining fuel will experience an equal
change in momentum in the opposite direction (
Astronauts can withstand forces up to about 10g for short periods and up to about 3g for sustained periods. Consequently there is a limit to the acceleration of the rocket during launching. The astronaut would experience a force of 10g when the rocket was accelerating at 88.2 ms^{2} (10 g = 1g weight + 9g acceleration). As the rocket rose the acceleration could be slightly more as the astronauts weight decreased due to the decrease in the Earth’s gravitational field.
Students learn to “analyse the forces involved in uniform
circular motion for a range of objects, including satellites orbiting the
Earth. ” (syllabus)
According to
F_{c }= mv^{2}/r Where F_{c} = centripetal force, m = mass, v = speed and r = radius
For a satellite orbiting the Earth the centripetal force is supplied by gravity, for an object being swung around on a string the tension in the string supplies the centripetal force and for a car turning a corner, the friction between the tyres and the road supply the centripetal force.
Students learn to “compare qualitatively low Earth and
geostationary orbits. ” (syllabus)
A satellite in a geostationary orbit remains above the same point on the Earth’s surface. It has a period of 24 hours and travels towards the east; the same direction as the Earth’s rotation. Since it also orbits the Earth’s centre of gravity it follows a path directly above the equator. A low Earth orbit is closer to the Earth’s surface and has a shorter period.
Students learn to “define the term orbital velocity and
the quantitative and qualitative relationship between orbital velocity, the
gravitational constant, mass of the satellite and the radius of the orbit using
Kepler’s Law of Periods. ” (syllabus)
Orbital velocity refers to the speed of an object in uniform circular motion about a central point.
For a satellite in uniform circular motion around the Earth the centripetal force is supplied by the gravitational force of attraction; i.e. mv^{2}/r = Gmm_{E} /r^{2}
Which becomes v^{2} = Gm_{E} /r
This shows the relationship between the orbital velocity, gravitational constant and radius of orbit. Note that the orbital speed is independent of the mass of the satellite but is proportional to the square root of the mass of the Earth
The orbital velocity can be related to the period by;
speed = distance/time
orbital velocity = circumference / period
v = 2pr /T v^{2} = 4p^{2}r^{2} / T^{2} = Gm_{E} /r
T^{2} = 4p^{2} r^{3 }/G m_{E}
This is in the form of Kepler’s Law of Periods that stated that the square of the period is proportional to the cube of the mean distance from the central body.
Note that this assumes that the mass of the satellite is negligible compared to the central body. This is true for satellites orbiting the Earth or for planets orbiting the Sun but for systems where the orbiting mass is significant such as binary stars, then the equation becomes T^{2} = (4p^{2}/G{m_{1} + m_{2}}) r^{3}
A good applet dealing with Kepler’s Laws but possibly a little off this topic:
http://www.physics.nwu.edu/ugrad/vpl/mechanics/planets.html
Students learn to “account for the orbital decay of
satellites in low Earth orbit. ” (syllabus)
There are still minute amounts of atmosphere at the height of a low Earth orbit so these satellites are subjected to atmospheric drag. Over a period of a few years they will suffer orbital decay whereby their speed gradually decreases due to the friction with the lowdensity atmosphere and they gradually spiral back towards the Earth’s surface.
Students learn to “discuss issues associated with safe
reentry into the Earth’s atmosphere and landing on the Earth’s surface. ”
(syllabus)
As the spacecraft reenters the Earth’s atmosphere it has to decelerate at a rate that will not cause harm to the occupants and also not produce sufficient friction to heat the spacecraft to dangerous levels. This is achieved largely by the spacecraft reentering the atmosphere at the correct angle. Even with the correct angle of reentry there is still a lot of heat generated. This makes the choice of the correct insulating tiles critical. Also the heat generated causes the atoms in the air surrounding the spacecraft to become ionised and prevent communication, as radio waves cannot penetrate the layer of ionised atoms.
Students learn to “identify that there is an optimum angle for safe reentry for a manned spacecraft into the Earth’s atmosphere and the consequences of failing to achieve this angle. ” (syllabus)
The optimum angle for a spacecraft’s reentry into the Earth’s atmosphere is
6.2^{o} +/ 1^{o}
If the spacecraft enters too steeply it will decelerate at too great a rate and produce forces that are dangerous to the occupants. Also, because of the greater friction, the heat produced will be greater and this can be fatal to the occupants. If a spacecraft reenters the Earth’s atmosphere at an angle less than the optimum angle it will bounce off the atmosphere and go into a large elliptical orbit.
Students: “Solve problems and analyse information to
calculate the actual velocity of a projectile from its horizontal and vertical
components using:
1. Calculate the vertical and horizontal components of the velocity for an object traveling at
(i)
60
ms^{1} at 30^{o} to the horizontal
(ii)
30
ms^{1} at 45^{o} to the horizontal
(iii)
40
ms^{1} at 40^{o} to the vertical
Calculate the
(i)
time
of flight for each object
(ii)
horizontal
distance traveled by each object
(iii)
distance
apart that the two objects will land
(i)
How
high did the ball go?
(ii)
How
far did the ball go?
Calculate the
(i)
vertical
component of the initial velocity
(ii)
horizontal
component of the initial velocity
(iii)
maximum
height reached
(iv)
time
to reach maximum height
(v)
velocity
at maximum height
(vi)
time
to fall from maximum height
(vii)
total
time of flight
(viii)
range
(ix)
speed
with which the bag hit the ground
(x)
angle
at which the bag hit the ground
ANSWERS:
1. (i) v_{y}
= 30 ms^{1} v_{x} = 52
ms^{1} (ii) v_{y} = 21.2 ms^{1} v_{x} = 21.2 ms^{1}
(iii) v_{y} = 30.6 ms^{1} v_{x} = 25.7 ms^{1}
2. (i) 5.0 s for each (ii) 20m, 70m
(iii) 50m
3. 59.3 ms^{1} at 82^{o}15’
to horizontal
4. (i) 8.43m (ii) 40.1m
5. (i) 5.2 ms^{1} (ii) 3 ms^{1}
(iii) 13.34m above ground (iv) 0.53s
(v) 3 ms^{1} horizontally. (vi) 1.65s
(vii) 2.18s (viii) 6.54m (ix) 16.45ms^{1} (x) 79^{o}29’ to
horizontal.
6. No. Height at fence = 1.865m
7. 15.92 ms^{1}
8. 23.8 ms^{1}
Students: “perform a firsthand investigation, gather
information and analyse data to calculate the initial and final velocity,
maximum height reached, range and time of flight of a projectile for a range of
situations by using simulations, data loggers and computer analysis. ”
(syllabus)
Try this computer simulation. It’s good fun:
http://jersey.uoregon.edu/vlab/Cannon/index.html
Projectile
Motion Experiment
Richard and Amanda used a video camera to study projectile motion. They drew a grid on a blackboard and threw a ball in front of the grid while filming with the video camera. They had researched T.V. operation and found that the picture is formed from two scans of an electron gun that scans at 50 Hz. i.e. 25 frames per second.
They played the video and paused at each frame to determine the position of the ball at 1/25 second intervals. They then plotted the vertical and horizontal displacements of the ball at 1/25 second intervals on a graph and a copy of that graph is reproduced below.
Graph: Vertical v^{s} horizontal displacement
Questions
1. What is the time of flight of the ball?
1 second
2. What feature of the graph shows that the horizontal component of the velocity is constant?
The points on the graph have equal horizontal spacings.
3. Calculate the horizontal component of the velocity.
v_{x} = s_{x}/t = 2.5/1 = 2.5 ms^{1}
4. What feature of the graph shows that the vertical component of the motion is accelerating?
The vertical spacings on the graph decrease until the ball reaches maximum height (deceleration) then increase.
5. Calculate the initial vertical component of the velocity.
Max. height = 2.6m (from graph) v_{y}^{2} = u_{y}^{2} + 2a_{y}s_{y}
0 = u_{y}^{2} + 2(9.8) 2.6 u_{y}^{2} = 50.96 u_{y} = 7.14 ms^{1}
6. Calculate the speed at the beginning of its motion past the grid.
u^{2} = u_{x}^{2} + u_{y}^{2} = 2.5^{2} + 7.14^{2} u = 7.56 ms^{1}
7. Calculate the angle at which the ball was travelling at the beginning of its motion past the grid.
Tan q = 7.14/2.5 q = 70^{o}42’
8. What feature of the graph shows that the vertical component of the velocity is zero at maximum height?
The direction of the motion is given by a tangent to the curve at that point. At maximum height the tangent is horizontal indicating that the vertical component of the velocity is zero.
9.What is the acceleration of the ball at maximum height?
9.8 ms^{2} vertically down
10. What is the direction of the acceleration of the ball when its horizontal displacement is 1.0 m?
Vertically down (the acceleration is g throughout)
11. What would happen to the time of flight if the ball was thrown at a steeper (more vertical) angle at the same speed?
The time of flight would increase.
12. How would the motion of the ball be different if it was thrown with the same speed and at the same angle on the Moon where gravity is less than it is on Earth?
The maximum height would be greater, the time of flight would be greater and the range would be greater.
Students: “Identify data sources, gather, analyse and
present information on the contribution of one of the following to the
development of space exploration: Tsiolkovsky, Oberth, Goddard,
EsnaultPelterie, O’Neill or von Braun. ” (syllabus)
(i) Konstantin Tsiolkovski: Russian scientist – In 1883 he demonstrated the reaction principle and proposed that is feasible for rockets to operate in the vacuum of space. In 1903 he designed a threestage rocket that used liquid hydrogen and liquid oxygen as a fuel. He was a theoretical scientist and never actually built a rocket.
http://www.hq.nasa.gov/office/pao/History/sputnik/kon.html
http://www.informatics.org/museum/tsiol.html
http://www.spaceline.org/history/21.html
(ii) Herman Oberth: Romanian born German. Theoretical scientist. Wrote doctorial thesis “ By Rocketry to Space” and also the book “ The Road to Space Travel”
http://www.100yearsofflight.com/history/oberth.html
http://www.astronautix.com/astros/oberth.htm
http://inventors.about.com/library/inventors/blrocketOberth.htm
http://spaceline.org/history/25.html
(iii) Robert
Goddard:
http://inventors.about.com/library/inventors/blgoddard.htm
http://spaceline.org/history/22.html
(iv) Roberts EsnaultPelterie: Wrote Astronautics and Astronautics Complement. Suggested that rockets could be used for long range ballistic missiles and employed by the French army to develop these missiles.
(v) Wernher von Braun: German rocket scientist
who helped develop the V1 and V2 rocket bombs which were used against
http://inventors.about.com/library/inventors/blvonBraun1.htm
Gerard K O’Neill (1927 – 1992)
In 1965 he began teaching the introductory
physics course at
Gerard O'Neill (1969) From UNSW notes
Students: “Solve problems and analyse information to
calculate centripetal force acting on a satellite undergoing uniform circular
motion about the Earth using: F = mv^{2 }/r ” (syllabus)
Exercise:
A satellite of mass 800 kg is orbiting the Earth at a distance of 10 000 km from its centre and travelling at 6326 ms^{1}.
(i) Calculate the centripetal force acting on the satellite.
(ii) Calculate the value of “g” at the height of the satellite.
Answer:
(i) F = mv^{2}/r = 800 x 6326^{2} / 10^{7} = 3201 N
(ii) g = F/m = 3201/800 = 4.0 ms^{2}
Exercise: A space station of mass 10 tonnes is orbiting the Earth at a height of 8.07 x 10^{6}m above its centre and has a period of 2.0 hours.
(i) Calculate the orbital speed of the space station.
(ii) Calculate the centripetal force acting on the space station.
(iii) Calculate the value of “g” at the height of the space station.
(iv) Since the value of “g” is not zero, explain how the astronauts inside the space station can appear weightless.
Answer:
(i) v = 2pr/T = 2p 8.07x10^{6}/ 7200 = 7042 ms^{1}
(ii) F = mv^{2}/r = 10000 x 7042^{2} / 8.07 x 10^{6} = 61457 N
(iii) g = F/m = 61457/10000 = 6.15 ms^{2}
(iv) The space station is falling around the Earth and the astronauts are falling at the same rate as the space station. Since there is no reaction force exerted by the space station on the astronauts they appear to be weightless.
Students: “Solve problems and analyse information using:
r^{3} / T^{2} = GM / 4p^{2}
” (syllabus)
1. Consider a satellite of mass m travelling at speed v in uniform circular motion about the Earth at a distance r from the Earth’s centre.
(i) What is the centripetal force acting on the satellite?
F_{cent. }= mv^{2} / r
(ii) What is the gravitational force acting on the satellite?
F_{grav.} = G mM / r^{2}
(iii) What is the period of the satellite in terms of v and r?
T = 2pr / v
(iv) Rearrange the above formulae to show:
r^{3} / T^{2} = GM / 4p^{2}
The gravitational force supplies the centripetal force so they are equal.
mv^{2} / r = GmM / r^{2} v^{2} = GM / r
but v = 2pr / T so 4pr^{2} / T^{2} = GM / r
r^{3} / T^{2} = GM / 4p^{2}
Answer:
Radius of orbit = radius of Earth + height = 6380km + 3620km = 10 000 km
= 10 000 000m
T^{2} = 4p^{2}r^{3}/ Gm = 4p^{2} (10^{7})^{ 3} / (6.67 x 10^{11} x 6 x 10^{24}) = 9.864 x 10^{7}
T = 9932s =165 min =2hr 45 min
Students learn to “describe a gravitational field in the
region surrounding a massive object in terms of its effects on other masses in
it. ” (syllabus)
A gravitational field is a region where a mass experiences a force. It occurs in a region surrounding a mass and means that there is a mutual force of attraction between any two masses. From a relativistic point of view gravity can be described as the warping of space around a mass but this treatment is beyond the scope of the HSC course.
Students learn to “define
Every object in the Universe attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres of mass.
F a m_{1}m_{2}/d^{2}
While
F = Gm_{1}m_{2}/d^{2}
Students learn to “discuss the importance of
Satellites are kept
in circular motion around the Earth by the gravitational attraction between the
satellite and Earth that provides the centripetal force necessary to constantly
change the satellite’s direction.
mv^{2} / r = Gmm_{E} / r^{2}
Students learn to “identify that a slingshot effect can
be provided by planets for space probes. ” (syllabus)
As a spacecraft approaches a planet it is captured by the gravity of the planet. The spacecraft attains the orbital speed of the planet around the Sun as some of the momentum of the planet is transferred to the spacecraft. As the mass of the planet is huge compared to the spacecraft, any slowing down of the planet is undetectable. As the spacecraft swings around the planet and returns to space it retains the speed it had perpendicular to the path of the planet’s orbital rotation. However it has also attained the velocity of the planets orbital rotation. Vector addition of these two components results in a greatly increased speed.
Students: “Present information and use available evidence
to discuss the factors affecting the strength of the gravitational force. ”
(syllabus)
According to
The greater the masses the greater the gravitational force.
The smaller the distance between the objects the greater the gravitational force.
The gravitational force is also proportional to the Universal gravitational constant. If the constant had a higher value then two objects would have a greater force of attraction between them.
Students: “Solve problems and analyse information using:
F = Gm_{1}m_{2}/d^{2} ” (syllabus)
The mass of the Earth is 6 x 10^{24 }kg while that of the Sun is 2 x 10^{30} kg. The distance between the Earth and the Sun is 1.5 x 10^{11} m. Calculate the gravitational force of attraction between the Earth and the Sun.
F = Gm_{1}m_{2}/d^{2} = 6.67 x 10^{11} x 2 x 10^{30} x 6 x 10^{24 }/ (1.5 x 10^{11})^{2} = 3.6 x 10^{22} N
Calculate the force of attraction between two 60 kg people standing 1.0m apart.
F = Gm_{1}m_{2}/d^{2} = 6.67 x 10^{11} x 60 x 60^{ }/ 1^{2} = 2.4 x 10^{7} N.
Students learn to “outline the features of the aether
model for the transmission of light. ” (syllabus)
Experiments by Young, Fresnel and others indicated that light is a wave. Scientists of the time thought that a medium was necessary to transmit waves so the aether was proposed as the medium that carried waves through space. The aether was thought to be transparent and fill all space. It permeated (went through) all matter and was permeable to all matter (all matter could go through it). It had high elasticity to facilitate the propagation of light waves.
Students learn to “describe and evaluate the
MichelsonMorley attempt to measure the relative velocity of the Earth through
the aether. ” (syllabus)
Michelson & Morley attempted to show that the speed of light was different when travelling parallel to the aether compared to when it was travelling perpendicular to the aether.
They shone monochromatic light onto a half silvered mirror, inclined at 45^{o} to the light, so that about half of the light passed through the mirror, and the other half was reflected at 90^{o} to the first. The two beams travelled approximately the same distance and were reflected back by plane mirrors at right angles to the beams. The two beams were recombined and formed an interference pattern of alternating bright and dark bands.
Michelson & Morley rotated their apparatus, expecting to see changes in the interference pattern as the beam that was originally travelling parallel to the aether was now travelling perpendicular to the aether and viceversa. No such change was detected, which indicated that the aether, if it existed, had no effect on the speed of light, and in all probability did not exist.
Students learn to “discuss the role of the
MichelsonMorley experiments in making determinations about competing theories.
” (syllabus)
Competing
theories of relativity illustrate the contribution.
Galilean relativity states that motion is relative and cannot be detected without reference to an external point, If the aether was stationary in space then the speed of light parallel to the aether should be different from the speed of light perpendicular to it.
By showing that the aether was superfluous to the speed of light it was possible for Einstein to state that the speed of light is constant and is independent of the motion of the source or observer. This led to his special and general theories of relativity.
Students learn to “outline the nature of inertial frames
of reference. ” (syllabus)
An inertial frame of reference is a nonaccelerating frame
of reference and
Students learn to “discuss the principal of relativity. ”
(syllabus)
On the basis of his experiments with falling bodies, Galileo
proposed that the laws of physics are the same whether the body is at rest or
in uniform motion.
Students learn to “describe the significance of
Einstein’s assumption of the constancy of the speed of light. ” (syllabus)
It explains the null result of the MichelsonMorley experiment and did away with the need for an aether. It also paved the way for Einstein’s special theory of relativity.
Students learn to “identify that if c is constant then
space and time become relative. ” (syllabus)
Observers of motion near the speed of light must observe different lengths and different times such that the speed of light = length/time = constant.
Students learn to “discuss the concept that length
standards are defined in terms of time in contrast to the original metre
standard. ” (syllabus)
The metre was originally defined as 1/ 10 000 000 the
distance between the equator and North Pole on a line through
Students learn to “explain qualitatively and quantitatively
the consequence of special relativity in relation to:

the relativity of simultaneity

the equivalence of mass and energy

length contraction

time dilation

mass dilation ” (syllabus)
The relativity of simultaneity
Events that appear simultaneous to one observer will not necessarily appear simultaneous to another observer.
The equivalence of mass and energy
Mass and energy are interrelated; mass can be converted to energy and energy can be converted to mass.
Exercise:
How much energy can be obtained from the complete conversion of one gram of matter?
E = mc^{2}
= 0.001 x (3x10^{8})^{2}
= 9.0 x 10^{13}J
Length contraction
The length of a moving object becomes shorter relative to a stationary observer.
Time dilation.
The passage of time on a moving object passes more slowly to a stationary observer outside the object.
Mass dilation
The mass of a moving object becomes greater, relative to a stationary observer.
Students learn to “discuss the implications of mass
increase, time dilation and length contraction for space travel. ” (syllabus)
Time dilation and length contraction mean that it may be possible in the future to reach distant stars by travelling at speeds close to that of light. While many years would have passed relative to an observer on Earth, the occupants of the spaceship will have aged only slightly because in their frame of reference, stellar distances become shorter and time travels more slowly.
Students: “gather and process information to interpret
the results of the MichelsonMorley experiment. ” (syllabus)
Michelson & Morley rotated their apparatus, expecting to see changes in the interference pattern as the beam that was originally travelling parallel to the aether was now travelling perpendicular to the aether and viceversa. No such change was detected, which indicated that the aether, if it existed, had no effect on the speed of light, and in all probability did not exist.
Students: “perform an investigation to help distinguish
between noninertial and inertial frames of reference. ” (syllabus)
An inertial frame of reference can be detected by using a pendulum i.e. a mass on the end of a light string. Hold the pendulum. If it hangs vertically then it is in an inertial frame of reference. If it hangs at an angle to the vertical then the frame of reference is accelerating and is noninertial.
Students: “analyse and interpret some of Einstein’s
thought experiments involving mirrors and trains and discuss the relationship
between thought and reality. ” (syllabus)
Suppose A & B are equidistant from observer O on a train moving with velocity v. Suppose also that A & B have mirrors so that they reflect a flash of light made by observer O. Observer O will see the flash reflected simultaneously from A & B so that from O’s frame of reference the two reflections are simultaneous.
For an outside observer O`, because the train has moved forward in the time it takes the light to reach A & B, the reflection from A will travel a shorter distance than the reflection from B so from O`’s frame of reference the events are not simultaneous; A’s reflection will occur before B’s.
Students: “analyse information to discuss the
relationship between theory and the evidence supporting it, using Einstein’s
predictions based on relativity that were made many years before evidence was
available to support it. ” (syllabus)
Einstein predicted the bending of light as it passed close to massive objects. The bending of light around the Sun was demonstrated during an eclipse of the Sun in 1919. Light from stars behind the Sun was bent so that the stars were visible even though they should have been obscured.
Students: “Solve problems and analyse information using:
E = mc^{2}
l_{v}
= l_{0} \/(1 – v^{2}/c^{2})
t_{v}
= t_{0} /{\/(1 – v^{2}/c^{2})}
m_{v}
= m_{0} /{\/(1 – v^{2}/c^{2})} ” (syllabus)
1. The power output of the Sun is around 3.9 x 10^{26} watts. How much mass is converted to energy each second in the Sun?
Power output = 3.9 x 10^{26} watts =3.9 x 10^{26} Js^{1}
3.9 x 10^{26} = mc^{2} = m x 9 x 10^{16} m = 3.9 x 10^{26} /9 x 10^{16}
= 4.3 x 10^{9} kgs^{1} = 4.3 million tonnes/second
2. A spacecraft travelled from Earth to Beta Centauri, 390 light years away, at a speed of 0.999c. When it reached Beta Centauri it immediately turned around and returned to Earth. During the journey to Beta Centauri and back again, how much time has passed relative to
(i) people on Earth ?
390 x 2/ 0.999 = 780.8 years
(ii) people on the spacecraft?
t_{v}
= t_{o} / \/(1 – v^{2}/c^{2}) = 780.8 / \/(1  0 .999^{2}) = 34.9 years
3. The spaceship Aychesee was speeding away from the Earth at a speed of 0.98c. The spaceship was 10.0m long when measured on Earth before launch.
(i) What was the length of the spaceship as measured by an occupant of the spaceship? (10.0m)
(ii) What was the length of the spaceship as measured by someone on earth?
L = 10.0 \/(1 – 0.98^{2}c^{2}/c^{2}) = 1.99m
(iii) How long would it take to watch a onehour video on the spaceship as measured by an occupant of the spaceship? (1.0 hour)
(iv) How long would it take to watch a onehour video on the spaceship as measured by someone on Earth?
t = 1.0/ \/(1 – 0.98^{2}c^{2}/c^{2}) = 5.03h
4. The rest mass of an electron is 9.1 x 10^{31} kg. What is the mass of an electron moving at 0.90c?
m_{v} = m_{o }/ \/(1 – v^{2}/c^{2}) = 9.1 x 10^{31} / \/(1 – 0.9^{2})
= 2.09 x 10^{30} kg