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. Ep = -G m1m2/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. 

Ep = -Gm1m2 / r         where Ep = potential energy, G = universal gravitational constant, m1 and m2 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 computer-assisted 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.walter-fendt.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 5o) 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 T2. Determine the gradient of the graph by drawing the line of best fit and multiply by 4p2. This will give the value of “g”. l = T2g / 4p2.

 

            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 = Gmmp/rp2 where mp and rp are the mass and radius of the planet respectively and G is the universal gravitational constant. Equating we get mg = Gmmp/rp2

or g = Gmp/rp2

 

Example: Given that the Universal gravitational constant , G = 6.67 x 10-11 Nm2kg-2 and the mass of the Earth is 6.0 x 1024 kg and its radius is 6.38 x 106 m, calculate the acceleration due to gravity on the Earth.

Solution: g = GmE/rE2  = 6.67 x 10-11 x 6.0 x 1024/(6.38 x 106)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 Nm2kg-2

Planet

Mass (kg)

Radius (m)

“g” (ms-2)

Mercury

3.35 x 1023

2.4 x 106

 

Venus

4.9 x 1024

5.6 x 106

 

Mars

6.44 x 1023

3.4 x 106

 

Jupiter

1.9 x 1027

7.2 x 107

 

Saturn

5.7 x 1026

6.0 x 107

 

Uranus

8.7 x 1025

2.6 x 107

 

Neptune

1.0 x 1026

2.5 x 107

 

The Moon

7.4 x 1022

1.75 x 106

 

 

Values of “g” (ms-2): Mercury 3.88; Venus 10.42; Mars 3.72; Jupiter 24.45; Saturn 10.56; Uranus 8.58; Neptune 10.67; Moon 1.61

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

 

 

 

 

Neptune

 

 

 

 

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

Neptune

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.

 

  • vy = uy + ayt                            where  uy = initial  vertical velocity

vy = final vertical velocity

  • Dy = uyt + ½ ayt2                                Dy =  vertical displacement

ay =  vertical acceleration

  • vy2 = uy2 + 2ayDy                                t = time

 

 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 vy = v sin q and the horizontal component vx = v cos q

 

Since there is no horizontal acceleration the horizontal distance travelled is given by Dx = vxt       

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. Ek + Ep >0  ½ mv2 – GmmE/rE > 0 v2 > 2GmE /rE

 

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

 

vescape =    \/(2 x 6.67 x 10-11 x 6 x 1024 / 6.38 x 106)    

= 11200.6 ms-1            = 11.2 kms-1

 

Students learn to “outline Newton’s concept of escape velocity. ” (syllabus)

 

Newton noticed that the faster an object was thrown horizontally from the top of a cliff or building, the greater was its range. He deduced that if an object was thrown fast enough it would fall right around the Earth, i.e. go into orbit.

Gravity would supply the centripetal force so that mg = mv2/r  i.e. g = v2/r  or v2 = rg

At sea level v2 = 6.38 x 106 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 v2 = 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 (Newton’s third law).

 

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 Newton’s first law an object in motion will continue to move at the same speed and in the same direction unless acted on by a force. An object that is travelling in a circular path requires a force to change its direction even if the speed remains unchanged. This force acts towards the centre of the circle and is called centripetal force.

Fc = mv2/r        Where Fc = 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 geo-stationary orbits. ” (syllabus)

 A satellite in a geo-stationary 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. mv2/r = GmmE /r2

Which becomes v2 = GmE /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        v2 = 4p2r2 / T2 = GmE /r 

T2 = 4p2 r3 /G mE

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 T2 = (4p2/G{m1 + m2}) r3

 

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 low-density atmosphere and they gradually spiral back towards the Earth’s surface.

 

Students learn to “discuss issues associated with safe re-entry into the Earth’s atmosphere and landing on the Earth’s surface. ” (syllabus)

 

As the spacecraft re-enters 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 re-entering the atmosphere at the correct angle. Even with the correct angle of re-entry 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 re-entry 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 re-entry into the Earth’s atmosphere is

6.2o  +/- 1o

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 re-enters 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:

  • vx2 = ux2
  • v = u + at
  • vy2 = uy2 + 2ayDy
  • Dx = uxt
  • Dy = uyt + ½ ayt2 ” (syllabus)

 

PROJECTILE MOTION – REVISION QUESTIONS

 

1.      Calculate the vertical and horizontal components of the velocity for an object traveling at

(i)                 60 ms-1 at 30o to the horizontal

(ii)               30 ms-1 at 45o to the horizontal

(iii)             40 ms-1 at 40o to the vertical

 

  1. Two objects are thrown simultaneously from the top of a cliff 122.5 m high. The first is thrown horizontally at a speed of 4.0 ms-1 and the second is thrown horizontally at a speed of 14.0 ms-1.

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

 

  1. A ball is thrown horizontally from the top of a cliff 176.4 m high at a speed of 8.0 ms-1. Calculate the velocity (vector quantity) with which it hits the ground below.

 

  1. A football player kicked a ball at 20 ms-1 at an angle of 40o to the horizontal.

(i)                 How high did the ball go?

(ii)               How far did the ball go?

 

  1. A student standing on a balcony 12.0 m high threw their school bag up over the edge at an angle of 60o to the horizontal at a speed of 6.0 ms-1.

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

 

  1. Junior threw a ball from a height of 1.0m towards a fence 2.0m high. If he threw the ball upwards at an angle of 50o to the horizontal at 6.0 ms-1 at a distance of 1.0m from the fence, will the ball clear the fence? Show calculations to justify your answer.

 

  1. A squash player hit a ball just above the ground at 30o up from the horizontal. The ball hit the wall of the court 1.5 m above the ground and 3.0 m in front of where it was hit. Calculate the speed with which the player hit the ball.

 

  1. A ball is thrown up at 300 to the horizontal from ground level and hits the ground 50 metres away. Calculate the speed with which the ball was thrown.

 

ANSWERS: 1. (i) vy = 30 ms-1  vx = 52 ms-1 (ii) vy = 21.2 ms-1  vx =  21.2 ms-1

(iii) vy = 30.6 ms-1  vx = 25.7 ms-1

2. (i) 5.0 s for each (ii) 20m, 70m (iii) 50m

3. 59.3 ms-1 at 82o15’ 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) 79o29’ to horizontal.

6. No. Height at fence = 1.865m

7. 15.92 ms-1

8. 23.8 ms-1

 

Students: “perform a first-hand 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 vs 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.

 

vx = sx/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)       vy2 = uy2 + 2aysy

0 = uy2 + 2(-9.8) 2.6    uy2 = 50.96                  uy = 7.14 ms-1

 

6.         Calculate the speed at the beginning of its motion past the grid.

 

u2 = ux2 + uy2   = 2.52 + 7.142  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 = 70o42’

 

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, Esnault-Pelterie, 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 three-stage 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.kiosek.com/oberth/

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: U.S. rocket scientist. He was the first to demonstrate that a rocket will work in a vacuum and developed rockets that operated on liquid fuels. He was the first to fire a rocket faster than sound.

 

http://inventors.about.com/library/inventors/blgoddard.htm

http://spaceline.org/history/22.html

 

(iv)       Roberts Esnault-Pelterie: 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 England in the second world war. Von Braun was in a group that surrendered to the U.S at the end of the war and after the war joined the U.S. rocket development programme that ultimately developed the Saturn V rocket that was used in the first Moon landing.

 

                        http://inventors.about.com/library/inventors/blvonBraun1.htm

 

 

Gerard K O’Neill (1927 – 1992)

In 1965 he began teaching the introductory physics course at Princeton University. To make the course more relevant he related the course to the current (1965) Apollo mission and emphasised the application of physics to people and objects orbiting the Moon. He investigated the colonisation of space and designed a mass driver, a device that can economically move materials into orbit from the Moon or an asteroid. In the mid 1970s he published “The High Frontier” where he outlines plans for human settlement beyond the Earth. In 1978 he founded the Space Studies Institute to establish a “conceptual, theoretical and technical foundation for the large-scale colonisation of space.”

Gerard O'Neill (1969)  From UNSW notes

    • American.
    • Conceived plans for space colonies at stable Lagrangian points between Earth-Moon (L5) - wrote book on this called "The High Frontier".
    • Devised concept of "mass-driver" and "electromagnetic slingshot" to accelerate cargoes into orbit from the Moon (eg after mining) and towards catcher ships.

http://www.ssi.org/obit.html

 

Students: “Solve problems and analyse information to calculate centripetal force acting on a satellite undergoing uniform circular motion about the Earth using: F = mv2 /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 = mv2/r           = 800 x 63262 / 107     = 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 106m 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 8.07x106/ 7200   =  7042 ms-1

(ii)        F = mv2/r         = 10000 x 70422 / 8.07 x 106  = 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: r3 / T2 = GM / 4p2 ” (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?

Fcent. =  mv2 / r

(ii)               What is the gravitational force acting on the satellite?

 Fgrav. = G mM / r2

(iii)             What is the period of the satellite in terms of v and r?

T = 2pr / v

(iv)             Rearrange the above formulae to show:

r3 / T2 = GM / 4p2

The gravitational force supplies the centripetal force so they are equal.

                  mv2 / r = GmM / r2            v2 = GM / r

                  but v = 2pr / T      so         4pr2 / T2 = GM / r

r3 / T2 = GM / 4p2

  1. Calculate the period of a satellite that orbits the Earth at a height of 3620 km above the Earth’s surface.

Answer:

Radius of orbit = radius of Earth + height = 6380km + 3620km = 10 000 km

= 10 000 000m

T2 = 4p2r3/ Gm            = 4p2 (107) 3 / (6.67 x 10-11 x 6 x 1024) = 9.864 x 107

T = 9932s =165 min =2hr 45 min

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. The Solar System is held together by gravity. ” (syllabus)

 

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 Newton’s Law of Universal Gravitation: F = Gm1m2/d2 ” (syllabus)

 

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 m1m2/d2

While Newton showed the proportionality, it was Henry Cavendish who experimentally determined the constant of proportionality ,G, in 1798.

F = Gm1m2/d2

 

Students learn to “discuss the importance of Newton’s Law of Universal Gravitation in understanding and calculating the motion of satellites. ” (syllabus)

 

 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. Newton’s Law of Universal Gravitation provides a quantitative measure of the gravitational force of attraction between the two bodies and the centripetal force gives a quantitative value for the force necessary to keep the body in circular motion. The motion of satellites can be analysed by equating these two forces.

mv2 / r = GmmE / r2

 

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 Newton’s Universal Law of Gravitation, 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.

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 = Gm1m2/d2 ” (syllabus)

 

The mass of the Earth is 6 x 1024 kg while that of the Sun is 2 x 1030 kg. The distance between the Earth and the Sun is 1.5 x 1011 m. Calculate the gravitational force of attraction between the Earth and the Sun.

 

F = Gm1m2/d2  = 6.67 x 10-11 x 2 x 1030 x 6 x 1024 / (1.5 x 1011)2        = 3.6 x 1022 N

 

Calculate the force of attraction between two 60 kg people standing 1.0m apart.

 

F = Gm1m2/d2  = 6.67 x 10-11 x 60 x 60 / 12     = 2.4 x 10-7 N.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. Current and emerging understanding about time and space has been dependent upon earlier models of the transmission of light. ” (syllabus)

 

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 Michelson-Morley 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 45o to the light, so that about half of the light passed through the mirror, and the other half was reflected at 90o 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 vice-versa. 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 Michelson-Morley 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 non-accelerating frame of reference and Newton’s laws hold. A non-inertial frame of reference is accelerating and Newton’s laws do not hold.

 

 

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. Newton proposed that no measurements conducted within an inertial frame of reference are capable of determining whether the frame is at rest or moving with constant velocity. These were restated by Einstein, who also stated that the speed of light in a vacuum is constant and is independent of the motion of the observer and source. This led to his proposal of firstly his special theory of relativity and later his general theory of relativity.

 

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 Michelson-Morley 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 Paris on the circumference of the Earth. This definition was cumbersome and accurate metres were hard to reproduce. The current definition is in terms of the distance travelled by light in an interval of time (The same way that the light year is defined). This has the advantage that scientists in any well-equipped laboratory throughout the world can reproduce it. An accurate method of recording time was needed and the second is now defined as the time for 9 192 631 770 oscillations of the caesium-133 atom. Since 1983 the metre has been defined as the distance travelled by light in a vacuum in 1 / 299 792 458 second.

 

 

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 inter-related; 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 = mc2           = 0.001 x (3x108)2       = 9.0 x 1013J

 

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 Michelson-Morley 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 vice-versa. 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 non-inertial 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 non-inertial.

 

 

 

 

 

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 = mc2

 

            lv = l0 \/(1 – v2/c2)

 

            tv = t0 /{\/(1 – v2/c2)}

 

            mv = m0 /{\/(1 – v2/c2)} ” (syllabus)

 

1.                  The power output of the Sun is around 3.9 x 1026 watts. How much mass is converted to energy each second in the Sun?

 

Power output = 3.9 x 1026 watts =3.9 x 1026 Js-1

3.9 x 1026 = mc2 = m x 9 x 1016           m = 3.9 x 1026 /9 x 1016

 = 4.3 x 109 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?

 

tv = to / \/(1 – v2/c2) = 780.8 / \/(1 - 0 .9992)     = 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 – 0.982c2/c2)        = 1.99m

(iii)             How long would it take to watch a one-hour video on the spaceship as measured by an occupant of the spaceship?  (1.0 hour)

(iv)             How long would it take to watch a one-hour video on the spaceship as measured by someone on Earth?

      t = 1.0/ \/(1 – 0.982c2/c2) =  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?

 

mv = mo / \/(1 – v2/c2) = 9.1 x 10-31 / \/(1 – 0.92)

= 2.09 x 10-30 kg

 

 

 

 

 

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