TOPIC ONE: SPACE (THEORY)
1. The Earth has a gravitational field that exerts a force on objects both on it and around it.
Define weight as the force on an object due to a gravitational field
- Weight is defined as the force on a mass due to the gravitational field of a large celestial body, such as the Earth.
- It is a force and therefore a vector quantity.
- Newton’s Second Law states F ma and hence: W mg Where: W = weight (N) m = mass (kg) g = acceleration due to gravity (m/s^2)
Explain that a change in gravitational potential energy is related to work done
- To have a change in gravitational potential of an object within a gravitational field, work is done. This means that if you move an object in a gravitational field, work is done on that object in order for it to move.
- If an object is moved against the gravitational field, positive work is done and if the object is moved with the gravitational field, negative work is done.
- For objects within a gravitational field, gravitational potential energy is given by the formula: 1 Ep mgh Where: Ep = gravitational potential energy (J) m = mass (kg), g = acceleration due to gravity (m/s^2), h = height (m).
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
- Gravitational potential energy is defined as the work done to move an object from a very large distance away to a point within a gravitational field.
- Due to the inverse relationship, (Constant/ r), Ep will only become zero when r is at a very large distance. Thus if its initial potential energy is zero then its final potential energy must be negative.
- Therefore gravitational potential energy is given by the formula: GMm Ep r Where: Ep = gravitational potential energy (J) G = 6.67 x10^-11 M = mass of central body (kg) m = mass of object (kg) r = distance separating from the centre of each mass (m)
- The change in potential energy of a mass m is defined as the work done to move the mass m from its initial distance Ri to the centre of a body of mass M to its final position Rf. This is expressed as:
1 1 E GmM p ri rf Where: Ep = change in gravitational potential energy (J) G = 6.67 x10^-11 M = mass of central body (kg) m = mass of object (kg) Ri = initial distance separating from the centre of each mass (m) Rf = final distance separating from the centre of each mass (m)
2 2. Many factors have to be taken into account to maintain a stable orbit and return to earth.
Describe the trajectory of an object undergoing projectile motion within the Earth’s gravitational field in terms of horizontal and vertical components
- The trajectory of any projectile is parabolic (ignoring air resistance).
- The velocity of the projectile continuously changes throughout its trajectory.
- The motion of a projectile can be split into two independent components, the horizontal (x) and the vertical (y) component.
- The horizontal (x) component is not subject to the acceleration due to gravity and therefore the velocity stays constant throughout its motion.
- The vertical (y) component is subject to the acceleration due to gravity and therefore the velocity changes throughout its motion.
- Adding these two vectors together, the initial or launch velocity of the projectile can be calculated via the use of Pythagoras theorem, and the launch angle can be calculated via the use of trigonometry.
Describe Galileo’s analysis of projectile motion
- The motion of a projectile can be split into two independent components, the horizontal (x) and the vertical (y) component.
3 The horizontal (x) component is not subject to the acceleration due to gravity and therefore the velocity stays constant throughout its motion.
The vertical (y) component is subject to the acceleration due to gravity and therefore the velocity changes throughout its motion.
- Galileo was responsible for stating that the trajectory of a projectile was parabolic. (ignoring air resistance)
- Galileo’s analysis of projectile motion led him to consider reference frames. This is what all measurements are referred to.
- Galileo supported the use of mathematics and experimentation in explaining natural phenomena.
Explain the concept of escape velocity in terms of the: i. Universal gravitational constant ii. Mass and radius of the planet
- By considering the kinetic and gravitational potential energy of a projectile, it can be shown mathematically that the escape velocity is only dependant upon the universal gravitational constant, the mass and the radius of the planet.
- Mathematical proof:
(At a distance of infinity both Ek and Ep = 0, similarly due to the principal of conservation of momentum Ek and Ep must also equal to zero at the planet’s surface. )
Ek Ep 0 1 _ GMm mv 2 0 2 r 1 GMm mv 2 2 r 2GM v r
Where: V = escape velocity (m/s) G = 6.67 x10^-11 M = mass of central body (kg) r = radius of the orbit from the centre of the central body (m)
4 - Therefore the escape velocity is only dependant upon the universal gravitational constant, the mass and the radius of the planet.
- It can be seen from the equation that escape velocity is not affected by the mass of the object meaning that the same escape velocity applies for all masses. Also the greater the mass of the planet the greater the escape velocity required while the greater the radius the smaller escape velocity required.
Outline Newton’s concept of escape velocity.
- In a thought experiment, Newton imagined that projectiles being launched horizontally from a very high above the surface above of the Earth will fall down to earth following a parabolic path.
- As the projectiles are given higher and higher velocities, they reach further and further around the earth’s curved surface.
- If the projectile is launched at just the right velocity, it will follow the curve of the Earth’s surface and orbit the earth.
- However if the speed of the projectile is equal or greater than the escape velocity, the projectile will fly off into space and never return to earth.
Identify why the term g-forces is used to explain the forces acting on an astronaut during launch
- During a launch an astronaut is subject to many forces such as thrust. The term ‘g’ force is a term used to express a person’s apparent weight (weight experienced at that instant) over the person’s true weight (mass x acceleration due to gravity).
- The astronaut feels an apparent weight of mg+ma and the true weight is 9.8m.
- Since g force is equal to apparent weight over true weight, therefore G-force mg ma 9.8m g a 9.8 Where: a = acceleration of rocket (m/s^2) g = acceleration due to gravity (m/s^2)
5 - By using this formula, the term g-forces can be used to explain the forces acting on an astronaut during a launch.
- Further a force of 5g is equivalent to 5 times the acceleration due to gravity. n o i t a r e l e c c a Max g-forces
Discuss the effect1st on the 2 earth’snd orbital 3rd motion and it’s its rotational motion on the launch of a rocket stage stage stage
- The Earth is a moving platform with two different motions which can be utilized in a rocket launch to gain a boost in velocity and therefore conserve fuel.
- The first is the orbital motion, Earth revolves around the Sun at 107,000km/h relative to the Sun and the second is its rotational motion, Earth rotates once on its axis per day so that a point on the Equator has a rotational velocity of approximately 1,700km/h relative to the Sun.
- A rocket heading into orbit is launched to the east at the equator to receive a velocity boost from the Earth’s rotational motion.
- The flight of a rocket heading into space is timed so that it can head out in the direction of the Earth’s rotational and orbital motion to receive a velocity boost; this is an inconvenience as if the rocket misses the opportunity it may have to wait for a long time before the rocket can be launched again.
6 Analyse the changing acceleration of a rocket launch in terms of: i. Law of conservation of momentum
- The Law of conservation of momentum states that the total momentum of any group of objects remains the same unless outside forces act on the objects.
- The Law of conservation of momentum applies during a launch. Rocket engines generate thrust by burning fuel and expelling the resulting gases. Conservation of momentum means that as the gases move in one way, the rocket moves the other. (Momentum before the launch is zero; hence the momentum after the launch is also zero. The gases carry the momentum in one direction, and so the rocket carries an equal momentum in the opposite direction up). This can be expressed mathematically as:
PROCKET PGAS 0
PROCKET PGAS
(Ft) ROCKET (Ft)GAS
FROCKET FGAS
Thus, the rocket is forcing a large volume of gases backwards behind it which in turn propels the rocket forward according to Newton’s 3rd Law.
- As fuel is consumed and the gases expelled, the mass of the system decreases. Therefore if thrust is constant as according to the law of conservation of momentum, the acceleration will increase as mass decreases.
- Note: 7 Impulse = (force) x (time) = Ft Momentum (P) = (mass) x (velocity) = mv
Ft = ΔP Ft = mv – mu Ft = m (v - u) F = m (v - u) / t
Analyse the changing acceleration of a rocket launch in terms of: ii. Forces experienced by an astronaut
- Two forces act upon an astronaut during launch. The upwards thrust (T) as well as the downward weight (mg). A simple expression for acceleration of a rocket that is launched directly upwards: can be derived from Newton’s 2nd law: F ma F a m T mg a m Where:
a = acceleration of rocket (m/s^2) T= thrust (N) m = mass of rocker (kg) g = acceleration due to gravity (m/s^2)
- As the rocket rises, the acceleration increases which in turn increases the forces experienced by the astronauts. This is because of the following factors:
1. As fuel is consumed and the gases expelled, the mass of the system decreases. According to the equation T mg a , acceleration is proportional to the thrust and inversely proportional to the mass, as the mass m decreases, the acceleration increases.
2. The acceleration due to gravity decreases when altitude increases.
3. The air resistance decreases as air becomes thinner at higher altitude.
- Variations in the forces and g-forces experienced by astronauts can be accounted from the sequential shutdowns during a multistage rocket. The above factors can also account for the variations in the forces experienced by the astronauts.
Analyse the forces involved in uniform circular motion for a range of objects including satellites orbiting the earth .
8 - When an object is undergoing uniform circular motion:
It is moving at constant speed The magnitude of its velocity remains the same The direction of its velocity is always changing, so it is continuously accelerating.
- There are two types of circular motions: angular (or orbital) and tangential (or linear)
- Angular velocity is the rate at which the angle is changing.
It is expressed as [w=θ/t] where:
w is the angular velocity in radian per second (rad-1) θ: is the angle travelled in radians (rad) t: is time taken in seconds (s)
One revolution/period is equal to 2pie radians, thus w = 2pie/t and the time is t = 2pie/ w.
- Tangential velocity: is the rate at which distance is changing. Tangential velocity = change in distance / change in time = circumference of circle/ period That is v -= 2pier/t s since w = 2pie/ t in this case therefore
Tangential velocity can be expressed as [v=wr] where: v= linear velocity (ms-1) w= angular velocity (rads-1) r = radius of circle (m)
- Centripetal acceleration is when an object moves in uniform circular motion, the magnitude of the velocity remains constant but its direction is always changing and so the object is therefore continuously accelerating. The direction of this acceleration is always towards the centre because this is the direction in which the force must be applied to keep the object moving in a circle. The vectors are always perpendicular at every point.
It is expressed as[Ac= v^2/r] or [Ac=w^2r] where: Ac= centripetal acceleration (ms-2) v= velocity (ms-1) r= radius (m)
- Centripetal force is the force needed to keep an object moving in uniform circular motion. Like centripetal acceleration, centripetal force is also directed towards the centre of the circle. An easy way to remember this is centripetal means “centre seeking”.
It is expressed as F = ma = mw^2r = mv^2/r
- Contextual info:
A radian is defined as the able subtended at the centre of a circle by an arc which is equal in length to its radius.
The size of the angle in radians is the ratio of the length of the arc, l, to the length of the radius, r, of the circle. It is expressed as θ = l / r
Theatre radian/ 2pie = theatre degrees / 360 degrees Therefore theatre radian= 2pie x (theatre degrees / 360 degrees) Theatre degrees= 360 degrees x (Theatre radian/ 2pie) 9 - Examples on above theory
1.Convert the following into radians per second: i) 250rpm ii) 25000rpm
Ai) knowing that one revolution is equal to 2 pie radians thus, 250rpm = 250 x 2pie therefore 250rps = (250 x 2pie)/60 = 26.18 rad/s Aii) knowing that one revolution is equal to 2 pie radians thus, 25000rpm = 25000 x 2pie x 60 therefore 25000rps = (25000 x 2pie)/60 = 2618 rad/s
2. Convert the following to revolutions per minute (rpm): i) 26.18 rad/s ii) 2618 rad/s
Ai) This is the opposite method to converting rpm to rads. We know that there are 60 seconds in one minute and 2pie radians in one circle. Thus 1 rpm = (26.18 x 60)/2pie.
3. Calculate the angle in radians subtended at the centre of the circle of: i) Radius 5cm by an arc of length 5cm ii) Radius 2cm by an arc of length 4cm
Ai) θ = l / r = 5/5 = 1rad Aii) θ = l / r = 4/2 = 2 rad
4. Calculate the angular (orbital) velocity of an object which moves in a circle through an angle of: i) pie radians in 5s ii) 270 degrees in 6s
Ai) w=θ/t = pie/ 5 Aii) w=θ/t = (3pie/2)/ 6 = pie/ 4
5. Calculate the tangential speed (linear speed) of an object that moves in a circle of: i) radius 2m at an angular velocity of 5rad/s ii) radius 6m at an angular velocity of pie/3rad/s
Ai) v=wr= 5 x 2 = 10m/s Aii) v=wr= pie/3 x 6 = 2pie
6. Calculate the centripetal acceleration…… 7. Calculate the centripetal force….
10 - Different objects in circular motion:
Circular motion Source of centripetal force
Ball on a string whirled in a circle Tension in the string
Car driving around a corner Friction between the tyres and the road
Satellite orbiting the Earth Gravitational attraction between the Earth and the satellite
11 Compare qualitatively low Earth and geo-stationary orbits
- There are two main types of satellites orbiting the earth Low earth and Geostationary.
Low Earth satellites Geostationary satellites
- Generally in orbit higher than 250km to avoid - Generally in orbit at an altitude at which the atmospheric drag, and lower than approximately period of the orbit is the same as the Earth. 1000km, which is the altitude at which the van Allen’s radiation belt begins. - This is at an altitude of about 35 800 km above the equatorial plane which is where it will remain - The space shuttle utilises a low earth orbit stationary over a fixed point. (Altitude is somewhere between 250km and 400km determined from Kepler’s 3rd Law and subtracting
12 depending on the mission. the radius of the earth- 6400km)
- At 250km, a low earth satellite takes 90 minutes - Period is the same as the Earth which is 24hours. to complete one orbit around the earth.
- Used for spying missions. - Used for modern communication and weather.
- Other satellites include geosynchronous which are satellites that have a period identical to the earth which is 24 hours, however they do not orbit in the equatorial plane and therefore they trace a ‘figure 8’ every 24 hours.
Define the term orbital velocity and the qualitative and quantitative relationship between orbital velocity, the gravitational constant, mass of the satellite and the radius of the orbit using Kepler’s law of periods
- Orbital velocity is the velocity required to maintain an object in orbit.
- By deriving the equation of orbital velocity, the qualitative and quantitative relationship between orbital velocity, the gravitational constant, mass of the satellite and the radius of the orbit can be seen.
- Orbital velocity is derived by equating the centripetal force of an object in orbit with the gravitational force between the two bodies, as the centripetal force is the gravitational force.
Fc Fg mv 2 GMm r r 2 GM V r
Where: V = orbital velocity (m/s) G = 6.67 x10^-11 M = mass of central body (kg) r = radius of the orbit from the centre of the central body (m)
- Orbital velocity can also be expressed as: 2nr V (V = displacement/ T) T Where: V = velocity (m/s) r = radius of the orbit from the centre of the central body (m) T = time for one orbit(seconds)
- It can be seen from the above equations that orbital velocity is independent of the mass of the satellite.
13 Account for the orbital decay of satellites in low earth orbit
- All satellites in low earth orbits experience atmospheric drag that will eventually decay (lose altitude) their orbit and limit their lifetimes. As a satellite slows, it loses altitude and begins a slow descent. As it descends, it encounters higher air density and therefore higher drag, in turn speeding up the process. The satellite eventually collides with the earth’s atmosphere.
Discuss issues associated with safe re-entry into the earth’s atmosphere and landing on the earth’s surface
- There are 4 main factors involved in safe re-entry, they are:
1. Heating effect:
- The heat is generated as the spacecraft contacts the earth’s atmosphere at high velocity due to the friction with air molecules. To lessen the heat, space craft are designed with a heat shield made of ablative material that would burn up to disperse heat on the spacecraft.
- The shape of the spacecraft will influence the spread and dispersion of the heat that the spacecraft experiences. Blunt noses are preferred over sharp noses as the heat is spread out over a large surface area rather than on one particular point.
2. G-forces:
- The human body is relatively unaffected by high speeds. However changes in that is, acceleration, can affect the human body.
- To minimise the effects of g forces on astronauts, several approaches have been implemented:
Reclining the astronaut and applying these forces perpendicular to the astronaut’s long axis.
Increasing the re-entry time and slowing the rate of descent.
- G-forces have two man effects on the body:
If the acceleration is in the direction of the persons head they may experience a ‘blackout’ as the blood rushes to the feet.
If the acceleration is towards the feet, they may experience a ‘red out’ where the blood rushes to their head and retina.
3. Angle of re-entry:
- The re – entry angle is critical in maintaining g-forces within a safe limit. The optimum angle of re-entry is between 5.2 – 7.2 degrees to the horizon.
If the angle of entry is too steep, the space craft may burn up on re-entry and/or be subject to massive g- forces. However if the angle is too shallow, the space craft may bounce off the atmosphere and will go back into space.
4. Reaching the earth’s surface:
- The spacecraft must reach the Earth’ surface in a safe manner. There are several methods to achieve this. 14 - One method is by first using parachutes to slow down the space capsule to lessen the impact when the capsule falls into the ocean which is then collected.
- Another method used currently, is that the spacecraft lands on an air strip just like aeroplanes do.
Identify that there is an optimum angle of re-entry for a manned spacecraft into the earth’s atmosphere and the consequences of failing to achieve this angle
- The re – entry angle is critical in maintaining g-forces within a safe limit. The optimum angle of re-entry is between 5.2 – 7.2 degrees to the horizon.
- If the angle of entry is too steep, the space craft may burn up on re-entry and/or be subject to massive g- forces. However if the angle is too shallow, the space craft may bounce off the atmosphere and back out into space.
3. The solar system is held together by gravity.
Describe a gravitational field in the region surrounding a massive object in terms of its effects on other masses in it
- A gravitational field is a field in which any mass within the gravitational field will experience a gravitational force. Since the force of gravity will act on any mass near the Earth, a gravitational field exists around the earth. The gravitational field is directed towards the Earth’s centre; therefore any mass on the earth’s surface will experience a downward force of gravity.
- On a planetary scale, say the moon and the earth, the moon and the earth will have their own gravitational fields, and are directed towards their centres, however the earth will have a larger gravitational field than the moon. If illustrated there would be a null point where there is no gravitational field. The gravitational field’s strength is indicated by field lines, the closer the field lines, the stronger the gravitational field is.
Define Newton’s Law of Universal Gravitation
- Newton’s Law of Universal Gravitation states that every mass in the universe is attracted to every other mass in the universe by a force of gravitation.
- This force of gravitation between any two masses is proportional to the product of the two masses and inversely proportional to the square of the distance between their centres. This force is exerted equally on both masses.
- Mathematically this force of gravitation between any two masses is represented as : GMm F r 2
Where:
15 F = the force experienced on the object m2 (N) M= the mass of the central object (kg) m= the mass of the orbiting object (kg) G= the universal gravitational constant r = distance of m from centre of M (m)
- Using Newton’s formula of Universal Gravitation and weight force (gravitational force on a mass while Newton’s formula of Universal Gravitation is the force between any two masses) formula a formula for gravity can be derived.
GMm mg r 2 GM g r 2 Where: g = acceleration due to gravity (m/s^2) G= the universal gravitational constant M= the mass of the central object (kg) r = radius of planet
Discuss the importance of Newton’s Law of Universal Gravitation in understanding and calculating the motion of satellites
- Satellites orbiting the earth via uniform circular motion must have a force to keep the satellite in orbit. In this case the centripetal force is Newton’s Law of Universal Gravitation.
- Therefore equating the centripetal force of an object in orbit with Newton’s Law of Universal Gravitation, we get:
Fc Fg mv 2 GMm r r 2 GM V r
- From this formula, the velocity required for the satellite to stay in orbit can be calculated which can be affected by the central mass (M) and radius of the central mass (R). This leads to a better understanding in the motion of satellites.
- Newton’s law combined with Kepler's 3rd law of gives us this expression:
A satellite orbiting the Earth is carrying out a uniform circular motion. The net force acting on the satellite is the centripetal force. The Earth’s gravitational attraction provides this centripetal force in keeping the satellite in orbiting to Newton’s Law Universal Gravitation attraction. Equating these 2 formulas, Kepler’s third law can be derived.
16 Fc Fg ma mg a g GM 2 r r 2 2 2 GM r .. T r 2 4 2 r GM T 2 r 2 r 3 GM T 2 4 2
Or preferable method, equating the two escape velocity methods, Kepler’s third law can be derived.
2r GM T r 4 2 r 2 GM T 2 r r 3 GM T 2 4 2
Where: r = radius of the orbit from the centre of the central body (m) T = time for one orbit (seconds) G = 6.67 x10^-11 M = mass of central body (kg)
Identify that a slingshot effect can be provided by planets for space probes
- As the probe approaches the planet it is accelerated by the planet’s gravitational attraction, causing it to speed up relative to the planet. (By Newton’s 3rd law the planet will also experience a force slowing it down, its mass however, is so much greater than that of the probe that its velocity decrease is negligible)
- As the probe passes the planet, its speed is reduced (relative to the planet). Relative to the sun however, its speed has increased, the probe picks up the angular momentum from the planet (which loses an equal amount of momentum due law of conservation of angular momentum).
- The sling shot effect is used to increase the velocity and/or change the direction of the satellite which conserves fuel.
17 4. Current and emerging understanding about time and space has been dependent upon earlier models of the transmission of light
Outline the features of the aether model for the transmission of light
- The aether was the proposed medium required for light waves to propagate.
- The Aether also had special properties:
It filled all of space, had low density and was perfectly transparent. Permeated all matter and yet was completely permeable to material objects. Had great elasticity to support and propagate the light waves.
Describe and evaluate the Michelson-Morley attempt to measure the relative velocity of the earth through the aether
Context:
- If the aether existed, the Earth should be moving through the aether, meaning that the earth should experience a flow of ‘aether wind’.
The experiment:
- Aim: In 1887 two United States scientists A.A Michelson and E.W Morley attempted to measure the relative motion between the earth and the aether. - How: They used the phenomenon of the interference of light (interferometer) to measure minute distances which indicated an interference change. - The experiment was conducted at different times to ensure that they did not just test at a time where the ether and the Earth were moving at the same speed.
The Methodology:
1. Light was sent from S and split into two perpendicular beams by the half silvered mirror M.
2. The beam M1 travelled across the aether while, whilst beam M2 travelled against and with the aether.
3. These two beams are then reflected back by the mirrors A and B and are recombined in the observer’s eye where an interference pattern would be observed.
18 4. When the entire apparatus was rotated 90 degrees, a change in the interference pattern was expected (since the beams A and B were now interchanged and the idea being that the aether wind would affect the path of at least one of the light beams causing a different interference pattern to appear.), no change in the interference pattern was observed.
5. This null result meant that the relative motion of the earth to the aether was not detectable.
Several reasons were given for the experiments failure even after many attempts:
- It was suggested that the earth ‘carried with it the aether’ so that there was no relative motion. - Lorentz suggested that the apparatus used contracted in the direction of the motion
Another way of looking at the Michelson-Morley experiment is through this boat analogy, In this case the current represents the ‘aether wind’ and the two boats A and B represent the two beams A and B.
19 Discuss the role of the Michelson-Morley experiment in making determinations about competing theories
- The Michelson-Morley experiment set out to measure the relative motion between the earth and the aether. The fact that a null result was found showed the ether hypothesis to be invalid
- The Michelson-Morley experiment provided supporting evidence for Einstein’s theory allowing a choice to be made between two conflicting theories – one requiring an ether and one that did not
- Almost twenty years after the Michelson-Morley experiments, Einstein proposed the theory of relativity, in which the aether model was not needed. The theory of relativity produced its own set of predictions, not all of which were testable at that time. As technology has improved, the predictions have been tested and found to be correct.
- The choice for scientists was as follows: continue to follow a theory for which no predictions proved true (aether) OR follow an alternative theory for which prediction does prove true (relativity).
Outline the nature of inertial frames of reference
20 - An inertial frame of reference is one that is moving with a constant velocity or is at rest (the two conditions being indistinguishable). In such reference frames, Newton’s law of motion holds.
- A non-inertial frame of reference is one that is accelerating
- Frames of reference are objects or coordinates systems with respect to which we take measurements.
Discuss the principle of relativity
- The study of motion began with Galileo. Galilean Relativity states that the laws of mechanics are the same for a body at rest and a body moving at constant velocity.
- In addition Newtonian relativity states that all steady motion is relative and cannot be detected without reference to an outside point. This means that it is impossible to do any mechanical experiments within an inertial frame of reference that can distinguish whether the frame is at rest or moving with constant velocity unless with reference to another frame of reference’s motion.
For example if you were in space in a spacecraft and if all the windows are closed, there is no way of you knowing how fast you are travelling or if you are moving at all. This is the principle of relativity, that is, there is no way of detecting inertial motion without reference to an outside point. So in our example, you can only detect the motion of the spacecraft you are in when you open the windows and look outside and see the Earth getting smaller and the moon becoming larger. To detect your motion you are referring to an outside reference point.
Two points to be reinforced:
- The principle of relativity only applies for non-accelerating motion or stationary motion. - This principle states that within an inertial frame of reference you cannot perform any mechanical experiment or observation that would reveal to you whether you were moving with uniform velocity or standing still.
Describe the significance of Einstein’s assumption of the constancy of the speed of light
- In 1905 Albert Einstein proposed that the speed of light is constant and is independent of the speed of the source or the observer.
- This proposal explained the null result of the Michelson-Morley experiment and showed that the aether was not necessary. As a consequence of this, there was no such thing as an absolute frame of reference (aether), all inertial frames are equivalent meaning that all motion is relative.
- Nothing can travel faster than the speed of light
- This assumption that the speed of light is constant and is independent of the speed of the source or the observer allowed Einstein to propose his theories of time dilation, length contraction and mass dilation.
Identify that if c is constant then space and time become relative
- In Newtonian physics, space and speed can be relative terms, but time is an absolute and fundamental quantity. Einstein radically altered the assumptions of Newtonian physics so that now the speed of light is absolute, and space and time are both relative quantities that depend upon the motion of the observer. This means that time taken for an event to occur, the measured length of an object and the mass of an object
21 entirely depends upon the observer’s velocity. Since Speed = d/t, and the speed of light is constant then distance and time must change.
Discuss the concept that length standards are defined in terms of time in contrast to the original metre standard
- Measurement is the process of comparing quantities such as length, mass or time to a selected standard and expressing the measured quantity as some factor of that standard.
- The standard length is the metre. This was originally to be one ten millionth of the distance between the equator and the North Pole along the meridian passing through Paris. This distance was marked on a platinum iridium rod and became the standard metre which was distributed all over the world. Following advanced in the accurate measurement of the wavelength of light, this definition was changed to one defined in terms of the wavelength of time light emitted by the element krypton -86 when excited in a discharge tube.
- Since 1960, our standard of distance has been defined in terms of time and velocity as follows: The metre is the distance travelled by light in a vacuum in the fraction 1/229792458 of a second.
- This modern definition takes advantage of the constancy of the speed of light, as well as the capability technology has given us to measure time and the speed of light with great precision. This means that the metre is the same for any frame of reference, meaning that it is unaffected by length contraction and time dilation.
Explain qualitatively and quantitatively the consequence of special relativity in relation to:
o the relativity of simultaneity o mass dilation o time dilation o length contraction o the equivalence between mass and energy
the relativity of simultaneity
- Simultaneous events in one frame of reference are not necessarily observed to be simultaneous in a different frame of reference. This can be explained with the following thought experiment involving a moving train and two observers, one observer ‘Alice’ who is in the middle of the train and the other “Bob” who is outside the train.
22 - When a flash of light is emitted from Alice, she sees the light beams hit the two ends of the train carriage at the same time. This is because the speed of light is constant irrelevant of the frame of references motion.
- However Bob will see things differently as the train is moving and he is stationary while Alice and the train are in the same frame of reference. Bob will see the light reach the back of the train before it hits the front of the train, this is because the back of the train has moved closer to the light source and the front of the train has moved away from the light source, remembering that he also observes the light to travel at c.
- In summary in Alice’s frame of reference these events appear to be simultaneous, but in Bob’s frame of reference they do not appear to be simultaneous.
mass dilation
- As you speed up an object the energy you use in order to accelerate it turns into kinetic energy which results in the acceleration of the object.
- But what happens at relativistic speeds? Well at relativistic speeds, the energy that is used to accelerate the object, does not accelerate anymore because the object cannot reach the speed of light, so this energy is not turning into kinetic energy as the object is not gaining much speed. Instead this energy you use to accelerate the object turns into mass and the object becomes heavier. This is mass dilation, which in turn makes it harder to accelerate the object even further. So to accelerate to the speed of light an infinite amount of energy is required.
mo mv v 2 1 c 2
23 Where: Mv = the mass for a ‘moving’ object Mo= the mass for the object when it is stationary v = velocity of the object c = the speed of light
time dilation
- Time in a moving frame of reference appears to go slower relative to a stationary observer.
to tv v 2 1 c 2
Where: Tv= the observed time for a stationary observer T0=the time for an observer traveling in the frame – To is called the proper time (this is the time measured by an observer present at the same location as the events that indicate the start and end of an event) v = velocity of the object c = the speed of light
- Time dilation can be explained by looking at an experiment that investigates how two different observers in different frames of reference measure time relative to a light clock.
- As shown in the diagram below, two observers Alice and Bob use the fact that velocity = distance/time to measure ‘one tick of time’. They decide that one ‘tick of time’ will be the time taken for light to travel from a source to a mirror and then be reflected back to the source. Alice and the apparatus are in a moving train and Bob is an outside stationary observer relative to the moving train.
- Rearranging speed = distance/ time we get time = distance/speed. So for Alice ‘one tick of time’ will be
. The light travels straight up and then down at the speed of light.
- However for Bob who is the stationary observer relative to the moving train, the light has to take a slightly longer path. This means that the distance will be greater and so the time will be greater. From Pythagoras’
theorem we see that . And so time passes slower for Bob’s frame of reference than for Alice’s frame of reference. This effect is time dilation.
24 length contraction
- The length of a moving object appears to contract in the direction of motion of its frame relative to a stationary observer.
v 2 l l 1 v o c 2 Where: Lv=the moving length Lo= the ‘rest’ length (that is, the length as measured by an observer at rest with respect to the rod) v = velocity of the object c = the speed of light
INCLUDE THOUGHT EXPERIMENT IF POSSIBLE
the equivalence between mass and energy
- If force is applied to an object then work is done on it, that is energy is given to the object. This energy would take in the form of kinetic energy and it increases as the object speeds up but at near light speeds the object doesn’t speed up and instead the object acquires extra mass (mass dilation). Einstein made an inference here and stated that the mass of the object contained this extra energy. Relativity results in a new definition of energy as follows:
2 E Ek mc Where: E= total energy Ek= kinetic energy m= mass c= speed of light
- When an object is stationary, it has no kinetic energy but still has some energy due to its mass. This is called its mass energy or rest energy and is given by:
E mc 2 Where: E= total rest energy (J) m= mass (kg) c= speed of light (3x10^8ms-1)
Discuss the implications of mass increase, time dilation and length contraction for space travel
- Mass increase means that spacecraft cannot travel at or above the speed of light. It also means that the faster the spacecraft gets, the heavier it will be and thus the more force needed to accelerate it. This means that the spacecraft cannot simply allow constant acceleration to increase the speed at a constant rate for as
25 long as they want. This limits the speeds spacecraft can travel at, and it limits them to low speeds relative to c.
- However time dilation and length contraction help for space travel. If relativistic speeds can be achieved then the length needed to travel will become less and as time slows people can travel further in their lifetime, however if they return to Earth people would have aged much more.
- For length contraction, it is best to use Einstein’s Relativity Postulate that states, the laws of physics are the same for observers in all inertial reference frames. No frame is preferred. This means that the spaceship speeding though space at a constant velocity could equally say that they in the spaceship are stationary and the rest of the space is speeding past the craft. In this respect, the space will contract in length and hence the spaceship will have less length to travel in order to reach the destination.
INCLUDE WORKING OUT FROM JACARANDA PAGE 90
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