Space Questions (plus answers) (1 Viewer)

[MuffinMan]

1. Outline the features of the aether model and the reasons that scientists believed that it needed to exist

Spoiler

Aether model – property (reason)
- fill space (light travelled everywhere)
- be stationary in space (light travelled in straight lines. If the aether was in motion, its movement would change the path of light travelling through it.)
- be transparent (we can’t see it)
- permeate all matter (light travelled everywhere)
- have an extremely low density (it can’t be detected)
- have great elasticity (transfer of energy over long distances requires the medium to have high elasticity otherwise significant amounts of energy will be “lost” to the particles in the medium

2. List the supposed features of the aether

Spoiler

– fill space
– be stationary in space
– be transparent
– permeate all matter
– have an extremely low density
– have great elasticity

3a. Identify the objective of the Michelson-Morley experiment

Spoiler

The purpose of this experiment was to detect the aether wind.

3b. Construct a diagram showing the paths of the light rays in the Michelson-Morley experiment

See surfing physics...sorry couldnt attatch it file was too big

3c. Write a one paragraph description on how the apparatus worked

Spoiler

The equipment used by Michelson and Morley (an interferometer) reflected light from a common source in two directions and then back to the observation point. If the aether existed, then the light rays travelling with and against should interfere with each other more than the rays travelling at right angles to the aether’s motion.

3d. The experiment had a very definitive result. Explain the result

Spoiler

The result of this experiment was a null result. This means that it did not approve or disapprove the hypothesis; in this case, it did not approve or disapprove the existence of the aether.

4. Evaluate the success of the Michelson-Morley experiment in proving or disapproving the aether model. Outline the result

Spoiler

Although it did not approve or disapprove the aether model, with hindsight, the result of the Michelson-Morley experiment has been able to help scientists in the twentieth century to reject the aether model and accept Einstein’s relativity. In this sense, it has been an important experiment in helping others to decide between competing theories, along with comparative success of relativity experiments.

It is important to note, however, that it did not sway scientific belief at that time. Aether supporters saw the null result as an indication that their model needed improvement. Einstein, although apparently aware of the Michelson-Morley result, was not influenced by it and was unconcerned with proposed aether model modifications. He was approaching the problem from an entirely different direction

5. Outline the essential aspects of inertial frame of reference and identify a method to distinguish between an inertial and non-inertial frame of reference

Spoiler

A frame that is not accelerating is called an inertial frame of reference. Motion can not be detected from an inertial frame of reference.

Motion is detectable from a non-inertial frame of reference – one which is accelerating. The principle of relativity does not hold in a non-inertial frame of reference.

Experiment: Inertial and Non-Inertial Frames of Reference

Aim: The purpose of the experiment was to distinguish between inertial and non-inertial frames of reference whilst on a train.

Method: A passenger travelling on a train records how they determined whether or not they detect the train moving and/or accelerating.

Results: Travelling at a constant speed change in position could not be felt (except for the rocking of the train) and was only detectable by looking outside the window of the train.

When the train was accelerating, the passenger is noted to have felt a force pushing them backwards. When the train was coming to rest, a force pushing forward was noted.

Conclusion: The experiment was successful in determining the difference between inertial and non-inertial frames of reference.

Discussion: While travelling at a constant speed, it was not possible to detect motion without the aid of an external point of reference (i.e. stationary objects outside the train as viewed from the window). This is therefore an inertial frame of reference.

While the train is accelerating/decelerating forces were felt by the passenger. This is therefore a non-inertial frame of reference. Thus the results prove Einstein’s theory of relativity.

6. You are in a spaceship heading, you think, in free motion towards Pluto; however you are far from any reference point to check your progress. Suddenly a comet approaches you and overtakes you, heading in the same direction. Identify which of the following interpretations of the events are correct and distinguish them

a) the comet is travelling towards Pluto, travelling at a high speed
b) your spaceship is stationary but the comet is heading towards Pluto
c) the comet is stationary and you are travelling away from Pluto
d) you are both travelling away from Pluto, but you have a higher speed

Spoiler

a could be right
u know that u are moving
so its not b
and space ships dont go backwards
u know that u are travelling forwards
ut if the comet is faster slower than u and
u are moving away from pluto then u are not moving
backwards
u hav to see the comet in front of u and go
backwards
so its not d, but a

whoever has jacaranda can you help me finish the rest
this took me ages and im only up to q6
sorry

 

[physician]

who is just so nice and deserves to be thanked time and time again for their more than generous contribution to the physics forums???

Spoiler

.................. HSC_sUcKsSsS... well done dude...

 

[skyby]

umm thanks

 

[MuffinMan]

who is just so nice and deserves to be thanked time and time again for their more than generous contribution to the physics forums???

Spoiler

.................. HSC_sUcKsSsS... well done dude...

thanks i'll try to do the questions tomorrow
i still have to do ideas to implementation and chem syllabuses
*sigh*

 

[MuffinMan]

7. Identify any method you can use to tell if your spacecraft, far from any planet or star, is standing still or travelling with an uniform velocity. What is the name given to this idea?

Spoiler

You can’t tell, inertial frame of reference

8. In classical physics, time was assumed to be absolute and the maximum velocity of interaction assumed infinite. Explain how these same concepts are viewed in the theory of relativity

Spoiler

Constant speed of light – c, holds in all reference frames
Space-time continuum – three dimensions of space + time dimension
The principle of relativity states that it is not possible to detect uniform velocity motion while within a frame of reference without referring to another frame of reference. Classical physics established the principle for mechanics but not for optics. Einstein included optics by extending the principle to include all laws of physics. Time becomes a relative term once this is accepted that the speed of light is an absolute term. Distance (or space) is also a relative term.

9. A traveller in a very fast train holds a mirror at arm’s length and looks at his reflections. Both he and the observer outside see the same velocity of light for the traveller’s image. Since velocity = distance/time, describe what has happened to the length of the traveller’s arm, and the time taken for the reflection to return, as seen by the outside observer.

Spoiler

Let the train’s speed = 0.5c (very fast speed)

Length contraction

L_V = L_O √ (1 – [ V² / C²) ]

= L_O √ (1- [ 0.5² / 1 ]

Since √ (1 – [0.5² / 1]) < 1
L_V < L_O

Therefore the traveller’s arm is shorter as seen by the outside observer

t_V = t_O / √ (1- [v² / c²

t_V = t_O / √ (1- [ {0.5c}² / c² ])

t_V = t_O / √ (1-0.25)

Since √ (1-0.25) < 1

Therefore t_V > t_O

According to the time dilation equation, time would always be perceived longer for the outside observer.

10a. Compare the current definition of the metre with the original metre standard

Spoiler

The standard of length (1875) was the distance between two lines one metre apart on a platinum-iridium alloy bar. With the realisation that the length of an object changes as the speed changes, a new definition was needed

10b. Evaulate light-time definition, such as the metre and the light year

Spoiler

Metre – because the speed of light is constant, our new definition of a metre is taken as the distance light travels in 1/299 792 458 of a second (speed of light, c = 299 792 458 ms^-1

Light year – the distance light travels in one year

(3600 x 24 x 365 x c)
Where c = speed of light in ms^-1

11. Identify the technologies required to support the current definition of the metre

Spoiler

very fast and accurate computation and analysis equipment. Add more if you wish

12. Complete the table of distances shown below. You will need the following conversion factors.
1 light year = 0.3066 parsecs = 63 240 AU = 9.4605 x 10¹² km
1 parsec = 3.2616 light years = 206 265 AU = 30.857 x 10¹² km
1 astronomical unit (AU) = 1.5813 x 10^-5 light years = 1.496 x 10⁸km

Star Distance(L/Y) Distance(parsecs) Distance(km)

Canopus 23
Rigel 900
Arcturus 10
Hadar 2.0857 x 1015

Spoiler

Canopus (L/Y) = 23 x 3.2616 = 75.0168 L/Y
Canopus (km) = 23 x 30.857 x 10¹² km

Rigel (parsecs) = 900 x 0.3066 = 275.94 parsecs
Rigel (km) = 900 x 9.4605 x 10¹² = 8.51445 x 10¹⁵ km

Arcturus (L/Y) = 10 x 3.2616 = 32.616 light years
Arcturus (km) = 10 x 30.857 x 10¹² = 3.0857 x 10¹⁴

Hadar (L/Y) = 3.0857 x 10¹⁵ / 9.4605 x 10¹² = 326.166931 light years
Hadar (parsecs) = 326.166931 x 0.3066 = 100.0027081 parsecs

 

[Libbster]

Good work HSC_sUcKsSsS!!! However, with the answer to 3a wasn't the aim to measure the relative velocity of the Earth through the aether? (according to the syllabus )

 

[sunjet]

yes it was.

 

[Abtari]

3c. Write a one paragraph description on how the apparatus worked

Spoiler

The equipment used by Michelson and Morley (an interferometer) reflected light from a common source in two directions and then back to the observation point. If the aether existed, then the light rays travelling with and against should interfere with each other more than the rays travelling at right angles to the aether’s motion.

i dont understand this. can some one pls explain.

why would the rays travelling with and against each other interfere more with each other than the rays travelling orthogonally? oh, and what do you mean by "the rays" - there is only ONE ray travelling with and against the supposed aether - there is only ONE ray travelling orthogonally to the supposed aether. they are reflected, yes, but its still only ONE. so i am a bit confused.

thanks heaps.

 

[speed2]

<<6. You are in a spaceship heading, you think, in free motion towards Pluto; however you are far from any reference point to check your progress. Suddenly a comet approaches you and overtakes you, heading in the same direction. Identify which of the following interpretations of the events are correct and distinguish them

a) the comet is travelling towards Pluto, travelling at a high speed
b) your spaceship is stationary but the comet is heading towards Pluto
c) the comet is stationary and you are travelling away from Pluto
d) you are both travelling away from Pluto, but you have a higher speed>>

if there isnt a reference point how can any of these be correct???

 

[Shael]

3c. Write a one paragraph description on how the apparatus worked

Spoiler

The equipment used by Michelson and Morley (an interferometer) reflected light from a common source in two directions and then back to the observation point. If the aether existed, then the light rays travelling with and against should interfere with each other more than the rays travelling at right angles to the aether’s motion.

i dont understand this. can some one pls explain.

why would the rays travelling with and against each other interfere more with each other than the rays travelling orthogonally? oh, and what do you mean by "the rays" - there is only ONE ray travelling with and against the supposed aether - there is only ONE ray travelling orthogonally to the supposed aether. they are reflected, yes, but its still only ONE. so i am a bit confused.

thanks heaps.

How to explain... well basically:

The Aether is like an invisible wind that scientists thought effected the speed of light, similar to how normal wind blowing effects a car driving through it causing friction and thus a slow down of speed of the car.
So Michelson and Morley carried out an experiment to see if there was such a wind that slowed the speed of light. They did this by shooting a beam of light into mirrors that would split the light beam into to two beams perpendicular to each other. One beam was went across the Aether, the other into the Aether. The idea was that the beam of light shot into the direction of the Aether would be reflected back after the other beam and once and for all prove that there was an Aether wind. However, the experiment failed many times with all sorts of changes (rotating the system, different times of the year, and altitudes). The failure was shown using an interferometer that displays interference patterns and the result was that there were no interference patterns (beams stayed in phase). Therefore both split beams returned in phase (at the same time).

Heres a little diagram I drew up to help explain:

http://darknight.bluewolf72.com/mmexp.jpg

 

[eleco]

"The equipment used by Michelson and Morley (an interferometer) reflected light from a common source in two directions and then back to the observation point. If the aether existed, then the light rays travelling with and against should interfere with each other more than the rays travelling at right angles to the aether’s motion. "

Are you sure this is correct. They used a half silvered mirror to split a single beam in two, one beam was reflect at 90 degrees to the other beam that was transmitted through the mirror... The interferometer was used as a detector nothing else... or am i wrong

 

[velox]

Yeh that seems wrong.

 

[Shael]

Thats right eleco, and I mentioned that in my explanation above and with the diagram.

 

[eleco]

Yeh that seems wrong.

Are you sure??

 

[velox]

i was agreeing with you................

 

[Bokky]

tV = tO / √ (1- [v2 / c2

tV = tO / √ (1- [ {0.5c}2 / c2 ])

tV = tO / √ (1-0.25)

^ One thing i dont understand about that is why the 0.5 gets squared when the c^2 is cancelled out with the c^2 on the bottom so 0.5c^2/c^2 should be 0.5 therefore (1-(0.5) ) doesnt it? can someone explain it to me?

 

[wanton-wonton]

6. You are in a spaceship heading, you think, in free motion towards Pluto; however you are far from any reference point to check your progress. Suddenly a comet approaches you and overtakes you, heading in the same direction. Identify which of the following interpretations of the events are correct and distinguish them

a) the comet is travelling towards Pluto, travelling at a high speed
b) your spaceship is stationary but the comet is heading towards Pluto
c) the comet is stationary and you are travelling away from Pluto
d) you are both travelling away from Pluto, but you have a higher speed

Spoiler

a could be right
u know that u are moving
so its not b
and space ships dont go backwards
u know that u are travelling forwards
ut if the comet is faster slower than u and
u are moving away from pluto then u are not moving
backwards
u hav to see the comet in front of u and go
backwards
so its not d, but a

whoever has jacaranda can you help me finish the rest
this took me ages and im only up to q6
sorry

Where did you get that question? You said 'You know that you're moving'. How do you know that when there's nothing outside the spaceship to refer to? For all you know, you could be stationary or moving at a constant velocity.

 

[richz]

wth

the eqn is tv = to / root(1-v^2/c^2)

so u have to square the 0.5

 

[Shael]

tV = tO / √ (1- [v2 / c2

tV = tO / √ (1- [ {0.5c}2 / c2 ])

tV = tO / √ (1-0.25)

^ One thing i dont understand about that is why the 0.5 gets squared when the c^2 is cancelled out with the c^2 on the bottom so 0.5c^2/c^2 should be 0.5 therefore (1-(0.5) ) doesnt it? can someone explain it to me?

0.5c means the velocity of the object is in terms of c, or a fraction of c which is what 0.5c is. And this is exactly the same as ( v^2 / c^2 ) which is a fraction c. Which is also the same as ( v / c )^2. So if we already have what fraction of c the object is travelling at it simply becomes ( fraction of c )^2 and in this case that is
( 0.5 )^2.

Hope that helps.

 

[Bokky]

ah yeah, i get it now, u have to square everything in the brackets so u get 0.25c^2 and the c^2 cancel out and u get 0.25. lol silly question but o well i get it now