The Universe

1. The Universe Revision

2. Gravity and mass

3. What’s the difference between mass and weight? • Mass is how much matter (material) is in an object and is measured in kg. • Weight is the force of gravity acting on an object (of known mass) and is measured in N kg-1.

4. This law states that there is a force of attraction between any two objects in the universe. The size of the force is proportional to the product of the masses of the two objects, and inversely proportional to the square of the distance between them. m1 and m2 are the masses of the two objects, and r is the distance between them. G = gravitational constant = 6.67 x 10-11 Nm2kg-2 Newton’s Inverse Square Law of Gravitation

5. What is the force of attraction between two pupils of average mass (60 kg) sitting 1.5 metres apart? F = Gm1m2 r2 = 6.67 x 10-11 x 60 x 60 1.52 = 1.07 x 10-7 N Example 1

6. Value of r • It is important to realise that the value for r, the distance between two masses, is the distance between the centre of the two masses. i.e. if we consider the force between the Earth and the Moon then the value of r is the distance from the centre of the Earth to the centre of the Moon.

7. Taking the radius of the Earth to be 6.4 x 106 m, find the force of attraction on a 250kg satellite that is orbiting at a height of 36 000km above the Earth. (mass of Earth = 6.0  1024 kg) Example 2 • This question should be broken down into two parts. • First of all, find the distance, r, between the two objects. • Use Newton’s Universal Law

8. 1. r = radius of Earth + the height above the Earth = 6.4 x 106 + 36 000 x 103 = 4.24 x 107 m 2. F = Gm1m2 r2 = 6.67 x 10-11 x 6.0  1024 x 250 (4.24 x 107)2 =55.7N Example 2

9. 2012 Revised Higher

10. 2013 Revised HigherOpen-ended question

11. Special Relativity

12. Time dilation • Imagine a lamp which sends a pulse of light at the same time as producing a click. • The light is reflected from a mirror, at a known distance, D, from the lamp. • When it arrives back at the lamp it produces a second click. • The total time will be: t = 2D c

13. Time dilation • Now imagine that the two lamps are moving at an identical horizontal velocity. • To an observer moving with the lamps nothing will have changed. • However, if there is a stationary observer watching the lamps move he will see the pulses of light take a different path and move a longer distance, 2h.

14. Time dilation • The time between clicks in this case will be: • t = 2h c • Therefore, time will be different for two observers watching an identical system (as h is clearly bigger than D).

15. What is meant by time dilation? • The time observed in a moving system will always be greater than that measured in the stationary frame of reference. • Time dilation is a difference in a time interval as measured by a stationary observer and a moving observer. • ie a stationary observer will record a greater time than a moving observer for the same journey travelling at speeds close to the speed of light.

16. Equation for time dilation • t’ = t . √1 – v2 c2 • t’ = time reference for the stationary observer • t = time reference for the moving observer • v = velocity of moving observer • c = 3 x 108 ms-1 • NB: v is often given as a unit of c i.e. 0.7c. In this case v = 0.7 and c = 1

17. Example 1 • A spacecraft leaves Earth and travels at a constant speed of 0.6c to its destination. An astronaut on board records a flight time of 5 days. • Calculate the time taken for the journey as measured by an observer on Earth. • t’ = t . √1 – v2 c2 • t’ = 5 . √1 – 0.62 12 • t’ = 6.25 days

18. Example 2* • A rocket leaves a planet and travels at a constant speed of 0.8c to a destination. An observer on the planet records a time of 20h. • Calculate the time taken for the journey as measured by the astronaut on board. • t’ = t . √1 – v2 c2 • 20 = t . √1 – 0.82 12 • 20 x √1 – 0.82 = t 12 • t = 20 x (0.6) • t = 12 h

19. 2012 Revised Higher C

20. 2014 Revised Higher B

21. The Doppler Effect

22. What is the Doppler effect? • The Doppler effect is the change in frequency you notice when a source of sound waves is moving relative to you. • When the source moves towards you, more waves reach you per second and the frequency is increased. • If the source moves away from you, fewer waves reach you per second and the frequency is decreased. • Doppler Shift Demonstrator

23. Calculating the frequency Moving towardsthe source • The observed frequency, fo, is higher: • fo = fs v . (v - vs) • fs = frequency of source • v = speed of sound (approx 340ms-1) • vs = speed of source • Towards = Take away

24. Calculating the frequency Moving awayfrom the source • The observed frequency, fo, is lower: • fo = fs v . (v + vs) • Away = Add

25. Example 1 • What is the frequency heard by a person driving at 15 ms-1toward a blowing factory whistle (f = 800 hz) if the speed of sound in air is 340 ms-1? • fo = fs v . (v - vs) = 800 340 . (340-15) = 800 x 1.04 fo = 837 Hz

26. Example 2 • What frequency would he hear after passing the factory if he continues at the same speed? • fo = fs v . (v + vs) = 800 340 . (340+15) = 800 x 0.931 fo = 745 Hz

27. 2013 Revised Higher C

28. Redshift

29. Background information • White light (light from galaxies and stars) is broken up into all the colours of the rainbow • RedOrange Yellow GreenBlue Indigo Violet • Longer λshorterλ • All the colours have different wavelengths

30. What is redshift? • Redshift (also known as Doppler shift) is how much the frequency of light from a far away object has moved toward the red end of the spectrum. • It is a measure of how much the ‘apparent’ wavelength of light has been increased. • It has the symbol Z and can be calculated using the following equation: • Z = λo – λr it canalso expressed as: Z = λo - 1 λr λr • λo = the wavelength observed • λr = the wavelength at rest

31. What is a blueshift? • When we use the equation for redshift, we can sometimes end up with a –ve value. • This means the object is moving closer to you and is said to be blueshifted. • It is a measure of how much the ‘apparent’ wavelength of light has been decreased.

32. Redshift and velocity • We can also work out the redshift if we know the velocity that the body is moving at (for slow moving galaxies): • Z = v c

33. Wavelengths • With a redshift, moving away, the wavelength increases. • With a blueshift, moving towards, the wavelength decreases.

34. Example 1 • Light from a distant galaxy is found to contain the spectral lines of hydrogen. The light causing one of these lines has (an observed) measured wavelength of 466 nm. When the same line is observed (at rest) from a hydrogen source on Earth it has a wavelength of 434 nm. (a) Calculate the Doppler shift, z, for this galaxy. (b) Calculate the speed at which the galaxy is moving relative to the Earth. (c) In which direction, towards or away from the Earth, is the galaxy moving? (a) Z = λo – λr λr = 466 - 434 434 Z = 0.074

35. Example 1 • Z = v c 0.074 = v . 3 x 108 v = 2.21 x 107 ms-1 (c) Z is positive therefore galaxy is moving away

36. Example 2 • A distant star is travelling directly away from the Earth at a speed of 2·4 × 107 ms1. (a) Calculate the value of z for this star. • A hydrogen line in the spectrum of light from this star is measured to be 443 nm. Calculate the wavelength of this line when it observed from a hydrogen source on the Earth. (a) Z = v / c = 2.4 x 107 / 3 x 108 = 0.08 • Z = λo - 1 λr 0.08 = (443x10-9) – 1 λr 0.08 + 1 = (443x10-9) λr λr =(443x10-9) 0.08 + 1 λr = 410 x 10-9 m / 410 nm

37. Revised Higher 2013 A

38. Revised Higher 2014 Qu: 25(b)

39. Revised Higher 2014 Qu: 25(b)

40. Specimen Paper Qu: 6(a)

41. Solution

42. Hubble’s law

43. Hubble’s Law • The astronomer Edwin Hubble noticed in the 1920s that the light from some distant galaxies was shifted towards the red end of the spectrum. • The size of the shift was the same for all elements coming from the galaxies. • This shift was due to the galaxies moving away from Earth at speed.

44. The bigger the shift the faster the galaxy moves • Hubble found that the further away a galaxy was the faster it was travelling. • The relationship between the distance and speed of a galaxy is known as Hubble’s Law: v = Ho d • Ho = Hubble’s constant = 2.3 x 10-18 s-1

45. The value of Ho = 2.3 x 10-18 s-1 is given in data sheet (and is the value you would use in an exam) but can vary as more accurate measurements are made. The gradient of the line in a graph of speed v distance of galaxies provides a value for Hubble’s constant. Hubble’s Constant

46. Example 1 • What is the speed of a galaxy relative to Earth that is at an approximate distance of 4.10 × 1023 m from Earth? • v = Ho d • v = 2.3 x 10-18 x 4.10 x 1023 • v = 9.43 x 105 ms-1

47. What is a light year? • Sometimes distances can be given in light years. • One light year is the distance travelled by light in one year. • It can be calculated as follows using d = vt: • 3 x 108 (speed of light) x 365 (days) x 24 (hours) x 60 (mins) x 60 (s) • One light year = 9.46 x 1015 m

48. Specimen Paper Qu: 6(b)

49. 2012 Revised Higher E

50. v = Ho d 5.5 x 105 = 2.3 x 10-18 x d d = 5.5 x 105 2.3 x 10-18 d = 2.39 x 1023 m Revised Higher (specimen paper)