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Key Areas covered Hubble’s law shows the relationship between the recession velocity of a galaxy and its distance from us. Hubble’s law allows us to estimate.

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Presentation on theme: "Key Areas covered Hubble’s law shows the relationship between the recession velocity of a galaxy and its distance from us. Hubble’s law allows us to estimate."— Presentation transcript:

1 Key Areas covered Hubble’s law shows the relationship between the recession velocity of a galaxy and its distance from us. Hubble’s law allows us to estimate the age of the Universe.

2 What we will do today Use Hubble’s law to calculate how far away a galaxy is. Describe how we can use Hubble’s law to estimate the age of the Universe.

3 Hubble’s law

4 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.

5 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 s-1

6 Hubble’s Constant The value of
Ho = 2.3 x 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. As you can see from the graph, v and d are directly proportional

7 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 x 4.10 x 1023 v = 9.43 x 105 ms-1

8 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

9 Specimen Paper Qu: 6(b)

10 2012 Revised Higher E

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

12 Questions You can now attempt the questions on Hubble’s Law in your class jotter Note the different value for Hubble’s law here – this should be changed to 2.3 x10-18 s-1(not 2.4 as in the booklet)

13 Questions 3. 1·63 × 105 5·44 × 10−4 4·35 × 106 1·45 × 10−2 7·39 × 1023
5·67 × 10−3 9·61 × 1023 7·36 × 10−3 4. (a) 7·37 × 10−2 (b) 2·21 × 107 ms−1 (c) Away 5. (a) 1·49 × 107 ms−1 (b) 6·48 × 1024 m 6. (a) 2·22 × 107 ms−1 (b) 9·65 × 1024 m 7. (a) 8 × 10−2 (b) 410 nm 8. (a) 3·0 × 107 ms−1 (b) 1·3 × m, 1·38 × 109 light years 9. 2·18 × 107 ms−1 10. (a) Teacher Check – use v=fλ (b) 4·55 × 1014 Hz 11. (a) 1·23 × 107 ms−1 (b) 565 million light years 12. 1·83 × 106 ms−1 13. 3·24 × 10−12 m


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