Astrophysics for Cambridge IGCSE Physics

Reading cue. The 1.1 × 10¹¹ m is measured from the centre of the star, so it is the orbit radius, not the distance travelled; 1.9 × 10⁷ s is the time for one full lap of the dashed circle.

1. aCalculate Calculate the distance the planet travels in going once round the dashed circle in Fig. 3.1.

   recall2 marks

   Check answer · 2 marks

   1. distance = 2πr with r = 1.1 × 10¹¹ m (1)
   2. distance = 6.9 × 10¹¹ m (1)

2. bDetermine Light travels at 3.0 × 10⁸ m/s. Determine, in minutes, the time light takes to travel from the star to the planet at the position drawn in Fig. 3.1.

   routine3 marks

   Check answer · 3 marks

   1. t = d / v = 1.1 × 10¹¹ / 3.0 × 10⁸ (1)
   2. t = 367 s (1)
   3. t = 6.1 minutes (1)

3. cExplain The arrow at the planet in Fig. 3.1 shows the direction in which it is travelling. Explain how the direction of the star's gravitational pull on the planet is related to that arrow, and explain why this pull changes the planet's direction without changing its speed.

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   Check answer · 4 marks

   1. the gravitational force acts from the planet towards the centre of the star (1)
   2. it is therefore at right angles to the arrow showing the direction of travel (1)
   3. so the force has no part of it along the direction of motion, and does no work on the planet (1)
   4. it changes only the direction of the velocity, so the planet keeps a constant speed while continually turning (1)

4. dExplain The orbit is drawn as a dashed circle, but the planet's path is described only as nearly circular. Explain how the planet's distance and speed would vary round a slightly squashed orbit, and explain what this means for any single speed calculated from Fig. 3.1.

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   Check answer · 4 marks

   1. the distance from the star would vary round the orbit rather than staying at 1.1 × 10¹¹ m (1)
   2. the planet moves fastest at the point of the orbit closest to the star (1)
   3. and slowest at the point furthest from the star (1)
   4. a value found from circumference ÷ period is therefore only an average speed for the whole orbit (1)

Reading cue. Both strips share one wavelength scale increasing to the right, and the labelled gap is the distance between two positions of one line — it is not itself a wavelength read off the scale.

1. aIdentify Identify which of the two strips in Fig. 6.1 was recorded in the laboratory, and state what must be true of the two lines being compared for the labelled gap to have any meaning.

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   Check answer · 2 marks

   1. the upper strip is the laboratory spectrum (1)
   2. both lines must come from the same element / be the same spectral line (1)

2. bCalculate The marked line lies at 486 nm in the laboratory strip and the same line lies at 500 nm in the galaxy strip. Calculate the increase in wavelength labelled on Fig. 6.1, and calculate it as a percentage of the laboratory wavelength.

   routine3 marks

   Check answer · 3 marks

   1. increase = 500 − 486 = 14 nm (1)
   2. percentage = 14 / 486 × 100 (1)
   3. = 2.9 % (1)

3. cExplain Light from a second galaxy shows the same line shifted twice as far to the right along the scale as the shift drawn in Fig. 6.1. Explain what this tells you about the second galaxy, in terms of both its speed and its distance.

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   Check answer · 4 marks

   1. a larger shift towards longer wavelengths means a greater speed of recession (1)
   2. the second galaxy is moving away about twice as fast (1)
   3. speed of recession is proportional to distance (1)
   4. so the second galaxy is roughly twice as far away as the one drawn (1)

4. dSuggest A star within our own Galaxy gives the same line at almost exactly the laboratory position, with a shift far too small to draw on Fig. 6.1. Suggest what this shows about the star, and suggest why such a measurement is of little use for finding a value of the Hubble constant.

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   Check answer · 4 marks

   1. the star is moving away from us very slowly, or hardly at all (1)
   2. it is extremely close compared with distant galaxies, and is held within our own Galaxy rather than being carried apart by the expansion (1)
   3. a shift that small cannot be measured accurately, so the speed found from it is very uncertain (1)
   4. dividing a very uncertain speed by a very small distance would give a value of the constant with an enormous uncertainty (1)

Reading cue. This is a plan view and not to scale: check what each dashed circle is drawn around before you name it, and note that both arrows go the same way round.

1. aIdentify Identify what each of the two dashed circles in Fig. 7.1 represents, and state the body at the centre of each one.

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   Check answer · 2 marks

   1. the large circle is the path of the Earth, centred on the Sun (1)
   2. the small circle is the path of the Moon, centred on the Earth (1)

2. bExplain The Sun is the only source of light drawn in Fig. 7.1. Explain why an observer on Earth sees the whole of the Moon's lit face when the Moon reaches the point of the small circle furthest from the Sun, and identify where on that circle the Moon is almost invisible.

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   Check answer · 3 marks

   1. the Moon is seen only by the sunlight it reflects (1)
   2. at the far point the lit half of the Moon faces the Earth, so the whole lit disc is seen (1)
   3. on the side of the small circle nearest the Sun, the lit half faces away from the Earth, so almost none of it can be seen (1)

3. cDetermine The Moon takes 27 days to travel once round the small dashed circle and the Earth spins once on its axis in 24 hours, both in the directions shown by the arrows. Determine the angle the Moon moves round its orbit in one day, and hence determine the extra time the Earth must spin each day before the observer faces the Moon again.

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   Show a hint

   In one day the Earth must turn through 360° plus the extra angle the Moon has moved on round its own circle.

   Check answer · 4 marks

   1. angle moved in one day = 360 / 27 = 13.3° (1)
   2. after one complete spin the Earth must turn a further 13.3° to point at the Moon again (1)
   3. extra time = 13.3 / 360 × 24 hours (1)
   4. = 0.89 hours, about 53 minutes (1)

4. dSuggest Fig. 7.1 is drawn not to scale. Suggest two ways in which a true scale drawing would look different, and suggest why the diagram is drawn as it is.

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   Check answer · 3 marks

   1. the Moon's orbit would be very much smaller compared with the Earth's orbit (1)
   2. the Sun would be drawn very much larger than the Earth, and both would be far too small to see beside orbits of this size (1)
   3. the diagram is drawn out of scale so that the directions of travel and the relative positions of the three bodies can all be shown on one page (1)

Reading cue. Each axis is labelled 'quantity divided by a power of ten', so the numbers along it are not the measured values — convert every reading into the axis units before plotting or reading off.

1. aState A galaxy is 8.0 × 10²⁴ m away and is receding at 1.8 × 10⁷ m/s. State the number to be plotted on each axis of the grid in Fig. 10.1 for this galaxy.

   recall2 marks

   Check answer · 2 marks

   1. plotted at 8.0 on the distance axis (1)
   2. 1.8 × 10⁷ m/s = 18 × 10⁶ m/s, so plotted at 18 on the speed axis (1)

2. bExplain Explain why the straight line of best fit drawn on the grid in Fig. 10.1 must be taken through the origin, and state what it would mean if one plotted point lay well above that line.

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   Check answer · 3 marks

   1. a galaxy at zero distance from us would have zero speed of recession (1)
   2. so the line must pass through the point (0, 0) (1)
   3. a point well above the line is a galaxy receding faster than the trend for its distance, from a measurement error or from its own motion within a cluster (1)

3. cDetermine A straight line of best fit through the origin is drawn on Fig. 10.1 and reaches the right-hand edge of the grid at the point d = 10, v = 22. Determine the speed of recession of a galaxy 3.5 × 10²⁴ m away, in m/s.

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   Check answer · 3 marks

   1. gradient in plotted units = 22 / 10 = 2.2 (1)
   2. reading at d = 3.5 gives v = 3.5 × 2.2 = 7.7 in plotted units (1)
   3. v = 7.7 × 10⁶ m/s (1)

4. dExplain A fifth galaxy is measured at a distance of 2.4 × 10²⁵ m, receding at 5.0 × 10⁷ m/s. Explain why it cannot be plotted on the grid in Fig. 10.1, state whether it follows the same relationship as the line of best fit in the previous part, and describe the change needed to each axis so that all five galaxies can be shown.

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   Check answer · 5 marks

   1. it would be plotted at 24 on the distance axis, which stops at 10 (1)
   2. it would be plotted at 50 on the speed axis, which stops at 25 (1)
   3. v / d = 50 / 24 = 2.1 in the plotted units, close to the gradient of 2.2, so it does follow the same relationship (1)
   4. extend the distance axis to at least 25, for example a numbered gridline every 2.5 (1)
   5. extend the speed axis to at least 50, for example a numbered gridline every 10, keeping both scales linear so the points still spread across the grid (1)