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Eclipse geometry for A Level

A ready-to-teach lesson on the geometry of eclipses: why the Moon can only just cover the Sun. A 5-minute starter, a 35-minute main activity on angular size, small-angle approximations and the Moon’s shadow, an extension and full worked answers. Use it any time, or when an eclipse is in the news.

Level
A Level Maths (Year 12 and Year 13)
Time
40 minutes, plus a 10-minute extension
Topics
Radians and small-angle approximations (Year 13); Trigonometry; Similar triangles and ratio; Percentage error
Equipment
The starter is non-calculator. A calculator is needed for the main activity.

Download student sheet (PDF)Answers (PDF)

Suggested timings

PartTimeWhat
Starter5 minQuick questions on the board
Main: task A12 minAngular size
Main: task B11 minTotal or annular?
Main: task C12 minThe Moon’s shadow
Extension10 minFast finishers or homework

Starter (5 minutes)

No calculator.

  1. Convert 30° to radians.
  2. Using sin θ ≈ θ, estimate sin 0.01.
  3. Find the arc length of a sector of radius 5 and angle 0.3 radians.
  4. Using cos θ ≈ 1 − θ2/2, estimate cos 0.2.

Main activity (35 minutes)

Task A: Angular size (12 min)

A ball of diameter D at distance d fills an angle θ = 2 arctan(D/(2d)). Take: Sun diameter 1 400 000 km at 150 000 000 km; Moon diameter 3500 km at 380 000 km.

  1. Find the Sun’s angular size in degrees, to 3 significant figures.
  2. Using tan θ ≈ θ, show that θ ≈ D/d. Find the Moon’s angular size in radians and in degrees, to 3 significant figures.
  3. Find the percentage error of the approximation D/d compared with the exact 2 arctan(D/(2d)) for the Moon, to 2 significant figures.

Task B: Total or annular? (11 min)

Take the Moon’s distance as 360 000 km when nearer and 405 000 km when further away. The eclipse can be total only when the Moon looks bigger than the Sun.

  1. Find the Moon’s angular size at 360 000 km and at 405 000 km, in degrees to 3 significant figures.
  2. Find the distance at which the Moon exactly covers the Sun.
  3. Taking the Moon’s distance to vary evenly between 360 000 and 405 000 km, what fraction of that range allows a total eclipse?

Task C: The Moon’s shadow (12 min)

The Moon’s umbra is a cone whose tip is where the edges of the Sun and Moon line up. Take the Sun as 150 000 000 km from the Moon and ignore the size of the Earth.

  1. Find the length L of the shadow cone, to 3 significant figures.
  2. Find the half-angle of the shadow cone, in radians, to 3 significant figures.
  3. Find the diameter of the shadow 360 000 km behind the Moon, to 3 significant figures.

Extension (10 minutes)

Seeing the Sun safely.

  1. A pinhole projector makes an image of the Sun on a screen s metres behind the pinhole. Find s for an image 2 cm across, to 3 significant figures.

For teachers

Teacher notes and full worked answers
  • Safety first: never look at the Sun directly, even during an eclipse. A pinhole projector (extension) is a safe way to see it.
  • Every physical value is a rounded value given in the question; the real values vary.
  • Task A (c) shows why small-angle approximations are so useful in astronomy: the error is far smaller than the uncertainty in the data.

Starter

  1. π/6
    • 30 × π/180
  2. 0.01
    • θ is small and in radians.
  3. 1.5
    • 5 × 0.3 = 1.5
  4. 0.98
    • 1 − 0.04/2 = 0.98

Task A: Angular size

  1. 0.535°
    • 2 arctan(700 000/150 000 000) = 0.5347…°
  2. 0.00921 rad = 0.528°
    • tan(θ/2) = D/(2d), and for small θ, θ/2 ≈ D/(2d).
    • 3500/380 000 = 0.009210…; × 180/π = 0.5277…°
  3. 0.00071%
    • Exact: 0.00921046…; approximate: 0.00921053…
    • (0.00921053 − 0.00921046)/0.00921046 × 100 = 0.000707…%

Task B: Total or annular?

  1. 0.557° and 0.495°
    • 2 arctan(1750/360 000) = 0.5570…°
    • 2 arctan(1750/405 000) = 0.4951…°
  2. 375 000 km
    • Equal angles: 3500/d = 1 400 000/150 000 000, so d = 375 000 km.
  3. 1/3
    • Total when d < 375 000: (375 000 − 360 000)/(405 000 − 360 000) = 15 000/45 000 = 1/3.

Task C: The Moon’s shadow

  1. 376 000 km
    • L/3500 = (L + 150 000 000)/1 400 000 gives L = 525 000 000 000/1 396 500 = 375 939.8…
  2. 0.00465 rad
    • tan α = 1750/L = 0.004655…, and α ≈ tan α for small angles: 0.00465.
  3. 148 km
    • 3500 × (1 − 360 000/375 939.8…) = 148.4…

Extension

  1. 2.14 m
    • Image/s = 1 400 000/150 000 000, so s = 0.02 × 150 000 000/1 400 000 = 2.142…

The same answers are in the answers PDF. Every answer was recomputed by computer before publishing.

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