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Eclipse geometry for IB Maths SL and HL

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, total and annular eclipses and the Moon’s shadow, an extension and full worked answers. Use it any time, or when an eclipse is in the news.

Level
IB Maths SL and HL (AA and AI)
Time
40 minutes, plus a 10-minute extension
Topics
Radians and arc length; Right-angled trigonometry; Similar triangles; Approximation and percentage error
Equipment
The starter is non-calculator. A GDC 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 B12 minTotal or annular?
Main: task C11 minThe Moon’s shadow
Extension10 minFast finishers or homework

Starter (5 minutes)

No GDC.

  1. Convert 0.5° to radians, in terms of π.
  2. Find the arc length of a sector of radius 10 and angle 0.2 radians.
  3. At the same moment, a 1.8 m person casts a 2.4 m shadow and a tree casts a 12 m shadow. How tall is the tree?
  4. Write down tan(π/4).

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 these rounded values: Sun diameter 1 400 000 km at 150 000 000 km; Moon diameter 3500 km at 380 000 km.

  1. Find the angle the Sun fills, in degrees, to 3 significant figures.
  2. Find the angle the Moon fills, in degrees, to 3 significant figures.
  3. For small angles, θ ≈ D/d radians. Use this to find the Moon’s angle in radians and in degrees, each to 3 significant figures. Is the approximation good?

Task B: Total or annular? (12 min)

The Moon’s distance varies. Take 360 000 km when it is nearer and 405 000 km when it is further away. If the Moon looks bigger than the Sun, an eclipse can be total; if smaller, the eclipse is annular (a ring of Sun is left).

  1. Find the Moon’s angular size when it is nearer, in degrees to 3 significant figures. Can the eclipse be total?
  2. Find the Moon’s angular size when it is further away, to 3 significant figures. What kind of eclipse happens?
  3. Use similar triangles to find the distance at which the Moon exactly covers the Sun.

Task C: The Moon’s shadow (11 min)

The Moon’s dark shadow (the umbra) is a cone. Its tip is where the edges of the Sun and Moon line up. Take the Sun to be 150 000 000 km from the Moon, and ignore the size of the Earth.

  1. Let L be the length of the shadow cone behind the Moon. Explain why L/3500 = (L + 150 000 000)/1 400 000 and find L to 3 significant figures.
  2. With the Moon 380 000 km away, by how much does the cone fail to reach us? Give 3 significant figures.
  3. With the Moon 360 000 km away the cone reaches us. 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 in a card makes an image of the Sun on a screen 1 m behind it. Using the rounded values above, find the diameter of the image in millimetres, 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 C is the key idea: the Moon’s shadow only just reaches the Earth, which is why total eclipses are seen only along a narrow path.

Starter

  1. π/360
    • 0.5 × π/180 = π/360
  2. 2
    • rθ = 10 × 0.2 = 2
  3. 9 m
    • 12 × 1.8/2.4 = 9
  4. 1
    • tan 45° = 1

Task A: Angular size

  1. 0.535°
    • 2 arctan(700 000/150 000 000) = 0.5347…°
  2. 0.528°
    • 2 arctan(1750/380 000) = 0.5277…°
  3. 0.00921 rad = 0.528°: very good
    • 3500/380 000 = 0.009210… rad
    • × 180/π = 0.5277…°: the same as the exact value to 3 s.f.

Task B: Total or annular?

  1. 0.557°: yes, it is bigger than the Sun’s 0.535°
    • 2 arctan(1750/360 000) = 0.5570…°
  2. 0.495°: annular
    • 2 arctan(1750/405 000) = 0.4951…°, smaller than the Sun’s.
  3. 375 000 km
    • d/3500 = 150 000 000/1 400 000
    • d = 3500 × 150 000 000/1 400 000 = 375 000 km

Task C: The Moon’s shadow

  1. 376 000 km
    • Similar triangles with a common tip: diameter ÷ distance from the tip is the same for the Moon and the Sun.
    • 1 400 000L = 3500L + 525 000 000 000, so L = 525 000 000 000/1 396 500 = 375 939.8…
  2. 4060 km
    • 380 000 − 375 939.8 = 4060.2…
  3. 148 km
    • The diameter shrinks linearly from 3500 km at the Moon to 0 at the tip.
    • 3500 × (1 − 360 000/375 939.8…) = 148.4…

Extension

  1. 9.33 mm
    • Image/1000 mm = 1 400 000/150 000 000
    • Image = 1000 × 1 400 000/150 000 000 = 9.333… mm

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

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