Solutions — Distance on Earth’s Surface

Fluency Same meridian, 10°N and 50°N.

θ = 50° − 10° = 40°
d = (π × 6400 × 40) / 180 ≈ 4 468 km

Fluency On equator at 45°W and 75°E.

θ = 45° + 75° = 120° (opposite sides of prime meridian)
d = (π × 6400 × 120) / 180 = (π × 6400 × 2) / 3 ≈ 13 404 km

Fluency Convert 30° arc.

(a) d = (π × 6400 × 30) / 180 = (π × 6400) / 6 ≈ 3 351 km
(b) 30° = 30 × 60 = 1 800 nautical miles; 1 800 × 1.852 ≈ 3 334 km ✓ (consistent)

Fluency At 50°N, 40° longitude separation.

r = 6400 × cos(50°) = 6400 × 0.6428 = 4114 km
d = (π × 4114 × 40) / 180 ≈ 2 872 km

Understanding Cairns (17°S) to Port Moresby (9°S), same meridian approximation.

θ = 17° − 9° = 8° (both south, closer to equator is Port Moresby)
d = (π × 6400 × 8) / 180 ≈ 893 km

Understanding At 60°S, 90° longitude separation.

(a) Distance along parallel (small circle):
r = 6400 × cos(60°) = 6400 × 0.5 = 3200 km
d = (π × 3200 × 90) / 180 = π × 3200 / 2 = π × 1600 ≈ 5 027 km

(b) Note: (60°S, 0°) to (60°S, 90°E) are NOT on the same meridian, so we cannot directly apply the meridian formula for great circle distance. The great circle distance would require more advanced spherical trigonometry. At this level, the parallel route (5 027 km) is what’s calculated.

Understanding Express 3000 km as degrees.

d = (πRθ) / 180 → θ = (180d) / (πR)
θ = (180 × 3000) / (π × 6400) = 540 000 / 20 106 ≈ 26.86°

Understanding Ship east along 30°S from 20°E to 80°E.

θ = 80° − 20° = 60°, φ = 30°
r = 6400 × cos(30°) = 6400 × 0.8660 = 5542 km
d = (π × 5542 × 60) / 180 = (π × 5542) / 3 ≈ 5 801 km

This is not a great circle route. The ship is travelling along a parallel of latitude (a small circle), which is not the shortest path.

Problem Solving London (51.5°N, 0°) to Moscow (55.8°N, 37.6°E), same meridian approximation.

Using same meridian approximation: θ = 55.8° − 51.5° = 4.3°
d = (π × 6400 × 4.3) / 180 ≈ 480 km

Note: This is only the north-south component. The actual great circle distance from London to Moscow is approximately 2 500 km because Moscow is significantly east as well as slightly north. The same-meridian approximation is poor here due to the large longitude difference (37.6°).

Problem Solving (40°N, 30°E) to (40°N, 90°E) by two routes.

(a) Along parallel of latitude (40°N):
Δλ = 90° − 30° = 60°
r = 6400 × cos(40°) = 6400 × 0.7660 = 4902 km
d = (π × 4902 × 60) / 180 = (π × 4902) / 3 ≈ 5 132 km

(b) Via equator (three great circle segments):
Leg 1: (40°N, 30°E) to equator (0°, 30°E): θ = 40° → d = (π × 6400 × 40)/180 ≈ 4 468 km
Leg 2: (0°, 30°E) to (0°, 90°E): θ = 60° → d = (π × 6400 × 60)/180 ≈ 6 702 km
Leg 3: (0°, 90°E) to (40°N, 90°E): θ = 40° → d ≈ 4 468 km
Total via equator: 4 468 + 6 702 + 4 468 ≈ 15 638 km

The parallel route (5 132 km) is shorter because it stays close to both cities. The detour via the equator is much longer. Note: the true great circle route between these two points is even shorter than the parallel route (≈ 4 800 km).