Question texta) Which proposition correctly states Euler’s fixed point theorem? The most general displacement of a Answer 1 Question 6[select: , rigid body, fluid] over the surface of Answer 2 Question 6[select: , a sphere, an ellipsoid] can be regarded as a rotation about Answer 3 Question 6[select: , 1 axis which passes, 2 axes which pass, 3 axes which pass] through the center of that Answer 4 Question 6[select: , ellipsoid, sphere]b) Which of the following surface features can be used as a reference frame for determining the absolute plate motion? Answer 5 Question 6[select: , Oceanic Trenches, Magnetic North Pole, Geographical North, Hawaiian Archipelago]c) U Chicago lies at a latitude of 41.79oN. Calculate the linear velocity of the U Chicago due to the rotation of the Earth around the North Pole axis. Radius of the Earth is 6371 km and a day has a period of exactly 24 hours. Give your answer to 3 significant figures in m/s. Speed = Answer 6 Question 6[input] m/sd) What is the maximum speed of a point at a depth of 11 km inside the Earth about the North-South pole axis of the Earth? Give your answer to three significant figures in m/s. Maximum speed = Answer 7 Question 6[input] m/s

Question

BlackTom AI · Accepted Answer

Angular rotation rate of the Earth: ω = 2π / (24×3600 s) = 2π / 86400 s = 7.272205×10^−5 s^−1.

1) Relevant concepts
- Linear (tangential) speed due to planetary rotation is v = ω r⊥, where r⊥ is the perpendicular distance from the rotation axis (i.e., R cos(latitude) for a surface point).
- For an interior point at depth d, its radius from Earth's center is R' = R − d; the maximum speed about the pole is at the equator so r⊥ = R'.

2) Complete step-by-step reasoning and intermediate calculations

(c) U Chicago at latitude 41.79°:
- Earth radius R = 6371 km = 6,371,000 m.
- Perpendicular radius to axis: r⊥ = R cos(41.79°).
  Compute cos(41.79°) ≈ 0.74392 (sufficient precision for the final 3 s.f.).
  r⊥ = 6,371,000 m × 0.74392 = 4,742,283 m (approx).
- Linear speed: v = ω r⊥ = 7.272205×10^−5 s^−1 × 4,742,283 m = 344.98 m/s.
- Round to three significant figures: v = 345 m/s.

(d) Maximum speed at depth 11 km:
- Radius at depth d = 11 km: R' = 6371 − 11 = 6360 km = 6,360,000 m.
- Maximum (equatorial) perpendicular radius r⊥ = R' (cos = 1).
- v_max = ω R' = 7.272205×10^−5 s^−1 × 6,360,000 m = 462.999… m/s.
- Round to three significant figures: v_max = 463 m/s.

3) Why these answers are correct
- The formulas used follow directly from rigid rotation: linear speed = angular speed × perpendicular radius. Numerical intermediate values (ω, r⊥) are shown and the final rounding uses three significant figures as requested.

4) Brief note on the multiple-choice parts (beyond the numerical answers)
- Euler’s fixed point theorem: the most general displacement of a rigid body on a sphere is a rotation about one axis through the center of the sphere. For absolute plate motion, hotspot tracks (e.g., the Hawaiian archipelago) provide a useful reference frame.

Conclusion: The calculated speeds, rounded to three significant figures, are 345 m/s (U Chicago) and 463 m/s (maximum at 11 km depth), matching the provided answers.

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