Problem Statement
A ball is dropped from height $h = 100\,\text{m}$ at latitude $\lambda = 45°N$. Find the eastward Coriolis deflection.
Given Information
- $h = 100\,\text{m}$
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Identify given quantities and set up the problem: Coriolis deflection of a freely falling body to the east:
Step 2 — Apply the relevant physical law or equation: $$x_{east} = \frac{1}{3}g t^3 \omega\cos\lambda$$
Step 3 — Solve algebraically for the unknown: Time of fall: $t = \sqrt{2h/g} = \sqrt{200/9.8} = 4.52\,\text{s}$
Step 4 — Substitute numerical values with units: $$x = \frac{1}{3}\times9.8\times(4.52)^3\times7.27\times10^{-5}\times\cos45°$$
$$= \frac{1}{3}\times9.8\times92.2\times7.27\times10^{-5}\times0.707$$
$$= \frac{9.8\times92.2\times5.14\times10^{-5}}{3} \approx \frac{0.0464}{3} \approx \boxed{15\,\text{mm}}$$
Step 5 — Compute and check the result: Small but measurable — Coriolis deflection of falling bodies was a classical test for Earth’s rotation.
Worked Calculation
$$x_{east} = \frac{1}{3}g t^3 \omega\cos\lambda$$
$$x = \frac{1}{3}\times9.8\times(4.52)^3\times7.27\times10^{-5}\times\cos45°$$
$$= \frac{1}{3}\times9.8\times92.2\times7.27\times10^{-5}\times0.707$$
Coriolis deflection of a freely falling body to the east:
$$x_{east} = \frac{1}{3}g t^3 \omega\cos\lambda$$
Time of fall: $t = \sqrt{2h/g} = \sqrt{200/9.8} = 4.52\,\text{s}$
$$x = \frac{1}{3}\times9.8\times(4.52)^3\times7.27\times10^{-5}\times\cos45°$$
$$= \frac{1}{3}\times9.8\times92.2\times7.27\times10^{-5}\times0.707$$
$$= \frac{9.8\times92.2\times5.14\times10^{-5}}{3} \approx \frac{0.0464}{3} \approx \boxed{15\,\text{mm}}$$
Small but measurable — Coriolis deflection of falling bodies was a classical test for Earth’s rotation.
Answer
$$= \frac{9.8\times92.2\times5.14\times10^{-5}}{3} \approx \frac{0.0464}{3} \approx \boxed{15\,\text{mm}}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
Leave a Reply