Question

Acceleration due to Coriolis force of a water parcel at a location P (67°E, 20°N) moving with a speed of 0.35 m/s is .......... \(\times 10^{-5}\) m/s\(^2\). (Round off to two decimal places)
[Assume the angular velocity of the Earth is \(7.3 \times 10^{-5}\) s\(^{-1}\))]

Hint:

The Coriolis acceleration depends on the speed of the object, the angular velocity of the Earth, and the sine of the latitude. It is strongest near the poles and weakest at the equator.

Answer 1

The Coriolis acceleration is given by the formula: \[ a_C = 2 \cdot \Omega \cdot v \cdot \sin(\phi) \] where:
- \( \Omega = 7.3 \times 10^{-5} \, {s}^{-1} \) is the angular velocity of the Earth,
- \( v = 0.35 \, {m/s} \) is the speed of the water parcel,
- \( \phi = 20^\circ \) is the latitude of location P.
Substitute these values into the formula: \[ a_C = 2 \times 7.3 \times 10^{-5} \times 0.35 \times \sin(20^\circ) \] Using \( \sin(20^\circ) \approx 0.342 \), we get: \[ a_C = 2 \times 7.3 \times 10^{-5} \times 0.35 \times 0.342 = 1.74 \times 10^{-5} \, {m/s}^2 \] Thus, the Coriolis acceleration is 1.72 to 1.78 \(\times 10^{-5}\) m/s\(^2\).