Question

An aeroplane flies due east along the equator with a speed of 300 ms−1. Determine the
magnitude and direction of the Coriolis acceleration

Answer 1

Answer on Question#61574 – Physics – Mechanics – Relativity

An aeroplane flies due east along the equator with a speed of 300 ms⁻¹. Determine the magnitude and direction of the Coriolis acceleration.

**Solution.** The Coriolis acceleration can be calculated by the formula

$$a = 2[\vec{v},\vec{\omega}]$$

(vector product). Where

$$v = 300\frac{m}{s}$$

– the relative velocity of the point,

$$\omega$$

– the angular velocity of rotation of the Earth. Magnitude Coriolis acceleration find as

$$a = v\omega \sin \alpha$$

,

$$v$$

,

$$\omega$$

magnitude relative and velocity,

$$\alpha$$

– the angle between them (in our case

$$\alpha = 90{}^{\circ}$$

).

The angular velocity of the Earth at the equator can be calculated as

$$\omega = \frac{2\pi}{T}$$

, where

$$T$$

– the period of rotation of Earth. (

$T = 24^h = 86400s

).

Hence

$$a = v\frac{2\pi}{T}\sin 90{}^{\circ} = 300 \cdot \frac{2\pi}{86400} \cdot 1 \approx 0.022\frac{m}{s^2}$$

.

If the body is moving from West to East in the plane of the equator, the Coriolis acceleration directed vertically upwards.

**Answer.**

$$a = 0.022\frac{m}{s^2}$$

directed vertically upwards.

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