Question

The magnitude of Coriolis acceleration (expressed in terms of \( g \), the acceleration due to Earth's gravity) for a particle of mass \( m \) moving with a velocity \( 10^4 \) cm/s horizontally along Earth's surface at the North Pole is about:

• A. \( 1.5g \)
• B. \( g \)
• C. \( 0.0015g \)
• D. \( 0.15g \)

Hint:

Coriolis acceleration depends on latitude and velocity, being maximum at the poles.

Answer 1

The Correct Option is C

Coriolis acceleration is given by:
\[ a_c = 2 \omega v \sin \theta \] where \( \omega = 7.29 \times 10^{-5} \) rad/s (Earth's angular velocity), \( v = 10^4 \) cm/s, and \( \theta = 90^\circ \) (at the North Pole, \( \sin 90^\circ = 1 \)).
\[ a_c = 2 \times (7.29 \times 10^{-5}) \times (10^2) \] \[ a_c = 0.0015g \]