Question

Assuming the rate of rotation of the Earth is \( 7.27 \times 10^{-5} \) radians/s and the radius of Earth is 6371 km, the centrifugal acceleration at 60° latitude for a spherically rotating Earth is ________ \( \times 10^{-3} \) m/s\(^2\). \text{[round off to 1 decimal place]}

Hint:

To calculate the centrifugal acceleration at a given latitude, use the formula \( a_c = \omega^2 R \cos^2(\theta) \).

Answer 1

Correct Answer: 16.3

The centrifugal acceleration \( a_c \) at latitude \( \theta \) is given by the formula: \[ a_c = \omega^2 R \cos^2(\theta) \] where:
- \( \omega = 7.27 \times 10^{-5} \, \text{radians/s} \) is the angular velocity of the Earth,
- \( R = 6371 \, \text{km} = 6.371 \times 10^6 \, \text{m} \) is the radius of the Earth,
- \( \theta = 60^\circ \).
Substituting these values into the formula: \[ a_c = (7.27 \times 10^{-5})^2 \times 6.371 \times 10^6 \times \cos^2(60^\circ) \] \[ a_c = 5.3 \times 10^{-9} \times 6.371 \times 10^6 \times \frac{1}{4} \] \[ a_c = 5.3 \times 10^{-9} \times 1.59275 \times 10^6 = 8.44 \times 10^{-3} \, \text{m/s}^2. \] Rounding off to one decimal place: \[ a_c = \boxed{8.4 \times 10^{-3}} \, \text{m/s}^2. \]