Question

If the angular velocity of earth's spin is increased such that the bodies at the equator start floating, the duration of the day would be approximately: [Take \(g = 10\) ms⁻², \(R = 6400 \times 10^3\) m, \(\pi = 3.14\)]

• A. does not change
• B. 1200 minutes
• C. 60 minutes
• D. 84 minutes

Hint:

This period (\(\approx 84\) min) is also the orbital period of a satellite orbiting very close to the Earth's surface.

Answer 1

The Correct Option is D

Step 1: For bodies to float at the equator, the effective gravity \(g'\) must be zero: \[g' = g - R\omega^2 = 0 \implies \omega = \sqrt{\frac{g}{R}}\]
Step 2: The duration of the day is the time period \(T\): \[T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{R}{g}}\]
Step 3: Substitute values: \[T = 2 \times 3.14 \times \sqrt{\frac{6400 \times 10^3}{10}} = 6.28 \times \sqrt{640000} = 6.28 \times 800 = 5024 \text{ seconds}\]
Step 4: Convert to minutes: \[T_{min} = \frac{5024}{60} \approx 83.73 \text{ minutes} \approx 84 \text{ minutes}\]