Determine the speed with which the Earth has to rotate so that the wei

1.	Determine the speed with which the Earth has to rotate so that the weight of a person on the equator is 2/3rd of the current weight. Assume radius of Earth is 6.4 × 106 m.
Answer» Let m be the mass of the person, then, Original weight = m^_g and new weight = m^_g' Given that \(\frac {{mg}'}{mg} \) = \(\frac {2}{3}\) ⇒ g' = \(\frac{2}{3}\)g We know that, g' = g - Rω² (for equatorial plane) \(\frac{2}{3}\) g = g - Rω² ⇒ ω = \(\sqrt\frac{g}{3R}\) ω = \(\sqrt \frac {9.8}{3 \times 6.4 \times 10^6}\) ω = 5.104 × 10^-4 rad s^-1.

Determine the speed with which the Earth has to rotate so that the weight of a person on the equator is 2/3rd of the current weight. Assume radius of Earth is 6.4 × 106 m.

Answer»

Let m be the mass of the person, then,

Original weight = m^_g

and new weight = m^_g'

Given that \(\frac {{mg}'}{mg} \) = \(\frac {2}{3}\)

⇒ g' = \(\frac{2}{3}\)g

We know that,

g' = g - Rω² (for equatorial plane)

\(\frac{2}{3}\) g = g - Rω²

⇒ ω = \(\sqrt\frac{g}{3R}\)

ω = \(\sqrt \frac {9.8}{3 \times 6.4 \times 10^6}\)

ω = 5.104 × 10^-4 rad s^-1.

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Determine the speed with which the Earth has to rotate so that the weight of a person on the equator is 2/3rd of the current weight. Assume radius of Earth is 6.4 × 106 m.

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