Question:
What should be the velocity of rotation of earth due to rotation about its own axis so that the weight of a person becomes (3)/(5) of the present weight at the equator. Equatorial radius of earth is 6400km.
What should be the velocity of rotation of earth due to rotation about its own axis so that the weight of a person becomes (3)/(5) of the present weight at the equator. Equatorial radius of earth is 6400km.
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At the equator, effective gravity is reduced due to centrifugal force:
gₑff = g - ω² R
Always use this relation when Earth’s rotation affects weight.
Updated On: Mar 19, 2026
- \(8.7 \times 10^{-4}\,\text{rad/s}\)
- \(7.8 \times 10^{-4}\,\text{rad/s}\)
- \(6.7 \times 10^{-4}\,\text{rad/s}\)
- 7.4 × 10⁻3rad/s
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The Correct Option is B
Solution and Explanation
Step 1: At the equator, effective weight of a person is W = m(g - ω² R)
Step 2: Given that the new weight becomes (3)/(5) of the original weight: m(g - ω² R) = (3)/(5)mg
Step 3: Simplifying: g - ω² R = (3)/(5)g ω² R = (2)/(5)g
Step 4: Substituting g = 9.8m/s² and R = 6400 × 10³m: ω² = (2 × 9.8)/(5 × 6.4 × 10⁶) ω² = 6.125 × 10⁻7
Step 5: Taking square root: ω = 7.8 × 10⁻4rad/s
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