Question

The speed with which the earth would have to rotate about its axis so that a person on the equator would weigh \(\frac{3}{5}\)th as much as at present is (\(g\) = gravitational acceleration, \(R\) = equatorial radius of the earth.)

• A. \(\sqrt{\frac{3}{5}gR}\)
• B. \(\sqrt{\frac{2g}{5R}}\)
• C. \(\sqrt{\frac{3g}{5R}}\)
• D. \(\sqrt{\frac{5R}{2g}}\)

Hint:

At the equator, effective gravity is \(g'=g-\omega^2R\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Because of rotation, the apparent weight at the equator is reduced by the centripetal term. \(g'=g-\omega^2R\).

Step 2: Key Formula or Approach
We want \(g'=\dfrac35g\).

Step 3: Detailed Explanation
\[ g-\omega^2R=\frac35g \Rightarrow \omega^2R=\frac25g \]
\[ \omega=\sqrt{\frac{2g}{5R}} \]

Final Answer:
The required angular speed is \(\sqrt{\frac{2g}{5R}}\), option (B). \[ \boxed{\sqrt{\dfrac{2g}{5R}}\ \text{(B)}} \]