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: Use force balance
Weight $=mg-m\omega^2R$ at the equator, so the fractional drop is $\omega^2R/g$.

Step 2: Match
The weight is $\tfrac35$ of normal, so the drop is $\tfrac25$: $\omega^2R/g=\tfrac25$, $\omega=\sqrt{2g/5R}$, option (B).

Final Answer:
$\omega=\sqrt{2g/(5R)}$, option (B). \[ \boxed{\sqrt{\dfrac{2g}{5R}}} \]