Question

The value by which free fall acceleration differs from acceleration due to gravity at the equator is:
 

• A. \(+\omega^2 R\)
• B. \(-\omega^2 R\)
• C. \(+2\omega^2 R\)
• D. \(-2\omega^2 R\)

Hint:

At the equator: centripetal effect is maximum (\(\omega^2 R \approx 0.034\) m/s\(^2\)), so \(g_{\text{eff}}\) is minimum (\(\approx 9.78\) m/s\(^2\)).
At the poles: \(\phi = 90^\circ\), \(\cos\phi = 0\), so centripetal effect is zero and \(g_{\text{eff}} = g\) (\(\approx 9.83\) m/s\(^2\)).

Answer 1

The Correct Option is B

Concept: On a rotating Earth, an observer experiences an effective (free-fall) gravitational acceleration \(g_{\text{eff}}\) that differs from the true gravitational acceleration \(g\) due to the centripetal acceleration arising from Earth's rotation:
• True gravitational acceleration: \(g\) (directed toward Earth's centre)
• Centripetal acceleration (directed outward from rotation axis): \(\omega^2 r\), where \(r\) is the perpendicular distance from the rotation axis.
• Effective (free-fall) acceleration at latitude \(\phi\): \(g_{\text{eff}} = g - \omega^2 R \cos\phi\)

Step 1: Apply the general formula at the equator.
At the equator, latitude \(\phi = 0^\circ\) and \(\cos\phi = 1\), so: \[ g_{\text{eff}} = g - \omega^2 R \]

Step 2: Find the difference between free-fall and gravitational acceleration.
\[ g_{\text{eff}} - g = (g - \omega^2 R) - g = -\omega^2 R \]

Step 3: Interpret the result.
The negative sign confirms that free-fall acceleration is smaller than \(g\) at the equator, as the centripetal term acts radially outward, opposing gravity: \[ \boxed{g_{\text{eff}} - g = -\omega^2 R} \;\Rightarrow\; Option (2). \]