Question

Earth is assumed to be a sphere of radius $R$. If '$g_{\phi}$' is value of effective acceleration due to gravity at latitude $30^{\circ}$ and '$g$' is the value at equator, then the value of $|g - g_{\phi}|$ is ($\omega$ is angular velocity of rotation of earth, $\cos 30^{\circ} = \frac{\sqrt{3{2}$ )}

• A. \frac{1}{4} \omega^2 R
• B. \frac{3}{4} \omega^2 R
• C. \omega^2 R
• D. \frac{1}{2} \omega^2 R

Hint:

Remember that effective gravity decreases as you move from the poles to the equator due to rotation.

Answer 1

The Correct Option is A

Step 1: The formula for effective acceleration due to gravity at a latitude $\phi$ is: \[ g_{\phi} = g_0 - \omega^2 R \cos^2 \phi \] where $g_0$ is the acceleration due to gravity without rotation.
Step 2: At the equator, the latitude $\phi = 0^{\circ}$. Thus, the effective gravity $g$ is: \[ g = g_0 - \omega^2 R \cos^2 0^{\circ} = g_0 - \omega^2 R \]
Step 3: At latitude $\phi = 30^{\circ}$, the effective gravity $g_{\phi}$ is: \[ g_{\phi} = g_0 - \omega^2 R \cos^2 30^{\circ} \] \[ g_{\phi} = g_0 - \omega^2 R \left( \frac{\sqrt{3{2} \right)^2 = g_0 - \frac{3}{4} \omega^2 R \]
Step 4: Find the absolute difference $|g - g_{\phi}|$: \[ |g - g_{\phi}| = |(g_0 - \omega^2 R) - (g_0 - \frac{3}{4} \omega^2 R)| \] \[ |g - g_{\phi}| = |-\omega^2 R + \frac{3}{4} \omega^2 R| = |-\frac{1}{4} \omega^2 R| = \frac{1}{4} \omega^2 R \]