MHT CET · Physics · Gravitation

Consider earth to be a sphere of radius \(R_e\) rotating about its own axis with angular speed \(\omega\). If \(g_{\mathrm{E}}\) and \(g_{\mathrm{P}}\) are the accelerations due to gravity at the equator and the poles respectively, then \(\left(g_{\mathrm{P}}-g_{\mathrm{E}}\right)\) is given by
\(\left[\cos \left(0^{\circ}\right)=\sin \left(\frac{\pi}{2}\right)=1, \sin \left(0^{\circ}\right)=\cos \left(\frac{\pi}{2}\right)=0\right]\)

A \(\frac{R_{\mathrm{E}}}{\omega^2}\)
B \(R_{\mathrm{E}} \omega^2\)
C \(R_{\mathrm{E}} \omega\)
D \(R_{\mathrm{E}}^2 \omega^2\)

Verified Solution

Answer & Solution

Correct Answer

(B) \(R_{\mathrm{E}} \omega^2\)

Step-by-step Solution

Detailed explanation

Acceleration due to gravity at a place of latitude \(\lambda\) due to the rotation of earth is
\(g^{\prime}=g-\mathrm{R}_{\mathrm{E}} \omega^2\left(\cos ^2 \lambda\right)\)
At equator, \(\lambda=0^{\circ}, \cos 0^{\circ}=1\)
\(\Rightarrow g_E=g-\mathrm{R}_{\mathrm{E}} \omega^2\)
At poles, \(\lambda=90^{\circ}, \cos (\lambda)=0\)
\(\begin{aligned} & \therefore g^{\prime}=g_{\mathrm{P}}=g \\ & \therefore g_{\mathrm{P}}-g_{\mathrm{E}}=g-\left(g-R_E \omega^2\right)=R_E \omega^2\end{aligned}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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