Understanding the effect of Earth's rotation on gravity.
Due to the rotation of the Earth, the effective acceleration due to gravity is given by: \[ g_{\text{eff}} = g - \omega^2 R \cos^2 \theta \] Where \( \theta \) is the angle made with the equator.
At the poles, where \( \theta = 90^\circ \), the change in gravity is zero because \( \cos(90^\circ) = 0 \). This shows no effect on the poles. For the equator, where \( \theta = 0^\circ \), the change in gravity is maximum: \[ g_{\text{eff}} = g - \omega^2 R \] Thus, the change in gravity is maximum at the equator and zero at the poles.
This contradicts Statement II, but Statement I is correct.
The acceleration due to gravity at a height of 6400 km from the surface of the earth is \(2.5 \, \text{ms}^{-2}\). The acceleration due to gravity at a height of 12800 km from the surface of the earth is (Radius of the earth = 6400 km)
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: