Acceleration due to gravity at the surface of a planet is given by \( g = \frac{4}{3} \pi G R \rho \). Therefore, \( \frac{g_1}{g_2} = \frac{R_1}{R_2} \times \frac{\rho_1}{\rho_2} = \frac{2}{3} \times \frac{3}{2} = 1 \).

Since acceleration due to gravity varies with height as \(g' = g \frac{R^2}{(R+h)^2}\), setting \(g' = g/9\) gives \(R+h = 3R\), which simplifies to \(h = 2R\).

Since \(g \propto \rho R\), on a planet with twice the radius and same density, \(g' = 2g\). Pendulum time period decreases (\(T_p = 2\pi\sqrt{l/g}\)), so clock P ticks faster. Spring clock is unaffected.

Since \(g = \frac{GM}{R^2}\), differentiating gives \(\frac{dg}{g} = -2 \frac{dR}{R}\). A -1% change in R leads to a +2% change in g.

The effective gravity is \(g' = g - \omega^2 R \cos^2 \theta\). At the poles, \(\theta = 90^\circ\), so \(g' = g\), which is independent of the earth's angular velocity.

Inside the earth, the acceleration due to gravity varies linearly with distance: \(g' = \frac{gr}{R}\). For \(r = R/2\), we have \(g' = \frac{g}{2}\).

The acceleration due to gravity at depth \(d\) is \(g' = g \left(1 - \frac{d}{R}\right)\). Setting \(g' = g/4\), we get \(1 - \frac{d}{R} = \frac{1}{4}\) which gives \(d = \frac{3R}{4}\).

Acceleration due to gravity on the surface of earth is given by \(g = \frac{G M_e}{R_e^2}\). Hence, it depends on the mass of the earth, not on the mass or size of the falling body.

Inside a solid sphere, the acceleration due to gravity at a distance \(r\) from the center is \(g' = \frac{g r}{R}\). For \(r = R/2\), we have \(g' = \frac{g (R/2)}{R} = g/2\).

The variation of gravity with height is \(g' = g \left(\frac{R}{R+h}\right)^2\). Given \(g' = 0.0625 g = \frac{1}{16} g\), we get \(\frac{R}{R+h} = \frac{1}{4} ⇒ R+h = 4R ⇒ h = 3R ⇒ \frac{h}{R} = 3\).