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.

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 \(d = \frac{3R}{4}\).

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, \(g\) increases linearly from center to surface: \(g(r) \propto r\). Outside, it decreases: \(g(r) \propto 1/r^2\). Thus, \(g\) is maximum at the surface of the earth.

In free fall under gravity (neglecting air resistance), all bodies experience the same acceleration due to gravity \(g\), regardless of their mass or density.

The error in force is \(\Delta F = m(g_1 - g_2) \approx m \frac{2g}{R} h\). Since \(g = \frac{4}{3} \pi G \rho R\), we have \(\frac{2g}{R} = \frac{8}{3} \pi G \rho\). Therefore, \(\Delta F = \frac{8}{3} \pi G \rho m h\).

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 \(g' = \frac{g (R/2)}{R} = g/2\).