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 \(h = 2R\).

As a satellite moves to a lower orbit, total energy decreases (becomes more negative) due to dissipation, but kinetic energy increases, so speed increases. The speed of a satellite is not universally constant.

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.

The distance between centers is \(2r\). Force between any two is \(F = \frac{GM^2}{4r^2}\). Since the angle between the forces is \(60^\circ\), the resultant force is \(F_{\text{net}} = \sqrt{3}F = \frac{\sqrt{3} GM^2}{4r^2}\).

The acceleration due to gravity is \(g = \frac{GM_e}{R_e^2}\). This depends on the mass of the Earth, not on the properties of the body itself.

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

Due to the symmetrical distribution of mass in a circular ring, the gravitational field at the center is zero. Therefore, the force on any mass placed at the center is zero.

According to shell theorem, the gravitational field inside a uniform spherical shell is zero at all points. Thus, the force acting on the point mass is zero.

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