Using energy conservation in the center-of-mass frame: \(\frac{1}{2} \mu v_{\text{rel}}^2 = \frac{GMm}{d}\), where \(\mu = \frac{Mm}{M+m}\) is the reduced mass. Substituting \(\mu\) yields \(v_{\text{rel}} = \sqrt{\frac{2G(M+m)}{d}}\).

The centripetal force is provided by the gravitational force: \(m \omega^2 R = \frac{k}{R^n} ⇒ \omega^2 \propto \frac{1}{R^{n+1}}\). Since \(T = \frac{2\pi}{\omega}\), we get \(T^2 \propto R^{n+1} ⇒ T \propto R^{\frac{n+1}{2}}\).

The increase in potential energy is \(\Delta U = \frac{mgh}{1 + h/R}\). Since \(h = nR\), we get \(\Delta U = \frac{mg(nR)}{1 + n} = \left(\frac{n}{n+1}\right) mgR\).

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

Due to symmetry, the gravitational forces exerted by all symmetric parts of the ring at the center cancel each other out, resulting in a net force of zero.

The gravitational field inside a uniform spherical shell is zero everywhere. Therefore, the gravitational force on any mass kept inside the shell is zero.

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 \(\frac{R}{R+h} = \frac{1}{4} ⇒ R+h = 4R ⇒ h = 3R ⇒ \frac{h}{R} = 3\).

The acceleration due to gravity at depth \(d\) is \(g' = g \left(1 - \frac{d}{R}\right)\). For \(g' = \frac{g}{4}\), we have \(1 - \frac{d}{R} = \frac{1}{4} ⇒ \frac{d}{R} = \frac{3}{4} ⇒ d = \frac{3R}{4}\).

Since \(g = \frac{GM}{R^2}\), for a small percentage change: \(\frac{\Delta g}{g} \times 100 = -2 \frac{\Delta R}{R} \times 100\). Since \(R\) decreases by \(1%\), \(g\) increases by \(2%\).