When both \( r_1 \) and \( r_2 \) are greater than \( R \) (i.e., both are outside the sphere), the gravitational force at a distance \( r \) from the center of a uniform sphere is given by:

\[
F = \frac{G M}{r^2}
\]

So, the forces \( F_1 \) and \( F_2 \) at distances \( r_1 \) and \( r_2 \) from the center are:

\[
F_1 = \frac{G M}{r_1^2}
\]
\[
F_2 = \frac{G M}{r_2^2}
\]

Now, taking the ratio \( \frac{F_1}{F_2} \):

\[
\frac{F_1}{F_2} = \frac{\frac{G M}{r_1^2}}{\frac{G M}{r_2^2}} = \frac{r_2^2}{r_1^2}
\]

Thus, the ratio of the gravitational forces is:

\[
\frac{F_1}{F_2} = \left( \frac{r_2}{r_1} \right)^2
\]

To find the dimensions of the gravitational constant \( G \), use Newton's law of gravitation:

\[
F = \frac{G M_1 M_2}{r^2}
\]

Where:
- \( F \) is force (with dimensions \( [M L T^{-2}] \)),
- \( M_1 \) and \( M_2 \) are masses (with dimensions \( [M] \)),
- \( r \) is distance (with dimensions \( [L] \)).

Rearranging for \( G \):

\[
G = \frac{F r^2}{M_1 M_2}
\]

Substitute the dimensions:

\[
G = \frac{[M L T^{-2}] [L^2]}{[M][M]}
\]

Simplify:

\[
G = [M^{-1} L^3 T^{-2}]
\]

Thus, the dimensions of \( G \) are \( [M^{-1} L^3 T^{-2}] \).

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.

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.

Formula of gravitational force is \(F = \frac{G m_1 m_2}{r^2}\). If masses and distance are doubled: \(F' = \frac{G(2m_1)(2m_2)}{(2r)^2} = \frac{4 G m_1 m_2}{4 r^2} = F\). Thus, the force remains unchanged.

The distance between the centers of any two touching spheres is \(2R\). The gravitational force between any two is \(F = \frac{GM^2}{(2R)^2} = \frac{GM^2}{4R^2}\). The angle between the two forces acting on one sphere is \(60^\circ\). Net force is \(F_{\text{net}} = \sqrt{3}F = \frac{\sqrt{3}GM^2}{4R^2}\).

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 centripetal force is \(F = m\omega^2 r = m\frac{4\pi^2}{T^2} r \propto \frac{r}{T^2}\). Given \(F \propto r^{-5/2}\), we get \(\frac{r}{T^2} \propto r^{-5/2} \Rightarrow T^2 \propto r^{3.5}\).