$$F = \frac{{GM\left( {3m} \right)}}{{{d^2}}} = \frac{{3GMm}}{{{d^2}}}.$$

$${E_P} = \frac{{GM}}{{{{\left( {3a} \right)}^2}}} + \frac{{G\left( {2M} \right)}}{{{{\left( {3a} \right)}^2}}} = \frac{{GM}}{{3{a^2}}}$$

For a part on the surface of a spherical uniform charge distribution the whole mass acts as a point mass kept at the centre.

Gravitational force will be due to $${{M_1}}$$ only.

According to Kepler's third law $${T^2} \propto {r^3}$$
where, $$T =$$  Time period of revolution
$$r =$$  Semi major axis
$$\eqalign{ & \therefore \frac{{T_A^2}}{{T_B^2}} = \frac{{r_A^3}}{{r_B^3}} \cr & \therefore \frac{{{r_A}}}{{{r_B}}} = {\left( {\frac{{{T_A}}}{{{T_B}}}} \right)^{\frac{2}{3}}} = {\left( 8 \right)^{\frac{2}{3}}} = {2^{3 \times \frac{2}{3}}} = 4 \cr & {\text{or}}\,\,{r_A} = 4{r_B} \cr} $$

$$V = - \int_\infty ^x E dx = \int_\infty ^x K {x^{ - 3}}dx = \frac{K}{{2{x^2}}}.$$

At the equator, $$g' = g - R{\omega ^2}$$
As $$\omega $$ increases, $${g'}$$ decreases and hence weight decreases.
At the pole, $$g' = g.$$
So weight remain unchanged.

$${U_i} = - \frac{{GMm}}{R} = $$    Initial potential energy of the system.
$${U_f} = - \frac{{GMm}}{{2R}} = $$    Final P.E. of the system.

$$\eqalign{ & \therefore \Delta U = {U_f} - {U_i} \cr & = - GMm\left[ {\frac{1}{{2R}} - \frac{1}{R}} \right] = \frac{{GMm}}{{2R}}\,.....(i) \cr & {\text{But }}g = \frac{{GM}}{{{R^2}}} \cr & \therefore GM = g{R^2}\,.....(ii) \cr & {\text{From (i) and (ii)}} \cr & \Delta U = \frac{{g{R^2}m}}{{2R}} = \frac{{mgR}}{2} \cr} $$

Gravitational potential at $$P,{V_p} = \frac{{ - GM}}{{\sqrt 5 R}}$$
Gravitational potential at $$O,{V_O} = \frac{{ - GM}}{R}$$

By work energy theorem,
$$\eqalign{ & W = \Delta K \Rightarrow m\left[ {{V_p} - {V_O}} \right] = \frac{1}{2}m{v^2} \cr & m\left[ {\frac{{GM}}{R} - \frac{{GM}}{{\sqrt 5 R}}} \right] = \frac{1}{2}m{v^2}\,{\text{or}}\,V = \sqrt {\frac{{2GM}}{R}\left( {1 - \frac{1}{{\sqrt 5 }}} \right)} \cr} $$

It is given that, acceleration due to gravity on planet $$A$$ is 9 times the acceleration due to gravity on planet $$B$$ i.e.
$${g_A} = 9{g_B}\,......\left( {\text{i}} \right)$$
From third equation of motion, $${v^2} = 2gh$$
At planet $$A,$$ $${h_A} = \frac{{{v^2}}}{{2{g_A}}}\,......\left( {{\text{ii}}} \right)$$
At planet $$B,$$ $${h_B} = \frac{{{v^2}}}{{2{g_B}}}\,......\left( {{\text{iii}}} \right)$$
Dividing Eq. (ii) by Eq. (i), we have
$$\frac{{{h_A}}}{{{h_B}}} = \frac{{{g_B}}}{{{g_A}}}$$
From Eq. (i), $${g_A} = 9{g_B}$$
$$\eqalign{ & \therefore \frac{{{h_A}}}{{{h_B}}} = \frac{{{g_B}}}{{9{g_B}}} = \frac{1}{9} \cr & {\text{or}}\,\,{h_B} = 9{h_A} = 9 \times 2 = 18\;m\,\,\left( {\because {h_A} = 2\;m} \right) \cr} $$

Here, centripetal force will be given by the gravitational force between the two particles.

$$\eqalign{ & \frac{{G{m^2}}}{{{{\left( {2R} \right)}^2}}} = m{\omega ^2}R \cr & \Rightarrow \frac{{Gm}}{{4{R^3}}} = {\omega ^2} \Rightarrow \omega = \sqrt {\frac{{Gm}}{{4{R^3}}}} \cr} $$
If the velocity of the two particles with respect to the centre of gravity is $$v$$ then $$v = \omega R$$
$$v = \sqrt {\frac{{Gm}}{{4{R^3}}}} \times R = \sqrt {\frac{{Gm}}{{4R}}} $$