The gravitational force acting on both the masses is the same. We know that
Force $$=$$ mass × acceleration.

For same force, $${\text{acceleration}} \propto \frac{1}{{{\text{mass}}}}$$
$$\therefore \frac{{{a_{5M}}}}{{{a_M}}} = \frac{M}{{5M}} = \frac{1}{5}\,.....(i)$$
Let $$t$$ be the time taken for the two masses to collide and $${x_{5M}},\,\,{x_M}$$   be the distance travelled by the mass $$5\,M$$  and $$M$$ respectively.
For mass $$5\,M$$
$$\eqalign{ & u = 0,\,\,\,S = {x_{5M}},\,\,\,\,t = t,\,\,\,a = {a_{5M}} \cr & \therefore S = ut + \frac{1}{2}a{t^2} \cr & \Rightarrow {x_{5M}} = \frac{1}{2}{a_{5M}}{t^2}\,.....\,(ii) \cr} $$
For mass $$M$$
$$\eqalign{ & u = 0,\,\,\,s = {x_M},\,\,\,\,t = t,\,\,\,a = {a_M} \cr & \therefore s = ut + \frac{1}{2}a{t^2} \cr & \Rightarrow {x_M} = \frac{1}{2}{a_M}{t^2}\,.....\,(iii) \cr} $$
Dividing (ii) by (iii)
$$\eqalign{ & \frac{{{x_{5M}}}}{{{x_M}}} = \frac{{\frac{1}{2}{a_{5M}}{t^2}}}{{\frac{1}{2}{a_M}{t^2}}} = \frac{{{a_{5M}}}}{{{a_M}}} = \frac{1}{5}\,\,\,\,\,\,\,\,\,\left[ {{\text{From }}(i)} \right] \cr & \therefore 5{x_{5M}} = {x_M}\,.....(iv) \cr} $$
From the figure it is clear that
$${x_{5M}} + {x_M} = 9R\,.....(v)$$
Where $$O$$ is the point where the two spheres collide.
From (iv) and (v)
$$\eqalign{ & \frac{{{x_M}}}{5} + {x_M} = 9R \cr & \therefore 6{x_M} = 45R \cr & \therefore {x_M} = \frac{{45}}{6}R = 7.5R \cr} $$

According to Kepler’s law of planetary motion $${T^2} \propto {R^3}$$
$$\therefore {T_2} = {T_{_1}}{\left( {\frac{{{R_2}}}{{{R_1}}}} \right)^{\frac{3}{2}}} = 5 \times {\left[ {\frac{{4R}}{R}} \right]^{\frac{3}{2}}} = 5 \times {2^3} = 40\,hour$$

From Kepler's third law $${T^2} \propto {r^3}$$
\[\left[ {\begin{array}{*{20}{c}} {{\rm{where,}}\,T = {\rm{time \,period \,of \,satellite}}}\\ {r = {\rm{radius\, of \,elliptical\, orbit}}\left( {{\rm{semi \,major \,axis}}} \right)} \end{array}} \right]\]
Hence, $$T_1^2 \propto r_1^3\,{\text{and}}\,\,T_2^2 \propto r_2^3$$
$$\eqalign{ & {\text{So,}}\,\,\frac{{T_2^2}}{{T_1^2}} = \frac{{r_2^3}}{{r_1^3}} = \frac{{{{\left( {3R} \right)}^3}}}{{{{\left( {6R} \right)}^3}}} \cr & {\text{or}}\,\,\frac{{T_2^2}}{{T_1^2}} = \frac{1}{8} \cr & T_2^2 = \frac{1}{8}T_1^2 \Rightarrow {T_2} = \frac{{24}}{{2\sqrt 2 }} = 6\sqrt 2 \,h \cr} $$

Acceleration due to gravity,
$$\eqalign{ & g \propto r\,\left( {{\text{if}}\,r < {R_E}} \right)\,{\text{and}} \cr & g \propto \frac{1}{{{r^2}}}\left( {{\text{if}}\,r < {R_E}} \right) \cr} $$

Gravitational force provides the necessary centripetal force.
$$\eqalign{ & \therefore \frac{{m{v^2}}}{{\left( {R + x} \right)}} = \frac{{GmM}}{{{{\left( {R + x} \right)}^2}}}\,\,{\text{also}}\,\,g = \frac{{GM}}{{{R^2}}} \cr & \therefore \frac{{m{v^2}}}{{\left( {R + x} \right)}} = m\left( {\frac{{GM}}{{{R^2}}}} \right)\frac{{{R^2}}}{{{{\left( {R + x} \right)}^2}}}\frac{{n!}}{{r!\left( {n - r} \right)!}} \cr & \therefore \frac{{m{v^2}}}{{\left( {R + x} \right)}} = mg\frac{{{R^2}}}{{{{\left( {R + x} \right)}^2}}} \cr & \therefore {v^2} = \frac{{g{R^2}}}{{R + x}} \Rightarrow v = {\left( {\frac{{g{R^2}}}{{R + x}}} \right)^{\frac{1}{2}}} \cr} $$

Various regions of spherical shell attract the point mass inside it in various directions. These forces cancel each other completely. Therefore the gravitational force on the particle is zero.

Potential at the given point = Potential at the point due to the shell + Potential due to the particle
$$ = - \frac{{GM}}{a} - \frac{{2GM}}{a} = - \frac{{3GM}}{a}$$

$$\eqalign{ & V = 2\left[ { - \frac{{G \times 1}}{1} - \frac{{G \times 1}}{2} - \frac{{G \times 1}}{4} - \frac{{G \times 1}}{8}..} \right] \cr & = - 2G\left[ {1 + \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ..} \right] \cr & = - 2G\left[ {\frac{1}{{1 - \frac{1}{2}}}} \right] = - 4G \cr & \therefore {\text{Magnitude}}\,{\text{of}}\,V = 4G \cr} $$

According to question, gravitational force between two objects $$F = \frac{k}{R}$$
In equilibrium, the gravitational force provides the required centripetal force to the particle
$$\therefore \frac{{m{v^2}}}{R} = \frac{k}{R}$$
Hence, $$v \propto {R^0}$$

$${W_{AB}} = m\left( {{V_B} - {V_A}} \right);{\text{But}}\,{V_A} = {V_B}\,{\text{and}}\,{\text{so}}\,{W_{AB}} = 0.$$