At a platform at a height $$h,$$
Escape energy = binding energy of sphere
$$\eqalign{ & {\text{or}}\,\,\frac{1}{2}m{\left( {f{v_e}} \right)^2} = \frac{{GMm}}{{R + h}} \cr & {\text{or}}\,\,f{v_e} = \sqrt {\frac{{2GM}}{{R + h}}} = \sqrt {\frac{{2GM}}{2R}} \,.......\left( {\text{i}} \right)\left( {\because h = R} \right) \cr} $$
But at surface of the earth, $${v_e} = \sqrt {\frac{{2GM}}{R}} \,......\left( {{\text{ii}}} \right)$$
Dividing Eq. (ii) by Eq. (i).
Hence, $$\frac{{f{v_e}}}{{{v_e}}} = \frac{{\sqrt {\frac{{GM}}{R}} }}{{\sqrt {\frac{{2GM}}{R}} }} \Rightarrow f = \frac{1}{{\sqrt 2 }}$$

According to Kepler’s law, the areal velocity of a planet around the sun always remains constant.
SCD : $${A_1} - {t_1}\,\left( {{\text{areal velocity constant}}} \right)$$
SAB : $${A_2} - {t_2}$$
$$\eqalign{ & \frac{{{A_1}}}{{{t_1}}} = \frac{{{A_2}}}{{{t_2}}},{t_1} = {t_2}.\frac{{{A_1}}}{{{A_2}}},\,\,\left( {{\text{given}}\,{A_1} = 2{A_2}} \right) \cr & = {t_2}.\frac{{2{A_2}}}{{{A_2}}} \cr & \therefore {t_1} = 2{t_2} \cr} $$

$$\eqalign{ & \therefore U = \frac{{ - GmM}}{{2R}} + \frac{{GmM}}{R}; \cr & \Delta U = \frac{{GmM}}{{2R}} \cr & {\text{Now}}\,\,\frac{{GM}}{{{R^2}}} = g; \cr & \therefore \frac{{GM}}{R} = gR \cr & \therefore \Delta U = \frac{1}{2}mgR \cr} $$

Apply Kepler's second law. The line joining the sun to the planet sweeps out equal areas in equal time interval i.e. areal velocity is constant.
$$\eqalign{ & \frac{{dA}}{{dt}} = {\text{constant}} \cr & {\text{or}}\,\,\frac{{{A_1}}}{{{t_1}}} = \frac{{{A_2}}}{{{t_2}}} \cr} $$
where, $${{A_1}} =$$  area under $$SCD$$
$${{A_2}} =$$  area under $$ABS$$
$$\eqalign{ & \Rightarrow {t_1} = \frac{{{A_1}}}{{{A_2}}}{t_2} \cr & {\text{Given,}}\,\,{A_1} = 2{A_2} \cr & \therefore {t_1} = 2{t_2} \cr} $$

$${m_1} = \frac{4}{3}\pi {R^3}\rho ,{m_2} = \frac{4}{3}\pi {\left( {2R} \right)^3}\rho ,$$       distance between the centres of the two spherical objects, $$r = 3R.$$
$$\eqalign{ & F = \frac{{G{M_1}\;{m_2}}}{{{r^2}}} \cr & = G\left( {\frac{4}{3}\pi {R^3}\rho } \right)\left( {8 \times \frac{4}{3}\pi {R^3}\rho } \right) \times \frac{1}{{{{\left( {3R} \right)}^2}}} \cr & = \frac{{128}}{{81}}G{\pi ^2}{R^4}{\rho ^2} \cr} $$

Let us consider a circular elemental area of radius $$x$$ and thickness $$dx.$$  The area of the shaded portion \[ = 2\pi xdx.\]
Let $$dm$$  be the mass of the shaded portion.

$$\eqalign{ & \therefore \frac{{Mass}}{{Area}} = \frac{M}{{\pi \left( {4{R^2}} \right) - \pi {{\left( {3R} \right)}^2}}} = \frac{{dm}}{{2\pi xdx}} \cr & \therefore dm = \frac{{2M}}{{7{R^2}}}xdx \cr} $$
The gravitational potential of the mass $$dm$$  at $$P$$ is
$$\eqalign{ & dV = \frac{{ - G\,dm}}{{\sqrt {{{\left( {4R} \right)}^2} + {x^2}} }} = - \frac{G}{{\sqrt {16{R^2} + {x^2}} }} \times \frac{{2M}}{{7{R^2}}}xdx \cr & = \frac{{ - 2GM}}{{7{R^2}}}\frac{{xdx}}{{\sqrt {16{R^2} + {x^2}} }}\,.....(1) \cr} $$
suppose $$16{R^2} + {x^2} = {t^2}$$
$$ \Rightarrow 2xdx = 2tdt\,\,\,\,\, \Rightarrow xdx = tdt$$
Also for $$x=3R,\,\,\,\,t=5R$$     and for $$x = 4R,\,\,t = 4\sqrt 2 R$$
On integrating equation (1), taking the above limits, we get:
$$\eqalign{ & V = - \int\limits_{5R}^{4\sqrt 2 R} {\frac{{2GM}}{{7{R^2}}}dt = \frac{{ - 2GM}}{{7{R^2}}}\left[ t \right]_{5R}^{4\sqrt 2 R}} \cr & = \frac{{ - 2GM}}{{7{R^2}}}\left[ {4\sqrt 2 R - 5R} \right] \cr & \Rightarrow V = \frac{{ - 2GM}}{{7R}}\left( {4\sqrt 2 - 5} \right) \cr & {\text{Now }}\frac{{{W_{P\,\infty }}}}{1} = {V_ \propto } - {V_P} = - {V_P}\,\,\,\,\,\,\,\,\left[ {\because {V_ \propto } = 0} \right] \cr & \therefore {W_{P\,\infty }} = \frac{{2GM}}{{7R}}\left( {4\sqrt 2 - 5} \right) \cr} $$

$$\eqalign{ & \therefore G\frac{{4 \times 1}}{{{x^2}}} = G\frac{{9 \times 1}}{{{{\left( {60 - x} \right)}^2}}} \cr & {\text{or}}\,\,\frac{2}{3} = \frac{x}{{\left( {60 - x} \right)}} \Rightarrow x = 24\,cm \cr} $$

Let the gravitational field at $$P,$$ distant $$x$$ from mass $$m,$$  be zero.
$$\therefore \frac{{GM}}{{{x^2}}} = \frac{{4Gm}}{{{{\left( {r - x} \right)}^2}}}\,\,\, \Rightarrow x = \frac{r}{3}$$

Gravitational potential a $$P,\,\,V = - \frac{{Gm}}{{\frac{r}{3}}} - \frac{{4Gm}}{{\frac{{2r}}{3}}} = - \frac{{9Gm}}{r}$$

According to Kepler's law: $$\frac{{T_1^2}}{{T_2^2}} = \frac{{R_1^3}}{{R_2^3}}$$
Here,
$$\eqalign{ & {T_1} = 365{\text{ days}}\,{\text{; }}\,\,{T_2} = ?{\text{;}}\,\,\,\,{R_1} = R\,{\text{; }}\,\,{R_2} = \frac{R}{2} \cr & \Rightarrow {T_2} = {T_1}{\left( {\frac{{{R_2}}}{{{R_1}}}} \right)^{\frac{3}{2}}} = 365{\left[ {\frac{{\frac{R}{2}}}{R}} \right]^{\frac{3}{2}}} = 129{\text{ days}} \cr} $$

Energy at surface of the earth = energy at maximum height
or $$\left( {K + U} \right)$$  at the earth's surface $$= \left( {K + U} \right)$$   at maximum height
$$\therefore \frac{1}{2}m{u^2} - \frac{{GMm}}{R} = \frac{1}{2}m \times {\left( 0 \right)^2} - \frac{{GMm}}{{R + h}}$$
At maximum height it has only potential energy
or $$\frac{1}{2}m{u^2} = \frac{{GMm}}{R} - \frac{{GMm}}{{R + R}}\,\,\left( {\because h = R} \right)$$
$$\eqalign{ & {\text{or}}\,\,{u^2} = \frac{{2GM}}{R} - \frac{{2GM}}{{2R}} \cr & {\text{or}}\,\,{u^2} = \frac{{GM}}{R} \cr & \therefore u = \sqrt {\frac{{GM}}{R}} \cr} $$
Alternative
The expression for the speed with which a body should be projected so as to reach a height $$h$$ is $$u = \sqrt {\frac{{2gh}}{{1 + \left( {\frac{h}{R}} \right)}}} $$
Here, $$h = R$$  (given)
$$u = \sqrt {\frac{{2gR}}{{1 + \left( {\frac{R}{R}} \right)}}} = \sqrt {\frac{{2 \times \frac{{GM}}{{{R^2}}} \times R}}{2}} = \sqrt {\frac{{GM}}{R}} $$