$$\eqalign{ & {E_{{\text{earth}}}} = {E_{{\text{moon}}}} \cr & \Rightarrow \frac{{GM}}{{{x^2}}} = \frac{{\frac{{GM}}{{81}}}}{{{{\left( {60R - x} \right)}^2}}} \cr & \Rightarrow \frac{1}{x} = \frac{1}{{9\left( {60R - x} \right)}} \cr & \Rightarrow x = 54\,R\,{\text{from centre of earth}}{\text{.}} \cr} $$

When two satellites of the earth are moving in same orbit, then time period of both are equal.
From Kepler’s third law $${T^2} \propto {r^3}$$
Time period is independent of mass, hence their time periods will be equal.
The potential energy and kinetic energy are mass dependent, hence the potential energy and kinetic energy of satellites are not equal.
But, if they are orbiting in a same orbit, then they have equal orbital speed.

From the principle of conserving angular momentum, we have
$$mvR = mv'r\,.....\left( {\text{i}} \right)$$
[$${v'}$$ = speed when spaceship is just touching the plane]
From conserving of energy, we have
$$\frac{1}{2}m{v^2} = \frac{1}{2}m{{v'}^2} - \frac{{GMm}}{r}\,......\left( {{\text{ii}}} \right)$$
Solving Eqs. (i) and (ii), we get
$$R = \frac{r}{v}{\left[ {{v^2} + \frac{{2GM}}{r}} \right]^{\frac{1}{2}}}$$

$$\eqalign{ & {E_1} = - \frac{{G{M_e}m}}{{{R_e} + \frac{{{R_e}}}{2}}} - \left( { - \frac{{G{M_e}m}}{{{R_e}}}} \right) = \frac{{G{M_e}m}}{{3{R_e}}} \cr & {E_2} = \frac{1}{2}mv_0^2 = \frac{1}{2}m.\frac{{G{M_e}}}{{R + \frac{{{R_e}}}{2}}} = \frac{{G{M_e}m}}{{3{R_e}}} \cr & \therefore \frac{{{E_1}}}{{{E_2}}} = 1:1 \cr} $$

Energy required $$=$$ (Potential energy of the Earth $$-$$ mass system when mass is at distance $$3R$$  ) $$-$$ (Potential energy of the Earth $$-$$ mass system when mass is at distance $$2R$$  )
$$\eqalign{ & = \frac{{ - GMm}}{{3R}} - \left( {\frac{{ - GMm}}{{2R}}} \right) \cr & = \frac{{ - GMm}}{{3R}} + \frac{{GMm}}{{2R}} \cr & = \frac{{ - 2GMm + 3GMm}}{{6R}} \cr & = \frac{{GMm}}{{6R}} \cr & \cr} $$

Applying energy conservation
$$\eqalign{ & k + \left( { - \frac{{GMm}}{R}} \right) = \frac{1}{2}mv_0^2 + \left( { - \frac{{GMm}}{{R + h}}} \right) \cr & k - \frac{{GMm}}{R} = \frac{1}{2}m + \left( {\frac{{Gm}}{{R + 2R}}} \right) - \frac{{GMm}}{{R + 2R}} \cr & \therefore k = \frac{{5GMm}}{{6R}} \cr} $$

In horizontal direction
$$\eqalign{ & {\text{Net}}\,{\text{force}} = \frac{{G\sqrt 3 mm}}{{12{d^2}}}\cos {30^ \circ } - \frac{{G{m^2}}}{{4{d^2}}}\cos {60^ \circ } \cr & = \frac{{G{m^2}}}{{8{d^2}}} - \frac{{G{m^2}}}{{8{d^2}}} = 0 \cr} $$
In vertical direction,
$$\eqalign{ & {\text{Net}}\,{\text{force}} = \frac{{G\sqrt 3 {m^2}}}{{12{d^2}}}\cos {60^ \circ } + \frac{{G\sqrt 3 {m^2}}}{{3{d^2}}} + \frac{{G{m^2}}}{{4d}}\cos {30^ \circ } \cr & = \frac{{\sqrt 3 G{m^2}}}{{24{d^2}}} + \frac{{\sqrt 3 G{m^2}}}{{3{d^2}}} + \frac{{\sqrt 3 G{m^2}}}{{8{d^2}}} \cr & = \frac{{\sqrt 3 G{m^2}}}{{{d^2}}}\left[ {\frac{{1 + 8 + 3}}{{24}}} \right] = \frac{{\sqrt 3 G{m^2}}}{{2{d^2}}}\,{\text{along}}\,SQ. \cr} $$

Since, the escape velocity of earth can be given as
$$\eqalign{ & {v_e} = \sqrt {2gR} = R\sqrt {\frac{8}{3}\pi G\rho } \,\,\left[ {\rho = {\text{density of earth}}} \right] \cr & \Rightarrow {v_e} = R\sqrt {\frac{8}{3}\pi G\rho } \,......\left( {\text{i}} \right) \cr} $$
As it is given that the radius and mean density of planet are twice as that of earth. So, escape velocity at planet will be
$${v_p} = 2R\sqrt {\frac{8}{3}\pi G2\rho } \,......\left( {{\text{ii}}} \right)$$
Divide, Eq. (i) by Eq. (ii), we get
$$\eqalign{ & \frac{{{v_e}}}{{{v_p}}} = \frac{{R\sqrt {\frac{8}{3}\pi G\rho } }}{{2R\sqrt {\frac{8}{3}\pi G2\rho } }} \cr & \Rightarrow \frac{{{v_e}}}{{{v_p}}} = \frac{1}{{2\sqrt 2 }} \cr} $$

The ball, when dropped from the orbiting satellite will not reach the surface of the earth. When ball is dropped from the satellite, the ball also starts moving with the same speed due to inertia. As the orbit of a satellite does not depend upon its mass, the ball continues to move along with the satellite in the same orbit.

When rocket accelerates upward with acceleration $$a,$$ then effective acceleration of rocket is $$\left( {g + a} \right).$$
As, $$T = 2\pi \sqrt {\frac{l}{g}} = 2\pi \sqrt {\left( {\frac{l}{{g + a}}} \right)} $$
Hence, period of oscillation of seconds pendulum decreases when the rocket moves up with uniform acceleration.