For the black hole, the escape speed is more than $$c$$ (speed of light). We should compare the escape speed with the $$c$$ (Note that the escape speed should be at least just greater than $$c$$).
$$\eqalign{ & {V_e} = \sqrt {\frac{{2GM}}{{R'}}} \,\,\left[ {R' = {\text{New radius of the earth}}} \right] \cr & c = \sqrt {\frac{{2GM}}{{R'}}} \left[ {{v_e} \approx c} \right] \Rightarrow {c^2} = 2\frac{{GM}}{{R'}} \cr & R' = \frac{{2GM}}{{{c^2}}} = \frac{{2 \times 6.67 \times {{10}^{ - 11}} \times 6 \times {{10}^{24}}}}{{9 \times {{10}^{16}}}} \cr & = \frac{{4 \times 6.67}}{3} \times {10^{ - 3}} = 8.89 \times {10^{ - 3}} \cr & = 0.889 \times {10^{ - 2}} \cr & \simeq {10^{ - 2}}\;m \cr} $$

$$\eqalign{ & 2F\cos {45^ \circ } + F' = \frac{{M{v^2}}}{R}\left( {{\text{From figure}}} \right) \cr & {\text{Where }}F = \frac{{G{M^2}}}{{{{\left( {\sqrt 2 R} \right)}^2}}}{\text{ and }}F' = \frac{{G{M^2}}}{{4{R^2}}} \cr} $$

$$\eqalign{ & \Rightarrow \frac{{2 \times G{M^2}}}{{\sqrt 2 {{\left( {R\sqrt 2 } \right)}^2}}} + \frac{{G{M^2}}}{{4{R^2}}} = \frac{{M{v^2}}}{R} \cr & \Rightarrow \frac{{G{M^2}}}{R}\left[ {\frac{1}{4} + \frac{1}{{\sqrt 2 }}} \right] = M{v^2} \cr & \therefore v = \sqrt {\frac{{Gm}}{R}\left( {\frac{{\sqrt 2 + 4}}{{4\sqrt 2 }}} \right)} = \frac{1}{2}\sqrt {\frac{{Gm}}{R}\left( {1 + 2\sqrt 2 } \right)} \cr} $$

Escape velocity on surface of the earth is given by
$${v_e} = \sqrt {2g{R_e}} = \sqrt {\frac{{2G{M_e}}}{{{R_e}}}} \,\,\left( {\because g = \frac{{G{M_e}}}{{R_e^2}}} \right)$$
where, $${M_e} =$$  mass of earth
$${R_e} =$$  radius of the earth
$$G =$$  gravitational constant
$$\therefore {v_e} \propto \sqrt {\frac{{{M_e}}}{{{R_e}}}} $$
If $${v_p}$$ is escape velocity from the surface of the planet, then $$\frac{{{v_e}}}{{{v_p}}} = \sqrt {\frac{{{M_e}}}{{{R_e}}}} \times \sqrt {\frac{{{R_p}}}{{{M_p}}}} $$
where, $${M_p}$$ is mass of the planet and $${R_p}$$ is radius of the planet.
but $${R_p} = 3{R_e}\,\,\left( {{\text{given}}} \right)$$
and $${M_p} = 3{M_e}$$
$$\eqalign{ & \therefore \frac{{{v_e}}}{{{v_p}}} = \sqrt {\frac{{{M_e}}}{{{R_e}}}} \times \sqrt {\frac{{3{R_e}}}{{3{M_e}}}} = \frac{1}{1} = 1 \cr & {\text{or}}\,\,{v_p} = {v_e} \cr} $$

$$\eqalign{ & - \frac{{G{M_e}m}}{{10{R_e}}} + \frac{1}{2}mv_i^2 = - \frac{{G{M_e}m}}{{{R_e}}} + \frac{1}{2}mv_f^2 \cr & \therefore v_f^2 = v_i^2 + \frac{{2G{M_e}}}{{{R_e}}}\left( {1 - \frac{1}{{10}}} \right). \cr} $$

$${\text{As}}\,\,g = \frac{{{\text{ }}GM{\text{ }}}}{{{R^2}}}$$
when radius is reduced by $$1\% $$  then
$$\eqalign{ & R' = R - \frac{R}{{100}} = .99\,R, \cr & {\text{so,}}\,g' = \frac{{GM}}{{{{\left( {.99\,R} \right)}^2}}} = 1.02\frac{{GM}}{{{R^2}}} \cr & \Rightarrow \% \,{\text{increase}}\, = 2\% \cr} $$

$${v_e} = \sqrt {2gR} $$
The escape velocity is independent of the angle at which the body is projected.

When the planet moves in circular or elliptical orbit, then torque is always acting parallel to displacement or velocity. So, angular momentum is conserved. When the field is attractive, then potential energy is given by $$U = - \frac{{GMm}}{R}$$
Negative sign shows that the potential energy is due to attractive gravitational force. Kinetic energy changes as velocity increases when distance is less. Hence, option (C) is correct.

K.E. of the person = P.E. at the maximum height
$$\eqalign{ & \frac{1}{2}m{v^2} = m\frac{{GM}}{{{R^2}}}h. \cr & \therefore m\frac{{GM}}{{{R^2}}}.h = mG\frac{{M'}}{{{{R'}^2}}}h' \cr & {\text{Given}}\,\rho = \frac{4}{5}p',\frac{R}{{R'}} = \frac{2}{3} \cr & \therefore mG.\frac{4}{3}\frac{{\pi {R^3}\rho }}{{{R^2}}}.h = mG.\frac{4}{3}\frac{{\pi {{R'}^3}\rho '}}{{{{R'}^2}}}h' \cr & \Rightarrow \frac{2}{3}R'.\frac{4}{5}\rho '1.5 = R'\rho 'h' \cr & \therefore h' = 0.8\;m. \cr} $$

In the space, there is no external gravity. Due to masses of the astronauts, there will be small gravitational attractive force between them. Thus, these astronauts will move towards each other.

In $$\Delta AOB:\cos {60^ \circ } = \frac{R}{{OB}} \Rightarrow OB = 2R$$
Here gravitational force will provide the required centripetal force.
Hence, $$\frac{{GMm}}{{{{\left( {OB} \right)}^2}}} = m\left( {OB} \right){\omega ^2}$$
$$ \Rightarrow \omega = \sqrt {\frac{{GM}}{{{{\left( {OB} \right)}^3}}}} = \sqrt {\frac{{GM}}{{{{\left( {2R} \right)}^3}}}} \Rightarrow \omega = \sqrt {\frac{{GM}}{{8{R^3}}}} $$