(3)

The escape velocity of Earth,

$v=\sqrt{\frac{2G{M}_e}{R_e}}=\sqrt{\frac{2G\times 9M_e}{16R_e}}$                        $(\because \begin{array}{c}{M}_p=9M_e\\ R_p=16R_e\end{array})$

$=\frac{3}{4}\sqrt{\frac{2G{M}_e}{R_e}} \Rightarrow v_p=\frac{3}{4}v$                    (using above eqn.)

(4)

The velocity at infinite distance when speed of projection and escape velocity are given, $v_{\infty }^2=v^2-v_e^2$

where $v$ is projection velocity 

$v_e$ is escape velocity. Given, $v=2v_e$

So,  $v_{\infty }=\sqrt{{\left(2v_e\right)}^2-v_e^2}$ $;$

        $v_{\infty }=\sqrt{3}{v}_e=11.2\sqrt{3}$ $\mathrm{km}/s$

(1)

The formula of escape velocity is

     $v=\sqrt{\frac{2GM}{R}}$

where, $G$ = gravitational constant, $M$ = mass and $R$ = radius.

Mass, $M=\frac{4}{3}\pi R^3\rho$ where, $\rho$ is the density of the planet

        $v=\sqrt{\frac{2\times G}{R}\times \frac{4}{3}\pi R^3\rho }$

       $v=\sqrt{\frac{8G\pi }{3}{R}^2\rho }$                                             ...(i)

Let the escape velocity on planet is $v\text{'}$.

For planet, $R\text{'}=4R,\rho \text{'}=\rho$

       $v\text{'}=\sqrt{\frac{8G\pi }{3}R{\text{'}}^2\rho \text{'}}$                                        ...(ii)

Divide equation (i) by (ii), we get

So,  $\frac{v\text{'}}{v}=\sqrt{\frac{R{\text{'}}^2}{R^2}\times \frac{\rho \text{'}}{\rho }}$

        $\frac{v\text{'}}{v}=\sqrt{\frac{16R^2}{R^2}\times \frac{\rho }{\rho }}=\sqrt{16}=4$

        $v\text{'}=4v$

(1)

The particle is fired vertically upwards from the Earth’s surface with a velocity $v$ and reaches a height $h$.

Energy of the particle at the surface of the Earth is

         $E_i=\frac{1}{2}m{v}^2-\frac{G{M}_{E}m}{R_E}$

Energy of the particle at a height $h$

         $E_f=-\frac{G{M}_{E}m}{R_E+h} \text{(∵at height }h,\text{ velocity of the particle is zero)}$

According to law of conservation of energy,

      $E_i=E_f$

      $\frac{1}{2}m{v}^2-\frac{G{M}_{E}m}{R_E}=-\frac{G{M}_{E}m}{R_E+h}$

       $\frac{1}{2}m{v}^2=G{M}_{E}m[\frac{{\displaystyle 1}}{{\displaystyle R_E}}-\frac{{\displaystyle 1}}{{\displaystyle R_E+h}}]=\frac{G{M}_{E}mh}{\left(R_E\right)(R_E+h)}$

       $\frac{1}{2}m{v}^2=\frac{mgh{R}_E}{R_E+h}$                   ...(i)                 $(\because g=\frac{G{M}_E}{R_E^2})$

As per question,

       $v=k{v}_e=k\sqrt{2g{R}_E}$                 ...(ii)                  $(\because v_e=V_e=\sqrt{2g{R}_E})$

Using (i) and (ii), we get

        $h=\frac{R_E{k}^2}{1-k^2}$

If $R_E=R$, then $h=\frac{R{k}^2}{1-k^2}$

(4)

As escape velocity, $v=\sqrt{\frac{2GM}{R}}$

   $=\sqrt{\frac{2G}{R}·\frac{4\pi R^3}{3}\rho }=R\sqrt{\frac{8\pi G}{3}\rho }$

$\therefore$   $\frac{v_e}{v_p}=\frac{R_e}{R_p}\times \sqrt{\frac{{\rho }_e}{{\rho }_p}}=\frac{1}{2}\times \sqrt{\frac{1}{2}}=\frac{1}{2\sqrt{2}}$         $(\therefore R_p=2R_e\text{ and }{\rho }_p=2{\rho }_e)$

(3)

Light cannot escape from a black hole,

$v_e=c\Rightarrow \sqrt{\frac{2GM}{R}}=c \text{or} R=\frac{2GM}{c^2}$

$R=\frac{2\times 6.67\times {10}^{-11} {\text{Nm}}^2{\text{kg}}^{-2}\times 5.98\times {10}^{24} \text{kg}}{{{\displaystyle (}{\displaystyle 3}{\displaystyle \times }{\displaystyle 10}{\displaystyle {}^8}{\displaystyle }{\displaystyle \text{ms}}{\displaystyle {}^{-1}}{\displaystyle )}}^2}$

      $=8.86\times {10}^{-3}\text{m≃}{10}^{-2}\text{m}$