$$K.E. = \frac{1}{2}mv_e^2$$     where $${v_e} = $$  escape velocity $$ = \sqrt {2gR} $$
$$\therefore K.E. = \frac{1}{2}m \times 2gR = mgR$$

$$\eqalign{ & g = \frac{{GM}}{{{R^2}}} = \frac{{G\rho \times V}}{{{R^{^2}}}}\,\,\,\,\, \Rightarrow g = \frac{{G \times \rho \times \frac{4}{3}\pi {R^3}}}{{{R^2}}} \cr & g = \frac{4}{3}\rho \pi G.R{\text{ where }}\rho \to {\text{ average density}} \cr} $$

Angular momentum of the planet about $$S$$ is conserved. So, $$mvr = {\text{constant}}{\text{.}}$$
$$v$$ is maximum when $$r$$ is minimum.
So, $$v$$ is maximum at point $${P_4}.$$ Hence $$K.E.$$  is maximum at $${P_4}.$$

As, $${v_e} = \sqrt {2gR} = \sqrt {\frac{{2GM}}{R}} $$
Hence, escape velocity does not depend on the angle of projection. Escape velocity will remain same.

The appearance of weightlessness occurs in space when the gravitational attraction of the earth on a body in space is equal to the centripetal force.
$$\eqalign{ & \therefore \frac{{m{v^2}}}{r} = mg \cr & {\text{or}}\,\,v = \sqrt {rg} = \sqrt {20 \times 10} = 14.14\,m/s \cr} $$

Volume of removed sphere
$${V_{{\text{remo}}}} = \frac{4}{3}\pi {\left( {\frac{R}{2}} \right)^3} = \frac{4}{3}\pi {R^3}\left( {\frac{1}{8}} \right)$$
Volume of the sphere (remaining)
$${V_{{\text{remain}}}} = \frac{4}{3}\pi {R^3} - \frac{4}{3}\pi {R^3}\left( {\frac{1}{8}} \right) = \frac{4}{3}\pi {R^3}\left( {\frac{7}{8}} \right)$$
Therefore mass of sphere carved and remaining sphere are at respectively $$\frac{1}{8}M$$  and $$\frac{7}{8}M.$$
Therefore, gravitational force between these two sphere,
$$\eqalign{ & F = \frac{{GMm}}{{{r^2}}} = \frac{{G\frac{{7M}}{8} \times \frac{1}{8}M}}{{{{\left( {3R} \right)}^2}}} = \frac{7}{{64 \times 9}}\frac{{G{M^2}}}{{{R^2}}} \cr & \simeq \frac{{41}}{{3600}}\frac{{G{M^2}}}{{{R^2}}} \cr} $$

Suppose, the smaller body cover a distance $$x$$ before collision, then

$$\eqalign{ & Mx = 5M\left( {9R - x} \right) \cr & {\text{or}}\,\,x = 45R - 5x \cr & {\text{or}}\,\,x = \frac{{45R}}{6} = 7.5\,R \cr} $$

Note : The gravitational force of attraction between the stars will provide the necessary centripetal forces. In this case angular velocity of both stars is the same.
Therefore time period remains the same. $$\left( {\omega = \frac{{2\pi }}{T}} \right)$$

Orbital velocity of satellite is given by, $$v = \sqrt {\frac{{GM}}{r}} $$
Ratio of orbital velocities of $$A$$ and $$B$$ is given by,
$$\eqalign{ & \Rightarrow \frac{{{v_A}}}{{{v_B}}} = \sqrt {\frac{{{r_B}}}{{{r_A}}}} = \sqrt {\frac{R}{{4R}}} = \frac{1}{2} \cr & \therefore \frac{{{v_A}}}{{{v_B}}} = \frac{{3v}}{{{v_B}}} = \frac{1}{2} \cr & \therefore {v_B} = 6v \cr} $$

The required centripetal force of particle of mass $$'m\,'$$ to revolve in a circular path is provided by gravitational pull of the mass $$'M\,'$$ present in the sphere of radius $$'r\,'$$.
Therefore
$$\eqalign{ & \frac{{m{v^2}}}{r} = \frac{{GMm}}{{{r^2}}}\,\,\, \Rightarrow \frac{1}{2}m{v^2} = \frac{{GMm}}{{2r}}\,\,\, \Rightarrow K = \frac{{GMm}}{{2r}} \cr & \therefore M = \frac{{2Kr}}{{Gm}} \cr} $$

Differentiating the above equation w.r.t $$'r\,'$$ we get
$$\eqalign{ & \frac{{dM}}{{dr}} = \frac{{2K}}{{Gm}}{\text{ or }}dM = \frac{{2K}}{{Gm}}dr \cr & \therefore 4\pi {r^2}dr\rho = \frac{{2K}}{{Gm}}dr \cr & \therefore \rho = \frac{K}{{2\pi {r^2}mG}} \cr & \therefore \frac{\rho }{m} = \frac{K}{{2\pi {r^2}{m^2}G}} \cr} $$