Gravitational force is independent of medium. Hence, this will remain same.

Escape velocity, $${v_e} = \sqrt {2gR} = \sqrt {\frac{{2GM}}{R}} \,\,\,\,\,\, \Rightarrow {V_e} \propto {m^0}$$
Where $$M, \,R$$  are the mass and radius of the planet respectively. In this expression the mass of the body ($$m$$ ) is not present showing that the escape velocity is independent of the mass.

Gravitational flux through a closed surface is given by $$\int {\overrightarrow {{E_{g\,\,}}} \,\,\overrightarrow d S = - 4\pi GM} $$
where, $$M =$$   mass enclosed in the closed surface
This is relationship is is valid valid when when $$\left| {{E_g}} \right| \propto \frac{1}{{{r^2}}}.$$

$$\eqalign{ & v = \omega R \cr & g = {g_0} - {\omega ^2}R\left[ {\;g = {\text{at}}\,{\text{equator,}}\,{g_0} = {\text{at}}\,{\text{poles}}} \right] \cr & \frac{{{g_0}}}{2} = {g_0} - {\omega ^2}R;\,\,{\omega ^2}R = \frac{{{g_0}}}{2}; \cr & {v^2} = \frac{{{g_0}R}}{2} \cr & {v_e} = \sqrt {2\;{g_0}R} = \sqrt {4{v^2}} = 2v \cr} $$

$${R_p} = \frac{{{R_e}}}{{10}} = 6 \times {10^5}\,m$$
The mass of the wire $$ = {10^{ - 3}} \times 1.2 \times {10^5} = 120\,kg$$
Let $${g_{pM}}$$ be the acceleration due to gravity at point $$M$$ which is the mid point of the wire and is at a depth of $$\frac{{{R_p}}}{{10}}.$$
Let $${g_p}$$ be the acceleration due to gravity at the surface of the planet.
$$\eqalign{ & {g_p} = \frac{4}{3}\pi \rho G{R_P}\,;\,\,{g_e} = \frac{4}{3}\pi \rho G{R_E} \cr & \therefore \frac{{{g_p}}}{{{g_e}}} = \frac{{{R_p}}}{{{R_E}}} = \frac{1}{{10}} \cr & \therefore {g_p} = \frac{{10}}{{10}} = 1\,m{s^{ - 2}}\,\,\,\,\,\,\,{\text{and}} \cr & {g_{pM}} = {g_p}\left[ {1 - \frac{{\frac{{{R_p}}}{{10}}}}{{{R_p}}}} \right] = 1\left[ {1 - 0.1} \right] = 0.9\,m{s^{ - 2}} \cr} $$
$$\therefore $$ Force $$=$$ mass of wire × $${g_{pM}}$$ $$=$$ 120 × 0.9 $$=$$ 108 N

$$\eqalign{ & E = - \frac{{GMm}}{{2r}} \cr & - \frac{{dE}}{{dt}} = \frac{{GMm}}{{2r}}\frac{1}{{{r^2}}}\frac{{dr}}{{dt}} \cr & \int\limits_0^t {dt} = - \frac{{GMm}}{{2C}}\int\limits_r^R {\frac{{dr}}{{{r^2}}}} ;t = \frac{{GMm}}{{2C}}\left[ {\frac{1}{R} - \frac{1}{r}} \right] \cr} $$

Body will move tangentially to the original orbit with same velocity.

Gravitational potential at point $$\frac{a}{2}$$ distance from centre is given by, $$V = - \frac{{GM}}{a} - \frac{{GM}}{{\frac{a}{2}}} = - \frac{{3GM}}{a}$$

Mass of satellite does not affect time period
$$\eqalign{ & \frac{{{T_A}}}{{{T_B}}} = {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^{\frac{3}{2}}} = {\left( {\frac{r}{{2r}}} \right)^{\frac{3}{2}}} = {\left( {\frac{1}{8}} \right)^{\frac{1}{2}}} = \frac{1}{{2\sqrt 2 }} \cr & \therefore {T_A} = {T_B} = 1:2\sqrt 2 \cr} $$

Given, height of a satellite $$h = 0.25 \times {10^6}m$$
Earth’s radius, $${R_e} = 6.38 \times {10^6}m$$
For the satellite revolving around the earth, orbital velocity of the satellite
$$\eqalign{ & {v_0} = \sqrt {\frac{{G{M_e}}}{{{R_e}}}} = \sqrt {\frac{{G{M_e}}}{{{R_e}\left[ {1 + \frac{h}{{{R_e}}}} \right]}}} \cr & \Rightarrow {v_0} = \sqrt {\frac{{g{R_e}}}{{1 + \frac{h}{{{R_e}}}}}} \cr} $$
Substitutes the values of $$g,{R_e}$$   and $$h,$$ we get
$$\eqalign{ & {v_0} = \sqrt {60 \times {{10}^6}} m/s \cr & {v_0} = 7.76 \times {10^3}m/s \cr & = 7.76\,km/s \cr} $$