According to Kepler's third law, the square of the time period of revolution of a planet around the sun is directly proportional to the cube of semi major-axis of its elliptical orbit i.e.
$${T^2} \propto {r^3}$$
where, $$T =$$  time taken by the planet to go once around the sun.
$$r =$$  semi-major axis of the elliptical orbit
$$\therefore \frac{{T_1^2}}{{T_2^2}} = \frac{{{{\left( r \right)}^3}}}{{{{\left( {2r} \right)}^3}}} = \frac{1}{8} \Rightarrow \frac{{{T_1}}}{{{T_2}}} = \frac{1}{{2\sqrt 2 }}$$

For calculation of gravitational field intensity inside the cavity.

$$\eqalign{ & {{\vec I}_1} = \frac{{G\left( {\frac{4}{3}\pi R_1^3} \right)\rho \left( { - {{\vec r}_1}} \right)}}{{R_1^3}} \cr & {{\vec I}_2} = \frac{{G\left( {\frac{4}{3}\pi R_2^3} \right)\rho \left( { - {{\vec r}_2}} \right)}}{{R_2^3}} \cr & \vec I = {{\vec I}_1} - {{\vec I}_2}\,\,\left( {\vec I - {\text{intensity inside the cavity}}} \right) \cr & = \frac{4}{3}G\pi \rho \left[ { - {{\vec r}_1} + {{\vec r}_2}} \right] = \frac{4}{3}G\pi \rho \vec \ell \cr} $$

The components of acceleration are as shown $$a = {\vec a_r} + {\vec a_t}$$
The resultant of transverse and radial component of the acceleration is represented by $$\vec a$$

Let mass of smaller sphere (which has to be removed) is $$m$$
Radius $$ = \frac{R}{2}\,\left( {{\text{from figure}}} \right)$$
$$\frac{M}{{\frac{4}{3}\pi {R^3}}} = \frac{m}{{\frac{4}{3}\pi {{\left( {\frac{R}{2}} \right)}^3}}} \Rightarrow m = \frac{M}{8}$$
Mass of the left over part of the sphere
$$M' = M - \frac{M}{8} = \frac{7}{8}M$$
Therefore gravitational field due to the left over part of the sphere
$$ = \frac{{GM'}}{{{x^2}}} = \frac{7}{8}\frac{{GM}}{{{x^2}}}$$

Escape velocity $$ = v = \sqrt {\frac{{2GM}}{R}} $$
$$\eqalign{ & \Rightarrow {v^2} = \frac{{2GM}}{R}\,......\left( {\text{i}} \right) \cr & {v^2} = \left( {2GM} \right){R^{ - 1}} \cr} $$
Differentiating both sides, we get,
$$2v\frac{{dv}}{{dR}} = - \frac{{2GM}}{{{R^2}}} \Rightarrow v\frac{{dv}}{{dR}} = \frac{{GM}}{{{R^2}}}\,......\left( {{\text{ii}}} \right)$$
Dividing (ii) by (i),
$$\frac{1}{v}\frac{{dv}}{{dR}} = - \frac{1}{{2R}}$$
$$ \Rightarrow \left| {\frac{{dv}}{v}} \right| \times 100 = \frac{1}{2} \times 4\% = 2\% $$
$$\therefore $$ If the radius decreases by $$4\% ,$$ escape velocity will increase by $$2\% .$$

According to Newton's law of gravitation, the force between two spheres is given by,
$$F = \frac{{GMm}}{{{r^2}}}$$
From the relation, we can say the gravitational force does not depend on the medium between two spheres hence, it remains same, i.e. $$F.$$

Potential energy, $$U = - \frac{{G{M_e}m}}{{{R_e}}}$$
where,
$${M_e} =$$  mass of the earth
$$m =$$  mass of satellite
$${R_e} =$$  radius of the earth
$$G =$$  gravitational constant
or $$\,\left| U \right| = \frac{{G{M_e}m}}{{{R_e}}}$$
Kinetic energy, $$K = \frac{1}{2}\frac{{G{M_e}m}}{{{R_e}}}$$
Thus, $$\frac{K}{{\left| U \right|}} = \frac{1}{2}\frac{{G{M_e}m}}{{{R_e}}} \times \frac{{{R_e}}}{{G{M_e}m}} = \frac{1}{2}$$
NOTE
The total energy, $$E = K + U = - \frac{{G{M_e}m}}{{2r}}$$

This energy is constant and negative, i.e. the system is closed. To farther the satellite from the earth, the greater is its total energy.

$$\eqalign{ & \frac{1}{2}mV_e^2 - \frac{{G{M_e}m}}{{{R_e}}} - \frac{{G{M_e}m \times 3 \times {{10}^5}}}{{2.5 \times {{10}^4}{R_e}}} = 0 \cr & \frac{{V_e^2}}{2} = \frac{{G{M_e}}}{{{R_e}}}\left[ {1 + \frac{{3 \times {{10}^5}}}{{2.5 \times {{10}^4}}}} \right] \cr & {V_e} = \sqrt {13\left( {\frac{{2G{M_e}}}{{{R_e}}}} \right)} = \sqrt {13} \times 11.2 \approx 42 \cr} $$

Due to complete solid sphere, potential at point $$P$$
$$\eqalign{ & {V_{sphere}} = \frac{{ - GM}}{{2{R^3}}}\left[ {3{R^2} - {{\left( {\frac{R}{2}} \right)}^2}} \right] \cr & = \frac{{ - GM}}{{2{R^3}}}\left( {\frac{{11{R^2}}}{4}} \right) \cr & = - 11\frac{{GM}}{{8R}} \cr} $$

Due to cavity part potential at point P
$${V_{cavity}} = - \frac{3}{2}\frac{{\frac{{GM}}{8}}}{{\frac{R}{2}}} = - \frac{{3GM}}{{8R}}$$
So potential at the centre of cavity
$$ = {V_{sphere}} - {V_{cavity}} = - \frac{{11GM}}{{8R}} - \left( { - \frac{3}{8}\frac{{GM}}{R}} \right) = \frac{{ - GM}}{R}$$

$$\eqalign{ & g = \frac{{GM}}{{{R^2}}}{\text{ and }}g' = \frac{{GM}}{{{{\left( {0.99R} \right)}^2}}} \cr & \therefore \frac{{g'}}{g} = {\left( {\frac{R}{{0.99R}}} \right)^2}\,\,\, \Rightarrow g' > g \cr} $$