Acceleration due to gravity with height $$h$$ varies as $$g \propto \frac{1}{{{r^2}}}$$
(when $$r = R + h$$  ). Thus variation of $$g$$ and $$r$$ is a parabolic curve.

$$E = - \frac{{G{M_r}}}{{{R^3}}}\left( {{\text{if}}\,r < R} \right)\,\,{\text{and}}\,\,E = - \frac{{GM}}{{{r^2}}}\left( {{\text{if}}\,r \geqslant R} \right)$$

We know that
$$\eqalign{ & g = \frac{{GM}}{{{R^2}}} = \frac{{G\left( {\frac{4}{3}\pi {R^3}} \right)\rho }}{{{R^2}}} = \frac{4}{3}\pi GR\rho \cr & \frac{{g'}}{g} = \frac{{R'}}{R} = \frac{{3R}}{R} = 3 \cr & \therefore g' = 3\;g \cr} $$

$$g\left( {1 - \frac{d}{R}} \right) = g - {\omega ^2}R;d = \frac{{{\omega ^2}{R^2}}}{g}$$

Apply conservation of angular momentum.
From the law of conservation of angular momentum, $${L_1} = {L_2}$$
\[{\rm{So,}}\,\,m{r_1}{v_1} = m{r_2}{v_2}\,\,\left[ {\begin{array}{*{20}{c}} {{\rm{where,}}\,m = {\rm{mass \,the \,of \,planet}}}\\ {r = {\rm{radius\, of \,orbit}}}\\ {v{\rm{ }} = {\rm{ velocity \,of \,the \,planet}}} \end{array}} \right]\]
$${r_1}{v_1} = {r_2}{v_2} \Rightarrow \frac{{{v_1}}}{{{v_2}}} = \frac{{{r_2}}}{{{r_1}}}$$

Angular speed of earth = angular speed of geostationary satellite.
If $$\omega $$ is double, time period become half using $${T^2} \propto {{r'}^3}$$
$$r' = \frac{r}{{{4^{\frac{1}{3}}}}}$$

Acceleration due to gravity at a depth $$d$$ below the surface of the earth is given by
$${g_{{\text{depth}}}} = {g_{{\text{surface}}}}\left( {1 - \frac{d}{R}} \right)$$
Also, for a point at height $$h$$ above surface,
$${g_{{\text{height}}}} = {g_{{\text{surface}}}}\left[ {\frac{{{R^2}}}{{{{\left( {R + h} \right)}^2}}}} \right]$$
Therefore, we can say that value of $$g$$ increases from centre of maximum at the surface and then decreases as depicted in graph (B).

$$\eqalign{ & F = K{R^{ - n}} = MR{\omega ^2} \cr & \Rightarrow {\omega ^2} = K{R^{ - \left( {n + 1} \right)}}\,\,\,or,\,\,\omega = K{R^{\frac{{ - \left( {n + 1} \right)}}{2}}} \cr & \frac{{2\pi }}{T} \propto {R^{\frac{{ - \left( {n + 1} \right)}}{2}}} \cr & \therefore T \propto {R^{\frac{{ + \left( {n + 1} \right)}}{2}}} \cr} $$

Potential energy, $$U = - \frac{{GMm}}{r}$$
At the earth’s surface, $$r = R$$
$$\therefore {U_e} = - \frac{{GMm}}{R}$$
Now, if a body is taken to height $$h = 3R,$$  then the potential energy is given by
$$\eqalign{ & {U_h} = - \frac{{GMm}}{{R + h}}\,\,\left( {\because r = h + R} \right) \cr & = - \frac{{GMm}}{{4R}} \cr} $$
Thus, change in gravitational potential energy,
$$\eqalign{ & \Delta U = {U_h} - {U_e} \cr & = - \frac{{GMm}}{{4R}} - \left( { - \frac{{GMm}}{R}} \right) \cr & = - \frac{{GMm}}{{4R}} + \frac{{GMm}}{R} = \frac{3}{4}\frac{{GMm}}{R} \cr & \therefore \Delta U = \frac{3}{4}\frac{{g{R^2}m}}{R}\,\,\left( {\because GM = g{R^2}} \right) \cr & = \frac{3}{4}mgR \cr} $$

$${M_A} = \sigma 4\pi R_A^2,{M_B} = \sigma 4\pi R_B^2,$$
where $$\sigma $$ is surface density
$$\eqalign{ & {V_A} = \frac{{ - G{M_A}}}{{{R_A}}},{V_B} = \frac{{ - G{M_B}}}{{{R_B}}} \cr & \frac{{{V_A}}}{{{V_B}}} = \frac{{{M_A}}}{{{M_B}}}\frac{{{R_B}}}{{{R_A}}} = \frac{{\sigma 4\pi R_A^2}}{{\sigma 4\pi R_B^2}}\frac{{{R_B}}}{{{R_A}}} = \frac{{{R_A}}}{{{R_B}}} \cr & {\text{Given}}\,\,\frac{{{V_A}}}{{{V_B}}} = \frac{{{R_A}}}{{{R_B}}} = \frac{3}{4}\,\,{\text{then}}\,\,{R_B} = \frac{4}{3}{R_A} \cr} $$
for new shell of mass $$M$$ and radius $$R$$
$$\eqalign{ & M = {M_A} + {M_B} = \sigma 4\pi R_A^2 + \sigma 4\pi R_B^2 \cr & \sigma 4\pi {R^2} = \sigma 4\pi \left( {R_A^2 + R_B^2} \right) \cr} $$
then
$$\eqalign{ & \frac{V}{{{V_A}}} = \frac{M}{R}\frac{{{R_A}}}{{{R_B}}} = \frac{{\sigma 4\pi \left( {R_A^2 + R_B^2} \right)}}{{{{\left( {R_A^2 + R_B^2} \right)}^{\frac{1}{2}}}}} = \frac{{{R_A}}}{{\sigma 4\pi R_A^2}} \cr & = \frac{{\sqrt {R_A^2 + R_B^2} }}{{{R_A}}} = \frac{5}{3} \cr} $$