$$\eqalign{ & \frac{{M{v^2}}}{R} = \frac{{G{M^2}}}{{4{R^2}}} \Rightarrow M = \frac{{4R{v^2}}}{G} \cr & v = \frac{{2\pi R}}{T} \cr & R = \frac{{vT}}{{2\pi }} \cr & M = \frac{{{v^3}T2}}{{\pi G}} \cr} $$

$$\eqalign{ & {T^2} \propto {R^3}\left( {{\text{According}}\,{\text{to}}\,{\text{Keple's}}\,{\text{law}}} \right) \cr & T_1^2 \propto {\left( {{{10}^{13}}} \right)^3}\,\,{\text{and}}\,\,T_2^2 \propto {\left( {{{10}^{12}}} \right)^3} \cr & \therefore \frac{{T_1^2}}{{T_2^2}} = {\left( {10} \right)^3}\,\,{\text{or}}\,\,\frac{{{T_1}}}{{{T_2}}} = 10\sqrt {10} \cr} $$

The gravitational potential at the centre of uniform spherical shell is equal to the gravitational potential at the surface of shell i.e.,
$$V = \frac{{ - GM}}{a},$$   where $$a$$ is radius of spherical shell
Now, if the shell shrinks then its radius decrease then density increases, but mass is constant. so from above expression if a decreases, then $$V$$ increases.

Potential energy = Kinetic energy
i.e. $$mgh = \frac{1}{2}m{v^2} \Rightarrow v = \sqrt {2gh} $$
If $${h_1}$$ and $${h_2}$$ are the initial and final heights, then
$${v_1} = \sqrt {2g{h_1}} ,{v_2} = \sqrt {2g{h_2}} $$
Loss in velocity $$\Delta v = {v_1} - {v_2} = \sqrt {2g{h_1}} - \sqrt {2g{h_2}} $$
$$\therefore $$ Fractional loss in velocity $$ = \frac{{\Delta v}}{{{v_1}}} = \frac{{\sqrt {2g{h_1}} - \sqrt {2g{h_2}} }}{{\sqrt {2g{h_1}} }} = 1 - \sqrt {\frac{{{h_2}}}{{{h_1}}}} $$
Substituting the values, we have
$$\eqalign{ & \therefore \frac{{\Delta v}}{{{v_1}}} = 1 - \sqrt {\frac{{1.8}}{5}} = 1 - \sqrt {0.36} = 1 - 0.6 \cr & = 0.4 = \frac{2}{5} \cr} $$

$$\eqalign{ & g' = g\left( {1 - \frac{d}{R}} \right) \Rightarrow \frac{g}{n} = g\left( {1 - \frac{d}{R}} \right) \cr & \Rightarrow d = \left( {\frac{{n - 1}}{n}} \right)R \cr} $$

Apply Newton's gravitation law. According to Newton's law of gravitation force, $$F = \frac{{GMm}}{{{R^2}}}$$
Force on planet of mass $${M_p}$$ and body of mass $$m$$ is given by
$$\eqalign{ & F = \frac{{G{M_p}m}}{{{{\left( {\frac{{{D_p}}}{2}} \right)}^2}}}\,\,\left[ {{\text{where,}}\,{D_p} = {\text{diameter of planet and }}{R_p} = {\text{radius of planet}} = \frac{{{D_p}}}{2}} \right] \cr & F = \frac{{4G{M_p}m}}{{D_p^2}} \cr} $$
As we know that, $$F = ma$$
So, acceleration due to gravity $$a = \frac{F}{m} = \frac{{4G{M_p}}}{{D_p^2}}$$

The binding energy of sphere of mass $$m$$ (say) on the surface of the earth kept at rest is $$\frac{{G{M_e}m}}{{{R_e}}}.$$   To escape it from the earth's surface, this much energy in the form of kinetic energy is supplied to it.
So, $$\frac{1}{2}mv_e^2 = \frac{{G{M_e}m}}{{{R_e}}}$$
or $${v_e} = {\text{escape}}\,{\text{velocity}} = \sqrt {\frac{{2G{M_e}}}{{{R_e}}}} $$
where, $${R_e} =$$  radius of earth,
$${M_e} =$$  mass of the earth.

Gravitational potential at some height $$h$$ from the surface of the earth is given by
$$V = - \frac{{GM}}{{R + h}}\,......\left( {\text{i}} \right)$$
And acceleration due to gravity at some height $$h$$ from the earth surface can be given as
$$g' = \frac{{GM}}{{{{\left( {R + h} \right)}^2}}}\,......\left( {{\text{ii}}} \right)$$
From Eqs. (i) and (ii), we get
$$\eqalign{ & \frac{{\left| V \right|}}{{g'}} = \frac{{GM}}{{\left( {R + h} \right)}} \times \frac{{{{\left( {R + h} \right)}^2}}}{{GM}} \cr & \Rightarrow \frac{{\left| V \right|}}{{g'}} = R + h\,......\left( {{\text{iii}}} \right) \cr & \because V = - 5.4 \times {10^7}\;J\;k{g^{ - 2}}\,{\text{and}}\,\,g' = 6.0\;m{s^{ - 2}} \cr} $$
Radius of earth, $$R = 6400\,km.$$
Substitute these values in Eq. (iii), we get
$$\eqalign{ & \frac{{5.4 \times {{10}^7}}}{{6.0}} = R + h \Rightarrow 9 \times {10^6} = R + h \cr & \Rightarrow h = \left( {9 - 6.4} \right) \times {10^6} = 2.6 \times {10^6}\;m \cr & \Rightarrow h = 2600\,km \cr} $$

The gravitational field due to the ring at a distance $$\sqrt 3 r$$  is given by
$$E = \frac{{Gm\left( {\sqrt 3 r} \right)}}{{{{\left[ {{r^2} + {{\left( {\sqrt 3 r} \right)}^2}} \right]}^{\frac{3}{2}}}}} \Rightarrow E = \frac{{\sqrt 3 Gm}}{{8{r^2}}}$$

$$\because dF = \frac{{Gm\left( {\mu dx} \right)}}{{{x^2}}}$$

$$\eqalign{ & F = Gm\int {\left( {A + Bx} \right)} \frac{{dx}}{x} \cr & F = Gm\left[ {A\left( {\frac{1}{a} - \frac{1}{{a + L}}} \right) + BL} \right] \cr} $$