$$\eqalign{ & \frac{L}{{2m}} = \frac{{dA}}{{dt}}\,\left( {L = {\text{angular momentum}}} \right) \cr & \frac{{m{v_{\max }}{r_{\min }}}}{{2m}} = \frac{{dA}}{{dt}};{v_{\max }} = \frac{{\frac{{2dA}}{{dt}}}}{{{r_{\min }}}} = 40 \cr} $$

$$V$$ is the orbital velocity. If $${V_C}$$ is the escape velocity then $${V_e} = \sqrt 2 V.$$
The kinetic energy at the time of ejection $$KE = \frac{1}{2}mV_e^2 = \frac{1}{2}m{\left( {\sqrt 2 V} \right)^2} = m{V^2}$$

Gravitational force provides the necessary centripetal force
$$\eqalign{ & \therefore \frac{{m{v^2}}}{{\left( {R + x} \right)}} = \frac{{GmM}}{{{{\left( {R + x} \right)}^2}}}{\text{ also }}g = \frac{{GM}}{{{R^2}}} \cr & \therefore {v^2} = \frac{{g{R^2}}}{{R + x}} \Rightarrow v = {\left( {\frac{{g{R^2}}}{{R + x}}} \right)^{\frac{1}{2}}} \cr} $$

$$\eqalign{ & {\text{Gravitational field,}}\,\,I = \left( {5\hat i + 12\hat j} \right)N/kg \cr & I = - \frac{{dv}}{{dr}} \cr & v = - \left[ {\int\limits_0^x {{I_x}} dx + \int\limits_0^y {{I_y}} dy} \right] = - \left[ {{I_x}.x + {I_y}.y} \right] \cr & = - \left[ {5\left( {7 - 0} \right) + 12\left( { - 3 - 0} \right)} \right] \cr & = - \left[ {35 + \left( { - 36} \right)} \right] = 1\,J/kg \cr} $$
i.e., change in gravitational potential $$1\,J/kg.$$  Hence change in gravitational potential energy $$1\,J.$$

$$E = - \frac{{dV}}{{dx}} \Rightarrow {\text{if}}\,V = 0,E\,{\text{must}}\,{\text{be}}\,{\text{zero}}{\text{.}}$$

$$g = \frac{{GM}}{{{R^2}}}\,{\text{also}}\,M = d \times \frac{4}{3}\pi {R^3}$$
$$\therefore g = \frac{4}{3}\;d\pi R$$    at the surface of planet
$$\eqalign{ & {g_p} = \frac{4}{3}\left( {2d} \right)\pi R',{g_e} = \frac{4}{3}\left( d \right)\pi R \cr & {g_e} = {g_p} \Rightarrow dR = 2dR' \Rightarrow R' = \frac{R}{2} \cr} $$

Let satellite of mass $$m$$ be revolving closely around the earth of mass $$M$$ and radius $$R.$$
Total energy of satellite $$ = PE + KE = - \frac{{GMm}}{R} + \frac{1}{2}m{v^2}$$
$$\eqalign{ & = - \frac{{GMm}}{R} + \frac{m}{2}\frac{{GM}}{R}\,\,\left[ {{\text{as}}\,v = \sqrt {\frac{{GM}}{R}} } \right] \cr & = - \frac{{GMm}}{{2R}} \cr} $$
∴ Total energy of satellite $$ = - \frac{1}{2}m{v^2}$$

$$\eqalign{ & \frac{{g'}}{g} = {\left( {\frac{R}{{R + h}}} \right)^2} \Rightarrow \frac{1}{{100}} = {\left( {\frac{R}{{R + h}}} \right)^2} \cr & \Rightarrow h = 9R \cr} $$

$$\eqalign{ & g = {g_p} - R{\omega ^2}{\cos ^2}\lambda \cr & = {g_p} - {\omega ^2}R{\cos ^2}{60^ \circ } = {g_p} - \frac{1}{4}R{\omega ^2} \cr} $$

Escape velocity on earth (or any other planet) is defined as the minimum velocity with which the body has to be projected vertically upwards from the surface of the earth (or any other planet). So, that it just crosses the gravitational field of earth and never returns on its own. The escape velocity of the earth is given by
$${v_e} = \sqrt {\frac{{2GM}}{R}} = \sqrt {2gR} = \sqrt {\frac{{8\pi \rho G{R^2}}}{3}} $$
From above equation it is clear that value of escape velocity of a body does not depend upon the mass $$\left( m \right)$$ of the body and its angle of projection from the surface of the earth or the planet. So, escape velocity remains same.