Gravitation MCQ Questions & Answers in Basic Physics | Physics

The acceleration due to gravity on an object of mass $$m$$
$$g = \frac{F}{m}$$
but from Newton's law of gravitation $$F = \frac{{GMm}}{{{R^2}}}$$
where, $$M$$ is the mass of the earth and $$R$$ is the radius of the earth.
$$\eqalign{ & \therefore g = \frac{{\frac{{GMm}}{{{R^2}}}}}{m} = \frac{{GM}}{{{R^2}}}\,\,\left[ {\because {M_p} = \frac{4}{3}\pi R_p^3{P_p}\,\,{\text{and}}\,\,{M_e} = \frac{4}{3}\pi R_e^3{P_e}} \right] \cr & {\text{Given,}}\,\,{\rho _{{\text{planet}}}} = 2{\rho _{{\text{earth}}}} \cr & {\text{Also,}}\,\,{g_{{\text{planet}}}} = {g_{{\text{earth}}}} \cr & \frac{{G{M_p}}}{{R_p^2}} = \frac{{G{M_e}}}{{R_e^2}} \cr & {\text{As,}}\,{\text{Density}}\,\,\left( \rho \right) = \frac{{{\text{Mass}}\left( M \right)}}{{{\text{ Volume}}\left( V \right)}} \cr & {\text{So,}}\,\,\frac{{G \times \frac{4}{3}\pi R_p^3{\rho _p}}}{{R_p^2}} = \frac{{G \times \frac{4}{3}\pi R_e^3{\rho _e}}}{{R_e^2}} \cr & {\text{or}}\,\,{R_p}{\rho _p} = {R_e}{\rho _e} \cr & {\text{or}}\,\,{R_p} \times 2{\rho _e} = {R_e}{\rho _e} \cr & {\text{or}}\,\,{R_p} = \frac{{{R_e}}}{2} = \frac{R}{2} \cr} $$

Centripetal force = net gravitational force
$$\eqalign{ & \Rightarrow \frac{{mv_0^2}}{r} = 2F\cos {45^ \circ } + {F_1} = \frac{{2G{m^2}}}{{{{\left( {\sqrt 2 } \right)}^2}}}\frac{1}{{\sqrt 2 }} + \frac{{G{m^2}}}{{4{r^2}}} \cr & \frac{{mv_0^2}}{r} = \frac{{G{m^2}}}{{4{r^2}}}\left[ {2\sqrt 2 + 1} \right] \cr & \Rightarrow {v_0} = {\left[ {\frac{{Gm\left( {2\sqrt 2 + 1} \right)}}{{4r}}} \right]^{\frac{1}{2}}} \cr} $$

Mass of the rock = $$m$$ and height of rock from earth $$\left( h \right) = \frac{{{R_e}}}{2} = 3.2 \times {10^6}m$$
We know that gravitational potential energy of the rock rotating at height
$$h = - \frac{{G{M_e}m}}{{{R_e} + h}} = - \frac{2}{3}mg{R_e}\,\,\left( {{\text{where,}}\,G{M_e} = g{R_e}\,{\text{and}}\,h = {R_e}} \right)$$

Time period does not depend upon the mass of satellite

$$\eqalign{ & g = \frac{4}{3}\pi \rho GR \cr & \therefore \frac{{{g_1}}}{{{g_2}}} = \frac{{{R_1}{\rho _1}}}{{{R_2}{\rho _2}}} \cr} $$

$$\eqalign{ & W = m\left( {{V_2} - {V_1}} \right) \cr & {\text{when,}}\,{V_1} = - \left[ {\frac{{G{M_1}}}{a} + \frac{{G{M_2}}}{{\sqrt 2 a}}} \right], \cr & {V_2} = - \left[ {\frac{{G{M_2}}}{a} + \frac{{G{M_1}}}{{\sqrt 2 a}}} \right] \cr & \therefore W = \frac{{Gm\left( {{M_2} - {M_1}} \right)}}{{a\sqrt 2 }}\left( {\sqrt 2 - 1} \right). \cr} $$

$$\vec g' = - \frac{{g\vec r}}{R}\,{\text{for}}\,r \leqslant R\,{\text{and}}\,g' = \frac{g}{{{{\left( {1 + \frac{r}{R}} \right)}^2}}}\,{\text{for}}\,r \geqslant R$$
so option (A) is correct.

As we know that, force on satellite is only gravitational force which will always be towards the centre of earth. Thus, the acceleration of $$S$$ is always directed towards the centre of the earth.

$$\eqalign{ & {g_h} = {\text{Acceleration due to gravity at height }}h{\text{ above earth's surface}} \cr & = g{\left( {\frac{R}{{R + h}}} \right)^2} = g\left( {1 - \frac{{2h}}{R}} \right) \cr & {g_d} = {\text{Acceleration at depth }}d{\text{ below earths surface}} \cr & = g\left( {1 - \frac{d}{R}} \right) \cr & {\text{Given, when }}h = 1\,km,\,{g_d} = {g_h} \cr & {\text{or}}\,\,g\left( {1 - \frac{d}{R}} \right) = g\left( {1 - \frac{{2h}}{R}} \right) \cr & \Rightarrow d = 2h\,{\text{or}}\,\,d = 2\,km \cr} $$

The resulting gravitational potential,
$$\eqalign{ & V = - 2G\left[ {\frac{1}{1} + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...} \right] \cr & \Rightarrow V = - 2G\left[ {1 + \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ...} \right] \cr & \Rightarrow V = - 2G{\left( {1 + \frac{1}{2}} \right)^{ - 1}} \cr & \Rightarrow V = - \frac{{2G}}{{\left( {1 - \frac{1}{2}} \right)}} = - \frac{{2G}}{{\left( {\frac{1}{2}} \right)}} = - 4G \cr} $$