Gravitation MCQ Questions & Answers in Basic Physics | Physics

Variation of $$g$$ with altitude is, $${g_h} = g\left[ {1 - \frac{{2h}}{R}} \right];$$
variation of $$g$$ with depth is, $${g_d} = g\left[ {1 - \frac{d}{R}} \right]$$
Equating $${g_h}$$ and $${g_d},$$ we get $$d=2h$$

Inside a shell $$V = - \frac{{GM}}{R} = {\text{constant}}\,{\text{and}}\,E = 0$$
Outside the shell, $$V = - \frac{{GM}}{r}\,{\text{and}}\,E = \frac{{GM}}{{{r^2}}}$$
As $$r$$ increases, $$V$$ increases and $$E$$ decreases.

Note: A satellite revolving near the earth's surface has a time period of 84.6 min.
We know that as the height increases, the time period increases. Thus the time period of the spy satellite should be slightly greater than 84.6 minutes.
$$\therefore {T_s} = 2hr$$

Orbital velocity for close to earth orbits $$ = \sqrt {gR} .$$   Escape velocity required $$ = \sqrt {2gR} .$$
$$\% $$ Increase required $$ = \frac{{{v_e} - {v_0}}}{{{v_0}}} \times 100$$
$$\eqalign{ & = \frac{{\sqrt {gR} \left( {\sqrt 2 - 1} \right)}}{{\sqrt {gR} }} \times 100 \cr & = 41\% \cr} $$

Let the gravitational field be zero at a point distant $$x$$ from $${M_1}.$$
$$\eqalign{ & \frac{{G{M_1}}}{{{x^2}}} = \frac{{G{M_2}}}{{{{\left( {d - x} \right)}^2}}};\frac{x}{{d - x}} = \sqrt {\frac{{{M_1}}}{{{M_2}}}} \cr & x\sqrt {{M_2}} = \sqrt {{M_1}} d - x\sqrt {{M_1}} \cr & x\left[ {\sqrt {{M_1}} + \sqrt {{M_2}} } \right] = \sqrt {{M_1}} d \cr & x = \frac{{d\sqrt {{M_1}} }}{{\sqrt {{M_1}} + \sqrt {{M_2}} }},d - x = \frac{{d\sqrt {{M_2}} }}{{\sqrt {{M_1}} + \sqrt {{M_2}} }} \cr} $$
Potential at this point due to both the masses will be
$$\eqalign{ & - \frac{{G{M_1}}}{x} - \frac{{G{M_2}}}{{\left( {d - x} \right)}} \cr & = - G\left[ {\frac{{{M_1}\left( {\sqrt {{M_1}} + \sqrt {{M_2}} } \right)}}{{d\sqrt {{M_1}} }} + \frac{{{M_2}\left( {\sqrt {{M_1}} + \sqrt {{M_2}} } \right)}}{{d\sqrt {{M_2}} }}} \right] \cr & = - \frac{G}{d}{\left( {\sqrt {{M_1}} + \sqrt {{M_2}} } \right)^2} \cr & = - \frac{G}{d}\left( {{M_1} + {M_2} + 2\sqrt {{M_1}} \sqrt {{M_2}} } \right) \cr} $$

$$\eqalign{ & {v_e} = \sqrt {2{g_e}{R_e}} ;\,\,{v_m} = \sqrt {2{g_m}{R_m}} \cr & \frac{{{v_e}}}{{{v_m}}} = \sqrt {\frac{{{g_e}}}{{{g_m}}}\frac{{{R_e}}}{{{R_m}}}} = \sqrt {\frac{{{g_e}}}{{\frac{{{g_e}}}{6}}}\frac{{{R_e}}}{{\frac{{{R_e}}}{4}}}} = \sqrt {24} \cr} $$

$$\eqalign{ & x = \frac{{mg}}{k} = \frac{{GMm}}{{{R^2}}}k \Rightarrow x \propto \frac{1}{{{R^2}}} \Rightarrow {x_2} \cr & = {\left( {\frac{{6400}}{{790n}}} \right)^2}{x_1} = 0.79\,cm \cr} $$

we know that $$\frac{{g'}}{g} = \frac{{{R^2}}}{{{{\left( {R + h} \right)}^2}}}$$
$$\eqalign{ & \therefore \frac{{\frac{g}{9}}}{g} = {\left[ {\frac{R}{{R + h}}} \right]^2} \cr & \therefore h = 2\,R \cr} $$

According to the question, $$\frac{{GMm}}{{{{\left( {R + h} \right)}^2}}} = \frac{1}{{16}}\frac{{GMm}}{{{R^2}}}$$
where, $$m =$$  mass of the body and $$\frac{{GM}}{{{R^2}}} = $$   gravitational acceleration
$$\eqalign{ & \frac{1}{{{{\left( {R + h} \right)}^2}}} = \frac{1}{{16{R^2}}} \cr & {\text{or}}\,\,\frac{R}{{R + h}} = \frac{1}{4} \cr & {\text{or}}\,\,\frac{{R + h}}{R} = 4 \cr & h = 3R \cr} $$

$$\because $$ Total energy of a satellite at height $$h$$ is $$ = KE + PE = \frac{{GMm}}{{2\left( {R + h} \right)}} - \frac{{GMm}}{{\left( {R + h} \right)}}$$
$$\eqalign{ & = \frac{{ - GMm}}{{2\left( {R + h} \right)}} = \frac{{ - GMm{R^2}}}{{2{R^2}\left( {R + h} \right)}} \cr & = \frac{{ - m{g_0}{R^2}}}{{2\left( {R + h} \right)}}\,\,\left( {\because {g_0} = \frac{{GM}}{{{R^2}}}} \right) \cr} $$