Gravitation MCQ Questions & Answers in Basic Physics | Physics

$$\eqalign{ & \frac{{{{\left( {{v_e}} \right)}_p}}}{{{{\left( {{v_e}} \right)}_e}}} = \frac{{\sqrt {\frac{{2G{M_p}}}{{{R_p}}}} }}{{\sqrt {\frac{{2G{M_e}}}{{{R_e}}}} }} = \sqrt {\frac{{{M_p}}}{{{M_e}}} \times \frac{{{R_e}}}{{{R_p}}}} = \sqrt {\frac{{10{M_e}}}{{{M_e}}} \times \frac{{{R_e}}}{{\frac{{{R_e}}}{{10}}}}} = 10 \cr & \therefore {\left( {{v_e}} \right)_p} = 10 \times {\left( {{v_e}} \right)_e} = 10 \times 11 = 110\,km/s \cr} $$

Variation of acceleration due to gravity, $$g$$ with distance $$'d\,'$$ from centre of the earth
$$\eqalign{ & {\text{If }}d < R,\,\,\,g = \frac{{Gm}}{{{R^2}}}.d\,\,\,\,i.e.,\,\,g \propto d\left( {{\text{straight line}}} \right) \cr & {\text{If }}d = R,\,\,\,{g_s} = \frac{{Gm}}{{{R^2}}} \cr & {\text{If }}d > R,\,\,\,g = \frac{{Gm}}{{{d^2}}}\,\,\,\,\,\,\,\,\,\,i.e.,\,\,g \propto \frac{1}{{{d^2}}} \cr} $$

At a point inside a spherical shell, the value of gravitational intensity, $$I = 0.$$
If $$V = 0$$  then gravitational field is necessarily Zero.

As we know, gravitational potential $$\left( v \right)$$ and acceleration due to gravity $$\left( g \right)$$ with height
$$\eqalign{ & V = \frac{{ - GM}}{{R + h}} = - 5.4 \times {10^7}\,......\left( {\text{i}} \right) \cr & {\text{and}}\,\,g = \frac{{{\text{ }}GM{\text{ }}}}{{{{\left( {R + h} \right)}^2}}} = 6\,......\left( {{\text{ii}}} \right) \cr} $$
Dividing (i) by (ii)
$$\eqalign{ & \frac{{\frac{{ - GM}}{{R + h}}}}{{\frac{{GM}}{{{{\left( {R + h} \right)}^2}}}}} = \frac{{ - 5.4 \times {{10}^7}}}{6} \Rightarrow \frac{{5.4 \times {{10}^7}}}{{\left( {R + h} \right)}} = 6 \cr & \Rightarrow R + h = 9000\,km\,{\text{so,}}\,h = 2600\,km \cr} $$

Applying conservation of total mechanical energy principle
$$\eqalign{ & \frac{1}{2}m{v^2} = m{g_A}{h_A} = m{g_B}{h_B} \cr & \Rightarrow {g_A}{h_A} = {g_B}{h_B} \cr & \Rightarrow {h_B} = \left( {\frac{{{g_A}}}{{{g_B}}}} \right){h_A} = 9 \times 2 = 18\,m \cr} $$

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.
$$\therefore {T_s} = 2\,hr$$

$$\eqalign{ & \frac{{T_1^2}}{{T_2^2}} = \frac{{r_1^3}}{{r_2^3}} = {\left( 4 \right)^3} \cr & \therefore {T_2} = \frac{{{T_1}}}{8}. \cr} $$

$$\eqalign{ & \frac{{{g_p}}}{{{g_e}}} = \frac{2}{1};\frac{{{R_p}}}{{{R_e}}} = \frac{1}{2} \cr & {\text{So,}}\,\frac{{{M_p}}}{{{M_e}}} = 1:2 \cr} $$

Escape velocity on the earth's surface is given by $${v_{{\text{es}}}} = \sqrt {\frac{{2G{M_e}}}{{{R_e}}}} $$
where, $$G$$ is gravitational constant, $${M_e}$$ and $${R_e}$$ are the mass and radius of the earth respectively. By taking the ratios of two different cases
$$\eqalign{ & \therefore \frac{{{{v'}_{{\text{es}}}}}}{{{v_{{\text{es}}}}}} = \sqrt {\frac{{{{M'}_e}}}{{{M_e}}} \times \frac{{{R_e}}}{{{{R'}_e}}}} \cr & {\text{but}}\,\,{{M'}_e} = 2{M_e} \cr & {\text{and}}\,\,{{R'}_e} = \frac{{{R_e}}}{2} \cr & {v_{{\text{es}}}} = 11.2\,km/s \cr & \therefore \frac{{{{v'}_{{\text{es}}}}}}{{{V_{{\text{es}}}}}} = \sqrt {\frac{{2{M_e}}}{{{M_e}}} \times \frac{{{R_e}}}{{\frac{{{R_e}}}{2}}}} = \sqrt 4 = 2 \cr & \therefore {{v'}_{{\text{es}}}} = 2{v_{{\text{es}}}} = 2 \times 11.2 = 22.4\,km/s \cr} $$
NOTE
The escape velocity on moon's surface is only $$2.5\,km/s.$$   This is the basic fundamental on which, absence of atmosphere on moon can be explained.

we have, $$\frac{{m{v^2}}}{{R + x}} = \frac{{GmM}}{{{{\left( {R + x} \right)}^2}}}$$
$$x \,\,=$$   height of satellite from earth surface, $$ m\,\,=$$   mass of satellite
$$\eqalign{ & \Rightarrow {v^2} = \frac{{GM}}{{\left( {R + x} \right)}}\,\,or\,\, v = \sqrt {\frac{{GM}}{{R + x}}\,} \cr & T = \frac{{2\pi \left( {R + x} \right)}}{v} = \frac{{2\pi \left( {R + x} \right)}}{{\sqrt {\frac{{GM}}{{R + x}}\,} }} \cr} $$
which is independent of mass of satellite.