Gravitation MCQ Questions & Answers in Basic Physics | Physics

Gauss law for gravitation
$$\eqalign{ & \int {\vec g \cdot d\vec s = - {m_{{\text{in}}}} \cdot 4\pi G;g = \frac{{GM}}{{{R^2}}}} \cr & 2 \times \frac{{GM}}{{{R^2}}} \times A = \frac{M}{{\frac{4}{3}\pi {R^3}}}\left( {h \times A} \right) \times 4\pi G \cr & \Rightarrow h = \frac{{2R}}{3} \cr} $$

$$\eqalign{ & \frac{1}{2}mv_{\min }^2 - \frac{{GMm}}{{R\sqrt 2 }} \times 2 = 0 \cr & \Rightarrow {v_{\min }} = \sqrt {\frac{{2GM\sqrt 2 }}{R}} \cr} $$

According to Kepler's third law (or law of periods) the square of the time taken to complete the orbit (time period $$T$$) is proportional to the cube of the semi-major axis $$\left( r \right)$$ of the elliptical orbit i.e. $${T^2} \propto {r^3}$$
Here, $${r_1} = {10^{13}}m,\,{r_2} = {10^{12}}m$$
$$\eqalign{ & \therefore \frac{{T_1^2}}{{T_2^2}} = \frac{{r_1^3}}{{r_2^3}} = \frac{{{{\left( {{{10}^{13}}} \right)}^3}}}{{{{\left( {{{10}^{12}}} \right)}^3}}} \cr & {\text{or}}\,\,\frac{{T_1^2}}{{T_2^2}} = \frac{{{{10}^{39}}}}{{{{10}^{36}}}} = {10^3} \cr & {\text{or}}\,\,\frac{{{T_1}}}{{{T_2}}} = 10\sqrt {10} \cr} $$

The Gravitational field due to a thin spherical shell of radius $$R$$ at distance $$r.$$
$$E = \frac{{GM}}{{{r^2}}}\,\,\left( {{\text{If }}r > R} \right)$$
For $$r = R$$  i.e. on the surface of the shell
$$E = \frac{{GM}}{{{R^2}}}$$
For $$r < R$$  i.e. inside the shell $$E = 0$$

Compare the equation of escape velocity of earth and planet. Escape velocity is given by,
$${v_{es}} = \sqrt {\frac{{2G{M_e}}}{{{R_e}}}} $$
From a planet, $${{v'}_{es}} = \sqrt {\frac{{2G{M_p}}}{{{R_p}}}} $$
Therefore, $$\frac{{{{v'}_{es}}}}{{{v_{es}}}} = \sqrt {\frac{{2G{M_p}}}{{{R_p}}}} \times \sqrt {\frac{{{R_e}}}{{2G{M_e}}}} $$
It is given that,
mass of the planet = mass of the earth
i.e. $${M_p} = {M_e}$$
So, $$\frac{{{{v'}_{es}}}}{{{v_{es}}}} = \sqrt {\frac{{{R_e}}}{{{R_p}}}} \,......\left( {\text{i}} \right)$$
Given, $${R_p} = \frac{{{R_e}}}{4} \Rightarrow \frac{{{R_p}}}{{{R_e}}} = \frac{1}{4}\,\,{\text{and}}\,\,{v_{es{\text{ }}}} = 11.2\,km/s$$
Substituting in Eq. (i), we have
$$\frac{{{{v'}_{es}}}}{{11.2}} = \sqrt {\frac{4}{1}} = 2,{{v'}_{es}} = 11.2 \times 2 = 22.4\,km/s$$

Range of projectile $$R = \frac{{{u^2}\sin 2\theta }}{g}$$
if $$u$$ and $$\theta $$ are constant then $$R \propto \frac{1}{g}$$
$$\eqalign{ & \frac{{{R_m}}}{{{R_e}}} = \frac{{{g_e}}}{{{g_m}}} \Rightarrow \frac{{{R_m}}}{{{R_e}}} = \frac{1}{{0.2}} \cr & \Rightarrow {R_m} = \frac{{{R_e}}}{{0.2}} \Rightarrow {R_m} = 5{R_e} \cr} $$

From the property of ellipse, $$\frac{2}{R} = \frac{1}{{{r_1}}} + \frac{1}{{{r_2}}}$$

$$\eqalign{ & {\text{or}}\,\,\frac{2}{R} = \frac{{{r_1} + {r_2}}}{{{r_1}{r_2}}} \cr & \Rightarrow R = \frac{{2{r_1}{r_2}}}{{{r_1} + {r_2}}} \cr} $$

For $$h < < R,$$   the orbital velocity is $$\sqrt {gR} $$
Escape velocity $$ = \sqrt {2gR} $$
$$\therefore $$ The minimum increase in its orbital velocity
$$ = \sqrt {2gR} - \sqrt {gR} = \sqrt {gR} \left( {\sqrt 2 - 1} \right)$$

We know that intensity is negative gradient of potential,
i.e., $$I = - \left( {\frac{{dV}}{{dr}}} \right)$$   and as here $$I = - \left( {\frac{K}{r}} \right),$$   so,
$$\eqalign{ & \frac{{dV}}{{dr}} = \frac{K}{r},\,\,{\text{i}}{\text{.e}}{\text{.,}}\,\,\int d V = K\int {\frac{{dr}}{r}} \cr & {\text{or}}\,\,V - {V_0} = K\log \frac{r}{{{r_0}}} \cr & {\text{so}}\,\,V = K\log \frac{r}{{{r_0}}} + {V_0} \cr} $$

Let mass per unit light of wire, $$\lambda = \frac{m}{\ell }$$  and $$\pi r = \ell ,r = \frac{\ell }{\pi }.$$   mass of element, $$dm = \lambda \,rd\theta \,\,{\text{then}}\,\,dE = \frac{{Gdm}}{{{r^2}}}$$
$$\eqalign{ & \int\limits_0^\pi d E = \int\limits_0^\pi {\frac{{G\lambda rd\theta }}{{{r^2}}}} \left( {\widehat i\cos \theta + \widehat j\sin \theta } \right) \cr & E = \frac{{G\lambda }}{r}\left[ {\int\limits_0^\pi {\hat i} \cos \theta + \int\limits_0^\pi {\hat j} \sin \theta } \right] \cr & = \frac{{2G\lambda }}{r}\widehat j = \frac{{2GM}}{{\ell r}}\widehat j = \frac{{2Gm\pi }}{{{\ell ^2}}}\widehat j{\text{ }}\left( {{\text{along }}y - {\text{axis}}} \right) \cr} $$