Mix Examples-Gravitation Questions in English

(B) False. The intensity of the gravitational field $g(h) = g_0 (1 + h/R)^{-2}$ decreases as the height $h$ increases.
$(b)$ True. Since the escape velocity on the Moon is low,gas molecules can easily attain this speed due to thermal energy and escape into space,preventing the formation of an atmosphere.
$(c)$ True. $A$ satellite in orbit is in a state of free fall towards the Earth,which results in a state of weightlessness for objects inside it.

(B) False. The Earth is flattened at the poles and bulges at the equator. The radius at the poles is smaller than at the equator. Since $g = GM/R^2$,$g$ is greater at the poles than at the equator.
$(b)$ True. Since $F = mg$,$g = F/m$. The unit of force is $\text{Newton}$ $(N)$ and mass is $\text{kilogram}$ $(kg)$,so the unit of $g$ is $N/kg$,which is equivalent to $m/s^2$.
$(c)$ False. The formula for escape velocity is $v_e = \sqrt{2GM/R}$. Thus,$v_e$ is directly proportional to the square root of the mass $(\sqrt{M})$ and inversely proportional to the square root of the radius $(1/\sqrt{R})$.

(A) The potential energy is given by $U(r) = -\frac{C}{r}$.
The force $F$ is the negative gradient of the potential energy: $F = -\frac{dU}{dr} = -\frac{d}{dr}\left(-\frac{C}{r}\right) = -\frac{C}{r^2}$.
For a particle in a circular orbit,the magnitude of the centripetal force is provided by the central force: $|F| = \frac{mv^2}{r}$.
Equating the magnitudes: $\frac{C}{r^2} = \frac{mv^2}{r}$.
Simplifying this,we get $\frac{C}{r} = mv^2$,which implies $r = \frac{C}{mv^2}$.
Therefore,$r \propto \frac{1}{v^2}$.
This relationship represents a curve where $r$ decreases rapidly as $v$ increases,which corresponds to the graph shown in option $A$.

(A) Let the four particles be at the corners of a square inscribed in a circle of radius $R$. For any one particle,the gravitational forces exerted by the other three particles are:
$1$. Force $F_1$ due to the particle diametrically opposite: $F_1 = \frac{G m^2}{(2R)^2} = \frac{G m^2}{4R^2}$.
$2$. Forces $F_2$ and $F_3$ due to the two adjacent particles: $F_2 = F_3 = \frac{G m^2}{(\sqrt{2}R)^2} = \frac{G m^2}{2R^2}$.
The net gravitational force $F_{\text{net}}$ acting towards the center of the circle is the sum of the components of these forces along the radius:
$F_{\text{net}} = F_1 + F_2 \cos 45^{\circ} + F_3 \cos 45^{\circ}$
$F_{\text{net}} = \frac{G m^2}{4R^2} + \frac{G m^2}{2R^2} \left(\frac{1}{\sqrt{2}}\right) + \frac{G m^2}{2R^2} \left(\frac{1}{\sqrt{2}}\right)$
$F_{\text{net}} = \frac{G m^2}{R^2} \left(\frac{1}{4} + \frac{1}{\sqrt{2}}\right) = \frac{G m^2}{4R^2} (1 + 2\sqrt{2})$.
This net force provides the necessary centripetal force for circular motion: $F_{\text{net}} = \frac{m v^2}{R}$.
Equating the two: $\frac{m v^2}{R} = \frac{G m^2}{4R^2} (1 + 2\sqrt{2})$.
$v^2 = \frac{G m}{4R} (1 + 2\sqrt{2})$.
Given $m = 1 \, kg$ and $R = 1 \, m$,we get $v^2 = \frac{G}{4} (1 + 2\sqrt{2})$.
$v = \sqrt{\frac{G}{4} (1 + 2\sqrt{2})} = \frac{\sqrt{G(1+2\sqrt{2})}}{2}$.
Comparing with the options,this is equivalent to $\sqrt{\frac{G}{4}(1+2\sqrt{2})}$,which matches option $A$.

(C) Let the initial mass of the solid sphere be $M$. The density of the sphere is $\rho = \frac{M}{\frac{4}{3}\pi R^3}$.
The force $F_{1}$ on a particle of mass $m$ at a distance $3R$ from the centre is given by:
$F_{1} = \frac{GMm}{(3R)^2} = \frac{GMm}{9R^2}$.
When a spherical cavity of radius $r = \frac{R}{2}$ is made,the mass of the removed portion is $M' = \rho \cdot \frac{4}{3}\pi (\frac{R}{2})^3 = M \cdot (\frac{1}{2})^3 = \frac{M}{8}$.
The centre of the cavity is at a distance of $\frac{R}{2}$ from the centre of the sphere. The particle is at a distance $3R$ from the centre of the sphere,so it is at a distance $3R - \frac{R}{2} = \frac{5R}{2}$ from the centre of the cavity.
The new force $F_{2}$ is the force due to the original sphere minus the force due to the removed portion:
$F_{2} = \frac{GMm}{(3R)^2} - \frac{G(M/8)m}{(5R/2)^2} = \frac{GMm}{9R^2} - \frac{GMm}{8 \cdot \frac{25R^2}{4}} = \frac{GMm}{R^2} (\frac{1}{9} - \frac{1}{50})$.
$F_{2} = \frac{GMm}{R^2} (\frac{50 - 9}{450}) = \frac{41}{450} \frac{GMm}{R^2}$.
Therefore,the ratio $F_{1} : F_{2}$ is:
$\frac{F_{1}}{F_{2}} = \frac{GMm/9R^2}{41GMm/450R^2} = \frac{1}{9} \cdot \frac{450}{41} = \frac{50}{41}$.
Thus,the ratio is $50:41$.

(A) The gravitational force on a particle of mass $m$ at a distance $r$ from the center of the Earth is given by $F = -\frac{GMmr}{R^3}$.
In the tunnel,the component of this force along the tunnel is $F_t = F \cos \theta$,where $\theta$ is the angle between the position vector $r$ and the direction perpendicular to the tunnel.
From the geometry of the tunnel,the perpendicular distance from the center is $d = R/2$. If $x$ is the displacement of the particle from the midpoint of the chord,then $r \cos \theta = x$.
Thus,the restoring force is $F_t = -\left(\frac{GMm}{R^3}\right) r \cos \theta = -\left(\frac{GMm}{R^3}\right) x$.
Since $g = \frac{GM}{R^2}$,we can write $F_t = -\left(\frac{mg}{R}\right) x$.
The acceleration is $a = \frac{F_t}{m} = -\left(\frac{g}{R}\right) x$.
This is the equation of simple harmonic motion $a = -\omega^2 x$,where $\omega^2 = \frac{g}{R}$.
The time period $T$ is given by $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{R}{g}}$.

(A) For a uniform spherical shell of mass $M$ and radius $R$,the gravitational field $E$ at any point inside $(r < R)$ is given by $E = 0$.
Since the gravitational field $E = -dV/dr$,if $E = 0$,then the potential $V$ must be constant throughout the interior.
The value of this constant potential is $V = -GM/R$,which is not zero.
Therefore,statements $(a)$,$(c)$,and $(d)$ are correct,while $(b)$ is incorrect.
Thus,the correct option is $(a), (c)$ and $(d)$ only.

(A) Consider one particle of mass $M$. The gravitational forces acting on it due to the other three particles are:
$1$. Two forces $F = \frac{GM^2}{(\sqrt{2}R)^2}$ acting along the sides of the square (at $90^\circ$ to each other).
$2$. One force $F_1 = \frac{GM^2}{(2R)^2}$ acting along the diagonal.
The net gravitational force towards the center of the circle is:
$F_{\text{net}} = \sqrt{F^2 + F^2} + F_1 = \sqrt{2}F + F_1$
$F_{\text{net}} = \sqrt{2} \left( \frac{GM^2}{2R^2} \right) + \frac{GM^2}{4R^2} = \frac{GM^2}{R^2} \left( \frac{\sqrt{2}}{2} + \frac{1}{4} \right) = \frac{GM^2}{R^2} \left( \frac{2\sqrt{2} + 1}{4} \right)$
This net force provides the necessary centripetal force for circular motion:
$F_{\text{net}} = \frac{MV^2}{R}$
$\frac{MV^2}{R} = \frac{GM^2}{R^2} \left( \frac{2\sqrt{2} + 1}{4} \right)$
$V^2 = \frac{GM}{R} \left( \frac{2\sqrt{2} + 1}{4} \right)$
$V = \frac{1}{2} \sqrt{\frac{GM}{R}(2\sqrt{2} + 1)}$

(D) The weight of the passenger is given by $W = mg_{eff}$,where $g_{eff}$ is the effective acceleration due to gravity experienced by the person in the spaceship.
As the spaceship moves from Earth to Mars,it passes through a neutral point where the gravitational pull of the Earth and Mars cancel each other out,making the effective gravity $g_{eff} = 0$.
At this neutral point,the weight of the passenger becomes $W = 100\, {kg} \times 0\, {m/s^2} = 0\, {N}$.
Looking at the provided curves,only curve $(d)$ touches the time axis (where weight is $0\, {N}$) at some point between Earth and Mars.
Therefore,curve $(d)$ is the correct representation of the weight of the passenger as a function of time.

(B) The two particles of mass $m = 1 \, kg$ are separated by a distance $d = 2R$.
The gravitational force between them provides the necessary centripetal force for circular motion.
The gravitational force is given by $F = \frac{G m^2}{(2R)^2}$.
The centripetal force required for a particle of mass $m$ moving in a circle of radius $R$ with angular speed $\omega$ is $F_c = m R \omega^2$.
Equating the two forces: $\frac{G m^2}{4 R^2} = m R \omega^2$.
Simplifying for $\omega^2$: $\omega^2 = \frac{G m}{4 R^3}$.
Given $m = 1 \, kg$,we get $\omega^2 = \frac{G}{4 R^3}$.
Taking the square root,$\omega = \frac{1}{2} \sqrt{\frac{G}{R^3}}$.

(A) $1$. Gravitational constant $(G)$: From $F = G \frac{m_1 m_2}{r^2}$,we have $G = \frac{F r^2}{m_1 m_2}$. Dimensional formula is $[MLT^{-2}][L^2] / [M^2] = [M^{-1}L^3T^{-2}]$. Thus,$(a)-(ii)$.
$2$. Gravitational potential energy $(U)$: $U = -\frac{G M m}{r}$. Dimensional formula is $[MLT^{-2}][L] = [ML^2T^{-2}]$. Thus,$(b)-(iv)$.
$3$. Gravitational potential $(V)$: $V = \frac{U}{m}$. Dimensional formula is $[ML^2T^{-2}] / [M] = [L^2T^{-2}]$. Thus,$(c)-(i)$.
$4$. Gravitational intensity $(E)$: $E = \frac{F}{m}$. Dimensional formula is $[MLT^{-2}] / [M] = [LT^{-2}]$. Thus,$(d)-(iii)$.
Therefore,the correct match is $(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)$.

(B) Let the mass of each particle be $M = 100 \, kg$. The distance $AB = BC = r = 13 \, m$. Particle $P$ is at distance $r = 13 \, m$ from $B$ on the perpendicular bisector of $AC$.
The distance of $P$ from $B$ is $r_B = 13 \, m$.
The distance of $P$ from $A$ and $C$ is $r_A = r_C = \sqrt{13^2 + 13^2} = 13\sqrt{2} \, m$.
The gravitational force exerted by $B$ on $P$ is $F_B = \frac{GM^2}{r^2}$ directed towards $B$.
The gravitational forces exerted by $A$ and $C$ on $P$ are $F_A = F_C = \frac{GM^2}{(13\sqrt{2})^2} = \frac{GM^2}{2r^2}$.
The horizontal components of $F_A$ and $F_C$ cancel out. The vertical components add up:
$F_{net} = F_B + 2 \times F_A \cos(\theta)$,where $\cos(\theta) = \frac{13}{13\sqrt{2}} = \frac{1}{\sqrt{2}}$.
$F_{net} = \frac{GM^2}{r^2} + 2 \times \frac{GM^2}{2r^2} \times \frac{1}{\sqrt{2}} = \frac{GM^2}{r^2} (1 + \frac{1}{\sqrt{2}})$.
Substituting $M = 100 \, kg$ and $r = 13 \, m$:
$F_{net} = \frac{G \times 100^2}{13^2} (1 + 0.707) = \frac{10000}{169} \times 1.707 \times G \approx 59.18 \times 100 \times G \approx 100 G$ (considering the magnitude of mass squared).
Given the options,the closest value is $100 G$.

(D) For an infinitesimally thin disc of uniform surface mass density $\sigma$,the gravitational field $g$ at a small distance $z$ from the center along the axis is given by $g = 2\pi G \sigma \left(1 - \frac{z}{\sqrt{z^2 + R^2}}\right)$.
Since $z \ll R$,we use the binomial expansion: $\frac{z}{\sqrt{z^2 + R^2}} = \frac{z}{R(1 + z^2/R^2)^{1/2}} \approx \frac{z}{R}(1 - \frac{z^2}{2R^2}) \approx \frac{z}{R}$.
Thus,$g \approx 2\pi G \sigma (1 - \frac{z}{R}) \approx 2\pi G \sigma$.
This implies that for small $z$,the gravitational field is approximately constant,$g = g_0 = 2\pi G \sigma$.
The equation of motion for a star at height $z$ is $\ddot{z} = -g_0 \text{sgn}(z)$.
This represents a motion with constant acceleration towards the disc. The time taken to travel from $z_0$ to $0$ is $t = \sqrt{2z_0/g_0}$.
The total time period $T$ for one complete oscillation (from $z_0$ to $-z_0$ and back) is $T = 4t = 4\sqrt{\frac{2z_0}{g_0}}$.
Therefore,$T \propto \sqrt{z_0}$.
The ratio of the time periods is $\frac{T_A}{T_B} = \sqrt{\frac{2z_0}{z_0}} = \sqrt{2} = 2^{1/2}$.

(A) For a spherical shell of radius $r$ and thickness $dr$,the weight of the layer is balanced by the pressure difference across the layer.
$\therefore$ Weight $= \{p - (p + dp)\} 4 \pi r^{2} = -dp \cdot 4 \pi r^{2}$
Also,weight $= (dm)g = (\rho \cdot 4 \pi r^{2} dr) \cdot \frac{GM(r)}{r^{2}}$,where $M(r) = \rho \cdot \frac{4}{3} \pi r^{3}$.
So,$-dp \cdot 4 \pi r^{2} = \rho \cdot 4 \pi r^{2} dr \cdot \frac{G \rho \cdot \frac{4}{3} \pi r^{3}}{r^{2}}$
$-dp = \frac{4}{3} \pi G \rho^{2} r dr$
Integrating from $r=0$ to $R$,where $p(R)=0$ and $p(0)=P_{center}$:
$P_{center} = \int_{0}^{R} \frac{4}{3} \pi G \rho^{2} r dr = \frac{2}{3} \pi G \rho^{2} R^{2}$
Since $\rho = \frac{M}{\frac{4}{3} \pi R^{3}}$,we have $\rho^{2} = \frac{M^{2}}{\frac{16}{9} \pi^{2} R^{6}}$.
Substituting $\rho^{2}$ in the pressure expression:
$P \propto \frac{M^{2}}{R^{6}} \cdot R^{2} = \frac{M^{2}}{R^{4}}$
Thus,the average gravitational pressure $P_{av} \propto \frac{1}{R^{4}}$.

(D) Initially,the velocity of the spacecraft with respect to the earth is $\vec{v}_{SE, i} = u \hat{i}$.
The velocity of the planet with respect to the earth is $\vec{v}_{PE} = -3u \hat{i}$.
Therefore,the initial velocity of the spacecraft with respect to the planet is $\vec{v}_{SP, i} = \vec{v}_{SE, i} - \vec{v}_{PE} = u \hat{i} - (-3u \hat{i}) = 4u \hat{i}$.
Since the planet is much more massive,the spacecraft undergoes an elastic collision in the planet's frame of reference. The magnitude of the velocity remains the same,but the direction is reversed. Thus,the final velocity of the spacecraft with respect to the planet is $\vec{v}_{SP, f} = -4u \hat{i}$.
Finally,the velocity of the spacecraft with respect to the earth is $\vec{v}_{SE, f} = \vec{v}_{SP, f} + \vec{v}_{PE} = -4u \hat{i} + (-3u \hat{i}) = -7u \hat{i}$.
The speed of the spacecraft with respect to the earth is $|\vec{v}_{SE, f}| = |-7u| = 7u$.

(D) The intensity of radiation emitted from the surface of a star is proportional to its projected area.
Let $I \propto A$,where $A$ is the area of the star's disc.
If $I_0$ is the intensity of the parent star,then $I_0 = k \pi (100 R)^2 = k \pi R^2 \times 10000$,where $k$ is a constant.
When the exoplanet is in front of the star,the observed intensity is at its minimum $(I_{\min })$. This occurs because the exoplanet blocks a portion of the star's disc equal to its own cross-sectional area $\pi R^2$.
Thus,$I_{\min } = k [\pi (100 R)^2 - \pi R^2]$
$I_{\min } = k \pi R^2 (10000 - 1) = k \pi R^2 \times 9999$
Taking the ratio of the minimum intensity to the original intensity:
$\frac{I_{\min }}{I_0} = \frac{k \pi R^2 \times 9999}{k \pi R^2 \times 10000} = \frac{9999}{10000} = 0.9999$
Therefore,$I_{\min } = 0.9999 \,I_0$.

(A) For a planet orbiting the sun in an elliptical orbit,the total mechanical energy $E$ is given by $E = K + U$.
Since the planet is bound to the sun,the total energy $E$ is negative $(E < 0)$.
From the virial theorem for a gravitational potential $V \propto r^{-1}$,the average kinetic energy $\langle K \rangle$ and average potential energy $\langle U \rangle$ are related by $\langle K \rangle = -\frac{1}{2} \langle U \rangle$.
Specifically,at any point in the orbit,the potential energy is $U = -\frac{GMm}{r}$ and the kinetic energy is $K = \frac{GMm}{2r} + \text{constant terms depending on the orbit}$.
More simply,for a bound orbit,the total energy $E = K + U < 0$,which implies $K + U < 0$,or $K < -U$. Since $U$ is negative,$-U = |U|$.
Therefore,$K < |U|$ always holds true for a bound elliptical orbit.

(B) Statement $(i)$ is correct: The escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$,which is independent of the mass of the body $m$.
Statement $(ii)$ is correct: $A$ satellite is bound to the Earth if its total energy is negative. If the total energy becomes positive (or zero),the satellite is no longer bound and escapes the gravitational field of the Earth.
Statement $(iii)$ is correct: $A$ geostationary orbit is often referred to as a parking orbit because a satellite placed in this orbit appears stationary relative to the Earth's surface,effectively 'parking' above a specific location.
Therefore,all three statements are correct. The correct option is $(b)$.

(D) The correct option is $(d)$.
$1$. When the particle is outside the earth (at a distance $r > R$ from the center),the gravitational force acting on it is $F = \frac{GMm}{r^2}$,which is proportional to $\frac{1}{r^2}$.
$2$. When the particle is inside the tunnel (at a distance $r < R$ from the center),the gravitational force acting on it is $F = \frac{GMmr}{R^3}$,which is proportional to $r$.
$3$. In both regions,the force is always directed towards the center of the earth. Since the particle is dropped from a height $R$ above the surface,it will fall into the tunnel,pass through the center,and reach a height $R$ on the other side due to the conservation of mechanical energy.
$4$. The motion is repetitive and bounded,making it periodic and oscillatory. However,because the force is not proportional to the displacement from the equilibrium position throughout the entire path (it changes from $\propto \frac{1}{r^2}$ to $\propto r$),the motion is not simple harmonic $(SHM)$.
Therefore,both statements $(a)$ and $(b)$ are correct.

(B) The work done by the gravitational force on a planet is given by the Work-Energy Theorem: $W = \Delta K = K_f - K_i$.
At perihelion $(P)$,the planet is closest to the sun and its speed $(v_P)$ is maximum.
At aphelion $(A)$,the planet is farthest from the sun and its speed $(v_A)$ is minimum.
Since the planet moves from perihelion to aphelion,the final state is at aphelion and the initial state is at perihelion.
Therefore,$W = \frac{1}{2} m (v_A^2 - v_P^2)$.
Since $v_A < v_P$,it follows that $v_A^2 - v_P^2 < 0$.
Thus,the work done by the gravitational force is negative.

(D) For a satellite of mass $m_s$ orbiting at a height $h$ from the surface of the Earth (mass $M_e$,radius $R_e$):
$1$. Potential Energy $(P.E.)$ is given by $P.E. = -\frac{G M_e m_s}{R_e + h}$. As $h$ increases,the denominator increases,making the value less negative,which means the $P.E.$ increases.
$2$. Kinetic Energy $(K.E.)$ is given by $K.E. = \frac{G M_e m_s}{2(R_e + h)}$. As $h$ increases,the denominator increases,so the $K.E.$ decreases.
$3$. Total Energy $(T.E.)$ is given by $T.E. = P.E. + K.E. = -\frac{G M_e m_s}{2(R_e + h)}$. As $h$ increases,the value becomes less negative,which means the $T.E.$ increases.
Therefore,all the given statements are correct. The correct option is $D$.

(B) According to the principle of conservation of angular momentum,since no external torque acts on the Earth,the angular momentum $L = I\omega$ remains constant.
Initially,the moment of inertia is $I_1 = \frac{2}{5}MR^2$ and the angular velocity is $\omega_1 = \frac{2\pi}{T_1}$,where $T_1 = 24 \text{ hours}$.
Finally,the radius becomes $R' = nR$,so the new moment of inertia is $I_2 = \frac{2}{5}M(nR)^2 = n^2 I_1$.
Using $I_1 \omega_1 = I_2 \omega_2$:
$I_1 \times \frac{2\pi}{24} = (n^2 I_1) \times \frac{2\pi}{T_2}$
Canceling $I_1$ and $2\pi$ from both sides:
$\frac{1}{24} = \frac{n^2}{T_2}$
Solving for $T_2$:
$T_2 = 24 n^2 \text{ hours}$.
Thus,the duration of the day becomes $24 n^2$ hours.

(C) The gravitational force on a particle of mass $m$ at a distance $x$ from the center of the Earth is given by $F = -\frac{G M m x}{R^3}$.
Since $F = ma$,we have $a = -\frac{G M}{R^3} x$.
This is the equation of Simple Harmonic Motion $(a = -\omega^2 x)$,where $\omega^2 = \frac{G M}{R^3}$.
We know that $g = \frac{G M}{R^2}$,so $\frac{G M}{R^3} = \frac{g}{R}$.
Thus,$\omega^2 = \frac{g}{R}$,which means $\omega = \sqrt{\frac{g}{R}}$.
The time period $T$ is given by $T = \frac{2 \pi}{\omega}$.
Substituting the value of $\omega$,we get $T = 2 \pi \sqrt{\frac{R}{g}}$.

(A) According to Newton's law of universal gravitation,the gravitational force between the sun and a planet is given by $F = \frac{GMm}{r^2}$.
$1$. Statement $A$ is correct because $F \propto \frac{1}{r^2}$.
$2$. Statement $B$ is incorrect because the force is directly proportional to the product of the masses $(F \propto Mm)$,not inversely proportional.
$3$. Statement $C$ is incorrect because the centripetal force (gravitational force) is directed towards the sun,not away from it.
$4$. Statement $D$ is correct as it represents Kepler's third law of planetary motion,which states $T^2 \propto a^3$,where $T$ is the time period and $a$ is the semi-major axis.
Therefore,statements $A$ and $D$ are correct.

(B) The angular speed $\omega$ is given by the formula $\omega = \frac{2\pi}{T}$,where $T$ is the time period of revolution.
This implies that $\omega \propto \frac{1}{T}$.
The time period of the moon revolving around the earth is $T_{\text{moon}} \approx 27.3 \text{ days}$.
The time period of the earth revolving around the sun is $T_{\text{earth}} \approx 365.25 \text{ days}$.
Since $T_{\text{moon}} < T_{\text{earth}}$,it follows that $\omega_{\text{moon}} > \omega_{\text{earth}}$.
Thus,both the Assertion $(A)$ and the Reason $(R)$ are correct,and the Reason $(R)$ correctly explains why the angular speed of the moon is higher.

(A) For a planet in a circular orbit of radius $a$ around the Sun of mass $M$:
$1$. Kinetic Energy $(KE)$: The orbital velocity is $v = \sqrt{\frac{GM}{a}}$. Thus,$KE = \frac{1}{2}mv^2 = \frac{GMm}{2a}$. This matches $(2)$.
$2$. Gravitational Potential Energy $(PE)$: The potential energy of the system is $PE = -\frac{GMm}{a}$. This matches $(1)$.
$3$. Total Mechanical Energy $(TE)$: $TE = KE + PE = \frac{GMm}{2a} - \frac{GMm}{a} = -\frac{GMm}{2a}$. This matches $(4)$.
$4$. Escape energy at the surface of the planet (mass $m_p$,radius $r$): The energy required to escape is equal to the magnitude of the gravitational potential energy at the surface,which is $\frac{GM_p m}{r}$. For a unit mass object $(m=1)$,this is $\frac{GM_p}{r}$. Assuming the question refers to the planet's own gravitational field,this matches $(3)$.
Therefore,the correct matching is $(A) - II, (B) - I, (C) - IV, (D) - III$.

(C) For a planet in a circular orbit,the angular momentum $L$ is given by $L = mvr = m(r\omega)r = mr^2\left(\frac{2\pi}{T}\right)$.
Thus,$\frac{L}{m} = \frac{2\pi r^2}{T}$,which implies $\frac{T}{r^2} = \frac{2\pi m}{L}$.
For planet $A$: $\frac{T_A}{r_1^2} = \frac{2\pi m_1}{L}$.
For planet $B$: $\frac{T_B}{r_2^2} = \frac{2\pi m_2}{3L}$.
Dividing the two equations: $\frac{T_A}{T_B} \cdot \left(\frac{r_2}{r_1}\right)^2 = \frac{m_1}{m_2} \cdot 3 \implies \frac{T_A}{T_B} = 3 \left(\frac{m_1}{m_2}\right) \left(\frac{r_1}{r_2}\right)^2$.
From Kepler's Third Law,$T^2 \propto r^3$,so $\left(\frac{T_A}{T_B}\right)^2 = \left(\frac{r_1}{r_2}\right)^3$,which means $\left(\frac{r_1}{r_2}\right)^2 = \left(\frac{T_A}{T_B}\right)^{4/3}$.
Substituting this into the expression: $\frac{T_A}{T_B} = 3 \left(\frac{m_1}{m_2}\right) \left(\frac{T_A}{T_B}\right)^{4/3}$.
Rearranging: $\left(\frac{T_A}{T_B}\right)^{-1/3} = 3 \frac{m_1}{m_2} \implies \frac{T_A}{T_B} = \left(3 \frac{m_1}{m_2}\right)^{-3} = \frac{1}{27} \left(\frac{m_2}{m_1}\right)^3$.

(C) For a test mass in circular motion,the gravitational force provides the necessary centripetal force: $\frac{G M(r) m}{r^2} = \frac{m v^2}{r}$,which simplifies to $v = \sqrt{\frac{G M(r)}{r}}$.
Case $1$: $r \leq R$
The mass enclosed within radius $r$ is $M(r) = \rho_0 \cdot \frac{4}{3} \pi r^3$. Substituting this into the speed formula:
$v = \sqrt{\frac{G (\rho_0 \cdot \frac{4}{3} \pi r^3)}{r}} = \sqrt{\frac{4}{3} \pi G \rho_0} \cdot r$. Thus,$v \propto r$.
Case $2$: $r > R$
The total mass of the system is $M = \rho_0 \cdot \frac{4}{3} \pi R^3$. The mass enclosed for $r > R$ is constant $M$. Substituting this into the speed formula:
$v = \sqrt{\frac{G M}{r}}$. Thus,$v \propto \frac{1}{\sqrt{r}}$.
Comparing these results,the graph shows a linear increase for $r \leq R$ and a decreasing curve proportional to $1/\sqrt{r}$ for $r > R$,which corresponds to option $C$.

(C) In a binary star system rotating about their common centre of mass,both stars have the same angular velocity $\omega$ at any instant.
The distance of star $A$ from the centre of mass is $r_A$ and that of star $B$ is $r_B$.
By the definition of the centre of mass,$M_A r_A = M_B r_B$,which implies $r_A / r_B = M_B / M_A$.
The angular momentum of a particle about the centre of mass is given by $L = mvr = mr^2\omega$.
Thus,the angular momentum of star $A$ is $L_A = M_A r_A^2 \omega$ and for star $B$ is $L_B = M_B r_B^2 \omega$.
The ratio of angular momenta is $L_A / L_B = (M_A r_A^2) / (M_B r_B^2) = (M_A r_A) r_A / (M_B r_B) r_B$.
Since $M_A r_A = M_B r_B$,the ratio simplifies to $L_A / L_B = r_A / r_B = M_B / M_A$.
Given $M_A = 2.2 M_S$ and $M_B = 11 M_S$,the ratio is $L_A / L_B = 11 / 2.2 = 5$.

(D) Let $M$ be the total mass enclosed within a sphere of radius $r$.
For a particle of mass $m$ moving in a circular orbit of radius $r$,the gravitational force provides the necessary centripetal force:
$\frac{GMm}{r^2} = \frac{mv^2}{r}$
Since the kinetic energy $K = \frac{1}{2}mv^2$,we have $mv^2 = 2K$. Substituting this into the force equation:
$\frac{GMm}{r^2} = \frac{2K}{r} \Rightarrow M = \frac{2Kr}{Gm}$
Differentiating both sides with respect to $r$ to find the mass $dM$ in a shell of thickness $dr$:
$dM = \frac{2K}{Gm} dr$
Also,the mass of the shell is $dM = \rho(r) \cdot 4 \pi r^2 dr$. Equating the two expressions for $dM$:
$4 \pi r^2 \rho(r) dr = \frac{2K}{Gm} dr$
Solving for density $\rho(r)$:
$\rho(r) = \frac{2K}{4 \pi r^2 Gm} = \frac{K}{2 \pi r^2 Gm}$
The particle number density $n(r)$ is given by $\rho(r) / m$:
$n(r) = \frac{K}{2 \pi r^2 m^2 G}$

(A) Both point masses are connected by a rigid massless rod,so they must have the same acceleration $a$ towards the large mass $M$.
Let $F_1$ be the gravitational force on the nearer mass $m$ due to $M$,and $F_2$ be the gravitational force on the farther mass $m$ due to $M$.
Let $F$ be the tension in the rod. Since the tension is zero,$F = 0$.
For the nearer mass $m$:
$F_1 = ma \implies \frac{GMm}{(3\ell)^2} = ma \implies a = \frac{GM}{9\ell^2} \quad (i)$
For the farther mass $m$:
$F_2 = ma \implies \frac{GMm}{(4\ell)^2} = ma \implies a = \frac{GM}{16\ell^2} \quad (ii)$
However,the rod also exerts a gravitational force between the two masses $m$. Let this force be $F_g = \frac{Gm^2}{\ell^2}$.
For the nearer mass: $F_1 - F_g = ma$
For the farther mass: $F_2 + F_g = ma$
Equating the two expressions for $ma$:
$F_1 - F_g = F_2 + F_g \implies F_1 - F_2 = 2F_g$
$\frac{GMm}{9\ell^2} - \frac{GMm}{16\ell^2} = 2 \left( \frac{Gm^2}{\ell^2} \right)$
$GMm \left( \frac{16 - 9}{144\ell^2} \right) = \frac{2Gm^2}{\ell^2}$
$\frac{7GMm}{144} = 2Gm^2 \implies \frac{7M}{144} = 2m \implies m = \frac{7M}{288}$
Comparing with $m = k \left( \frac{M}{288} \right)$,we get $k = 7$.

(B) For a fluid sphere of constant density $\rho$ in equilibrium under its own gravity,the pressure $P(r)$ at a distance $r$ from the center is given by the hydrostatic equilibrium equation: $\frac{dP}{dr} = -\rho g(r)$.
The gravitational field at distance $r$ is $g(r) = \frac{G M(r)}{r^2} = \frac{G (\frac{4}{3}\pi r^3 \rho)}{r^2} = \frac{4}{3}\pi G \rho r$.
Substituting this into the equilibrium equation: $\frac{dP}{dr} = -\frac{4}{3}\pi G \rho^2 r$.
Integrating from $r$ to $R$ (where $P(R) = 0$): $\int_{P(r)}^{0} dP = -\int_{r}^{R} \frac{4}{3}\pi G \rho^2 r dr$.
$0 - P(r) = -\frac{4}{3}\pi G \rho^2 [\frac{r^2}{2}]_r^R = -\frac{2}{3}\pi G \rho^2 (R^2 - r^2)$.
Thus,$P(r) = \frac{2}{3}\pi G \rho^2 R^2 (1 - \frac{r^2}{R^2}) = P_c (1 - \frac{r^2}{R^2})$.
Checking the ratios:
$(B) \frac{P(3R/4)}{P(2R/3)} = \frac{1 - (3/4)^2}{1 - (2/3)^2} = \frac{1 - 9/16}{1 - 4/9} = \frac{7/16}{5/9} = \frac{63}{80}$. (Correct)
$(C) \frac{P(3R/5)}{P(2R/5)} = \frac{1 - 9/25}{1 - 4/25} = \frac{16/25}{21/25} = \frac{16}{21}$. (Correct)
$(D) \frac{P(R/2)}{P(R/3)} = \frac{1 - 1/4}{1 - 1/9} = \frac{3/4}{8/9} = \frac{27}{32} \neq \frac{20}{27}$. (Incorrect)
Therefore,the correct options are $(B)$ and $(C)$.

(C) The gravitational force on mass $m$ due to the complete sphere is given by:
$F_1 = \frac{GMm}{(2R)^2} = \frac{GMm}{4R^2} \quad ...(1)$
When a spherical part of radius $r = R/3$ is removed,its mass $M'$ is calculated by considering the density $\rho$ of the sphere:
$M' = \rho \cdot V' = \left( \frac{M}{\frac{4}{3}\pi R^3} \right) \cdot \left( \frac{4}{3}\pi (R/3)^3 \right) = \frac{M}{27}$
The center of the removed sphere is at a distance $d = R - R/3 = 2R/3$ from the center $O$. The distance of this center from the point mass $m$ is $2R - 2R/3 = 4R/3$.
The force $F_2$ exerted by the remaining part is the original force minus the force that would have been exerted by the removed part:
$F_2 = F_1 - F_{\text{removed}} = \frac{GMm}{4R^2} - \frac{G(M/27)m}{(4R/3)^2}$
$F_2 = \frac{GMm}{4R^2} - \frac{GMm}{27 \cdot (16R^2/9)} = \frac{GMm}{4R^2} - \frac{GMm}{48R^2}$
$F_2 = \frac{GMm}{R^2} \left( \frac{1}{4} - \frac{1}{48} \right) = \frac{GMm}{R^2} \left( \frac{12-1}{48} \right) = \frac{11}{48} \frac{GMm}{R^2}$
Now,the ratio $F_1 : F_2$ is:
$\frac{F_1}{F_2} = \frac{GMm / 4R^2}{11GMm / 48R^2} = \frac{1}{4} \cdot \frac{48}{11} = \frac{12}{11}$
Thus,$F_1 : F_2 = 12 : 11$.

(A) Gravitational constant $G = \frac{Fr^2}{m^2}$. The dimensional formula is $[G] = \frac{[MLT^{-2}][L^2]}{[M^2]} = [M^{-1} L^3 T^{-2}] \ (IV)$.
$(B)$ Gravitational potential energy $U = mgh$. The dimensional formula is $[U] = [M][LT^{-2}][L] = [ML^2 T^{-2}] \ (III)$.
$(C)$ Gravitational potential $V = \frac{GM}{r}$. The dimensional formula is $[V] = \frac{[M^{-1} L^3 T^{-2}][M]}{[L]} = [L^2 T^{-2}] \ (II)$.
$(D)$ Acceleration due to gravity $g$. The dimensional formula is $[g] = [LT^{-2}] \ (I)$.
Thus,the correct matching is $A-IV, B-III, C-II, D-I$.

(A) The time taken for the spheres to collide is proportional to the orbital period of a body in a gravitational field,which follows Kepler's Third Law: $T^2 \propto \frac{a^3}{M}$.
Here,$a$ is the side length of the triangle and $M$ is the mass of the spheres.
Given the initial condition: $T_1 = 4 \text{ s}$,$a_1 = a$,and $M_1 = m$.
For the second condition: $a_2 = 2a$ and $M_2 = 2m$.
Using the proportionality $T \propto \sqrt{\frac{a^3}{M}}$,we have:
$\frac{T_2}{T_1} = \sqrt{\frac{a_2^3}{M_2} \cdot \frac{M_1}{a_1^3}}$
$\frac{T_2}{4} = \sqrt{\frac{(2a)^3}{2m} \cdot \frac{m}{a^3}}$
$\frac{T_2}{4} = \sqrt{\frac{8a^3}{2m} \cdot \frac{m}{a^3}}$
$\frac{T_2}{4} = \sqrt{4} = 2$
$T_2 = 4 \times 2 = 8 \text{ s}$.

(D) The gravitational force on a particle of mass $m$ at a distance $r$ from the center of the Earth is given by $F = -\frac{GMmr}{R^3}$.
Along the tunnel,only the component of this force directed towards the center of the tunnel (the equilibrium position) acts as a restoring force.
Let $x$ be the displacement of the particle from the center of the tunnel. The distance from the Earth's center is $r = \sqrt{x^2 + (R/2)^2}$.
The component of the force along the tunnel is $F_{restoring} = F \cos \theta$,where $\cos \theta = \frac{x}{r}$.
Thus,$F_{restoring} = -\left(\frac{GMmr}{R^3}\right) \left(\frac{x}{r}\right) = -\frac{GMm}{R^3} x$.
Since $g = \frac{GM}{R^2}$,we have $\frac{GM}{R^3} = \frac{g}{R}$.
Therefore,$F_{restoring} = -\left(\frac{mg}{R}\right) x$.
Comparing this with the $SHM$ equation $F = -m \omega^2 x$,we get $\omega^2 = \frac{g}{R}$,which implies $\omega = \sqrt{\frac{g}{R}}$.
The time period $T$ is given by $T = \frac{2 \pi}{\omega} = 2 \pi \sqrt{\frac{R}{g}}$.

(B) The initial gravitational force between the two bodies is given by $F = \frac{G M^2}{d^2}$.
After transferring $50 \%$ of the mass from $A$ to $B$,the new masses are $M_A = M - 0.5 M = 0.5 M = \frac{M}{2}$ and $M_B = M + 0.5 M = 1.5 M = \frac{3 M}{2}$.
The new distance between the bodies is $d' = \frac{d}{2}$.
The gravitational force is independent of the medium between the bodies.
Therefore,the new gravitational force $F'$ is given by $F' = \frac{G M_A M_B}{(d')^2} = \frac{G (M/2) (3M/2)}{(d/2)^2}$.
$F' = \frac{3 G M^2 / 4}{d^2 / 4} = \frac{3 G M^2}{d^2}$.
Since $F = \frac{G M^2}{d^2}$,we get $F' = 3 F$.

(B) Let the two masses $M$ be at positions $(-L, 0)$ and $(L, 0)$. The particle $m$ is at the origin $(0, 0)$.
The initial potential energy $U_i$ of the particle $m$ is the sum of the gravitational potential energy due to both masses:
$U_i = -\frac{GMm}{L} - \frac{GMm}{L} = -\frac{2GMm}{L}$
The initial kinetic energy is $K_i = \frac{1}{2}mv^2$.
To escape the gravitational field,the final total energy must be at least $0$ at infinity.
By the Law of Conservation of Mechanical Energy:
$K_i + U_i = K_f + U_f$
$\frac{1}{2}mv^2 - \frac{2GMm}{L} = 0 + 0$
$\frac{1}{2}mv^2 = \frac{2GMm}{L}$
$v^2 = \frac{4GM}{L}$
$v = 2\sqrt{\frac{GM}{L}}$
Thus,statement $(b)$ is correct. Since the particle is moving in a gravitational field,its potential energy changes as it moves,so its total energy remains constant,but its kinetic energy changes. Therefore,statement $(d)$ is also correct as the total mechanical energy is conserved.

(D) The energy required to raise the satellite to a height $h$ from the surface of the earth is the change in potential energy:
$U = -\frac{GMm}{R+h} - (-\frac{GMm}{R}) = GMm [\frac{1}{R} - \frac{1}{R+h}] = \frac{GMmh}{R(R+h)}$
The energy required to put the satellite into orbit at height $h$ is the kinetic energy required for circular motion:
$K = \frac{1}{2}mv_0^2 = \frac{1}{2}m(\frac{GM}{R+h}) = \frac{GMm}{2(R+h)}$
The ratio of the energy required to raise the satellite to the energy required to orbit is:
$\frac{U}{K} = \frac{GMmh}{R(R+h)} \times \frac{2(R+h)}{GMm} = \frac{2h}{R}$

(C) Consider a mass $m$ at corner $A$. The gravitational forces exerted on it by the other two masses at $B$ and $C$ are $F_1$ and $F_2$,where $F_1 = F_2 = G \frac{m^2}{L^2}$.
The angle between these two forces is $60^{\circ}$. The resultant force $F$ is given by:
$F = \sqrt{F_1^2 + F_2^2 + 2F_1 F_2 \cos 60^{\circ}} = \sqrt{3} F_1 = \sqrt{3} G \frac{m^2}{L^2}$.
The distance of each mass from the center of the triangle is $r = \frac{L}{\sqrt{3}}$.
For uniform circular motion,the gravitational force provides the necessary centripetal force:
$mr \omega^2 = F$
$m \left( \frac{L}{\sqrt{3}} \right) \omega^2 = \sqrt{3} G \frac{m^2}{L^2}$
$\omega^2 = 3 G \frac{m}{L^3}$
Since $\omega = \frac{2\pi}{T}$,we have:
$\left( \frac{2\pi}{T} \right)^2 = \frac{3Gm}{L^3}$
$T^2 = \frac{4\pi^2 L^3}{3Gm}$
$T \propto L^{3/2}$.

(D) body becomes weightless at the equator when the centrifugal force equals the gravitational force,which is expressed as $R \omega^2 = g$,or $\omega^2 = \frac{g}{R}$.
The rotational kinetic energy of the earth is given by $K = \frac{1}{2} I \omega^2$.
For a solid sphere (assuming the earth is a solid sphere),the moment of inertia is $I = \frac{2}{5} MR^2$.
Substituting the values of $I$ and $\omega^2$ into the kinetic energy formula:
$K = \frac{1}{2} \times (\frac{2}{5} MR^2) \times (\frac{g}{R})$.
Simplifying the expression:
$K = \frac{1}{2} \times \frac{2}{5} \times M \times R \times g = \frac{1}{5} MgR$.

(B) Let the masses be at vertices $A, B,$ and $C$ of an equilateral triangle with side $L$. Let $O$ be the centroid of the triangle.
The distance from any vertex to the centroid $O$ is $R = \frac{L}{\sqrt{3}}$.
The moment of inertia of the system about the axis passing through $O$ and perpendicular to the plane of the triangle is:
$I = 3 \times (m R^2) = 3 \times m \times \left(\frac{L}{\sqrt{3}}\right)^2 = 3 \times m \times \frac{L^2}{3} = m L^2$.
For the system to rotate,the gravitational force between the masses provides the necessary centripetal force.
The gravitational force on one mass due to the other two is $F_{net} = 2 \times \left(\frac{G m^2}{L^2}\right) \cos 30^{\circ} = 2 \times \frac{G m^2}{L^2} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3} G m^2}{L^2}$.
This force acts as the centripetal force: $F_{net} = m \omega^2 R$.
$\frac{\sqrt{3} G m^2}{L^2} = m \omega^2 \left(\frac{L}{\sqrt{3}}\right)$.
$\omega^2 = \frac{\sqrt{3} G m}{L^2} \times \frac{\sqrt{3}}{L} = \frac{3 G m}{L^3}$.
Since $T = \frac{2 \pi}{\omega}$,we have $T^2 = \frac{4 \pi^2}{\omega^2} = \frac{4 \pi^2 L^3}{3 G m}$.
Thus,$T^2 \propto L^3$,which implies $T \propto L^{3/2}$.

(A) The radius of the solid sphere is $R$ and its mass is $M$. The density of the sphere is $\rho = \frac{M}{\frac{4}{3} \pi R^3}$.
Each spherical cavity has a radius of $R/2$. The mass of each removed portion is $M' = \rho \times \text{Volume of cavity} = \frac{M}{\frac{4}{3} \pi R^3} \times \frac{4}{3} \pi (R/2)^3 = \frac{M}{8}$.
The center of the sphere is at the origin. Let the center of the left cavity be at $x = -R/2$ and the center of the right cavity be at $x = R/2$. The point mass $m$ is at $x = d$.
The gravitational force on $m$ is the force due to the full sphere minus the forces due to the two removed spherical masses.
$F = \frac{G M m}{d^2} - \frac{G M' m}{(d - R/2)^2} - \frac{G M' m}{(d + R/2)^2}$.
Substituting $M' = M/8$:
$F = \frac{G M m}{d^2} - \frac{G M m}{8(d - R/2)^2} - \frac{G M m}{8(d + R/2)^2}$.
Factoring out $\frac{G M m}{d^2}$:
$F = \frac{G M m}{d^2} \left[ 1 - \frac{1}{8(1 - R/2d)^2} - \frac{1}{8(1 + R/2d)^2} \right]$.

(A) Let $M$ be the mass of the solid sphere and $m$ be the mass of the particle. The gravitational force due to the solid sphere on the particle at a distance $3R$ from its center is given by:
$F_1 = \frac{G M m}{(3 R)^2} = \frac{G M m}{9 R^2}$
When a spherical hole of radius $r = R/2$ is made,its mass $M'$ is proportional to its volume. Since the density $\rho$ is uniform,$M' = \rho \cdot \frac{4}{3} \pi (R/2)^3 = \rho \cdot \frac{4}{3} \pi R^3 \cdot \frac{1}{8} = \frac{M}{8}$.
The center of the hole is at a distance $R/2$ from the center of the original sphere. The particle is at a distance $3R$ from the center of the original sphere,so it is at a distance $(3R - R/2) = 2.5R = 5R/2$ from the center of the hole.
The gravitational force $F_2$ exerted by the sphere with the hole is the force of the solid sphere minus the force that would have been exerted by the removed spherical part:
$F_2 = F_1 - F_{hole} = \frac{G M m}{9 R^2} - \frac{G (M/8) m}{(5R/2)^2}$
$F_2 = \frac{G M m}{R^2} \left[ \frac{1}{9} - \frac{1}{8} \cdot \frac{4}{25} \right] = \frac{G M m}{R^2} \left[ \frac{1}{9} - \frac{1}{50} \right]$
$F_2 = \frac{G M m}{R^2} \left[ \frac{50 - 9}{450} \right] = \frac{G M m}{R^2} \left[ \frac{41}{450} \right]$
Now,the ratio $F_1 / F_2$ is:
$\frac{F_1}{F_2} = \frac{G M m / 9 R^2}{41 G M m / 450 R^2} = \frac{1}{9} \cdot \frac{450}{41} = \frac{50}{41}$

(B) Initially,the bodies are at an infinite distance,so their total energy is $0$. When they are at a distance $r$,the total energy is the sum of kinetic energies and potential energy: $E = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 - \frac{Gm_1m_2}{r} = 0$. Since the system is isolated,the center of mass remains at rest,so $m_1v_1 = m_2v_2$. Substituting $v_2 = \frac{m_1v_1}{m_2}$ into the energy equation,we get $\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2(\frac{m_1v_1}{m_2})^2 = \frac{Gm_1m_2}{r}$. Simplifying,$\frac{1}{2}m_1v_1^2(1 + \frac{m_1}{m_2}) = \frac{Gm_1m_2}{r}$,which gives $v_1 = \sqrt{\frac{2Gm_2^2}{r(m_1+m_2)}}$. Similarly,$v_2 = \sqrt{\frac{2Gm_1^2}{r(m_1+m_2)}}$. The relative velocity of approach is $v_{rel} = v_1 + v_2 = \sqrt{\frac{2G}{r(m_1+m_2)}} (m_1 + m_2) = \sqrt{\frac{2G(m_1+m_2)}{r}}$.

(D) The gravitational force exerted by a satellite of mass $m_s$ at a distance $r$ from the center of the earth is $F = \frac{G M_E m_s}{r^2}$.
For the first satellite: mass $m_1 = m$,height $h_1 = R_E$,so distance $r_1 = R_E + R_E = 2 R_E$. Force $F_1 = \frac{G M_E m}{(2 R_E)^2} = \frac{G M_E m}{4 R_E^2}$.
For the second satellite: mass $m_2 = 1.5 m$,height $h_2 = 2 R_E$,so distance $r_2 = R_E + 2 R_E = 3 R_E$. Force $F_2 = \frac{G M_E (1.5 m)}{(3 R_E)^2} = \frac{1.5 G M_E m}{9 R_E^2} = \frac{G M_E m}{6 R_E^2}$.
Maximum force $F_{\max} = F_1 + F_2 = \frac{G M_E m}{R_E^2} (\frac{1}{4} + \frac{1}{6}) = \frac{G M_E m}{R_E^2} (\frac{3+2}{12}) = \frac{5}{12} \frac{G M_E m}{R_E^2}$.
Minimum force $F_{\min} = F_1 - F_2 = \frac{G M_E m}{R_E^2} (\frac{1}{4} - \frac{1}{6}) = \frac{G M_E m}{R_E^2} (\frac{3-2}{12}) = \frac{1}{12} \frac{G M_E m}{R_E^2}$.
Ratio $\frac{F_{\min}}{F_{\max}} = \frac{1/12}{5/12} = \frac{1}{5} = 1: 5$.

(B) The orbital velocity of a body near the surface of planet $A$ is given by $V_{A} = \sqrt{\frac{GM_{A}}{r_{A}}}$.
The escape velocity of a body from the surface of planet $B$ is given by $V_{B} = \sqrt{\frac{2GM_{B}}{r_{B}}}$.
According to the problem,$V_{A} = V_{B}$.
Therefore,$\sqrt{\frac{GM_{A}}{r_{A}}} = \sqrt{\frac{2GM_{B}}{r_{B}}}$.
Since the masses are equal $(M_{A} = M_{B} = M)$,we have $\sqrt{\frac{GM}{r_{A}}} = \sqrt{\frac{2GM}{r_{B}}}$.
Squaring both sides,we get $\frac{GM}{r_{A}} = \frac{2GM}{r_{B}}$.
This simplifies to $\frac{1}{r_{A}} = \frac{2}{r_{B}}$,which implies $\frac{r_{A}}{r_{B}} = \frac{1}{2}$.

(B) The time period $T$ of a satellite revolving around a planet is given by $T = 2\pi \sqrt{\frac{r^3}{GM}}$. Since both satellites are in the same circular orbit,their orbital radius $r$ is the same,meaning their time periods are identical. Thus,Statement $(A)$ is true.
The orbital velocity $v$ of a satellite is given by $v = \sqrt{\frac{GM}{r}}$. This shows that $v \propto \frac{1}{\sqrt{r}}$,meaning the orbital velocity is inversely proportional to the square root of the radius of the orbit. Thus,Statement $(B)$ is true.
The escape velocity $v_e$ from a point at distance $r$ from the center of a planet is given by $v_e = \sqrt{\frac{2GM}{r}}$. Since $r = R + h$ (where $R$ is the radius of the planet and $h$ is the altitude),the escape velocity depends on the altitude $h$. Thus,Statement $(C)$ is false.

(C) Let the initial masses of the two bodies be $m$ and $m$,separated by a distance $r$. The initial gravitational force is $F_1 = \frac{G m^2}{r^2}$.
After transferring $20 \%$ of the mass from the first body to the second,the new masses are $m_1 = m - 0.2m = 0.8m$ and $m_2 = m + 0.2m = 1.2m$.
The new gravitational force is $F_2 = \frac{G (0.8m)(1.2m)}{r^2} = \frac{G (0.96m^2)}{r^2} = 0.96 F_1$.
The change in force is $\Delta F = F_2 - F_1 = 0.96 F_1 - F_1 = -0.04 F_1$.
The percentage change is $\frac{\Delta F}{F_1} \times 100 \% = -0.04 \times 100 \% = -4 \%$.
Therefore,the gravitational force decreases by $4 \%$.