Mix Examples-Gravitation Questions in English

(C) The apparent weight of a body in an accelerating frame is given by $W_{app} = m(g + a)$,where $a$ is the upward acceleration.
Initially,as the aeroplane takes off and gains altitude,it experiences upward acceleration,causing the reading of the spring balance to increase.
As the aeroplane reaches a high altitude,the acceleration becomes zero (or negligible) and the value of the acceleration due to gravity $g$ decreases with height $(g' = g(1 - 2h/R))$.
Therefore,the reading of the spring balance will first increase due to the upward acceleration and then decrease due to the reduction in the gravitational force at higher altitudes.

(A) For a spherically symmetric massive body such as a planet,the escape velocity $(V_e)$ at the surface is given by the formula: $V_e = \sqrt{\frac{2GM}{R}}$.
For a satellite in a circular orbit close to the Earth's surface,the orbital velocity $(V_0)$ is given by the formula: $V_0 = \sqrt{\frac{GM}{R}}$.
Dividing the two expressions,we get: $\frac{V_e}{V_0} = \frac{\sqrt{\frac{2GM}{R}}}{\sqrt{\frac{GM}{R}}} = \sqrt{2}$.
Therefore,the escape velocity is $\sqrt{2}$ times the orbital velocity.

(B) The $K.E.$ required for a satellite to escape from Earth's gravitational field is given by the escape energy: $E_e = \frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2 = \frac{GMm}{R}$.
The $K.E.$ required for a satellite to move in a circular orbit just above Earth's atmosphere is given by the orbital energy: $E_o = \frac{1}{2}mv_0^2 = \frac{1}{2}m(\sqrt{\frac{GM}{R}})^2 = \frac{GMm}{2R}$.
The ratio of these two energies is $\frac{E_e}{E_o} = \frac{GMm/R}{GMm/2R} = 2$.

(D) The orbital speed of a satellite is given by the formula $v = \sqrt{\frac{GM}{r}}$,where $G$ is the gravitational constant,$M$ is the mass of the Earth,and $r$ is the orbital radius.
From this relation,it is clear that $v \propto \frac{1}{\sqrt{r}}$.
Therefore,as the radius of the orbit $r$ increases,the orbital speed $v$ of the satellite decreases.
Statement $A$ is incorrect because a satellite can move in a stable orbit in any plane passing through the Earth's center (e.g.,polar orbits).
Statement $C$ is incorrect because one geostationary satellite covers only about one-third of the Earth's surface; three are required for global coverage.
However,in the context of standard physics problems of this type,statement $D$ is the most fundamentally incorrect physical relationship provided.

(D) $(i)$ For a satellite orbiting just above the Earth's surface,the time period is $T_{st} = 2\pi \sqrt{\frac{R}{g}} \approx 84.6 \; \text{min}$.
$(ii)$ For a mass oscillating in a tunnel through the Earth's diameter,the motion is simple harmonic with $T_{ma} = 2\pi \sqrt{\frac{R}{g}} \approx 84.6 \; \text{min}$.
$(iii)$ For a simple pendulum of length $l=R$ in a uniform field $g$,$T_{sp} = 2\pi \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{R}{g}} \approx 84.6 \; \text{min}$.
$(iv)$ For an infinite length simple pendulum in the Earth's real gravitational field,the period is $T_{is} = 2\pi \sqrt{\frac{R}{g}} \approx 84.6 \; \text{min}$.
Thus,all four periods are equal: ${T_{st}} = {T_{ma}} = {T_{sp}} = {T_{is}}$.

(A) Nicolaus Copernicus,a Polish astronomer,studied the sky from the top of a cathedral. In $1543$,he published his book 'De revolutionibus orbium coelestium' (On the Revolutions of the Heavenly Spheres). In this work,he proposed the heliocentric model,stating that the Earth revolves around the Sun once a year and also rotates around its own axis.

(D) The acceleration due to gravity is $g = \frac{GM}{R^2}$,escape velocity is $v_e = \sqrt{\frac{2GM}{R}}$,and gravitational potential energy is $U = -\frac{GMm}{R}$.
From these relations,we observe that $g \propto \frac{M}{R^2}$,$v_e \propto \sqrt{\frac{M}{R}}$,and $U \propto \frac{M}{R}$.
Let the change be $x = 0.5\% = 0.005$. If $M' = M(1+x)$ and $R' = R(1+x)$:
$1$. For $g$: $g' = \frac{G M(1+x)}{R^2(1+x)^2} = g(1+x)^{-1} \approx g(1-x)$. Thus,$g$ decreases by $0.5\%$.
$2$. For $v_e$: $v_e' = \sqrt{\frac{2G M(1+x)}{R(1+x)}} = \sqrt{\frac{2GM}{R}} = v_e$. Thus,escape velocity remains unchanged.
$3$. For $U$: $U' = -\frac{G M(1+x)m}{R(1+x)} = -\frac{GMm}{R} = U$. Thus,potential energy remains unchanged.
Therefore,all statements are correct.

(D) The gravitational potential $V$ at any point $P$ due to the solid sphere with cavities is given by the principle of superposition: $V = V_{\text{large sphere}} - V_{\text{cavity A}} - V_{\text{cavity B}}$.
$1$. At the origin $O(0, 0, 0)$,the gravitational field $\vec{E}$ due to the large sphere is zero. The fields due to the two cavities are equal in magnitude and opposite in direction $(\vec{E}_A = -\vec{E}_B)$,so the net field at the origin is zero. Thus,option $(A)$ is correct.
$2$. The gravitational potential at any point $(x, y, z)$ due to a sphere with a cavity is symmetric about the axis passing through the centres of the cavities. For the circle ${y^2} + {z^2} = R^2$ in the plane $x = 0$,the distance from any point on the circle to the centres $A(-2, 0, 0)$ and $B(2, 0, 0)$ is the same. Specifically,for any point $(0, y, z)$ on the circle,the distance to $A$ is $\sqrt{(-2-0)^2 + y^2 + z^2} = \sqrt{4 + R^2}$ and the distance to $B$ is $\sqrt{(2-0)^2 + y^2 + z^2} = \sqrt{4 + R^2}$. Since the distances are equal,the potentials due to the cavities are equal,and the potential due to the large sphere is constant on this circle. Thus,the total potential is constant on the circles ${y^2} + {z^2} = 4$ and ${y^2} + {z^2} = 36$. Options $(B)$ and $(C)$ are also correct.
Therefore,the correct option is $(D)$.

(D) The weight of the passenger is the net gravitational force acting on them. Let $M_e$ and $M_m$ be the masses of Earth and Mars,and $R_e$ and $R_m$ be their radii. The gravitational force $F$ at a distance $x$ from Earth along the path is given by $F = |F_e - F_m| = |\frac{G M_e m}{x^2} - \frac{G M_m m}{(d-x)^2}|$,where $d$ is the distance between Earth and Mars.
At the surface of Earth $(x = R_e)$,the weight is $W_e = m g_e = 60 \times 10 = 600 \, N$.
At the surface of Mars $(x = d - R_m)$,the weight is $W_m = m g_m = 60 \times 4 = 240 \, N$.
As the spaceship moves from Earth to Mars,the gravitational pull of Earth decreases and the gravitational pull of Mars increases. At some point between the two planets,the gravitational forces from Earth and Mars will be equal in magnitude,making the net gravitational force (weight) zero.
Since the spaceship moves with constant velocity,the distance $x$ is proportional to time $t$. Therefore,the graph of weight versus time must decrease from $600 \, N$,reach $0 \, N$ at some intermediate time,and then increase to $240 \, N$ at the arrival time $t_0$. Curve $D$ is the only one that shows the weight becoming zero and then increasing to $240 \, N$.

(C) For a satellite of mass $m$ orbiting a planet of mass $M$ at a distance $r$ from its center,the energies are given by:
Potential Energy: $U = -\frac{GMm}{r}$
Kinetic Energy: $K = \frac{GMm}{2r}$
Total Energy: $E = U + K = -\frac{GMm}{2r}$
From these expressions:
$1$. $U$ is always negative and its magnitude decreases as $r$ increases,approaching $0$ from below.
$2$. $K$ is always positive and decreases as $r$ increases,approaching $0$ from above.
$3$. $E$ is always negative and its magnitude decreases as $r$ increases,approaching $0$ from below.
$4$. Comparing magnitudes: $|U| = 2|E|$ and $K = |E|$.
Thus,$U$ and $E$ are negative curves,while $K$ is a positive curve. As $r \to \infty$,all energies approach $0$. The correct graph shows $K$ above the $r$-axis,and both $U$ and $E$ below the $r$-axis,with $U$ being more negative than $E$ (i.e.,$|U| > |E|$). This corresponds to the graph where $K$ is positive,$E$ is the top negative curve,and $U$ is the bottom negative curve.

(D) Binary stars are two stars that orbit around a common center of mass. By observing their orbital periods and the separation between them,astronomers can apply Kepler's laws and Newton's law of universal gravitation to determine the masses of the stars. Furthermore,the motion of binary stars provides a direct observational test for Newton's law of gravitation in a celestial context,confirming that the same gravitational laws apply to distant stars as they do to objects on Earth.

(B) The solar constant $S$ is defined as the solar energy incident per unit area per unit time at the distance of the Earth from the Sun.
According to the inverse square law,the intensity of radiation $I$ from a point source is inversely proportional to the square of the distance $r$ from the source,i.e.,$I \propto \frac{1}{r^2}$.
Let $S$ be the solar constant at Earth's distance $r_E = 1 \, AU$.
Let $S'$ be the solar constant at a distance $r_P = 5.3 \, AU$.
Then,$\frac{S'}{S} = \frac{r_E^2}{r_P^2} = \left(\frac{1}{5.3}\right)^2$.
Therefore,$S' = \frac{S}{(5.3)^2}$.

(D) As we move towards the centre of the Sun,the gravitational force exerted by the outer layers increases,leading to a significant increase in both density and pressure. Therefore,the correct option is $D$.

(A) The solar constant $S$ is inversely proportional to the square of the distance $R$ from the Sun,i.e.,$S \propto \frac{1}{R^2}$.
Given,$R_M = 0.4 \, R_E$,where $R_M$ is the distance of Mercury and $R_E$ is the distance of Earth.
The solar constant of Earth is $S_E = 2 \, cal/min \cdot cm^2$.
Using the relation $\frac{S_M}{S_E} = \frac{R_E^2}{R_M^2}$,we get:
$\frac{S_M}{2} = \frac{R_E^2}{(0.4 \, R_E)^2} = \frac{1}{0.16} = 6.25$.
Therefore,$S_M = 6.25 \times 2 = 12.5 \, cal/min \cdot cm^2$.

(A) Let the particle at the origin be $A(0,0)$,the particle at $(0.2, 0)$ be $C$,and the particle at $(0, 0.2)$ be $B$. All masses $m = 1 \, kg$.
The gravitational force exerted by particle $C$ on $A$ is:
$\vec{F}_{AC} = \frac{G m_A m_C}{r_{AC}^2} \hat{i} = \frac{6.67 \times 10^{-11} \times 1 \times 1}{(0.2)^2} \hat{i} = \frac{6.67 \times 10^{-11}}{0.04} \hat{i} = 1.6675 \times 10^{-9} \hat{i} \approx 1.67 \times 10^{-9} \hat{i} \, N$
The gravitational force exerted by particle $B$ on $A$ is:
$\vec{F}_{AB} = \frac{G m_A m_B}{r_{AB}^2} \hat{j} = \frac{6.67 \times 10^{-11} \times 1 \times 1}{(0.2)^2} \hat{j} = 1.67 \times 10^{-9} \hat{j} \, N$
The net force on particle $A$ is:
$\vec{F}_{net} = \vec{F}_{AC} + \vec{F}_{AB} = 1.67 \times 10^{-9} (\hat{i} + \hat{j}) \, N$

(C) Let the center of the square be $P$. The distance from each corner to the center $P$ is $x = \frac{a}{\sqrt{2}}$.
The gravitational forces exerted by the masses at corners $A, B, C, D$ on the mass $m$ at $P$ are:
$F_{PA} = \frac{G(m)(m)}{x^2} = \frac{Gm^2}{x^2} = F$
$F_{PB} = \frac{G(2m)(m)}{x^2} = 2F$
$F_{PC} = \frac{G(3m)(m)}{x^2} = 3F$
$F_{PD} = \frac{G(4m)(m)}{x^2} = 4F$
These forces act along the diagonals. The net force along the diagonal $AC$ is $F_{net, AC} = F_{PC} - F_{PA} = 3F - F = 2F$ (directed towards $C$).
The net force along the diagonal $BD$ is $F_{net, BD} = F_{PD} - F_{PB} = 4F - 2F = 2F$ (directed towards $D$).
The resultant force $F_{net}$ is the vector sum of these two perpendicular forces:
$F_{net} = \sqrt{(2F)^2 + (2F)^2} = 2\sqrt{2}F$
Substituting $F = \frac{Gm^2}{x^2} = \frac{Gm^2}{(a/\sqrt{2})^2} = \frac{2Gm^2}{a^2}$:
$F_{net} = 2\sqrt{2} \left( \frac{2Gm^2}{a^2} \right) = \frac{4\sqrt{2}Gm^2}{a^2}$

(A) The gravitational field $I$ at a distance $r$ along the axis of a ring of mass $m$ and radius $a$ is given by $I = \frac{Gmr}{(a^2 + r^2)^{3/2}}$.
The gravitational force $F$ exerted by the sphere of mass $M$ on the ring is $F = M \times I = \frac{GMmr}{(a^2 + r^2)^{3/2}}$.
Given $r = \sqrt{3}a$,substitute this into the formula:
$F = \frac{GMm(\sqrt{3}a)}{(a^2 + (\sqrt{3}a)^2)^{3/2}}$
$F = \frac{GMm\sqrt{3}a}{(a^2 + 3a^2)^{3/2}}$
$F = \frac{GMm\sqrt{3}a}{(4a^2)^{3/2}}$
$F = \frac{GMm\sqrt{3}a}{8a^3}$
$F = \frac{\sqrt{3}GMm}{8a^2}$

(C) The gravitational potential energy $U$ of a system of three particles is given by:
$U = -\frac{G m_1 m_2}{r_{12}} - \frac{G m_2 m_3}{r_{23}} - \frac{G m_1 m_3}{r_{13}}$
Given $m_1 = m_2 = m_3 = 100 \, g = 0.1 \, kg$ and $r_{12} = r_{23} = r_{13} = 20 \, cm = 0.2 \, m$.
Since all masses and distances are equal,$U = -3 \times \frac{G m^2}{r}$.
$U = -3 \times \frac{6.67 \times 10^{-11} \times (0.1)^2}{0.2}$
$U = -3 \times \frac{6.67 \times 10^{-11} \times 0.01}{0.2} = -3 \times 3.335 \times 10^{-12} = -1.00 \times 10^{-11} \, J$.
The work required to move the particles to infinity is $W = U_{\infty} - U_{initial} = 0 - (-1.00 \times 10^{-11} \, J) = 1.00 \times 10^{-11} \, J$.

(C) The gravitational force between the two particles provides the necessary centripetal force for circular motion.
The distance between the two particles is $d = 2R$.
The gravitational force is $F_g = \frac{G \cdot m \cdot m}{(2R)^2} = \frac{Gm^2}{4R^2}$.
The centripetal force required for a particle of mass $m$ moving with speed $v$ in a circle of radius $R$ is $F_c = \frac{mv^2}{R}$.
Equating the two forces: $\frac{mv^2}{R} = \frac{Gm^2}{4R^2}$.
Simplifying the equation: $v^2 = \frac{Gm^2}{4R^2} \cdot \frac{R}{m} = \frac{Gm}{4R}$.
Taking the square root: $v = \sqrt{\frac{Gm}{4R}} = \frac{1}{2}\sqrt{\frac{Gm}{R}}$.

(B) The angular momentum $L$ of the Earth remains conserved as no external torque acts on it.
$L = I\omega = \text{constant}$
Since $I = \frac{2}{5}MR^2$ and $\omega = \frac{2\pi}{T}$,we have:
$\frac{2}{5}MR^2 \times \frac{2\pi}{T} = \text{constant}$
This implies $T \propto R^2$.
Given the new radius $R_2 = \frac{R}{n}$,the new time period $T_2$ is related to the original time period $T_1 = 24 \ hr$ by:
$\frac{T_2}{T_1} = \left( \frac{R_2}{R_1} \right)^2 = \left( \frac{R/n}{R} \right)^2 = \frac{1}{n^2}$
Therefore,$T_2 = \frac{T_1}{n^2} = \frac{24}{n^2} \ hr$.

(B) Power is defined as the dot product of force and velocity: $P = \vec{F} \cdot \vec{v} = Fv \cos \theta$.
At the highest point,the velocity $v = 0$,so power $P = 0$.
During the motion,the gravitational force $\vec{F}$ is always directed towards the center of the earth.
As the body falls back,the velocity $\vec{v}$ is directed downwards,which is in the same direction as the gravitational force $\vec{F}$ (i.e.,$\theta = 0^\circ$).
Since $P = Fv \cos(0^\circ) = Fv$,the power is proportional to the speed $v$.
The speed $v$ of the body is maximum at the instant just before it hits the earth's surface.
Therefore,the power exerted by the gravitational force is greatest at the instant just before the body hits the earth.

(C) Given:
Mass of the particle $= M$
Mass of the spherical shell $= M$
Radius of the spherical shell $= a$
Let $O$ be the center of the spherical shell.
The gravitational potential at a point $P$ at a distance $r = a/2$ from the center due to the particle at the center is given by $V_1 = -\frac{GM}{r} = -\frac{GM}{a/2} = -\frac{2GM}{a}$.
The gravitational potential at any point inside a spherical shell is constant and equal to the potential at its surface,which is $V_2 = -\frac{GM}{a}$.
The total gravitational potential at point $P$ is $V = V_1 + V_2$.
$V = -\frac{2GM}{a} + \left( -\frac{GM}{a} \right) = -\frac{3GM}{a}$.
The magnitude of the gravitational potential is $|V| = \frac{3GM}{a}$.

(D) The gravitational field $F$ due to a thin spherical shell of mass $M$ and radius $R$ is given by:
$1$. Inside the shell,i.e.,for $r < R$:
The gravitational field $F = 0$.
$2$. On the surface of the shell,i.e.,for $r = R$:
The gravitational field $F = \frac{GM}{R^2}$.
$3$. Outside the shell,i.e.,for $r > R$:
The gravitational field $F = \frac{GM}{r^2}$.
Thus,for $r < R$,the field is zero,and for $r > R$,it follows an inverse-square law. The correct plot is the one that shows $F = 0$ for $r < R$ and a curve representing $F \propto 1/r^2$ for $r > R$.

(A) For a point inside the Earth,i.e.,$r < R$:
$E = -\frac{GM}{R^3}r$
Where $M$ and $R$ are the mass and radius of the Earth,respectively. The negative sign indicates that the field is directed towards the center.
At the center,$r = 0$,therefore $E = 0$.
For a point outside the Earth,i.e.,$r > R$:
$E = -\frac{GM}{r^2}$
On the surface of the Earth,i.e.,$r = R$:
$E = -\frac{GM}{R^2}$
The magnitude of the gravitational field increases linearly with distance $r$ inside the Earth and decreases inversely with the square of the distance $r$ outside the Earth. The graph representing this variation is shown in the provided solution image.

(C) The initial distance between the centres is $12R.$ Collision occurs when the distance between the centres is equal to the sum of their radii,which is $R + 2R = 3R.$
Therefore,the total distance covered by both bodies before collision is $12R - 3R = 9R.$
Let $x_1$ and $x_2$ be the distances covered by the bodies of mass $M$ and $5M$ respectively. Since there is no external force,the centre of mass of the system remains stationary. Thus,$M x_1 = 5M x_2$,which gives $x_1 = 5x_2$.
Given $x_1 + x_2 = 9R$,we substitute $x_1 = 5x_2$ to get $5x_2 + x_2 = 9R$,so $6x_2 = 9R$,which means $x_2 = 1.5R$.
Then,$x_1 = 5(1.5R) = 7.5R$.
The distance covered by the smaller body (mass $M$) is $7.5R$.

(A) Let $\omega$ be the angular velocity of the earth's rotation about its axis.
Let $g$ be the acceleration due to gravity at the poles.
Let $R$ be the radius of the earth.
The effective acceleration due to gravity at the equator is $g_{eq} = g - \omega^2 R$.
The acceleration due to gravity at a depth $d$ below the surface is $g_d = g(1 - \frac{d}{R})$.
According to the problem,the readings are the same,so $g_{eq} = g_d$.
$g - \omega^2 R = g(1 - \frac{d}{R})$
$g - \omega^2 R = g - \frac{gd}{R}$
$\omega^2 R = \frac{gd}{R}$
$d = \frac{\omega^2 R^2}{g}$

(B) Let the mass of the asteroid be $M$ and its radius be $R$. The density $\rho$ is given by $\rho = M / (\frac{4}{3} \pi R^3)$.
When a spherical hole of radius $r = R/2$ is excavated,the mass of the removed part is $m = \rho \cdot (\frac{4}{3} \pi r^3) = \rho \cdot (\frac{4}{3} \pi (R/2)^3) = \rho \cdot (\frac{4}{3} \pi R^3) / 8 = M/8$.
The gravitational acceleration at the point $P$ (just above the hole) due to the complete asteroid is $g_1 = GM/R^2$ directed towards the center.
The gravitational acceleration at point $P$ due to the removed sphere (hole) is $g_2 = G(M/8) / (R/2)^2 = (GM/8) / (R^2/4) = GM / 2R^2$ directed away from the center of the hole (towards the center of the asteroid).
Since the hole is removed,the net gravitational acceleration $g_{net}$ is the vector sum of the acceleration due to the complete sphere and the negative mass of the hole: $g_{net} = g_1 - g_2 = GM/R^2 - GM/2R^2 = GM/2R^2$.

(A) Outside the planet (at the surface),the gravitational force on the man is given by Newton's law of gravitation: $F_{\text{old}} = \frac{GMm}{R^2}$.
Inside the planet,the man is inside the spherical shell of mass $\frac{2M}{3}$. According to the shell theorem,the gravitational field due to a spherical shell at any point inside it is zero. Thus,the shell exerts no force on the man.
The man is at a distance $R$ from the point mass $\frac{M}{3}$ located at the center. The gravitational force exerted by this point mass is $F_{\text{new}} = \frac{G(\frac{M}{3})m}{R^2} = \frac{GMm}{3R^2}$.
The change in the force of gravity experienced by the man is $\Delta F = F_{\text{old}} - F_{\text{new}} = \frac{GMm}{R^2} - \frac{GMm}{3R^2} = \frac{2GMm}{3R^2}$.

(D) For a thin spherical shell of mass $M$ and radius $a$,the gravitational field inside the shell due to the shell itself is zero,and the gravitational potential inside is constant and equal to $-GM/a$.
However,there is an external particle of mass $M$ at a distance $a$ from the surface of the shell. The distance of this particle from the centre of the shell is $r = a + a = 2a$.
This external particle creates a non-zero gravitational field and a non-zero gravitational potential at all points inside the shell.
Therefore,inside the shell,the net gravitational field is equal to the field produced by the external particle,and the net gravitational potential is the sum of the constant potential due to the shell and the potential due to the external particle.
Thus,neither the net gravitational field nor the net gravitational potential is zero inside the shell.

(D) The gravitational potential $V$ at a distance $x$ from the centre of a ring of radius $R$ and mass $M$ is given by $V(x) = -\frac{GM}{\sqrt{x^2 + R^2}}$.
Applying the principle of conservation of mechanical energy between the initial position $(x = R)$ and the centre $(x = 0)$:
$K_i + U_i = K_f + U_f$
Since the particle starts from rest,$K_i = 0$.
$0 + m V(R) = \frac{1}{2}mv^2 + m V(0)$
$0 + m \left( -\frac{GM}{\sqrt{R^2 + R^2}} \right) = \frac{1}{2}mv^2 + m \left( -\frac{GM}{R} \right)$
$-\frac{GM}{\sqrt{2}R} = \frac{1}{2}v^2 - \frac{GM}{R}$
$\frac{1}{2}v^2 = \frac{GM}{R} - \frac{GM}{\sqrt{2}R} = \frac{GM}{R} \left( 1 - \frac{1}{\sqrt{2}} \right)$
$v^2 = \frac{2GM}{R} \left( 1 - \frac{1}{\sqrt{2}} \right)$
$v = \sqrt{2\left( 1 - \frac{1}{\sqrt{2}} \right) \frac{GM}{R}}$

(A) The two masses revolve about their common center of mass (c.m.).
Let the distance of mass $4m$ from the center of mass be $x$,and the distance of mass $m$ be $(d-x)$.
By the definition of the center of mass: $4m(x) = m(d-x) \Rightarrow 4x = d-x \Rightarrow 5x = d \Rightarrow x = d/5$.
Thus,the distance of mass $4m$ from the center of mass is $r_1 = d/5$,and the distance of mass $m$ is $r_2 = 4d/5$.
Both masses revolve with the same angular velocity $\omega$.
The kinetic energy of a particle in circular motion is $K = \frac{1}{2}mv^2 = \frac{1}{2}m(r\omega)^2 = \frac{1}{2}mr^2\omega^2$.
The ratio of kinetic energies is:
$\frac{K_{4m}}{K_m} = \frac{\frac{1}{2}(4m)r_1^2\omega^2}{\frac{1}{2}(m)r_2^2\omega^2} = \frac{4(d/5)^2}{(4d/5)^2} = \frac{4(d^2/25)}{16(d^2/25)} = \frac{4}{16} = \frac{1}{4}$.
Therefore,the ratio is $1:4$.

(A, C) The gravitational field inside a spherical cavity within a spherical planet is uniform and non-zero. This is a standard result derived from the superposition principle of gravitational fields.
Regarding option $C$,the total mechanical energy $E = K + U$. If $E = 0$,then $K = -U$. Since $U$ is negative in a gravitational field,$K$ must be positive. As the body moves away from the source,$U$ increases (becomes less negative) and $K$ decreases. At infinity,$U = 0$ and $K = 0$,meaning the body has just enough energy to reach infinity,which is the definition of escape velocity. Thus,a body with zero total mechanical energy will escape the gravitational field.

(D) The gravitational potential at the mid-point $P$ due to two fixed particles of mass $m$ is given by:
$V = V_1 + V_2 = -\frac{Gm}{l/2} - \frac{Gm}{l/2} = -\frac{4Gm}{l}$
The gravitational potential energy of the particle of mass $m_0$ at point $P$ is:
$U = m_0 V = -\frac{4Gmm_0}{l}$
For the particle to just escape to infinity,its total energy at the point of projection must be equal to its total energy at infinity,which is zero.
By the law of conservation of energy:
$K_i + U_i = K_f + U_f$
$\frac{1}{2} m_0 v^2 + \left( -\frac{4Gmm_0}{l} \right) = 0 + 0$
$\frac{1}{2} m_0 v^2 = \frac{4Gmm_0}{l}$
$v^2 = \frac{8Gm}{l}$
$v = \sqrt{\frac{8Gm}{l}} = 2 \sqrt{\frac{2Gm}{l}}$
Thus,the minimum velocity is $2 \sqrt{\frac{2Gm}{l}}.$

(D) By the principle of conservation of energy,the total mechanical energy at the initial position equals the total mechanical energy at the final position (where the ball comes to rest momentarily).
Initial energy $E_i = K_i + U_i = 0 - \frac{GMm}{2R}$.
Final energy $E_f = K_f + U_f = 0 + U_{grav, depth} + U_{spring}$.
The gravitational potential energy at a depth $d = R/2$ (i.e.,at distance $r = R/2$ from the center) is $U_{grav} = -\frac{GMm}{2R^3}(3R^2 - r^2) = -\frac{GMm}{2R^3}(3R^2 - (R/2)^2) = -\frac{GMm}{2R^3}(3R^2 - R^2/4) = -\frac{GMm}{2R^3}(\frac{11R^2}{4}) = -\frac{11GMm}{8R}$.
The spring is compressed by $x = R/2$,so $U_{spring} = \frac{1}{2}K(R/2)^2 = \frac{KR^2}{8}$.
Equating $E_i = E_f$: $-\frac{GMm}{2R} = -\frac{11GMm}{8R} + \frac{KR^2}{8}$.
$\frac{KR^2}{8} = \frac{11GMm}{8R} - \frac{4GMm}{8R} = \frac{7GMm}{8R}$.
Therefore,$K = \frac{7GMm}{R^3}$.

(B) The gravitational potential energy at a distance $r$ from the center of the Earth is $U = -\frac{GMm}{r}$.
Using the conservation of energy,the velocity $v$ of the particle just before hitting the surface (at distance $R$ from the center) when dropped from height $h$ is given by $\frac{1}{2}mv^2 = GMm(\frac{1}{R} - \frac{1}{R+h})$.
For the initial drop from height $h_1 = R$,the velocity of approach $v_1$ is $\frac{1}{2}mv_1^2 = GMm(\frac{1}{R} - \frac{1}{2R}) = \frac{GMm}{2R}$,so $v_1 = \sqrt{\frac{GM}{R}}$.
For the rebound to height $h_2 = R/2$,the velocity of separation $v_2$ is $\frac{1}{2}mv_2^2 = GMm(\frac{1}{R} - \frac{1}{R + R/2}) = GMm(\frac{1}{R} - \frac{2}{3R}) = \frac{GMm}{3R}$,so $v_2 = \sqrt{\frac{2GM}{3R}}$.
The coefficient of restitution $e$ is defined as $e = \frac{v_2}{v_1} = \frac{\sqrt{2GM/3R}}{\sqrt{GM/R}} = \sqrt{\frac{2}{3}}$.

(B) Let the orbital velocity of the satellite of mass $5M$ be $v_r$. The velocities of the pieces of mass $M$ and $4M$ after the explosion are $v_1$ and $v_4$ respectively.
Given that the mass $M$ travels in the same orbit but in the opposite direction,its velocity is $v_1 = -v_r$.
According to the law of conservation of linear momentum,the total momentum before the explosion is equal to the total momentum after the explosion:
$5M v_r = (4M) v_4 + (M)(-v_r)$
$5M v_r = 4M v_4 - M v_r$
$6M v_r = 4M v_4$
$v_4 = 1.5 v_r$
The escape velocity for a satellite in a circular orbit is $v_e = \sqrt{2} v_r \approx 1.414 v_r$.
Since $v_4 = 1.5 v_r > 1.414 v_r$,the velocity of the $4M$ mass exceeds the escape velocity.
Therefore,the mass $4M$ will follow an unbound orbit (hyperbolic trajectory).

(C) According to the Virial Theorem for a system bound by an inverse-square law force (like gravity),the time-averaged kinetic energy $\langle K \rangle$ and potential energy $\langle U \rangle$ are related by $\langle U \rangle = -2 \langle K \rangle.$
For a planet in an elliptical orbit,the total mechanical energy $E$ is constant and is given by $E = K + U = -\frac{GMm}{2a},$ where $a$ is the semi-major axis.
At any point in the orbit,the potential energy is $U = -\frac{GMm}{r}.$
Since the total energy $E = K + U = -\frac{GMm}{2a},$ we have $K = E - U = -\frac{GMm}{2a} - (-\frac{GMm}{r}) = GMm(\frac{1}{r} - \frac{1}{2a}).$
Comparing the magnitudes $|U| = \frac{GMm}{r}$ and $|K| = GMm|\frac{1}{r} - \frac{1}{2a}|,$ we observe that $|U| > |K|$ holds true for all points in the orbit,including $C$ and $D.$
Specifically,at any point $r$ in the orbit,since $r$ is always less than $2a$ (the major axis length),the condition $|U| > |K|$ is satisfied throughout the entire elliptical path.

(B) The acceleration due to gravity at a distance $x$ from the center of the Earth is given by $g_{x} = \frac{g}{R} x$,where $g$ is the acceleration due to gravity at the surface and $R$ is the radius of the Earth.
Since the force $F = -m g_{x} = -\left(\frac{mg}{R}\right) x$ is proportional to the displacement $x$ and directed towards the center,the motion of the ball is simple harmonic motion $(SHM)$.
The time period of this $SHM$ is $T = 2 \pi \sqrt{\frac{m}{k}}$,where $k = \frac{mg}{R}$.
Thus,$T = 2 \pi \sqrt{\frac{m}{mg/R}} = 2 \pi \sqrt{\frac{R}{g}}$.
Substituting $R \approx 6400 \text{ km}$ and $g \approx 9.8 \text{ m/s}^2$,we get $T \approx 84.6 \text{ min}$.
The time taken to reach the other end is half the time period,so $t = \frac{T}{2} = \frac{84.6}{2} = 42.3 \text{ min}$.

(D) From the geometry of the problem,consider a right-angled triangle formed by the center of the Earth,the satellite,and the point on the horizon visible from the satellite. The hypotenuse is $(R+h)$ and the side opposite to the angle $\theta$ is $R$. Thus,$\sin \theta = \frac{R}{R+h}$.
The colatitude is the angle measured from the pole. The satellite can communicate up to a certain latitude,which corresponds to a minimum colatitude of $\theta = \sin^{-1} \left( \frac{R}{R+h} \right)$.
The area of the spherical cap visible from the satellite is $A_{visible} = 2\pi R^2 (1 - \cos \alpha)$,where $\alpha$ is the angular radius of the visible region. In this geometry,$\cos \alpha = \frac{R}{R+h} = \sin \theta$. Thus,the visible area is $2\pi R^2 (1 - \sin \theta)$.
The area on Earth that is $NOT$ visible (escaped) from the satellite is the total surface area minus the visible area:
$A_{escaped} = 4\pi R^2 - 2\pi R^2 (1 - \sin \theta) = 2\pi R^2 (1 + \sin \theta).$
Therefore,both statements $(A)$ and $(C)$ are correct.

(D) When a satellite encounters air resistance,it loses mechanical energy,causing it to spiral inward towards the earth. As the orbital radius $r$ decreases,the potential energy becomes more negative,and the kinetic energy $K = \frac{GMm}{2r}$ increases.
Because the air resistance acts as a non-conservative force,it exerts a torque on the satellite,which causes the angular momentum $L = mvr$ to decrease over time.
Therefore,both the kinetic energy increases and the angular momentum decreases. Thus,option $(D)$ is correct.

(D) According to Kepler's Third Law,$T^2 \propto r^3$. Therefore,$T_1/T_2 = (R_1/R_2)^{3/2} = (R/4R)^{3/2} = (1/4)^{3/2} = 1/8$. Thus,option $A$ is correct.
For orbital velocity,$v = \sqrt{GM/r}$,so $v \propto 1/\sqrt{r}$. Thus,$v_1/v_2 = \sqrt{r_2/r_1} = \sqrt{4R/R} = 2/1$. Thus,option $B$ is correct.
Angular velocity $\omega = v/r = \sqrt{GM/r^3}$,so $\omega \propto r^{-3/2}$.
$\omega_1 = \sqrt{GM/R^3}$ and $\omega_2 = \sqrt{GM/(4R)^3} = \omega_1/8$.
The relative angular velocity $\omega_{rel} = |\omega_1 - \omega_2| = |\omega_1 - \omega_1/8| = 7\omega_1/8$.
However,the question asks for the ratio of angular velocities of $s_2$ w.r.t $s_1$ in a specific context. Given the options,$A$ and $B$ are clearly correct,making $D$ the intended answer.

(D) The orbital speed of a satellite is given by $v = \sqrt{\frac{GM}{r}}$. As $r$ decreases,the speed $v$ increases. Thus,for the smallest possible $r$,the speed is maximum.
The time period of a satellite is given by $T = 2\pi \sqrt{\frac{r^3}{GM}}$. As $r$ decreases,the time period $T$ decreases. Thus,for the smallest possible $r$,the time period is minimum.
The total energy of the satellite-earth system is given by $E = -\frac{GMm}{2r}$. As $r$ decreases,the value of $E$ becomes more negative,meaning the total energy is minimum.
Since all three statements are correct,the correct option is $D$.

(D) For any satellite to orbit the Earth,the center of the Earth must lie in the orbital plane of the satellite. Since the equator is a great circle passing through the center of the Earth,the orbital plane of any satellite must intersect the equatorial plane. Therefore,the satellite must pass over the equator at some point in its orbit.
The time period $T$ of a satellite is given by $T = 2\pi \sqrt{\frac{r}{g_{eff}}}$,where $r = R + h$ is the orbital radius.
Since the minimum height $h$ above the surface is $0$,the minimum orbital radius is $R$. Thus,the time period $T$ must satisfy $T = 2\pi \sqrt{\frac{R+h}{g}} > 2\pi \sqrt{\frac{R}{g}}$.
Both statements $(A)$ and $(B)$ are correct.

(C) Let the velocity of mass $m$ be $u$ and mass $2m$ be $u$ when they reach the center of the Earth. By conservation of energy, $u = \sqrt{gR} = \sqrt{\frac{GM}{R}}$.
At the center, the masses collide and stick. By conservation of linear momentum: $(2m)u - m(u) = (3m)v$, where $v$ is the velocity of the combined mass $3m$ after collision.
$mu = 3mv \Rightarrow v = u/3$.
The potential energy of the combined mass at distance $x$ from the center is $U(x) = \frac{1}{2} (3m) \omega^2 x^2$, where $\omega^2 = \frac{g}{R} = \frac{GM}{R^3}$.
Using conservation of energy for the combined mass: $\frac{1}{2} (3m) v^2 = \frac{1}{2} (3m) \omega^2 A^2$.
$v^2 = \omega^2 A^2 \Rightarrow A = \frac{v}{\omega} = \frac{u/3}{\sqrt{GM/R^3}} = \frac{\sqrt{GM/R}/3}{\sqrt{GM/R^3}} = \frac{R}{3}$.

(C) Consider one particle of mass $M$. It is acted upon by three other particles. Let the distance between adjacent particles be $a = R\sqrt{2}$ and the distance between opposite particles be $d = 2R$.
The gravitational force exerted by the two adjacent particles is $F = \frac{GM^2}{a^2} = \frac{GM^2}{2R^2}$. The components of these forces directed towards the center are $F \cos(45^{\circ})$ each.
The gravitational force exerted by the diagonally opposite particle is $F' = \frac{GM^2}{d^2} = \frac{GM^2}{(2R)^2} = \frac{GM^2}{4R^2}$.
The net centripetal force required for circular motion is provided by the sum of these forces directed towards the center:
$F_{net} = 2F \cos(45^{\circ}) + F' = \frac{Mv^2}{R}$
Substituting the values:
$2 \left( \frac{GM^2}{2R^2} \right) \frac{1}{\sqrt{2}} + \frac{GM^2}{4R^2} = \frac{Mv^2}{R}$
$\frac{GM^2}{R^2} \left( \frac{1}{\sqrt{2}} + \frac{1}{4} \right) = \frac{Mv^2}{R}$
$v^2 = \frac{GM}{R} \left( \frac{4 + \sqrt{2}}{4\sqrt{2}} \right) = \frac{GM}{R} \left( \frac{2\sqrt{2} + 1}{4} \right)$
$v = \frac{1}{2} \sqrt{\frac{GM}{R} (1 + 2\sqrt{2})}$

(B) The orbital velocity of a satellite at height $h$ is given by $v_o = \sqrt{\frac{GM}{R+h}}$. Since $h \ll R$,we can approximate $R+h \approx R$. Thus,$v_o \approx \sqrt{\frac{GM}{R}} = \sqrt{gR}$.
The escape velocity from the Earth's surface (or near the surface) is given by $v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}$.
The required increase in velocity $\Delta v$ is the difference between the escape velocity and the orbital velocity:
$\Delta v = v_e - v_o = \sqrt{2gR} - \sqrt{gR}$.
Factoring out $\sqrt{gR}$,we get:
$\Delta v = \sqrt{gR}(\sqrt{2} - 1)$.

(D) The variation of acceleration due to gravity $g$ with distance $d$ from the center of the earth is given by:
$1$. Inside the earth $(d < R)$:
$g = \frac{GM}{R^3} d$
Since $G, M,$ and $R$ are constants,$g \propto d$. This represents a straight line passing through the origin.
$2$. At the surface of the earth $(d = R)$:
$g = \frac{GM}{R^2} = g_s$ (maximum value).
$3$. Outside the earth $(d > R)$:
$g = \frac{GM}{d^2}$
Here,$g \propto \frac{1}{d^2}$. This represents a rectangular hyperbola.
Combining these,the graph shows a linear increase from the center to the surface,followed by a hyperbolic decrease as distance increases beyond the surface. This matches the graph in option $D$.