Motion of Satellites in Circular Orbits and Planets in Elliptical Orbits Questions in English

(C) For a geostationary orbit,the orbital period is $T = 24 \text{ hrs}$. The angular velocity of the satellite is $\omega_{s} = \frac{2\pi}{T} = \frac{2\pi}{24} = \frac{\pi}{12} \text{ rad/hr}$.
Since the Earth rotates from west to east with angular velocity $\omega_{e} = \frac{2\pi}{24} = \frac{\pi}{12} \text{ rad/hr}$,and the satellite moves in the opposite direction (east to west),their relative angular velocity is $\omega_{r} = \omega_{e} + \omega_{s} = \frac{\pi}{12} + \frac{\pi}{12} = \frac{\pi}{6} \text{ rad/hr}$.
The time interval between successive passes over the same point on the equator is the relative time period $T_{r} = \frac{2\pi}{\omega_{r}} = \frac{2\pi}{\pi/6} = 12 \text{ hrs}$.

(C) For a geostationary orbit,the angular velocity of the satellite must be equal to the angular velocity of the planet,$\omega_s = \omega_p$.
From Kepler's third law,the orbital radius $r$ and angular velocity $\omega$ are related by $G M = \omega^2 r^3$,which implies $r^3 \propto \omega^{-2}$ or $r \propto \omega^{-2/3}$.
Let the initial angular velocity be $\omega$ and the new angular velocity be $\omega' = 2\omega$.
Let the initial radius be $r$ and the new radius be $r'$.
Then,$\frac{r'}{r} = \left( \frac{\omega'}{\omega} \right)^{-2/3} = (2)^{-2/3} = \frac{1}{(2^2)^{1/3}} = \frac{1}{4^{1/3}}$.
Therefore,$r' = \frac{r}{4^{1/3}}$.

(B) The two particles of mass $m$ are separated by a distance $d = 2R$ because they move in a circle of radius $R$ around their common centre of mass.
The gravitational force between them is $F_G = \frac{G m^2}{(2R)^2} = \frac{G m^2}{4R^2}$.
This gravitational force provides the necessary centripetal force for each particle to move in a circle of radius $R$. The centripetal force is $F_c = \frac{m v^2}{R}$.
Equating the two forces: $\frac{m v^2}{R} = \frac{G m^2}{4R^2}$.
Solving for $v$: $v^2 = \frac{G m^2}{4R^2} \cdot \frac{R}{m} = \frac{Gm}{4R}$.
Therefore,$v = \sqrt{\frac{Gm}{4R}}$.

(A) The angular momentum of a satellite in an elliptical orbit is given by $L = mvr \sin \theta$. For a satellite of mass $m$ in an orbit with semi-major axis $a$,the angular momentum is $L = m \sqrt{GMa(1-e^2)}$,where $M$ is the mass of the planet and $e$ is the eccentricity.
Alternatively,for a circular orbit,$L = m \sqrt{GMr}$. For an elliptical orbit,the angular momentum is proportional to the square root of the semi-major axis $a$,i.e.,$L \propto \sqrt{a}$.
From the figure,the semi-major axis $a_A$ of satellite $A$ is the largest,followed by the semi-major axis $a_B$ of satellite $B$,and the semi-major axis $a_C$ of satellite $C$ is the smallest.
Therefore,$a_A > a_B > a_C$.
Since $L \propto \sqrt{a}$,it follows that $L_A > L_B > L_C$.

(A) The angular velocity of a satellite is given by $\omega = \sqrt{\frac{GM_e}{r^3}}$.
For the first satellite with radius $R$,$\omega_1 = \sqrt{\frac{GM_e}{R^3}}$.
For the second satellite with radius $4R$,$\omega_2 = \sqrt{\frac{GM_e}{(4R)^3}} = \sqrt{\frac{GM_e}{64R^3}} = \frac{1}{8}\sqrt{\frac{GM_e}{R^3}} = \frac{\omega_1}{8}$.
Since they start on the same radial line,they have maximum separation when the difference in their angular positions is $\pi$ radians.
Let $\theta_1 = \omega_1 t$ and $\theta_2 = \omega_2 t$. The condition for maximum separation is $\theta_1 - \theta_2 = \pi$.
Substituting the values: $(\omega_1 - \omega_2)t = \pi$.
$t = \frac{\pi}{\omega_1 - \omega_2} = \frac{\pi}{\omega_1 - \frac{\omega_1}{8}} = \frac{\pi}{\frac{7\omega_1}{8}} = \frac{8\pi}{7\omega_1}$.
Substituting $\omega_1 = \sqrt{\frac{GM_e}{R^3}}$,we get $t = \frac{8\pi}{7}\sqrt{\frac{R^3}{GM_e}}$.

(B) Consider two masses $m$ at distances $r_1 = R - l/2$ and $r_2 = R + l/2$ from the planet of mass $M$,where $R$ is the orbital radius of the center of mass and $l$ is the length of the rod.
For the rod to remain vertical,it must rotate with angular velocity $\omega = \sqrt{GM/R^3}$.
The gravitational force on the inner mass is $F_1 = GMm/(R - l/2)^2$ and on the outer mass is $F_2 = GMm/(R + l/2)^2$.
The required centripetal force for the inner mass is $m(R - l/2)\omega^2 = GMm(R - l/2)/R^3$ and for the outer mass is $m(R + l/2)\omega^2 = GMm(R + l/2)/R^3$.
The net force on the inner mass (directed towards the planet) is $F_1 - F_{c1} = GMm [1/(R - l/2)^2 - (R - l/2)/R^3] > 0$,meaning the inner mass wants to fall faster than the rod's rotation allows,pulling the rod inward.
The net force on the outer mass (directed away from the planet) is $F_{c2} - F_2 = GMm [(R + l/2)/R^3 - 1/(R + l/2)^2] > 0$,meaning the outer mass wants to fly away,pulling the rod outward.
Thus,the rod is in tension.
For stability,a small angular displacement $\theta$ creates a restoring torque due to the gravity gradient,which acts to return the rod to the vertical position. This is a classic gravity-gradient stabilization problem,and the vertical orientation is the stable equilibrium position.

(D) For a circular orbit at distance $R$,the orbital velocity is $v_0 = \sqrt{\frac{GM}{R}}$.
For the second satellite,the initial total energy $E$ is given by $E = KE + PE = \frac{1}{2}m(\frac{v_0}{2})^2 - \frac{GMm}{R}$.
Substituting $GM = v_0^2 R$,we get $E = \frac{1}{8}mv_0^2 - mv_0^2 = -\frac{7}{8}mv_0^2$.
By conservation of angular momentum at the launch point $(R)$ and the point of minimum distance $(R_1)$,where velocity is $v_1$: $m(\frac{v_0}{2})R = mv_1 R_1$,so $v_1 = \frac{v_0 R}{2 R_1}$.
By conservation of energy: $E = \frac{1}{2}mv_1^2 - \frac{GMm}{R_1}$.
$-\frac{7}{8}mv_0^2 = \frac{1}{2}m(\frac{v_0 R}{2 R_1})^2 - \frac{mv_0^2 R}{R_1}$.
Dividing by $mv_0^2$: $-\frac{7}{8} = \frac{R^2}{8 R_1^2} - \frac{R}{R_1}$.
Let $x = \frac{R}{R_1}$,then $-\frac{7}{8} = \frac{1}{8}x^2 - x$.
$x^2 - 8x + 7 = 0$.
$(x-7)(x-1) = 0$.
Since $R_1 < R$,$x = 7$,so $R_1 = \frac{R}{7}$.

(D) According to the law of conservation of angular momentum,the angular momentum of the satellite remains constant at all points in its orbit.
$L_a = L_b$
$m v_a r_a = m v_b r_b$
Given from the figure,the distance of point $a$ from the planet $P$ is $r_a = r$ and the distance of point $b$ from the planet $P$ is $r_b = 3r$.
Substituting these values into the equation:
$m v_a r = m v_b (3r)$
$v_a = 3 v_b$
Therefore,the ratio of the speed at point $a$ to the speed at point $b$ is:
$\frac{v_a}{v_b} = \frac{3}{1} = 3 : 1$

(D) The magnitude of the gravitational potential energy of a satellite in an orbit of radius $r$ is given by $E = \frac{GMm}{2r}$.
According to the problem,the energy is quantized such that $E_n = n \cdot k$,where $n$ is an integer and $k$ is a constant energy.
For the first orbit with radius $R$,we have $\frac{GMm}{2R} = n \cdot k$.
For the next successive orbit with radius $\frac{3R}{2}$,the energy must be $(n-1) \cdot k$ (since energy magnitude decreases as radius increases),so $\frac{GMm}{2(3R/2)} = (n-1) \cdot k$,which simplifies to $\frac{GMm}{3R} = (n-1) \cdot k$.
Dividing the two equations: $\frac{GMm/2R}{GMm/3R} = \frac{n}{n-1} \Rightarrow \frac{3}{2} = \frac{n}{n-1}$.
Solving for $n$: $3n - 3 = 2n \Rightarrow n = 3$.
Substituting $n=3$ back into the first equation: $\frac{GMm}{2R} = 3k \Rightarrow k = \frac{GMm}{6R}$.
The maximum radius $R_{\text{max}}$ corresponds to the lowest energy level,which is $n=1$.
Thus,$\frac{GMm}{2R_{\text{max}}} = 1 \cdot k = \frac{GMm}{6R}$.
Solving for $R_{\text{max}}$ gives $R_{\text{max}} = 3R$.

(A) For a circular orbit of radius $r = 2R$,the orbital speed $v$ is given by $v = \sqrt{\frac{GM}{2R}}$.
For the elliptical orbit,the semi-major axis $a$ is the average of the periapsis and apoapsis distances: $a = \frac{R + 2R}{2} = \frac{3R}{2}$.
Using the conservation of energy for the elliptical orbit at the apoapsis (distance $2R$): $\frac{1}{2}mv_0^2 - \frac{GMm}{2R} = -\frac{GMm}{2a} = -\frac{GMm}{3R}$.
Rearranging the energy equation: $\frac{1}{2}v_0^2 = \frac{GM}{2R} - \frac{GM}{3R} = \frac{GM}{6R}$.
Thus,$\frac{GM}{R} = 3v_0^2$.
Substituting this into the expression for the circular orbit speed: $v = \sqrt{\frac{GM}{2R}} = \sqrt{\frac{3v_0^2}{2}} = \sqrt{\frac{3}{2}} v_0$.

(D) The time period $T$ of a satellite revolving near the surface of a spherical body is given by $T = 2\pi \sqrt{\frac{R^3}{GM}}$.
Since the mass $M$ of the asteroid can be expressed in terms of its density $\rho$ and radius $R$ as $M = \rho \times \frac{4}{3} \pi R^3$,we substitute this into the formula:
$T = 2\pi \sqrt{\frac{R^3}{G (\rho \cdot \frac{4}{3} \pi R^3)}} = 2\pi \sqrt{\frac{3}{4 \pi G \rho}} = \sqrt{\frac{3\pi}{G\rho}}$.
This shows that the time period depends only on the density $\rho$ of the body.
Since the asteroid has the same density as Earth,the time period of the pebble revolving near its surface is the same as the time period of a satellite revolving near the surface of the Earth.
This value is approximately $84 \ \text{min}$,which is equal to $1 \ \text{hr} \ 24 \ \text{min}$.

(B) According to Kepler's second law,the angular momentum of the planet remains conserved as the gravitational force is a central force.
Applying the conservation of angular momentum at positions $A$ and $B$:
$L_A = L_B$
$m v_A r_A = m v_B r_B$
where $r_A = OA$ and $r_B = OB$.
Rearranging the terms to find the ratio of speeds:
$\frac{v_B}{v_A} = \frac{r_A}{r_B} = \frac{OA}{OB}$
Given that $\frac{OA}{OB} = x$,we get:
$\frac{v_B}{v_A} = x$

(B) For a planet of mass $m$ revolving in a circular orbit of radius $R$ with angular velocity $\omega$,the centripetal force is provided by the gravitational force.
Given that the gravitational force $F \propto R^{-5/2}$,we can write $F = k R^{-5/2}$ for some constant $k$.
The centripetal force is given by $F = m R \omega^2$.
Equating the two,we have $m R \omega^2 \propto R^{-5/2}$.
Since $\omega = \frac{2\pi}{T}$,we substitute this into the equation: $m R (\frac{2\pi}{T})^2 \propto R^{-5/2}$.
This simplifies to $\frac{R}{T^2} \propto R^{-5/2}$.
Rearranging for $T^2$,we get $T^2 \propto R \cdot R^{5/2} = R^{1 + 5/2} = R^{7/2}$.
Thus,$T^2 \propto R^{7/2}$.

(A) The speed of an object moving in a parabolic path at a distance $R$ from the center of a celestial body is equal to the escape velocity at that point.
$V_{p} = V_{e} = \sqrt{\frac{2GM}{R}}$
To convert the spaceship into a satellite in a circular orbit of radius $R$,its speed must be reduced to the orbital velocity $V_{0}$ at that distance.
$V_{0} = \sqrt{\frac{GM}{R}}$
The change in speed required is $\Delta V = V_{p} - V_{0}$.
Substituting the values,we get:
$\Delta V = \sqrt{\frac{2GM}{R}} - \sqrt{\frac{GM}{R}}$
$\Delta V = \sqrt{\frac{GM}{R}} (\sqrt{2} - 1)$

(C) The orbital velocity of a satellite at a height $h$ above the Earth's surface is given by $V_{0} = \sqrt{\frac{GM}{R+h}}$.
The escape velocity from the Earth's surface is given by $V_{e} = \sqrt{\frac{2GM}{R}}$.
According to the problem,the orbital speed is half the escape velocity: $V_{0} = \frac{V_{e}}{2}$.
Substituting the expressions: $\sqrt{\frac{GM}{R+h}} = \frac{1}{2} \sqrt{\frac{2GM}{R}}$.
Squaring both sides: $\frac{GM}{R+h} = \frac{1}{4} \times \frac{2GM}{R}$.
Simplifying the equation: $\frac{1}{R+h} = \frac{1}{2R}$.
Therefore,$R+h = 2R$,which gives $h = R$.

(A) The areal speed $v_a$ is defined as the rate of change of area swept by the radius vector,given by $v_a = \frac{dA}{dt} = \frac{1}{2} r v$.
Squaring both sides,we get $v_a^2 = \frac{1}{4} r^2 v^2$.
For a satellite in a circular orbit,the orbital velocity is $v = \sqrt{\frac{GM}{r}}$. Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$. Thus,$v^2 = \frac{gR^2}{r}$.
Substituting $v^2$ into the areal speed equation: $v_a^2 = \frac{1}{4} r^2 \left( \frac{gR^2}{r} \right) = \frac{1}{4} r g R^2$.
Solving for the orbital radius $r$: $r = \frac{4v_a^2}{gR^2}$.
The height $h$ from the surface of the Earth is $h = r - R$.
Therefore,$h = \frac{4v_a^2}{gR^2} - R$.

(D) According to the law of conservation of angular momentum,the angular momentum $L$ of the planet remains constant throughout its orbit.
Angular momentum is given by $L = I\omega = mr^2\omega$.
At the closest point (perihelion),$L = m r_{min}^2 \omega$.
At the farthest point (aphelion),$L = m r_{max}^2 \omega'$,where $\omega'$ is the angular velocity at the farthest distance.
Equating the two,we get $m r_{min}^2 \omega = m r_{max}^2 \omega'$.
Solving for $\omega'$,we get $\omega' = \frac{r_{min}^2}{r_{max}^2} \omega$.

(A) The orbital speed of a satellite is given by $v = \sqrt{\frac{GM}{r}}$.
This implies that $v \propto r^{-1/2}$.
Taking the logarithmic derivative,we get $\frac{\Delta v}{v} = -\frac{1}{2} \frac{\Delta r}{r}$.
Given that the radius is decreased by $2\%$,we have $\frac{\Delta r}{r} = -0.02$.
Substituting this value,$\frac{\Delta v}{v} = -\frac{1}{2} \times (-0.02) = 0.01$.
Therefore,the speed increases by $1\%$.

(C) The orbital velocity of a satellite at a distance $r$ from the center of the Earth is given by $v = \sqrt{\frac{GM}{r}}$.
For a satellite just above the Earth's surface,the orbital radius is $r_1 = R$,so $v = \sqrt{\frac{GM}{R}}$.
For a satellite at an altitude $h = R/2$ above the surface,the orbital radius is $r_2 = R + R/2 = \frac{3R}{2}$.
The new orbital velocity $v'$ is given by $v' = \sqrt{\frac{GM}{r_2}} = \sqrt{\frac{GM}{3R/2}} = \sqrt{\frac{2GM}{3R}}$.
Comparing the two velocities: $\frac{v'}{v} = \frac{\sqrt{\frac{2GM}{3R}}}{\sqrt{\frac{GM}{R}}} = \sqrt{\frac{2}{3}}$.
Therefore,$v' = \sqrt{2/3} v$.

(B) According to Kepler's third law of planetary motion,the square of the time period $T$ is proportional to the cube of the orbital radius $R$,i.e.,$T^2 \propto R^3$.
Given $T_1 = 2 \, days$,$R_1 = R$,$T_2 = 16 \, days$,and $R_2 = ?$.
Using the ratio: $\frac{T_2^2}{T_1^2} = \frac{R_2^3}{R_1^3} \Rightarrow \frac{16^2}{2^2} = \left(\frac{R_2}{R}\right)^3$.
$\frac{256}{4} = 64 = \left(\frac{R_2}{R}\right)^3$.
Taking the cube root on both sides: $\frac{R_2}{R} = 4$,so $R_2 = 4R$.
For orbital velocity $v = \sqrt{\frac{GM}{R}}$,we have $v \propto \frac{1}{\sqrt{R}}$.
Therefore,$\frac{v_2}{v_1} = \sqrt{\frac{R_1}{R_2}} = \sqrt{\frac{R}{4R}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Thus,$v_2 = \frac{v_1}{2} = \frac{v_0}{2}$.

(D) For a satellite of mass $m$ orbiting the Earth of mass $M_e$ in a circular orbit of radius $r$,the gravitational force provides the necessary centripetal force:
$\frac{GM_e m}{r^2} = \frac{mv^2}{r} \Rightarrow v = \sqrt{\frac{GM_e}{r}}$
$1$. Kinetic Energy $(K.E.)$: $K.E. = \frac{1}{2}mv^2 = \frac{GM_e m}{2r}$. Thus,$K.E. \propto \frac{1}{r}$.
$2$. Angular Momentum $(L)$: $L = mvr = m \sqrt{\frac{GM_e}{r}} \cdot r = m\sqrt{GM_e r}$. Thus,$L \propto r^{1/2}$.
$3$. Linear Momentum $(P)$: $P = mv = m\sqrt{\frac{GM_e}{r}}$. Thus,$P \propto r^{-1/2}$.
$4$. Frequency of revolution $(f)$: According to Kepler's Third Law,$T^2 \propto r^3$,so $T \propto r^{3/2}$. Since $f = \frac{1}{T}$,we have $f \propto r^{-3/2}$ or $f \propto \frac{1}{r^{3/2}}$.
Therefore,option $D$ is correct.

(B) The orbital angular momentum $L$ of a satellite of mass $m$ revolving in a circular orbit of radius $r$ is given by $L = mvr$.
Since the orbital velocity $v = \sqrt{\frac{GM}{r}}$,we substitute this into the expression for $L$:
$L = m \sqrt{\frac{GM}{r}} \cdot r = m \sqrt{GMr}$.
This shows that $L \propto \sqrt{r}$.
Let the initial angular momentum be $L_1 = L$ at distance $r_1 = r$.
Let the new angular momentum be $L_2$ at distance $r_2 = 4r$.
Using the proportionality $L \propto \sqrt{r}$,we have:
$\frac{L_2}{L_1} = \sqrt{\frac{r_2}{r_1}} = \sqrt{\frac{4r}{r}} = \sqrt{4} = 2$.
Therefore,$L_2 = 2L_1 = 2L$.

(B) For two bodies revolving around their common centre of mass,the gravitational force provides the necessary centripetal force for both particles.
Since the particles must always remain on opposite sides of the centre of mass to maintain the equilibrium of the system,they must complete their circular orbits in the same time period $T$.
Angular velocity is defined as $\omega = \frac{2\pi}{T}$.
Since both particles have the same time period $T$,their angular velocities must be equal,i.e.,$\omega_1 = \omega_2$.
This angular velocity is determined by the gravitational interaction between the two masses and is independent of the specific fraction $f$ in the context of the orbital motion dynamics.

(B) Let the satellite move with velocity $v$ in the tangential direction (say,along the $y$-axis) and the meteorite move with velocity $v$ towards the Earth (along the $x$-axis). After the perfectly inelastic collision,the combined mass $2M$ moves with a new velocity vector $\vec{V}$.
By conservation of linear momentum:
Along the $x$-axis: $Mv = (2M)v_x \Rightarrow v_x = v/2$.
Along the $y$-axis: $Mv = (2M)v_y \Rightarrow v_y = v/2$.
The magnitude of the new velocity is $V = \sqrt{v_x^2 + v_y^2} = \sqrt{(v/2)^2 + (v/2)^2} = v/\sqrt{2}$.
Since the satellite was in a circular orbit,its initial velocity was $v = \sqrt{GM/R}$. The new velocity $V = v/\sqrt{2} < v$. Also,the direction of the velocity is no longer purely tangential. $A$ change in both speed and direction at the same orbital radius $R$ results in an elliptical orbit.

(B) The kinetic energy of a satellite of mass $m$ in a circular orbit of radius $r$ around the Earth of mass $M$ is given by $K.E. = \frac{GMm}{2r}$.
For satellite $A$: mass $= m$,radius $= R$. Therefore,$K.E._A = \frac{GMm}{2R}$.
For satellite $B$: mass $= 2m$,radius $= 2R$. Therefore,$K.E._B = \frac{GM(2m)}{2(2R)} = \frac{GMm}{2R}$.
Taking the ratio: $\frac{K.E._A}{K.E._B} = \frac{GMm/2R}{GMm/2R} = 1$.

(D) The gravitational field $E$ at a distance $r$ is given by $E = \frac{GM(r)}{r^2}$,where $M(r)$ is the mass enclosed within radius $r$.
Given $\rho(r) = \frac{K}{r^2}$,the mass $M(r)$ is calculated as:
$M(r) = \int_0^r \rho(x) \cdot 4\pi x^2 dx = \int_0^r \frac{K}{x^2} \cdot 4\pi x^2 dx = \int_0^r 4\pi K dx = 4\pi Kr$.
Substituting $M(r)$ into the expression for the gravitational field:
$E = \frac{G(4\pi Kr)}{r^2} = \frac{4\pi GK}{r}$.
For a particle of mass $m$ in a circular orbit of radius $R$,the centripetal force is provided by the gravitational force:
$\frac{mV^2}{R} = mE = m \left( \frac{4\pi GK}{R} \right)$.
This simplifies to $V^2 = 4\pi GK$,which means the orbital speed $V$ is a constant.
The period $T$ of the orbit is given by $T = \frac{2\pi R}{V}$.
Since $V$ is constant,$T \propto R$,which implies $\frac{T}{R} = \text{constant}$.

(B) The orbital velocity $V$ of a satellite at a distance $r$ from the center of a planet is given by $\frac{mV^2}{r} = \frac{GMm}{r^2}$, which simplifies to $V = \sqrt{\frac{GM}{r}}$.
Here, $r = R + h = 2 \times 10^6 \, m + 20 \times 10^3 \, m = 2.02 \times 10^6 \, m$.
The time period $T_p$ of one revolution is $T_p = \frac{2\pi r}{V} = 2\pi \sqrt{\frac{r^3}{GM}}$.
The number of revolutions $n$ in time $T = 24 \, hours = 24 \times 3600 \, s$ is $n = \frac{T}{T_p} = \frac{T}{2\pi} \sqrt{\frac{GM}{r^3}}$.
Substituting the values: $n = \frac{24 \times 3600}{2 \times 3.14} \sqrt{\frac{6.67 \times 10^{-11} \times 8 \times 10^{22}}{(2.02 \times 10^6)^3}}$.
$n = \frac{86400}{6.28} \sqrt{\frac{53.36 \times 10^{11}}{8.2424 \times 10^{18}}} = 13758 \times \sqrt{6.47 \times 10^{-7}} = 13758 \times 8.04 \times 10^{-4} \approx 11.06$.
Thus, the number of complete revolutions is $11$.

(C) Consider three bodies of mass $M$ placed at the vertices of an equilateral triangle inscribed in a circle of radius $R$. Let $L$ be the side length of the equilateral triangle.
The distance from the center $O$ to any vertex is $R$. In an equilateral triangle,the distance from the center to a vertex is related to the side length $L$ by $R = \frac{L}{\sqrt{3}}$,which implies $L = \sqrt{3}R$.
The gravitational force exerted on one body by the other two bodies is the vector sum of two forces,each of magnitude $F = \frac{GM^2}{L^2}$. The angle between these two forces is $60^{\circ}$.
The resultant gravitational force $F_{net}$ is given by:
$F_{net} = 2F \cos(30^{\circ}) = 2 \left( \frac{GM^2}{L^2} \right) \frac{\sqrt{3}}{2} = \frac{\sqrt{3}GM^2}{L^2}$.
This resultant force provides the necessary centripetal force for circular motion:
$\frac{Mv^2}{R} = \frac{\sqrt{3}GM^2}{L^2}$.
Substituting $L^2 = 3R^2$:
$\frac{Mv^2}{R} = \frac{\sqrt{3}GM^2}{3R^2} = \frac{GM^2}{\sqrt{3}R^2}$.
Solving for $v$:
$v^2 = \frac{GM}{\sqrt{3}R} \implies v = \sqrt{\frac{GM}{\sqrt{3}R}}$.

(B) The gravitational force is given by $F = \frac{k}{R^n}$.
For a planet in a circular orbit,this force provides the necessary centripetal force: $\frac{mv^2}{R} = \frac{k}{R^n}$.
Solving for the orbital velocity $v$,we get $v^2 = \frac{k}{mR^{n-1}}$,which implies $v \propto R^{-(n-1)/2}$.
The time period $T$ of the orbit is given by $T = \frac{2\pi R}{v}$.
Substituting the expression for $v$,we get $T \propto \frac{R}{R^{-(n-1)/2}} = R^{1 + (n-1)/2} = R^{(2+n-1)/2} = R^{(n+1)/2}$.
Therefore,$T \propto R^{(n+1)/2}$.

(D) According to the law of conservation of angular momentum,the angular momentum $L$ of the planet remains constant throughout its orbit.
The angular momentum is given by $L = I\omega = mr^2\omega$,where $m$ is the mass of the planet,$r$ is the distance from the sun,and $\omega$ is the angular velocity.
At the closest point (perihelion),$L = mr_{min}^2\omega$.
At the farthest point (aphelion),$L = mr_{max}^2\omega'$,where $\omega'$ is the angular velocity at the farthest distance.
Since angular momentum is conserved,$mr_{min}^2\omega = mr_{max}^2\omega'$.
Solving for $\omega'$,we get $\omega' = \frac{r_{min}^2}{r_{max}^2}\omega$.

(D) $1$. For a satellite of mass $m$ at a distance $r$ from the center of the Earth,the total energy $E = -\frac{GMm}{2r}$. The energy required to escape is $E_{escape} = -E = \frac{GMm}{2r}$.
$2$. For a stationary object at the same height $r$,the potential energy is $U = -\frac{GMm}{r}$. The energy required to escape is $E'_{escape} = -U = \frac{GMm}{r}$.
$3$. Since $\frac{GMm}{r} > \frac{GMm}{2r}$,option $A$ is incorrect.
$4$. Total energy $E = K + U$. For a satellite,$K = -\frac{1}{2}U$. Thus,$E = -\frac{1}{2}U + U = \frac{1}{2}U$. Option $B$ is incorrect.
$5$. Sputnik $I$ was launched in $1957$,not $1950$. Option $C$ is incorrect.
$6$. The orbital speed of a geostationary satellite is $v = \sqrt{\frac{GM}{r}}$. For $r \approx 42200 \ km$,$v \approx 3.07 \times 10^3 \ m \ s^{-1}$. Option $D$ is correct.

(A) According to Newton's law of universal gravitation,the gravitational force $F$ between two point masses $m_1$ and $m_2$ separated by a distance $r$ is given by $F = \frac{G m_1 m_2}{r^2}$.
In a binary star system,the two stars revolve around their common centre of mass $(CM)$.
The distance between the two stars is the sum of their orbital radii from the centre of mass,which is $r = r_1 + r_2$.
Substituting this distance into the formula,we get the magnitude of the gravitational force as $F = \frac{G m_1 m_2}{(r_1 + r_2)^2}$.

(D) The gravitational force provides the necessary centripetal force for the circular orbit of the satellite: $\frac{GMm}{(R+h)^{2}} = m \omega^{2}(R+h)$.
Here,$G$ is the gravitational constant,$M$ is the mass of the Earth,$m$ is the mass of the satellite,$R$ is the radius of the Earth,and $h$ is the height of the satellite above the surface.
Simplifying for angular velocity $\omega$: $\omega = \sqrt{\frac{GM}{(R+h)^{3}}}$.
Since $\omega = \frac{2\pi}{T}$,the time period $T$ is given by $T = 2\pi \sqrt{\frac{(R+h)^{3}}{GM}}$.
The radius of the orbit is $r = R+h$.
Thus,$T = 2\pi \sqrt{\frac{r^{3}}{GM}}$.
From this expression,$T$ depends on the mass of the Earth $(M)$,the radius of the orbit $(r)$,and the height $(h)$ (since $r = R+h$). It does not depend on the mass of the satellite $(m)$.
Therefore,the correct factors are $(ii), (iii),$ and $(iv)$.

(C) Polar satellites are low-altitude satellites,typically orbiting at an altitude of $500 \, \text{km}$ to $800 \, \text{km}$ above the Earth's surface.
Therefore,the statement that a polar satellite is a high-altitude satellite is incorrect.
Option $(C)$ is the correct answer.

(A) The angular momentum $L$ of a satellite in a circular orbit of radius $r$ is given by $L = mvr$.
For a satellite in a circular orbit,the orbital velocity $v$ is given by $v = \sqrt{\frac{GM}{r}}$.
Substituting this into the expression for angular momentum,we get $L = m \sqrt{\frac{GM}{r}} \cdot r = m \sqrt{GMr}$.
When the radius is increased from $r$ to $r^{\prime} = 2r$,the new angular momentum $L^{\prime}$ becomes:
$L^{\prime} = m \sqrt{GM(2r)} = \sqrt{2} \cdot m \sqrt{GMr} = \sqrt{2} L$.
Since $\sqrt{2} > 1$,the angular momentum increases.

(A) The orbital velocity of a satellite is given by $v = \sqrt{\frac{GM}{R}} = \sqrt{GM} \cdot R^{-1/2}$.
Differentiating both sides with respect to $R$,we get:
$\frac{dv}{dR} = \sqrt{GM} \cdot (-\frac{1}{2}) R^{-3/2} = -\frac{1}{2} \frac{\sqrt{GM}}{R \sqrt{R}} = -\frac{1}{2R} \sqrt{\frac{GM}{R}}$.
For small changes,$\Delta v \approx \frac{dv}{dR} \Delta R$.
Since the radius decreases by $\Delta R$,the change in radius is $-\Delta R$.
Therefore,$\Delta v = (-\frac{1}{2R} \sqrt{\frac{GM}{R}}) \cdot (-\Delta R) = \frac{\Delta R}{2R} \sqrt{\frac{GM}{R}} = \frac{\Delta R}{2} \sqrt{\frac{GM}{R^3}}$.
Thus,the change in orbital velocity is $\frac{\Delta R}{2} \sqrt{\frac{GM}{R^3}}$.

(C) The orbital velocity of a satellite is given by $v_0 = \sqrt{\frac{GM}{r}}$,where $r$ is the distance from the center of the Earth.
This implies $v_0 \propto \frac{1}{\sqrt{r}}$.
For a satellite very close to the Earth's surface,the orbital radius is $r_1 \approx R$,and its velocity is $v_1 = v$.
For a satellite at an altitude of $3R$,the orbital radius is $r_2 = R + 3R = 4R$.
Using the ratio formula: $\frac{v_1}{v_2} = \sqrt{\frac{r_2}{r_1}}$.
Substituting the values: $\frac{v}{v_2} = \sqrt{\frac{4R}{R}} = \sqrt{4} = 2$.
Therefore,$v_2 = \frac{v}{2}$.

(A) According to Kepler's Third Law,the time period $T$ of a satellite is related to its orbital radius $r$ by $T \propto r^{3/2}$.
Here,the orbital radius $r$ is the distance from the center of the earth,so $r = R_{earth} + h$.
For the geostationary satellite: $r_1 = R + 5R = 6R$. Its time period $T_1 = 24$ hours.
For the second satellite: $r_2 = R + 2R = 3R$.
Using the ratio: $\frac{T_1}{T_2} = \left( \frac{r_1}{r_2} \right)^{3/2}$.
Substituting the values: $\frac{24}{T_2} = \left( \frac{6R}{3R} \right)^{3/2} = (2)^{3/2} = 2\sqrt{2}$.
Therefore,$T_2 = \frac{24}{2\sqrt{2}} = \frac{12}{\sqrt{2}} = 6\sqrt{2}$ hours.

(D) For a satellite of mass $m$ orbiting a planet of mass $M$ at a distance $r$:
Kinetic Energy $(KE)$ = $\frac{GMm}{2r}$
Potential Energy $(PE)$ = $-\frac{GMm}{r}$
Total Energy $(E)$ = $-\frac{GMm}{2r}$
If the radius $r$ is decreased:
$1$. Since $KE = \frac{GMm}{2r}$,as $r$ decreases,$KE$ increases.
$2$. Since $PE = -\frac{GMm}{r}$,as $r$ decreases,the magnitude $\frac{GMm}{r}$ increases,making the negative value more negative. Thus,$PE$ decreases.
Therefore,both statements $(A)$ and $(B)$ are incorrect,making option $(D)$ the correct choice.

(C) The orbital speed $v_0$ of a satellite at a distance $r = R + x$ from the center of the Earth is given by $v_0 = \sqrt{\frac{GM}{r}} = \sqrt{\frac{GM}{R + x}}$.
We know that the acceleration due to gravity on the surface of the Earth is $g = \frac{GM}{R^2}$,which implies $GM = gR^2$.
Substituting $GM = gR^2$ into the expression for $v_0$,we get:
$v_0 = \sqrt{\frac{gR^2}{R + x}} = \left( \frac{gR^2}{R + x} \right)^{1/2}$.
Thus,the correct option is $C$.

(C) For a planet revolving around the sun in a circular orbit,the gravitational force provides the necessary centripetal force.
$\frac{GMm}{r^2} = \frac{mv^2}{r}$
Solving for the orbital velocity $v$:
$v = \sqrt{\frac{GM}{r}}$
The angular momentum $L$ is given by the formula $L = mvr$.
Substituting the expression for $v$:
$L = m \left( \sqrt{\frac{GM}{r}} \right) r$
$L = m \sqrt{GM} \sqrt{r}$
Since $m$,$G$,and $M$ are constants,we find that $L \propto \sqrt{r}$.

(C) satellite in a parking orbit (geostationary orbit) moves with the same orbital velocity as the earth's rotation. When a body is dropped from such a satellite,it retains the same orbital velocity as the satellite at that instant. Therefore,the body will continue to move in the same orbit as the satellite. Consequently,it will appear stationary relative to a point on the earth's surface and will rotate about the earth's axis with a time period of $24$ hours.

(C) The orbital period $T$ of a satellite revolving in a circular orbit at a distance $r$ from the center of a planet of mass $M$ is given by the formula $T = 2\pi \sqrt{\frac{r^3}{GM}}$.
From this expression,it is clear that the period $T$ depends on the mass of the planet $M$ and the orbital radius $r$.
The orbital radius $r$ is related to the radius of the planet $R$ and the height $h$ above the surface $(r = R + h)$.
Therefore,the period depends on the mass of the planet and the radius of the planet.
However,the mass of the satellite $m$ does not appear in the formula,meaning the period is independent of the mass of the satellite.

(B) The orbital velocity $v_0$ of a satellite revolving around the earth at a distance $r$ from the center is given by the formula:
$v_0 = \sqrt{\frac{GM}{r}}$
From this expression,it is clear that the orbital velocity is inversely proportional to the square root of the orbital radius:
$v_0 \propto \frac{1}{\sqrt{r}}$
Given the radii of the two satellites are $r_1 = 9R$ and $r_2 = 16R$,the ratio of their speeds $v_1$ and $v_2$ is:
$\frac{v_1}{v_2} = \sqrt{\frac{r_2}{r_1}}$
Substituting the given values:
$\frac{v_1}{v_2} = \sqrt{\frac{16R}{9R}} = \sqrt{\frac{16}{9}} = \frac{4}{3}$
Therefore,the ratio of the speeds is $4/3$.

(D) The gravitational force acting on the satellite provides the necessary centripetal force for its circular orbit.
Let $M_e$ be the mass of the earth. The gravitational force is $F_g = \frac{G M_e m}{(R + x)^2}$.
The centripetal force required is $F_c = \frac{m v_0^2}{R + x}$.
Equating these,we get $\frac{G M_e m}{(R + x)^2} = \frac{m v_0^2}{R + x}$,which simplifies to $v_0^2 = \frac{G M_e}{R + x}$.
We know that the acceleration due to gravity on the earth's surface is $g = \frac{G M_e}{R^2}$,which implies $G M_e = g R^2$.
Substituting this into the expression for $v_0^2$,we get $v_0^2 = \frac{g R^2}{R + x}$.
Therefore,the orbital speed is $v_0 = \sqrt{\frac{g R^2}{R + x}} = \left( \frac{g R^2}{R + x} \right)^{1/2}$.

(C) The gravitational force exerted by the Sun on a planet acts along the line joining the Sun and the planet.
This force is a central force,meaning its torque about the Sun is zero.
According to the principle of conservation of angular momentum,if the net external torque acting on a system is zero,the angular momentum of the system remains constant.
Therefore,the motion of planets in the solar system is an example of the conservation of angular momentum.

(D) geostationary satellite orbits the Earth with a time period $T = 24 \, \text{hours} = 86400 \, \text{s}$.
Using the formula for the orbital radius $r$ of a satellite: $r = \left( \frac{G M T^2}{4 \pi^2} \right)^{1/3}$.
Given $GM = g R^2$,where $g = 9.8 \, \text{m/s}^2$ and $R = 6.4 \times 10^6 \, \text{m}$,the orbital radius $r$ is approximately $42200 \, \text{km}$.
The height $h$ above the Earth's surface is $h = r - R$.
$h = 42200 \, \text{km} - 6400 \, \text{km} = 35800 \, \text{km}$.
Rounding to the nearest standard value,the height is approximately $36000 \, \text{km}$.

(A) According to Kepler's Third Law of planetary motion,the square of the time period $(T)$ is proportional to the cube of the orbital radius $(r)$: $T^2 \propto r^3$.
The orbital radius $r$ is the distance from the center of the Earth,given by $r = R + h$,where $h$ is the height above the surface.
For the geostationary satellite $(S_1)$: $h_1 = 6R$,so $r_1 = R + 6R = 7R$. The time period $T_1 = 24 \text{ hours}$.
For the second satellite $(S_2)$: $h_2 = 2.5R$,so $r_2 = R + 2.5R = 3.5R$.
Using the ratio: $\left(\frac{T_2}{T_1}\right)^2 = \left(\frac{r_2}{r_1}\right)^3$.
$\left(\frac{T_2}{24}\right)^2 = \left(\frac{3.5R}{7R}\right)^3 = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$.
$\frac{T_2}{24} = \sqrt{\frac{1}{8}} = \frac{1}{2\sqrt{2}}$.
$T_2 = \frac{24}{2\sqrt{2}} = \frac{12}{\sqrt{2}} = 6\sqrt{2} \text{ hours}$.

(C) The $Assertion$ is correct because an astronaut in a satellite is in a state of free fall towards the Earth,resulting in weightlessness.
The $Reason$ is incorrect because a body in free fall $DOES$ experience gravity (it is the gravitational force that causes the acceleration $g$). The sensation of weightlessness arises because the body and the satellite are falling together with the same acceleration $g$,making the normal force (apparent weight) zero.
Therefore,the correct option is $C$.

(B) $1$. Apply energy conservation from the surface to height $R$ (distance $2R$ from the center):
$\frac{1}{2}mu^2 - \frac{GMm}{R} = \frac{1}{2}mv^2 - \frac{GMm}{2R}$
$\frac{1}{2}v^2 = \frac{1}{2}u^2 - \frac{GM}{2R} \Rightarrow v = \sqrt{u^2 - \frac{GM}{R}}$
$2$. At height $R$, the satellite (mass $m$) ejects a rocket (mass $m/10$). The remaining satellite (mass $9m/10$) enters a circular orbit at distance $2R$. The orbital velocity is $v_o = \sqrt{\frac{GM}{2R}}$.
$3$. By conservation of linear momentum in the radial and tangential directions:
Radial: $\frac{m}{10} v_r = m v = m \sqrt{u^2 - \frac{GM}{R}} \Rightarrow v_r = 10 \sqrt{u^2 - \frac{GM}{R}}$
Tangential: $\frac{m}{10} v_T = \frac{9m}{10} v_o = \frac{9m}{10} \sqrt{\frac{GM}{2R}} \Rightarrow v_T = 9 \sqrt{\frac{GM}{2R}}$
$4$. Kinetic energy of the rocket $(K_r = \frac{1}{2} (m/10) (v_r^2 + v_T^2))$:
$K_r = \frac{m}{20} \left( 100(u^2 - \frac{GM}{R}) + 81(\frac{GM}{2R}) \right)$
$K_r = \frac{m}{20} \left( 100u^2 - 100\frac{GM}{R} + 40.5\frac{GM}{R} \right) = \frac{m}{20} \left( 100u^2 - 59.5\frac{GM}{R} \right)$
$K_r = 5m \left( u^2 - 0.595 \frac{GM}{R} \right) = 5m \left( u^2 - \frac{119}{200} \frac{GM}{R} \right)$.