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

(B) The gravitational force provides the necessary centripetal force for the particle to move in a circular orbit.
For a circular orbit,the centripetal force is given by $F_c = \frac{mv^2}{R}$.
According to the problem,the gravitational force is $F_g \propto \frac{1}{R}$,which can be written as $F_g = \frac{K}{R}$ for some constant $K$.
Equating the two forces: $\frac{mv^2}{R} = \frac{K}{R}$.
Canceling $R$ from both sides,we get $mv^2 = K$.
Since $m$ and $K$ are constants,$v^2$ is constant,which means $v$ is constant.
Therefore,$v \propto R^0$.

(B) For a satellite orbiting a planet,the gravitational force acts as a central force directed towards the center of the planet.
Since the torque exerted by a central force about the center of the planet is zero $(\vec{\tau} = \vec{r} \times \vec{F} = 0)$,the angular momentum $(\vec{L})$ of the satellite remains constant according to the law of conservation of angular momentum.
Velocity,potential energy,and acceleration change depending on the distance from the planet in an elliptical orbit.

(B) According to Kepler's third law of planetary motion,the square of the time period $(T)$ of a satellite is directly proportional to the cube of the radius $(r)$ of its orbit.
Mathematically,$T^2 \propto r^3$ or $T \propto r^{3/2}$.
When the satellite is shifted towards the Earth,the orbital radius $(r)$ decreases.
Since $T \propto r^{3/2}$,a decrease in $r$ leads to a decrease in the time period $(T)$.

(C) In a satellite,both the chair and the person are in a state of free fall towards the Earth. Since both have the same acceleration (the acceleration due to gravity at that altitude),there is no relative motion between them. Consequently,the chair does not exert any contact force (normal force) on the person. Since the sensation of weight is due to the normal force exerted by a surface,the absence of this force leads to the feeling of weightlessness.

(C) The orbital speed of a satellite in a circular orbit is given by the formula $v = \sqrt{\frac{GM}{r}}$,where $G$ is the gravitational constant,$M$ is the mass of the planet,and $r$ is the radius of the orbit.
From this relation,we see that $v \propto \frac{1}{\sqrt{r}}$.
Therefore,the ratio of the speeds of satellites $A$ and $B$ is $\frac{v_B}{v_A} = \sqrt{\frac{r_A}{r_B}}$.
Given $r_A = 4R$ and $r_B = R$,we have $\frac{v_B}{v_A} = \sqrt{\frac{4R}{R}} = \sqrt{4} = 2$.
Given the speed of satellite $A$ is $v_A = 3v$,the speed of satellite $B$ is $v_B = 2 \times v_A = 2 \times 3v = 6v$.

(B) The gravitational force provides the necessary centripetal force for the satellite to orbit the Earth.
The gravitational force is given by $F_g = m g$,where $g = \frac{GM}{R^3}$ according to the modified condition.
Equating the centripetal force to the gravitational force:
$m \omega^2 R = m g$
$m \left( \frac{4\pi^2}{T^2} \right) R = m \left( \frac{GM}{R^3} \right)$
Simplifying the equation:
$\frac{4\pi^2}{T^2} R = \frac{GM}{R^3}$
$\frac{1}{T^2} = \frac{GM}{4\pi^2 R^4}$
Therefore,$T^2 \propto R^4$,which implies $T \propto R^2$.

(B) The orbital speed $v$ of a planet at a distance $r$ from the Sun is given by the formula $v = \sqrt{\frac{GM}{r}}$,where $G$ is the gravitational constant and $M$ is the mass of the Sun.
From this relation,we see that $v \propto \frac{1}{\sqrt{r}}$.
Since the orbital radius of Jupiter $(r_J)$ is greater than the orbital radius of Earth $(r_e)$,it follows that the orbital speed of Jupiter $(v_J)$ must be less than the orbital speed of Earth $(v_e)$.
Therefore,$v_J < v_e$.

(C) The earth moves around the sun in an elliptical path. According to the properties of an ellipse:
${r_1} = a(1 + e)$ (Aphelion distance)
${r_2} = a(1 - e)$ (Perihelion distance)
Adding these,we get ${r_1} + {r_2} = 2a$,so $a = \frac{{{r_1} + {r_2}}}{2}$.
Multiplying these,we get ${r_1}{r_2} = a^2(1 - e^2)$.
Since $b^2 = a^2(1 - e^2)$,we have ${r_1}{r_2} = b^2$.
The distance from the sun when the earth is perpendicular to the major axis is the semi-latus rectum $(l)$ of the ellipse.
The formula for the semi-latus rectum is $l = \frac{{{b^2}}}{a}$.
Substituting the values,$l = \frac{{{r_1}{r_2}}}{({r_1} + {r_2})/2} = \frac{{2{r_1}{r_2}}}{{{r_1} + {r_2}}}$.

(C) 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.,$T^2 \propto r^3$ or $T \propto r^{3/2}$.
For the first satellite,the distance from the center of the Earth is $r_1 = R$,where $R$ is the radius of the Earth. The time period is $T_1 = 83 \text{ minutes}$.
For the second satellite,the orbit is at a distance of $3R$ from the surface,so the distance from the center of the Earth is $r_2 = R + 3R = 4R$.
Using the ratio formula: $\frac{T_2}{T_1} = \left( \frac{r_2}{r_1} \right)^{3/2}$.
Substituting the values: $\frac{T_2}{83} = \left( \frac{4R}{R} \right)^{3/2} = (4)^{3/2}$.
Calculating the value: $(4)^{3/2} = (2^2)^{3/2} = 2^3 = 8$.
Therefore,$T_2 = 83 \times 8 = 664 \text{ minutes}$.

(A) The orbital velocity $v$ of a satellite at a distance $R_0$ from the centre of the Earth is given by $v = \sqrt{\frac{GM}{R_0}}$.
Angular momentum $L$ is defined as $L = mvr$.
Substituting the values,we get $L = m \times \sqrt{\frac{GM}{R_0}} \times R_0$.
Simplifying the expression,$L = m \sqrt{GM R_0}$.

(C) The gravitational force exerted by the Sun on a planet is a central force.
$A$ central force is always directed towards or away from a fixed point (the center of the Sun).
The torque $\vec{\tau}$ exerted by a central force about the center of force is given by $\vec{\tau} = \vec{r} \times \vec{F}$.
Since the force $\vec{F}$ is parallel to the position vector $\vec{r}$,the cross product $\vec{r} \times \vec{F} = 0$.
Because the torque is zero,the angular momentum $\vec{L}$ of the planet remains constant over time.
Therefore,the motion of planets in the solar system is governed by the law of conservation of angular momentum.

(B) According to Kepler's Third Law of planetary motion,the square of the time period $(T)$ is proportional to the cube of the semi-major axis $(r)$ of the orbit: $T^2 \propto r^3$.
This relationship can be expressed as $T = 2\pi \sqrt{\frac{r^3}{GM}}$,where $M$ is the mass of the central body (e.g.,Earth).
Note that the mass of the satellite $(m)$ does not appear in the formula for the time period.
Since the time period $(T)$ remains constant and the mass of the central body $(M)$ is unchanged,the orbital radius $(r)$ must also remain constant.
Therefore,the ratio of the orbital radii in the two cases is $1:1$.

(A) According to the law of conservation of angular momentum,the angular momentum of the comet remains constant throughout its orbit.
$L = mvr \sin(\theta) = \text{constant}$.
At the points of perihelion (nearest) and aphelion (farthest),the velocity vector is perpendicular to the radius vector,so $\theta = 90^{\circ}$.
Thus,$m v_{\text{near}} r_{\text{near}} = m v_{\text{far}} r_{\text{far}}$.
Given:
$r_{\text{near}} = 1.6 \times 10^{12} \, m$
$r_{\text{far}} = 8 \times 10^{12} \, m$
$v_{\text{near}} = 60 \, m/s$
Substituting these values into the equation:
$60 \times 1.6 \times 10^{12} = v_{\text{far}} \times 8 \times 10^{12}$
$v_{\text{far}} = \frac{60 \times 1.6}{8} = \frac{96}{8} = 12 \, m/s$.

(D) The time period $T$ of a satellite revolving around a planet is given by the formula: $T = 2\pi \sqrt{\frac{r^3}{GM}}$, where $r$ is the orbital radius, $G$ is the universal gravitational constant, and $M$ is the mass of the planet (Earth).
Note that the mass of the satellite (Moon) does not appear in this formula.
Since the time period is independent of the mass of the satellite, changing the mass of the Moon will have no effect on its orbital period.
Therefore, the period remains $29$ days.

(D) The orbital angular momentum $L$ of a satellite of mass $m$ revolving at a distance $r$ from the center of a planet of mass $M$ is given by $L = mvr$.
Since the orbital velocity $v = \sqrt{\frac{GM}{r}}$,we substitute this into the expression for $L$:
$L = m \sqrt{\frac{GM}{r}} \cdot r = m \sqrt{GMr}$.
From this expression,we can see that $L \propto \sqrt{r}$.
Let $L_1 = L$ at distance $r_1 = r$,and $L_2$ be the new angular momentum at distance $r_2 = 16r$.
Then,$\frac{L_2}{L_1} = \sqrt{\frac{r_2}{r_1}} = \sqrt{\frac{16r}{r}} = \sqrt{16} = 4$.
Therefore,$L_2 = 4L_1 = 4L$.

(A) For a moon orbiting a planet in a circular orbit,the gravitational force provides the necessary centripetal force.
Let $M$ be the mass of the planet and $m$ be the mass of the moon.
The gravitational force is $F_g = \frac{GMm}{R^2}$.
The centripetal force is $F_c = m\omega^2 R$,where $\omega = \frac{2\pi}{T}$.
Equating the two forces: $m\omega^2 R = \frac{GMm}{R^2}$.
Substituting $\omega = \frac{2\pi}{T}$: $m \left(\frac{2\pi}{T}\right)^2 R = \frac{GMm}{R^2}$.
Simplifying the equation: $\frac{4\pi^2}{T^2} R = \frac{GM}{R^2}$.
Solving for $M$: $M = \frac{4\pi^2 R^3}{GT^2} = 4\pi^2 R^3 G^{-1} T^{-2}$.

(C) The orbital velocity of a satellite revolving around the Earth at a distance $r$ from the center of the Earth is given by the formula: $v = \sqrt{\frac{GM}{r}}$,where $G$ is the universal gravitational constant and $M$ is the mass of the Earth.
From this relation,it is clear that the orbital velocity $v$ is inversely proportional to the square root of the radius $r$,i.e.,$v \propto \frac{1}{\sqrt{r}}$.
Given that $r_1 > r_2$,it follows that $\sqrt{r_1} > \sqrt{r_2}$.
Therefore,$\frac{1}{\sqrt{r_1}} < \frac{1}{\sqrt{r_2}}$,which implies $v_1 < v_2$.

(B) The centripetal force required for the circular motion of the planet is provided by the gravitational force of attraction.
The gravitational force is given as $F_g \propto R^{-5/2}$.
The centripetal force is given by $F_c = m \omega^2 R = m \left( \frac{2\pi}{T} \right)^2 R = \frac{4\pi^2 m R}{T^2}$.
Equating the two forces: $\frac{m R}{T^2} \propto R^{-5/2}$.
Rearranging for $T^2$: $\frac{1}{T^2} \propto \frac{R^{-5/2}}{R} = R^{-7/2}$.
Therefore,$T^2 \propto R^{7/2}$.

(A) The gravitational force exerted by the Earth on the satellite always acts towards the centre of the Earth. According to Newton's second law,$F = ma$,the acceleration of the satellite is always directed towards the centre of the Earth.
Since the gravitational force is a central force,the torque acting on the satellite about the centre of the Earth is zero. Therefore,the angular momentum $L$ of the satellite remains constant in both magnitude and direction.
According to the law of conservation of energy,in the absence of non-conservative forces,the total mechanical energy of the satellite remains constant throughout its orbit.
Since the distance $r$ of the satellite from the Earth varies in an elliptical orbit,the orbital velocity $v$ must change to conserve angular momentum $(L = mvr \sin \theta)$. Consequently,the linear momentum $p = mv$ is not constant.

(A) The gravitational force $F$ is given by $F \propto \frac{1}{R^n}$.
For a planet in a circular orbit,this force provides the necessary centripetal force:
$F = m\omega^2 R = m\left( \frac{4\pi^2}{T^2} \right) R$.
Equating the two expressions:
$m\left( \frac{4\pi^2}{T^2} \right) R \propto \frac{1}{R^n}$.
Since $m$ and $4\pi^2$ are constants,we have:
$\frac{R}{T^2} \propto \frac{1}{R^n}$.
Rearranging for $T^2$:
$T^2 \propto R \cdot R^n = R^{n+1}$.
Taking the square root on both sides:
$T \propto R^{\left( \frac{n+1}{2} \right)}$.

(C) According to Kepler's third law,the time period $T$ of a satellite in a circular orbit is related to the orbital radius $r$ by the relation $T \propto r^{3/2}$.
Taking the natural logarithm on both sides: $\ln T = \frac{3}{2} \ln r$.
Differentiating both sides,we get the fractional change: $\frac{\Delta T}{T} = \frac{3}{2} \frac{\Delta r}{r}$.
Given that the radius increases from $R$ to $(1.01)R$,the percentage change in radius is $\frac{\Delta r}{r} \times 100 = 1\%$.
Therefore,the percentage change in the time period is $\frac{\Delta T}{T} \times 100 = \frac{3}{2} \times (1\%) = 1.5\%$.
Thus,the period of the second satellite is larger by approximately $1.5\%$.

(C) 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.,$T^2 \propto r^3$.
For a geostationary satellite,$T_1 = 24\, h$ and $r_1 = 36000\, km$.
For a satellite orbiting near the Earth's surface,$r_2 \approx R_{\text{Earth}} = 6400\, km$.
Using the ratio formula: $\frac{T_2}{T_1} = \left( \frac{r_2}{r_1} \right)^{3/2}$.
Substituting the values: $T_2 = 24 \times \left( \frac{6400}{36000} \right)^{3/2}$.
$T_2 = 24 \times \left( \frac{64}{360} \right)^{3/2} = 24 \times \left( \frac{8}{18.97} \right)^3 \approx 24 \times (0.177)^{3/2} \approx 24 \times 0.075 \approx 1.8\, h$.
Rounding to the nearest integer,the time period is approximately $2\, hours$.

(A) The gravitational force between the planet and the satellite provides the necessary centripetal force for the circular motion.
$F_g = F_c$
$\frac{G M m}{r^2} = m r \omega^2$
Since the angular velocity $\omega = \frac{2\pi}{T}$,we substitute this into the equation:
$\frac{G M m}{r^2} = m r \left( \frac{2\pi}{T} \right)^2$
$\frac{G M}{r^2} = r \frac{4\pi^2}{T^2}$
Solving for $M$:
$M = \frac{4\pi^2 r^3}{G T^2}$

(B) The solar system consists of $8$ planets. Mercury is the closest planet to the Sun,followed by Venus,Earth,Mars,Jupiter,Saturn,Uranus,and Neptune. Therefore,the correct option is $B$.

(D) Let the masses of the two stars be $m_1 = m$ and $m_2 = 2m$,separated by a distance $l$.
The distance of the centre of mass from the star of mass $m$ is $r_1 = \frac{m_2 l}{m_1 + m_2} = \frac{2ml}{3m} = \frac{2l}{3}$.
The gravitational force between the stars provides the necessary centripetal force for rotation.
$\frac{G(m)(2m)}{l^2} = m \omega^2 r_1 = m \left(\frac{2\pi}{T}\right)^2 \left(\frac{2l}{3}\right)$.
$\frac{2Gm^2}{l^2} = m \frac{4\pi^2}{T^2} \frac{2l}{3}$.
$\frac{Gm}{l^2} = \frac{4\pi^2 l}{3T^2}$.
$T^2 = \frac{4\pi^2 l^3}{3Gm}$.
$T = \sqrt{\frac{4\pi^2 l^3}{3Gm}} = 2\pi \sqrt{\frac{l^3}{3Gm}}$.
Thus,$T \propto m^{-1/2}$.

(B) The mass $M$ of the galaxy can be calculated using the formula for orbital motion: $M = \frac{v^2 r}{G}$.
Given:
Speed $v = 250 \, km/s = 2.5 \times 10^5 \, m/s$.
Radius $r = 3 \times 10^4 \, \text{light years}$.
Since $1 \, \text{light year} \approx 9.46 \times 10^{15} \, m$, we have $r = 3 \times 10^4 \times 9.46 \times 10^{15} \, m \approx 2.84 \times 10^{20} \, m \approx 3 \times 10^{20} \, m$.
Gravitational constant $G = 6.67 \times 10^{-11} \, N \cdot m^2/kg^2$.
Substituting these values into the formula:
$M = \frac{(2.5 \times 10^5)^2 \times (3 \times 10^{20})}{6.67 \times 10^{-11}}$
$M = \frac{6.25 \times 10^{10} \times 3 \times 10^{20}}{6.67 \times 10^{-11}}$
$M = \frac{18.75 \times 10^{30}}{6.67 \times 10^{-11}} \approx 2.81 \times 10^{41} \, kg$.
Rounding to the nearest given option, we get $M \approx 3 \times 10^{41} \, kg$.

(B) For a binary star system, the sum of masses is given by Kepler's Third Law in units of solar masses, astronomical units, and years: $M_1 + M_2 = \frac{r^3}{T^2}$.
Given $r = 30 \text{ AU}$ and $T = 30 \text{ years}$, we have $M_1 + M_2 = \frac{30^3}{30^2} = 30 \text{ solar masses}$ .....$(i)$.
By the definition of the centre of mass, $M_1 r_1 = M_2 r_2$. Given that one star is $5$ times farther from the centre of mass than the other, let $r_2 = 5r_1$. Thus, $M_1 r_1 = M_2 (5r_1)$, which implies $M_1 = 5M_2$ .....$(ii)$.
Substituting $(ii)$ into $(i)$: $5M_2 + M_2 = 30 \Rightarrow 6M_2 = 30 \Rightarrow M_2 = 5 \text{ solar masses}$.
Then $M_1 = 5(5) = 25 \text{ solar masses}$.
Therefore, the masses are $25$ and $5$ solar masses.

(B) From the conservation of energy,the total energy at the perihelion and aphelion is equal:
$\frac{1}{2}mv_1^2 - \frac{GM_Sm}{r_1} = \frac{1}{2}mv_2^2 - \frac{GM_Sm}{r_2}$
Angular momentum is conserved,so $L = mv_1r_1 = mv_2r_2$,which implies $v_2 = v_1\frac{r_1}{r_2}$.
Substituting $v_2$ into the energy equation:
$\frac{1}{2}mv_1^2 - \frac{GM_Sm}{r_1} = \frac{1}{2}m\left(v_1\frac{r_1}{r_2}\right)^2 - \frac{GM_Sm}{r_2}$
Solving for $v_1$:
$v_1^2 \left(1 - \frac{r_1^2}{r_2^2}\right) = 2GM_S \left(\frac{1}{r_1} - \frac{1}{r_2}\right)$
$v_1^2 \left(\frac{r_2^2 - r_1^2}{r_2^2}\right) = 2GM_S \left(\frac{r_2 - r_1}{r_1r_2}\right)$
$v_1^2 \left(\frac{(r_2 - r_1)(r_2 + r_1)}{r_2^2}\right) = 2GM_S \left(\frac{r_2 - r_1}{r_1r_2}\right)$
$v_1 = \sqrt{\frac{2GM_Sr_2}{r_1(r_1 + r_2)}}$
Finally,the angular momentum $L = mv_1r_1 = m \sqrt{\frac{2GM_Sr_2}{r_1(r_1 + r_2)}} \cdot r_1 = \sqrt{\frac{2GM_Sm^2r_1r_2}{r_1 + r_2}}$.

(A) The relationship between areal velocity $(\frac{dA}{dt})$ and angular momentum $(L)$ for a satellite of mass $(m)$ is given by the formula: $\frac{dA}{dt} = \frac{L}{2m}$.
Given:
Mass $(m) = 500 \ kg$
Areal velocity $(\frac{dA}{dt}) = 4 \times 10^4 \ m^2s^{-1}$.
Rearranging the formula to solve for angular momentum $(L)$:
$L = 2m \times \frac{dA}{dt}$.
Substituting the given values:
$L = 2 \times 500 \times (4 \times 10^4)$.
$L = 1000 \times 4 \times 10^4$.
$L = 4 \times 10^7 \ kg \ m^2s^{-1}$.
Therefore,the angular momentum is $4 \times 10^7 \ kg \ m^2s^{-1}$.

(A) According to the law of conservation of angular momentum,the angular momentum of the planet remains constant at all points in its orbit.
$L = mvr \sin \theta = \text{constant}$.
At the perihelion (closest) and aphelion (farthest) points,the velocity vector is perpendicular to the radius vector,so $\theta = 90^\circ$.
Therefore,$m v_{\max} r_{\min} = m v_{\min} r_{\max}$.
Given:
$r_{\min} = 1.6 \times 10^{12} \ m$
$v_{\max} = 60 \ m/s$
$r_{\max} = 8 \times 10^{12} \ m$
Substituting the values:
$60 \times (1.6 \times 10^{12}) = v_{\min} \times (8 \times 10^{12})$
$v_{\min} = \frac{60 \times 1.6 \times 10^{12}}{8 \times 10^{12}}$
$v_{\min} = 60 \times 0.2 = 12 \ m/s$.

(D) According to Kepler's third law of planetary motion,the square of the time period $(T)$ is directly proportional to the cube of the orbital radius $(r)$: $T^2 \propto r^3$.
Therefore,the ratio of the time periods is given by $\frac{T_A}{T_B} = \left( \frac{r_A}{r_B} \right)^{3/2}$.
Given $r_A = r$ and $r_B = 2r$,we substitute these values into the formula:
$\frac{T_A}{T_B} = \left( \frac{r}{2r} \right)^{3/2} = \left( \frac{1}{2} \right)^{3/2} = \frac{1}{2^{3/2}} = \frac{1}{2\sqrt{2}}$.
Thus,the ratio of the time periods is $1 : 2\sqrt{2}$.

(C) According to the law of conservation of angular momentum,the angular momentum of the planet remains constant throughout its orbit.
At the closest point (perihelion),the velocity vector is perpendicular to the position vector,so the angular momentum is $L_1 = m v_1 d_1$.
At the farthest point (aphelion),the velocity vector is also perpendicular to the position vector,so the angular momentum is $L_2 = m v_2 d_2$.
Since $L_1 = L_2$,we have $m v_1 d_1 = m v_2 d_2$.
By simplifying this equation,we get $v_1 d_1 = v_2 d_2$.
Therefore,the velocity at the farthest point is $v_2 = \frac{d_1 v_1}{d_2}$.

(B) The orbital velocity $v$ of a satellite at a distance $r$ from the center of the Earth is given by the formula: $v = \sqrt{\frac{GM}{r}}$.
From this relation,we can see that $v \propto \frac{1}{\sqrt{r}}$.
Therefore,the ratio of the velocities of satellites $B$ and $A$ is: $\frac{v_B}{v_A} = \sqrt{\frac{r_A}{r_B}}$.
Given $r_A = 4R$,$r_B = R$,and $v_A = 3V$,we substitute these values into the equation:
$\frac{v_B}{3V} = \sqrt{\frac{4R}{R}} = \sqrt{4} = 2$.
Thus,$v_B = 2 \times 3V = 6V$.

(B) For a circular orbit,the gravitational force provides the necessary centripetal force.
$F = \frac{mv^2}{R}$
Given that $F \propto \frac{1}{R}$,we can write $F = \frac{k}{R}$ for some constant $k$.
Equating the two expressions:
$\frac{mv^2}{R} = \frac{k}{R}$
$mv^2 = k$
$v^2 = \frac{k}{m}$
Since $k$ and $m$ are constants,$v^2$ is constant,which means $v$ is independent of $R$.
Therefore,$v \propto R^0$.

(B) The time period $T$ of a satellite at a height $h$ above the Earth's surface is given by the formula:
$T = 2\pi \sqrt {\frac{{(R + h)^3}}{{GM}}}$
Given that the height $h = R$,we substitute this into the equation:
$T = 2\pi \sqrt {\frac{{(R + R)^3}}{{GM}}}$
Since the acceleration due to gravity at the surface is $g = \frac{GM}{R^2}$,we can write $GM = gR^2$:
$T = 2\pi \sqrt {\frac{{(2R)^3}}{{gR^2}}}$
$T = 2\pi \sqrt {\frac{{8R^3}}{{gR^2}}}$
$T = 2\pi \sqrt {\frac{{8R}}{g}}$
$T = 2\pi \cdot 2\sqrt{2} \sqrt {\frac{R}{g}}$
$T = 4\sqrt{2} \pi \sqrt {\frac{R}{g}}$

(D) The angular momentum $L$ of a satellite of mass $m$ orbiting a planet of mass $M$ at a distance $r$ is given by $L = mvr$.
Since the orbital velocity $v = \sqrt{\frac{GM}{r}}$,we have $L = m \sqrt{\frac{GM}{r}} \cdot r = \sqrt{m^2 GMr}$.
This shows that $L \propto \sqrt{r}$.
Given the initial distance $r_1 = r$ and final distance $r_2 = 16r$,we can write the ratio of angular momenta as:
$\frac{L_2}{L_1} = \sqrt{\frac{r_2}{r_1}} = \sqrt{\frac{16r}{r}} = \sqrt{16} = 4$.
Therefore,the new angular momentum $L_2 = 4L_1 = 4L$.

(B) The orbital speed of a satellite around the earth is given by $v = \sqrt{\frac{GM}{r}}$.
For satellite $A$,the radius is $r_A = 4R$ and the speed is $v_A = 3V$.
Thus,$v_A = \sqrt{\frac{GM}{4R}} = 3V$ ... $(i)$
For satellite $B$,the radius is $r_B = R$ and the speed is $v_B$.
Thus,$v_B = \sqrt{\frac{GM}{R}}$ ... (ii)
Dividing equation (ii) by equation $(i)$,we get:
$\frac{v_B}{v_A} = \frac{\sqrt{GM/R}}{\sqrt{GM/4R}} = \sqrt{\frac{4R}{R}} = \sqrt{4} = 2$.
Therefore,$v_B = 2 \times v_A = 2 \times 3V = 6V$.
The speed of satellite $B$ is $6V$.

(C) According to Kepler's second law of planetary motion,the areal velocity of a planet remains constant. This is equivalent to the conservation of angular momentum.
Since the gravitational force is a central force,the torque acting on the planet about the sun is zero.
Therefore,the angular momentum $L$ is conserved: $L = m v_1 d_1 = m v_2 d_2$.
Here,$m$ is the mass of the planet.
Canceling $m$ from both sides,we get $v_1 d_1 = v_2 d_2$.
Solving for $v_2$,we get $v_2 = \frac{d_1 v_1}{d_2}$.

(C) 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.,$T^2 \propto r^3$ or $T \propto r^{3/2}$.
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,$h_1 = 5R$,so $r_1 = R + 5R = 6R$. The time period $T_1 = 24 \, hr$.
For the second satellite,$h_2 = 2R$,so $r_2 = R + 2R = 3R$.
Using the ratio formula: $\frac{T_2}{T_1} = \left( \frac{r_2}{r_1} \right)^{3/2}$.
Substituting the values: $\frac{T_2}{24} = \left( \frac{3R}{6R} \right)^{3/2} = \left( \frac{1}{2} \right)^{3/2} = \frac{1}{2\sqrt{2}}$.
Therefore,$T_2 = \frac{24}{2\sqrt{2}} = \frac{12}{\sqrt{2}} = 6\sqrt{2} \, hr$.

(A) According to the law of conservation of angular momentum,the angular momentum of the planet remains constant throughout its orbit.
At the closest point (perihelion),the velocity vector is perpendicular to the position vector,so the angular momentum is $L_1 = m v_1 r_1$.
At the farthest point (aphelion),the velocity vector is also perpendicular to the position vector,so the angular momentum is $L_2 = m v_2 r_2$.
Since $L_1 = L_2$,we have $m v_1 r_1 = m v_2 r_2$.
Canceling the mass $m$ from both sides,we get $v_1 r_1 = v_2 r_2$.
Therefore,the ratio of the velocities is $\frac{v_1}{v_2} = \frac{r_2}{r_1}$.

(B) The orbital speed of a satellite is given by the formula $v_0 = \sqrt{\frac{GM}{R+h}}$.
Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$.
Substituting this into the orbital speed formula,we get $v_0 = \sqrt{\frac{gR^2}{R+h}} = R \sqrt{\frac{g}{R+h}}$.
Given values: $R = 6.38 \times 10^6 \, m$,$h = 0.25 \times 10^6 \, m$,and $g = 9.8 \, m s^{-2}$.
Total orbital radius $r = R + h = (6.38 + 0.25) \times 10^6 \, m = 6.63 \times 10^6 \, m$.
Now,$v_0 = \sqrt{\frac{9.8 \times (6.38 \times 10^6)^2}{6.63 \times 10^6}} = \sqrt{\frac{9.8 \times 40.7044 \times 10^{12}}{6.63 \times 10^6}} = \sqrt{60.16 \times 10^6} \approx 7.756 \times 10^3 \, m s^{-1}$.
Thus,$v_0 \approx 7.76 \, km s^{-1}$.

(A) Two heavenly bodies interacting via gravitational force form a binary star system.
According to the laws of mechanics,in a two-body system,both bodies revolve around their common centre of mass.
The gravitational force between them provides the necessary centripetal force for both bodies to maintain their respective orbits.
Therefore,the correct option is $A$.

(A) The time period $T$ of a satellite orbiting very close to a planet is given by the formula:
$T = 2\pi \sqrt{\frac{R^3}{GM}}$
Since the mass of the planet $M = \rho V = \rho \left( \frac{4}{3}\pi R^3 \right)$,we substitute this into the formula:
$T = 2\pi \sqrt{\frac{R^3}{G \cdot \rho \cdot \frac{4}{3}\pi R^3}}$
$T = 2\pi \sqrt{\frac{3}{4\pi G \rho}}$
From this expression,it is clear that the time period $T$ depends only on the density $\rho$ of the planet and is independent of the radius $R$ of the planet.
Since both planets have the same density $\rho$,the time period of the satellite will remain the same,i.e.,$T$.

(B) The orbital velocity of a satellite orbiting close to the Earth's surface is given by $v_o = \sqrt{gR_e}$.
If the velocity of the satellite exceeds this value,it will no longer remain in a stable circular orbit.
Specifically,if the velocity reaches the escape velocity $v_e = \sqrt{2gR_e}$,the satellite will escape the Earth's gravitational pull.
Therefore,for a stable circular orbit,the maximum velocity is limited by the condition where the gravitational force provides the necessary centripetal force.
Thus,the maximum possible velocity for a stable orbit is $v = \sqrt{gR_e}$.

(C) Inside an artificial satellite,both the satellite and the man are in a state of free fall towards the Earth. The gravitational force exerted by the Earth on the man provides the necessary centripetal force required for his circular motion in orbit. Since the man and the satellite are accelerating at the same rate towards the Earth,the normal reaction force exerted by the floor of the satellite on the man is zero. This state is perceived as weightlessness. Therefore,the gravitational force is equal to the centripetal force required for the orbit.

(D) The angular momentum $L$ of a particle in circular motion is given by $L = mvr$.
Since $v = r\omega$ and $\omega = \frac{2\pi}{T}$,we have $L = m(r\omega)r = m\omega r^2 = m \left( \frac{2\pi}{T} \right) r^2$.
Given: $m = 6 \times 10^{24} \ kg$,$r = 1.5 \times 10^8 \ km = 1.5 \times 10^{11} \ m$,$T = 3.14 \times 10^7 \ s$.
Substituting the values:
$L = \frac{6 \times 10^{24} \times 2 \times 3.14 \times (1.5 \times 10^{11})^2}{3.14 \times 10^7}$
$L = \frac{6 \times 10^{24} \times 2 \times 2.25 \times 10^{22}}{10^7}$
$L = 27 \times 10^{39} = 2.7 \times 10^{40} \ kg \ m^2/s$.

(B) The satellite is revolving around the earth because the required centripetal force is provided by the earth's gravitational pull.
If the gravitational force suddenly disappears,the satellite is no longer subject to any centripetal force.
According to Newton's first law of motion,an object in motion will continue to move in a straight line with constant velocity unless acted upon by an external force.
Therefore,the satellite will continue to move with its instantaneous velocity $V$ in a direction tangential to the circular path at the point where the force disappeared.

(D) The gravitational force provides the necessary centripetal force for the satellite's circular motion:
$\frac{GMm}{r^2} = m\omega_0^2 r$
From this,we can find the product of the gravitational constant and the mass of the planet:
$GM = \omega_0^2 r^3$
The acceleration due to gravity $g$ on the surface of the planet is given by:
$g = \frac{GM}{R^2}$
Substituting the value of $GM$ into the equation for $g$:
$g = \frac{\omega_0^2 r^3}{R^2}$