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

(A) The orbital speed of a planet at a distance $r$ from the sun is given by the formula: $v_{\text{orb}} = \sqrt{\frac{GM}{r}}$.
For planet $A$,the orbital speed is $v_A = v = \sqrt{\frac{GM}{r_A}}$.
For planet $B$,the orbital speed is $v_B = \sqrt{\frac{GM}{r_B}}$.
Substituting $r_B = 100 r_A$ into the expression for $v_B$:
$v_B = \sqrt{\frac{GM}{100 r_A}} = \frac{1}{\sqrt{100}} \sqrt{\frac{GM}{r_A}} = \frac{1}{10} \sqrt{\frac{GM}{r_A}}$.
Since $v = \sqrt{\frac{GM}{r_A}}$,we have $v_B = \frac{v}{10}$.

(C) From Kepler's third law,the time period $T$ is given by $T = 2 \pi \sqrt{\frac{r^3}{GM}}$.
Since the satellite is revolving close to the surface of the planet,we take the orbital radius $r = R$.
Thus,$T = 2 \pi \sqrt{\frac{R^3}{GM}} \dots (i)$.
The mass $M$ of the planet can be expressed in terms of its density $\rho$ and radius $R$ as $M = \text{Volume} \times \text{Density} = \frac{4}{3} \pi R^3 \rho \dots (ii)$.
Substituting the value of $M$ from equation $(ii)$ into equation $(i)$:
$T = 2 \pi \sqrt{\frac{R^3}{G \times \frac{4}{3} \pi R^3 \rho}}$
$T = 2 \pi \sqrt{\frac{3}{4 \pi G \rho}}$
$T = 2 \pi \times \frac{1}{2} \sqrt{\frac{3}{\pi G \rho}}$
$T = \sqrt{\frac{4 \pi^2 \times 3}{4 \pi G \rho}} = \sqrt{\frac{3 \pi}{\rho G}}$.

(D) For a planet to orbit a star in a circular path,the gravitational force provides the necessary centripetal force.
Given that the gravitational force $F_G \propto R^{-5/2}$.
The centripetal force is given by $F_c = m \omega^2 R$,where $\omega = \frac{2\pi}{T}$.
Thus,$F_c = m \left(\frac{4\pi^2}{T^2}\right) R$.
Equating the forces: $m \left(\frac{4\pi^2}{T^2}\right) R \propto R^{-5/2}$.
Since $m$ and $4\pi^2$ are constants,we have $\frac{R}{T^2} \propto R^{-5/2}$.
Rearranging for $T^2$: $T^2 \propto R \cdot R^{5/2} = R^{7/2}$.
Taking the square root on both sides: $T \propto R^{7/4}$.

(C) The gravitational force provides the necessary centripetal force for the satellite to revolve in a circular orbit:
$\frac{mv^2}{r} = \frac{GMm}{r^2}$
Solving for the orbital velocity $v$:
$v^2 = \frac{GM}{r} \implies v = \sqrt{\frac{GM}{r}}$
The angular momentum $L$ of the satellite is defined as $L = mvr$.
Substituting the expression for $v$:
$L = m \left( \sqrt{\frac{GM}{r}} \right) r$
$L = m \sqrt{GM} \cdot \sqrt{r} = m(GMr)^{1/2}$

(D) For a satellite orbiting close to the surface of a planet of radius $R$ and density $\rho$,the time period $T$ is given by $T = 2\pi \sqrt{\frac{R}{g}}$.
Since $g = \frac{GM}{R^2}$ and $M = \rho \cdot \frac{4}{3} \pi R^3$,we have $g = \frac{G \cdot \rho \cdot \frac{4}{3} \pi R^3}{R^2} = \frac{4}{3} G \pi \rho R$.
Substituting $g$ into the time period formula: $T = 2\pi \sqrt{\frac{R}{\frac{4}{3} G \pi \rho R}} = 2\pi \sqrt{\frac{3}{4 G \pi \rho}}$.
Thus,$T \propto \rho^{-1/2}$.

(D) Concept: Angular momentum $(L)$ is conserved because the gravitational force exerted by the sun on the planet is a central force, resulting in zero torque about the sun.
$L = mvr \sin(\theta) = \text{constant}$.
At the perihelion (the point closest to the sun), the distance $r$ is minimum.
Since $L = mvr$ (where $v$ is the orbital velocity perpendicular to the radius vector at perihelion), for a constant $L$, the velocity $v$ must be maximum when $r$ is minimum.
Kinetic energy $(K.E.)$ is given by $K.E. = \frac{1}{2}mv^2$.
Therefore, when the velocity $v$ is maximum, the kinetic energy is also maximum.
Looking at the diagram, point $C$ is the closest to the sun (perihelion).
Thus, the planet has the maximum kinetic energy at position $C$.

(A) For a satellite to orbit the Earth,the gravitational force provides the necessary centripetal force.
$m r \omega^{2} = \frac{G M m}{r^{2}}$
Here,$r$ is the orbital radius,$M$ is the mass of the Earth,and $\omega$ is the angular velocity.
Simplifying the equation,we get $r^{3} = \frac{G M}{\omega^{2}}$.
We know that the acceleration due to gravity on the Earth's surface is $g = \frac{G M}{R^{2}}$,which implies $G M = g R^{2}$.
Substituting $G M$ in the expression for $r^{3}$,we get $r^{3} = \frac{g R^{2}}{\omega^{2}}$.
Therefore,the radius of the orbit is $r = \left(\frac{g R^{2}}{\omega^{2}}\right)^{1 / 3}$.

(C) The orbital speed of a satellite is given by $v = \sqrt{\frac{GM}{r}}$,where $G$ is the gravitational constant,$M$ is the mass of the earth,and $r$ is the radius of the orbit.
Since both satellites are in the same orbit,$r$ is the same for both.
Therefore,the orbital speed $v$ is independent of the mass of the satellite.
Thus,$S_{1}$ and $S_{2}$ move with the same speed.
The time period is given by $T = \frac{2\pi r}{v}$,which is also independent of the mass of the satellite.
Potential energy $U = -\frac{GMm}{r}$ and kinetic energy $K = \frac{GMm}{2r}$ both depend on the mass $m$ of the satellite,so they are not equal for $S_{1}$ and $S_{2}$.

(A) The two particles of mass $m$ are moving in a circle of radius $r$. The distance between them is the diameter of the circle,which is $2r$.
The gravitational force between the two particles acts as the centripetal force required for circular motion.
The gravitational force is given by $F_g = \frac{G m m}{(2r)^2} = \frac{G m^2}{4r^2}$.
The centripetal force required for a particle of mass $m$ moving with speed $v$ in a circle of radius $r$ is $F_c = \frac{m v^2}{r}$.
Equating the two forces: $\frac{m v^2}{r} = \frac{G m^2}{4r^2}$.
Solving for $v^2$: $v^2 = \frac{G m}{4r}$.
Taking the square root,we get $v = \sqrt{\frac{G m}{4r}}$.

(A) The orbital speed $v$ 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 $v \propto \frac{1}{\sqrt{r}}$.
Let $v_A$ and $v_B$ be the speeds of satellites $A$ and $B$ with orbital radii $r_A = 4R$ and $r_B = R$ respectively.
Taking the ratio,we get $\frac{v_A}{v_B} = \sqrt{\frac{r_B}{r_A}}$.
Substituting the given values,$\frac{v_A}{6V} = \sqrt{\frac{R}{4R}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Therefore,$v_A = 6V \times \frac{1}{2} = 3V$.

(A) The orbital speed $v$ of a satellite revolving around a planet of mass $M$ at a distance $r$ is given by the formula $v = \sqrt{\frac{GM}{r}}$.
This implies that $v \propto \frac{1}{\sqrt{r}}$.
Given the radii of the orbits are $r_P = 3R$ and $r_Q = R$.
Let $v_P$ and $v_Q$ be the orbital speeds of satellites $P$ and $Q$ respectively.
Then,$\frac{v_Q}{v_P} = \sqrt{\frac{r_P}{r_Q}}$.
Substituting the given values: $\frac{v_Q}{2V} = \sqrt{\frac{3R}{R}} = \sqrt{3}$.
Therefore,$v_Q = 2V \times \sqrt{3} = 2\sqrt{3}V$.

(D) The energy required to raise a satellite to a height $h$ is the change in potential energy: $E_1 = U_f - U_i = -\frac{GMm}{R+h} - (-\frac{GMm}{R}) = GMm \left( \frac{1}{R} - \frac{1}{R+h} \right) = \frac{GMmh}{R(R+h)}$.
Using $g = \frac{GM}{R^2}$,we get $GM = gR^2$. Substituting this,$E_1 = \frac{gR^2mh}{R(R+h)} = \frac{mgh}{1 + h/R}$.
The energy required to put the satellite into orbit at height $h$ is the kinetic energy $E_2 = \frac{1}{2}mv_0^2$. Since $v_0 = \sqrt{\frac{GM}{R+h}}$,$E_2 = \frac{1}{2}m \left( \frac{GM}{R+h} \right) = \frac{GMm}{2(R+h)}$.
Substituting $GM = gR^2$,$E_2 = \frac{gR^2m}{2(R+h)} = \frac{mgR}{2(1 + h/R)}$.
Taking the ratio: $\frac{E_1}{E_2} = \frac{mgh / (1 + h/R)}{mgR / (2(1 + h/R))} = \frac{h}{R/2} = \frac{2h}{R}$.

(D) The total energy $E$ of a satellite of mass $m$ revolving in an orbit of radius $r$ around a planet of mass $M$ is given by $E = -\frac{GMm}{2r}$.
From this expression,we can see that $E \propto \frac{m}{r}$.
Given the mass ratio $\frac{m_A}{m_B} = \frac{3}{1}$ and the radius ratio $\frac{r_A}{r_B} = \frac{r}{4r} = \frac{1}{4}$.
Therefore,the ratio of the total energies is $\frac{E_A}{E_B} = \frac{m_A}{m_B} \times \frac{r_B}{r_A}$.
Substituting the given values: $\frac{E_A}{E_B} = \frac{3}{1} \times \frac{4r}{r} = \frac{12}{1}$.
Thus,the ratio is $12: 1$.

(B) Let $M_e$ be the mass of the earth,$m$ be the mass of the satellite,and $r = R_e + h$ be the orbital radius.
The potential energy $U$ of the satellite is given by $U = -\frac{GM_e m}{r}$.
The orbital velocity $v$ of the satellite is given by $v = \sqrt{\frac{GM_e}{r}}$.
The kinetic energy $K$ is given by $K = \frac{1}{2}mv^2 = \frac{1}{2}m \left(\frac{GM_e}{r}\right) = \frac{GM_e m}{2r}$.
Comparing the magnitudes,we have $K = \frac{1}{2} |U|$.
Therefore,the ratio of kinetic energy to the magnitude of potential energy is $K : |U| = 1 : 2$.

(C) The time period $T$ of a satellite orbiting just above the surface of a planet of radius $R$ is given by $T = 2 \pi \sqrt{\frac{R}{g}}$.
Squaring both sides,we get $T^2 = 4 \pi^2 \frac{R}{g}$.
We know that the acceleration due to gravity $g$ at the surface of a planet is given by $g = \frac{GM}{R^2}$.
Since the mass $M$ of the planet is $M = \text{Volume} \times \text{Density} = \frac{4}{3} \pi R^3 \rho$,we substitute $M$ into the expression for $g$:
$g = \frac{G}{R^2} \left( \frac{4}{3} \pi R^3 \rho \right) = \frac{4}{3} \pi \rho G R$.
Now,substitute this expression for $g$ into the equation for $T^2$:
$T^2 = 4 \pi^2 \frac{R}{\frac{4}{3} \pi \rho G R} = 4 \pi^2 \times \frac{3}{4 \pi \rho G} = \frac{3 \pi}{\rho G}$.
Rearranging the terms,we get $T^2 \rho = \frac{3 \pi}{G}$.

(B) The formula for the time period of a satellite at height $h$ above the Earth's surface is given by $T = 2 \pi \sqrt{\frac{(R+h)^3}{GM}}$.
Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$.
Substituting this into the formula,we get $T = 2 \pi \sqrt{\frac{(R+h)^3}{gR^2}}$.
Given that the height $h = R$,we substitute $h$ with $R$:
$T = 2 \pi \sqrt{\frac{(R+R)^3}{gR^2}} = 2 \pi \sqrt{\frac{(2R)^3}{gR^2}}$.
$T = 2 \pi \sqrt{\frac{8R^3}{gR^2}} = 2 \pi \sqrt{\frac{8R}{g}}$.
Simplifying the expression,we get $T = 2 \pi \cdot 2 \sqrt{\frac{2R}{g}} = 4 \pi \sqrt{\frac{2R}{g}}$.

(A) For a geostationary satellite,the gravitational force provides the necessary centripetal force for circular motion.
$m r \omega^2 = \frac{G M m}{r^2}$
Where $m$ is the mass of the satellite,$r$ is the orbital radius,$M$ is the mass of the Earth,and $\omega$ is the angular velocity.
Simplifying the equation: $\omega^2 = \frac{G M}{r^3}$.
We know that the acceleration due to gravity on the Earth's surface is $g = \frac{G M}{R^2}$,which implies $G M = g R^2$.
Substituting $G M$ in the expression for $\omega^2$:
$\omega^2 = \frac{g R^2}{r^3}$
Rearranging to solve for $r$:
$r^3 = \frac{g R^2}{\omega^2}$
$r = \left(\frac{g R^2}{\omega^2}\right)^{\frac{1}{3}}$

(D) The correct option is $D$.
Considering the force balance in the orbit:
$\frac{GMm}{r^2} = \frac{mv^2}{r}$
Here,the gravitational force provides the necessary centripetal force for circular motion.
Solving for the orbital velocity $v$:
$v^2 = \frac{GM}{r} \implies v = \sqrt{\frac{GM}{r}}$
The angular momentum $L$ of the satellite is given by the formula $L = mvr$.
Substituting the value of $v$:
$L = m \times \sqrt{\frac{GM}{r}} \times r$
$L = \sqrt{m^2 \times \frac{GM}{r} \times r^2}$
$L = \sqrt{GMm^2 r}$

(D) The orbital speed $v$ of a satellite at a distance $r$ from the center of a planet is given by the formula $v = \sqrt{\frac{GM}{r}}$.
From this relation,we can see that $v \propto \frac{1}{\sqrt{r}}$.
Given for satellite $A$: $r_A = 4R$ and $v_A = 3V$.
Given for satellite $B$: $r_B = R$.
Using the ratio: $\frac{v_B}{v_A} = \sqrt{\frac{r_A}{r_B}}$.
Substituting the values: $\frac{v_B}{3V} = \sqrt{\frac{4R}{R}} = \sqrt{4} = 2$.
Therefore,$v_B = 2 \times 3V = 6V$.
The correct option is $D$.

(C) Let the satellite revolve in a circular orbit of radius $r$ with a linear velocity $v$.
The centripetal force acting on the satellite is provided by the gravitational force due to the earth.
$\therefore F_{\text{centripetal}} = F_{\text{gravitational}}$
$\Rightarrow \frac{mv^2}{r} = \frac{GMm}{r^2}$
$\Rightarrow v = \sqrt{\frac{GM}{r}}$
The angular momentum $L$ of the satellite about the center of the orbit is given by $L = mvr$.
Substituting the value of $v$:
$L = m \left(\sqrt{\frac{GM}{r}}\right) r$
$L = m \sqrt{GM} \cdot \sqrt{r} = \sqrt{GMm^2r}$
$L = (GMm^2r)^{1/2}$

(D) For a geostationary satellite,the gravitational force provides the necessary centripetal force for its circular orbit.
Let $r$ be the radius of the orbit of the satellite and $M$ be the mass of the Earth.
The condition for circular motion is: $m \omega^2 r = \frac{GMm}{r^2}$.
Simplifying this,we get: $r^3 = \frac{GM}{\omega^2}$.
We know that the acceleration due to gravity at the Earth's surface is given by $g = \frac{GM}{R^2}$,which implies $GM = gR^2$.
Substituting $GM = gR^2$ into the expression for $r^3$,we get: $r^3 = \frac{gR^2}{\omega^2}$.
Therefore,the radius of the orbit is $r = \left[\frac{R^2 g}{\omega^2}\right]^{1/3}$.

(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 $r = R$,so the velocity is $V = \sqrt{\frac{GM}{R}}$.
For a satellite at an altitude $h = \frac{R}{2}$,the orbital radius is $r' = R + h = R + \frac{R}{2} = \frac{3R}{2}$.
The new orbital velocity $V'$ is given by $V' = \sqrt{\frac{GM}{r'}} = \sqrt{\frac{GM}{3R/2}} = \sqrt{\frac{2GM}{3R}}$.
Substituting $V = \sqrt{\frac{GM}{R}}$ into the expression,we get $V' = \sqrt{\frac{2}{3}} V$.

(B) The kinetic energy $(K.E.)$ of a satellite of mass $m$ orbiting a planet of mass $M$ at a distance $r$ from the center is given by $K.E. = \frac{GMm}{2r}$.
Here,$r$ is the distance from the center of the Earth,so $r = R_{earth} + h$.
For the first satellite at height $h_1 = R$,the orbital radius is $r_1 = R + R = 2R$.
For the second satellite at height $h_2 = 2R$,the orbital radius is $r_2 = R + 2R = 3R$.
Since both satellites have equal mass $m$,the kinetic energy is inversely proportional to the orbital radius: $K.E. \propto \frac{1}{r}$.
Therefore,the ratio of their kinetic energies is $\frac{(K.E.)_1}{(K.E.)_2} = \frac{r_2}{r_1} = \frac{3R}{2R} = \frac{3}{2}$.

(D) The motion of a satellite depends on its horizontal velocity $(v)$ relative to the critical velocity $(v_c)$ and escape velocity $(v_e)$.
If $v < v_c$,the satellite will fall back to Earth.
If $v = v_c$,the satellite will revolve in a circular orbit.
If $v_c < v < v_e$,the satellite will revolve in an elliptical orbit with the Earth at one of the foci.
If $v = v_e$,the satellite will escape the gravitational field of the Earth.
Therefore,if the horizontal velocity is greater than the critical velocity but less than the escape velocity,the satellite will revolve in an elliptical orbit.

(B) The critical velocity (orbital velocity) of a satellite revolving around the Earth at a distance '$r$' from the center is given by the formula:
$v = \sqrt{\frac{GM}{r}}$
where '$G$' is the gravitational constant and '$M$' is the mass of the Earth.
For satellite '$A$' with radius '$R$':
$v_{A} = \sqrt{\frac{GM}{R}}$
For satellite '$B$' with radius '$2R$':
$v_{B} = \sqrt{\frac{GM}{2R}}$
Taking the ratio of the two velocities:
$\frac{v_{A}}{v_{B}} = \frac{\sqrt{\frac{GM}{R}}}{\sqrt{\frac{GM}{2R}}} = \sqrt{\frac{GM}{R} \times \frac{2R}{GM}} = \sqrt{2}$
Therefore,the ratio $\frac{v_{A}}{v_{B}}$ is $\sqrt{2}: 1$.

(C) The orbital velocity of a satellite is given by $v = \sqrt{\frac{GM}{r}}$.
The time period of revolution is $T = \frac{2\pi r}{v} = 2\pi \sqrt{\frac{r^3}{GM}}$.
The frequency of revolution is $f = \frac{1}{T} = \frac{1}{2\pi} \sqrt{\frac{GM}{r^3}}$.
As observed,the frequency '$f$' depends only on the mass of the Earth '$M$' and the radius of the orbit '$r$'.
It is independent of the mass of the satellite '$m$'.
Therefore,the ratio of their frequencies is $1:1$.

(A) 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 gravitational constant,and $M$ is the mass of the planet.
From this expression,it is clear that the time period $T$ depends only on the mass of the planet $M$ and the orbital radius $r$.
It does not depend on the mass of the satellite $m$.
Therefore,the period of revolution is independent of the mass of the satellite.

(D) The kinetic energy $(K)$ of a satellite of mass $m$ revolving at a distance $r$ from the center of the Earth is given by $K = \frac{GMm}{2r}$.
Here,$r = R + h$,where $R$ is the radius of the Earth and $h$ is the height above the surface.
For satellite '$A$',$h_A = 2R$,so $r_A = R + 2R = 3R$.
For satellite '$B$',$h_B = 3R$,so $r_B = R + 3R = 4R$.
The ratio of kinetic energies is $\frac{K_A}{K_B} = \frac{r_B}{r_A} = \frac{4R}{3R} = \frac{4}{3}$.

(D) The orbital period $T$ of a satellite revolving around the Earth is given by the formula $T = 2 \pi \sqrt{\frac{r^3}{GM}}$,where $r$ is the orbital radius,$G$ is the gravitational constant,and $M$ is the mass of the Earth.
Since both satellites are revolving in the same orbit,their orbital radius $r$ is the same.
Because $T$ depends only on the orbital radius $r$ and the mass of the Earth $M$,and not on the mass of the satellite $m$,both satellites will have the same orbital period.
Therefore,the correct statement is that they have the same period.

(C) The time period $T$ of a satellite orbiting near the surface of a planet is given by $T = 2 \pi \sqrt{\frac{R^3}{GM}}$.
Here,$R$ is the radius of the planet and $M$ is its mass.
The mass $M$ of the planet can be expressed in terms of its density $\rho$ as $M = \rho \cdot V = \rho \cdot \frac{4}{3} \pi R^3$.
Substituting this value of $M$ into the time period formula:
$T = 2 \pi \sqrt{\frac{R^3}{G \cdot (\frac{4}{3} \pi R^3 \rho)}}$
$T = 2 \pi \sqrt{\frac{R^3 \cdot 3}{4 \pi G R^3 \rho}}$
$T = 2 \pi \sqrt{\frac{3}{4 \pi G \rho}}$
$T = \sqrt{\frac{4 \pi^2 \cdot 3}{4 \pi G \rho}}$
$T = \sqrt{\frac{3 \pi}{G \rho}}$

(D) We know that angular momentum $L = m v R$ ... $(i)$
For a planet revolving around the sun,the gravitational force provides the necessary centripetal force:
$\frac{m v^2}{R} = \frac{G M m}{R^2}$
$v^2 = \frac{G M}{R}$
$v = \sqrt{\frac{G M}{R}}$ ... (ii)
Substituting equation (ii) into $(i)$:
$L = m \times \sqrt{\frac{G M}{R}} \times R$
$L = m \sqrt{G M R}$
Since $m$,$G$,and $M$ are constants,we have:
$L \propto \sqrt{R}$
$L \propto R^{1/2}$
Comparing this with $L \propto R^n$,we get $n = 0.5$.

(B) According to Newton's laws of motion,an object moving in a circular path requires a centripetal force directed towards the center of the circle to maintain its motion.
Since the earth revolves around the sun in a nearly circular orbit,there must be a centripetal force acting on the earth directed towards the sun.
This force is provided by the gravitational attraction between the sun and the earth.
Therefore,the revolution of the earth around the sun is the evidence that such a force exists.

(C) For a particle moving in a circular orbit,the centripetal force is provided by the central attractive force.
Given $F = \frac{k}{r}$,the magnitude of the centripetal force is $F_c = m r \omega^2$.
Equating the two: $m r \omega^2 = \frac{k}{r}$.
Rearranging for angular velocity $\omega$: $\omega^2 = \frac{k}{m r^2}$.
Taking the square root: $\omega = \sqrt{\frac{k}{m}} \cdot \frac{1}{r}$.
Since the periodic time $T = \frac{2\pi}{\omega}$,we have $T = 2\pi \sqrt{\frac{m}{k}} \cdot r$.
Therefore,$T \propto r$.

(D) The angular momentum $L$ of a particle moving in a circular orbit is given by $L = mvr$,where $m$ is the mass,$v$ is the orbital velocity,and $r$ is the radius.
For a planet revolving around the sun,the gravitational force provides the necessary centripetal force:
$\frac{GMm}{R^2} = \frac{mv^2}{R}$
Solving for $v$,we get $v = \sqrt{\frac{GM}{R}}$.
Substituting this into the angular momentum formula:
$L = m \times \sqrt{\frac{GM}{R}} \times R$
$L = m \sqrt{GM} \times \sqrt{R}$
Since $m$,$G$,and $M$ are constants,we find that $L \propto \sqrt{R}$.
Therefore,the correct option is $D$.

(C) The angular momentum $L$ of a particle in circular motion is given by $L = mvr$,where $m$ is the mass of the earth,$v$ is its orbital velocity,and $r$ is the radius of the orbit $R$.
For a planet revolving around the sun,the centripetal force is provided by the gravitational force:
$\frac{mv^2}{R} = \frac{GMm}{R^2}$
Solving for $v$,we get $v = \sqrt{\frac{GM}{R}}$.
Substituting this into the expression for angular momentum:
$L = m \times \sqrt{\frac{GM}{R}} \times R$
$L = m \sqrt{GM} \times \frac{R}{\sqrt{R}}$
$L = m \sqrt{GM} \times \sqrt{R}$
Since $m$,$G$,and $M$ are constants,we find that $L \propto \sqrt{R}$.

(B) The orbital velocity $v$ of a satellite revolving around the Earth at a distance $r$ is given by $v = \sqrt{\frac{GM}{r}}$.
Angular momentum $L$ is defined as $L = mvr$.
Substituting the value of $v$ into the equation for $L$:
$L = m \left(\sqrt{\frac{GM}{r}}\right) r$
$L = m \sqrt{GM} \cdot \sqrt{r}$
$L = m \sqrt{GMr}$
$L = m(GMr)^{1/2}$.

(C) According to Kepler's third law of planetary motion,the square of the time period $T$ of a satellite is proportional to the cube of the orbital radius $R$,given by $T^2 \propto R^3$.
This implies $T = 2\pi \sqrt{\frac{R^3}{GM}}$,where $G$ is the gravitational constant and $M$ is the mass of the planet.
Note that the time period $T$ depends only on the radius of the orbit $R$ and the mass of the central planet $M$.
It is independent of the mass of the satellite $m$.
Since both satellites orbit the same planet at the same radius $R$,their time periods will be equal.
Therefore,the ratio of their periods of revolution is $1: 1$.

(C) In central force motion,the angular momentum of the body remains conserved.
Angular momentum is given by $L = \vec{r} \times \vec{p} = mvr \sin \theta$.
At points $A$ and $B$,the velocity vector is perpendicular to the position vector,so $\theta = 90^\circ$ and $\sin 90^\circ = 1$.
Thus,$L = mvr$.
Let $v_A$ and $v_B$ be the speeds of the Earth at $A$ and $B$,respectively,and $r_A = OA$ and $r_B = OB$ be the distances from the Sun.
By conservation of angular momentum:
$m v_A r_A = m v_B r_B$
$\Rightarrow \frac{v_A}{v_B} = \frac{r_B}{r_A} = \frac{OB}{OA}$
Given that $\frac{OB}{OA} = R$,we have $\frac{v_A}{v_B} = R$.
Therefore,the ratio of Earth's velocities at $A$ and $B$ is $R$.

(C) The period of revolution $T$ of a satellite orbiting very close to the Earth's surface is given by the formula:
$T = 2 \pi \sqrt{\frac{R_E}{g}}$
Given values:
$R_E = 6400 \ km = 6.4 \times 10^6 \ m$
$g = 10 \ ms^{-2}$
$\pi = 3.14$
Substituting these values into the formula:
$T = 2 \times 3.14 \times \sqrt{\frac{6.4 \times 10^6}{10}}$
$T = 6.28 \times \sqrt{6.4 \times 10^5} = 6.28 \times \sqrt{64 \times 10^4}$
$T = 6.28 \times 800 = 5024 \ s$
To convert the time into minutes:
$T = \frac{5024}{60} \ min \approx 83.73 \ min$
Therefore, the period of revolution is $83.73$ minutes.

(A) geostationary satellite orbits around the Earth above the equator such that it remains stationary as seen from the Earth.
This means the satellite completes one revolution around the Earth in the same time it takes for the Earth to complete one rotation about its axis.
The rotational period of the Earth is $24 \text{ h}$.
Therefore,the time period of a geostationary satellite is $24 \text{ h}$.

(D) The escape velocity from the surface of the earth is given by $v_e = \sqrt{\frac{2GM}{R}} = 11.2 \,km \,s^{-1}$.
The orbital velocity of a satellite at a height $h$ from the surface is given by $v_o = \sqrt{\frac{GM}{R+h}}$.
Given that the height $h = R$, the orbital velocity becomes $v_o = \sqrt{\frac{GM}{R+R}} = \sqrt{\frac{GM}{2R}}$.
We can express $v_o$ in terms of $v_e$ as follows:
$v_o = \sqrt{\frac{1}{4} \cdot \frac{2GM}{R}} = \frac{v_e}{2}$.
Substituting the value of $v_e = 11.2 \,km \,s^{-1}$:
$v_o = \frac{11.2}{2} = 5.6 \,km \,s^{-1}$.

(B) When the velocity of a body $(v)$ is greater than the orbital velocity $(v_o)$ but less than the escape velocity $(v_e)$,i.e.,$v_o < v < v_e$,the total energy of the body is negative.
For a negative total energy in a central gravitational field,the trajectory of the body is an ellipse with the center of the Earth at one of the foci.
If $v = v_e$,the total energy is zero,and the path becomes parabolic.
If $v > v_e$,the total energy is positive,and the path becomes hyperbolic.

(D) According to Kepler's Third Law,the time period $T$ of a satellite in a circular orbit of radius $a$ is given by $T = 2\pi \sqrt{\frac{a^3}{GM}}$,where $M$ is the mass of the planet.
We know that the acceleration due to gravity at the surface of the planet is $g = \frac{GM}{R^2}$,which implies $GM = gR^2$.
Substituting $GM$ into the time period formula: $T = 2\pi \sqrt{\frac{a^3}{gR^2}} = 2\pi a^{3/2} g^{-1/2} R^{-1}$.
Comparing this with the given expression $T \propto a^{3/2} g^x R^y$,we get $x = -1/2$ and $y = -1$.

(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 $(r)$:
$T^2 \propto r^3$
For the first satellite, the orbital radius is $r_1 = R$ (where $R$ is the radius of the planet).
For the second satellite, the height is $h = 3R$, so the orbital radius is $r_2 = R + 3R = 4R$.
Using the ratio:
$\left(\frac{T_2}{T_1}\right)^2 = \left(\frac{r_2}{r_1}\right)^3$
$\left(\frac{T_2}{T_1}\right)^2 = \left(\frac{4R}{R}\right)^3 = 4^3 = 64$
Taking the square root on both sides:
$\frac{T_2}{T_1} = \sqrt{64} = 8$
Given $T_1 = 80 \text{ minutes}$, we find:
$T_2 = 8 \times 80 \text{ minutes} = 640 \text{ minutes}$.

(B) According to the law of conservation of angular momentum,the angular momentum $(L)$ of a planet is constant.
$L = mvr \sin \theta$,where $\theta$ is the angle between the velocity vector and the radius vector.
However,the problem specifies the angle between the velocity vector and the major axis. Let $\phi$ be the angle between the velocity vector and the radius vector. The angular momentum is $L = mvr \sin \phi$.
From the geometry of an ellipse,the angle between the radius vector and the tangent (velocity vector) is related to the angle with the major axis. Given the standard approach for such problems where the angular momentum component perpendicular to the radius vector is $v \sin \phi$,the conservation law is $m v_A r_A \sin \phi_A = m v_B r_B \sin \phi_B$.
Given $r_A = 90 \times 10^6 \text{ km}$,$r_B = 60 \times 10^6 \text{ km}$,$\phi_A = 30^{\circ}$,and $\phi_B = 60^{\circ}$:
$\frac{v_A}{v_B} = \frac{r_B}{r_A} \times \frac{\sin \phi_B}{\sin \phi_A}$
$\frac{v_A}{v_B} = \frac{60 \times 10^6}{90 \times 10^6} \times \frac{\sin 60^{\circ}}{\sin 30^{\circ}}$
$\frac{v_A}{v_B} = \frac{2}{3} \times \frac{\sqrt{3}/2}{1/2} = \frac{2}{3} \times \sqrt{3} = \frac{2}{\sqrt{3}}$.

(A) For a stationary satellite,its orbital period $T$ must be equal to the rotational period of the planet about its axis. The angular velocity of the planet is $\omega = 2\pi / T$.
Since the angular velocity is halved $(\omega' = \omega / 2)$,the new period $T'$ becomes $2T$ because $T = 2\pi / \omega$.
According to Kepler's Third Law,the square of the orbital period is proportional to the cube of the orbital radius: $T^2 \propto r^3$.
Therefore,$(T'/T)^2 = (r'/r)^3$.
Substituting $T' = 2T$,we get $(2)^2 = (r'/r)^3$,which implies $4 = (r'/r)^3$.
Taking the cube root on both sides,$r'/r = 4^{1/3} = (2^2)^{1/3} = 2^{2/3}$.
Comparing this with $r'/r = 2^n$,we find $n = 2/3$.

(D) The orbital speed of a satellite at a distance $r$ from the center of the earth is given by $v = \sqrt{\frac{GM_E}{r}}$,where $r = R_E + h$.
For satellite $A$,the height is $h_A = 1.25 R_E$,so the distance from the center is $r_A = R_E + 1.25 R_E = 2.25 R_E$.
For satellite $B$,the height is $h_B = 19.25 R_E$,so the distance from the center is $r_B = R_E + 19.25 R_E = 20.25 R_E$.
The ratio of the orbital speeds is $\frac{v_A}{v_B} = \sqrt{\frac{r_B}{r_A}} = \sqrt{\frac{20.25 R_E}{2.25 R_E}} = \sqrt{\frac{2025}{225}} = \sqrt{9} = 3$.
Thus,the ratio is $3: 1$.

(B) satellite with a time period equal to the rotation of the Earth $(T = 24 \,h)$ is a geostationary satellite.
For a geostationary satellite,the orbital radius $r$ is given by $r = \left( \frac{T^2 GM_e}{4 \pi^2} \right)^{1/3}$.
Substituting the values $T = 86,400 \,s$,$G = 6.67 \times 10^{-11} \,Nm^2/kg^2$,and $M_e = 5.97 \times 10^{24} \,kg$,we get $r \approx 42,200 \,km$.
The altitude $h$ is given by $h = r - R_e$,where $R_e \approx 6,400 \,km$.
Thus,$h = 42,200 \,km - 6,400 \,km = 35,800 \,km$.
Therefore,the altitude is nearly $35,840 \,km$.

(B) Polar satellites orbit the Earth in a north-south direction as the Earth rotates beneath them. This allows them to scan the entire globe,making them ideal for remote sensing,meteorology,and environmental monitoring.