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

(A) Initially,the body of mass $m$ is moving in a circular orbit of radius $R$. So,it must be moving with orbital speed $v_{0} = \sqrt{\frac{GM}{R}}$.
After the collision,let the combined mass move with speed $v_{1}$.
Using the conservation of linear momentum:
$m v_{0} + \frac{m}{2} \left(\frac{v_{0}}{2}\right) = \left(m + \frac{m}{2}\right) v_{1}$
$m v_{0} + \frac{m v_{0}}{4} = \frac{3m}{2} v_{1}$
$\frac{5m v_{0}}{4} = \frac{3m}{2} v_{1}$
$v_{1} = \frac{5}{4} \times \frac{2}{3} v_{0} = \frac{5}{6} v_{0}$.
Since the speed after the collision $(v_{1} = \frac{5}{6} v_{0})$ is not equal to the orbital speed $(v_{0})$,the motion cannot be circular.
Since the velocity remains tangential,it cannot fall vertically towards the planet.
Also,the speed after the collision is less than the escape speed $(v_{e} = \sqrt{2} v_{0})$,so the body cannot escape the gravitational field.
Therefore,the combined body will move in an elliptical orbit around the planet.

(C) Given $k = 10^{-13} \; s^{2} \; m^{-3}$.
To convert $k$ into $d^{2} \; km^{-3}$:
$1 \; s = \frac{1}{24 \times 60 \times 60} \; d = \frac{1}{86400} \; d$
$1 \; m = 10^{-3} \; km$
Substituting these values into $k$:
$k = 10^{-13} \times (\frac{1}{86400})^{2} \; d^{2} \times (10^{-3})^{-3} \; km^{-3}$
$k = 10^{-13} \times \frac{1}{7.465 \times 10^{9}} \times 10^{9} \; d^{2} \; km^{-3}$
$k \approx 1.33 \times 10^{-14} \; d^{2} \; km^{-3}$
Now,using $T^{2} = k(R_{E}+h)^{3}$ where $R_{E}+h = 3.84 \times 10^{5} \; km$:
$T^{2} = (1.33 \times 10^{-14}) \times (3.84 \times 10^{5})^{3}$
$T^{2} = (1.33 \times 10^{-14}) \times (56.623 \times 10^{15})$
$T^{2} \approx 753.08$
$T = \sqrt{753.08} \approx 27.4 \; d$.
The closest option is $27.3 \; d$.

(N/A) Mass of our galaxy, $M = 2.5 \times 10^{11} \times (2.0 \times 10^{30} \; kg) = 5 \times 10^{41} \; kg$.
Radius of the orbit, $r = 50,000 \; ly = 5 \times 10^{4} \times 9.46 \times 10^{15} \; m = 4.73 \times 10^{20} \; m$.
Using Kepler's Third Law for the time period $T$ of a star revolving around the galactic centre:
$T = \sqrt{\frac{4 \pi^{2} r^{3}}{G M}}$.
Substituting the values:
$T = \sqrt{\frac{4 \times (3.14)^{2} \times (4.73 \times 10^{20})^{3}}{6.67 \times 10^{-11} \times 5 \times 10^{41}}}$.
$T = \sqrt{\frac{39.48 \times 105.82 \times 10^{60}}{33.35 \times 10^{30}}} = \sqrt{125.27 \times 10^{30}} \approx 1.12 \times 10^{16} \; s$.
Converting seconds to years:
$T = \frac{1.12 \times 10^{16}}{365.25 \times 24 \times 3600} \approx 3.55 \times 10^{8} \; \text{years}$.

(C, F) No. The linear speed changes as the distance from the Sun changes.
$(b)$ No. The angular speed changes according to Kepler's second law.
$(c)$ Yes. Since the gravitational force is a central force,the torque acting on the comet is zero,so angular momentum is conserved.
$(d)$ No. Kinetic energy changes as the linear speed changes.
$(e)$ No. Potential energy depends on the distance from the Sun,which varies in an elliptical orbit.
$(f)$ Yes. The total mechanical energy of the comet remains constant in the gravitational field of the Sun.

(N/A) $1$. Select two points $F_{1}$ and $F_{2}$.
$2$. Fix the ends of a string at $F_{1}$ and $F_{2}$.
$3$. Using the tip of a pencil,stretch the string taut and draw a curve by moving the pencil while keeping the string taut throughout the motion.
$4$. The resulting closed curve is called an ellipse.
$5$. For any point $T$ on the ellipse,the sum of the distances from $F_{1}$ and $F_{2}$ is constant. The points $F_{1}$ and $F_{2}$ are called the foci.
$6$. Join the points $F_{1}$ and $F_{2}$ and extend the line to intersect the ellipse at points $P$ and $A$. The midpoint of the line segment $PA$ is the centre of the ellipse,denoted by $O$.
$7$. The length $PO = AO$ is called the semi-major axis of the ellipse.

(A) According to Bohr's postulate, the angular momentum $L$ is given by $m v_n r_n = n h / 2 \pi$.
Given: mass $m = 10\; kg$, radius $r_n = 8000\; km = 8 \times 10^6\; m$, and time period $T = 2\; hours = 7200\; s$.
The orbital velocity $v_n$ is given by $v_n = 2 \pi r_n / T$.
Substituting $v_n$ into the angular momentum equation: $m (2 \pi r_n / T) r_n = n h / 2 \pi$.
Rearranging for the quantum number $n$: $n = (2 \pi r_n)^2 \times m / (T \times h)$.
Substituting the values: $n = (2 \pi \times 8 \times 10^6)^2 \times 10 / (7200 \times 6.63 \times 10^{-34})$.
$n = (25.13 \times 10^6)^2 \times 10 / (4.77 \times 10^{-30})$.
$n = 6.315 \times 10^{14} \times 10 / 4.77 \times 10^{-30} \approx 5.3 \times 10^{45}$.

(C) The moon revolves around the earth in a nearly circular orbit.
In circular motion,the gravitational force exerted by the earth on the moon acts as the centripetal force,which is directed towards the center of the orbit (radially inward).
The displacement of the moon at any instant is along the tangent to the circular path.
The angle between the force (centripetal force) and the displacement is $90^{\circ}$.
The work done $W$ is given by the formula $W = F \cdot s \cdot \cos(\theta)$,where $\theta$ is the angle between force and displacement.
Since $\theta = 90^{\circ}$,$\cos(90^{\circ}) = 0$.
Therefore,$W = F \cdot s \cdot 0 = 0$.
Thus,the work done by the earth's gravitational force on the moon is zero.

(N/A) body revolving around a planet in the direction of the planet's revolution is called a satellite.
The motion of a satellite is very similar to the motion of planets around the sun; hence,$Kepler's$ laws of planetary motion are equally applicable to them.
Their orbits around the earth are circular or elliptic.
There are two types of satellites:
$(i)$ Natural satellite: The moon is the natural satellite of the earth with a time period of approximately $27.3$ days. It is equal to the rotational period of the moon about its own axis. The planet Jupiter has many satellites.
$(ii)$ Artificial satellite: The first artificial satellite made by mankind was $Sputnik$,put into orbit around the earth by Russian scientists in $1957$. India ($19$ April $1975$) also successfully launched '$Aryabhatta$' and the '$INSAT$' series of satellites.
Satellites are used for scientific research,engineering,communication,weather forecasting,spying,military operations,geophysics,and meteorology.

(N/A) The velocity required by a satellite to revolve around the Earth in a given orbit is known as the orbital velocity of the satellite.
Consider a satellite of mass $m$ at a height $h$ from the surface of the Earth,revolving around the Earth as shown in the figure. Its distance from the center of the Earth is $r = R_E + h$. Let the orbital velocity of the satellite be $v_0$.
The gravitational force exerted on the satellite by the Earth is:
$F = \frac{G M_E m}{r^2}$
The necessary centripetal force for the circular motion of the satellite is provided by the Earth's gravitational force:
$F_c = \frac{m v_0^2}{r}$
Equating the centripetal force to the gravitational force:
$\frac{m v_0^2}{r} = \frac{G M_E m}{r^2}$
Simplifying the equation:
$v_0^2 = \frac{G M_E}{r}$
Substituting $r = R_E + h$:
$v_0 = \sqrt{\frac{G M_E}{R_E + h}}$
This equation indicates that as the height $h$ increases,the orbital velocity $v_0$ decreases.

(N/A) The orbital period $(T)$ of a satellite is the time taken to complete one full revolution around the Earth.
The orbital velocity of a satellite at a height $h$ above the Earth's surface is given by:
$v_{0} = \sqrt{\frac{GM_{E}}{R_{E}+h}}$ ... $(1)$
The orbital velocity is also defined as the ratio of the circumference of the orbit to the time period:
$v_{0} = \frac{2\pi(R_{E}+h)}{T}$ ... $(2)$
Equating $(1)$ and $(2)$:
$\sqrt{\frac{GM_{E}}{R_{E}+h}} = \frac{2\pi(R_{E}+h)}{T}$
Rearranging for $T$:
$T = \frac{2\pi(R_{E}+h)}{\sqrt{\frac{GM_{E}}{R_{E}+h}}}$
$T = 2\pi \sqrt{\frac{(R_{E}+h)^{3}}{GM_{E}}}$ ... $(3)$
Squaring both sides:
$T^{2} = \frac{4\pi^{2}}{GM_{E}}(R_{E}+h)^{3}$ ... $(4)$
Using the relation $g = \frac{GM_{E}}{(R_{E}+h)^{2}}$,we can express the period in terms of acceleration due to gravity $g$ at height $h$:
$T = 2\pi \sqrt{\frac{R_{E}+h}{g}}$ ... $(5)$
From equation $(4)$,if we let $r = R_{E}+h$,then $T^{2} = Kr^{3}$,where $K = \frac{4\pi^{2}}{GM_{E}}$. This confirms Kepler's third law of planetary motion,$T^{2} \propto r^{3}$.

(A) satellite revolving around the Earth moves in a nearly circular orbit.
For any object to move in a circular path,a centripetal force is required.
In the case of a satellite,this centripetal force is provided by the gravitational force of attraction between the Earth and the satellite.
Mathematically,$F_c = F_g$,where $F_c = \frac{mv^2}{r}$ and $F_g = \frac{GMm}{r^2}$.
Therefore,the gravitational force is responsible for the necessary centripetal force.

(N/A) The orbital velocity $v_o$ of a satellite at a distance $r$ from the center of the Earth is given by $v_o = \sqrt{\frac{GM}{r}}$.
For a satellite revolving very close to the surface of the Earth,the distance $r$ is approximately equal to the radius of the Earth $R_e$ (i.e.,$r \approx R_e$).
Substituting $r = R_e$ into the formula,we get $v_o = \sqrt{\frac{GM}{R_e}}$.
Since the acceleration due to gravity $g$ at the surface of the Earth is given by $g = \frac{GM}{R_e^2}$,we can write $GM = gR_e^2$.
Substituting this into the orbital velocity equation,we get $v_o = \sqrt{\frac{gR_e^2}{R_e}} = \sqrt{gR_e}$.

(N/A) The periodic time $(T)$ of a satellite is defined as the total time taken by the satellite to complete one full revolution around the central body (such as a planet or star) in its orbit.
Mathematically,for a satellite at a distance $(r)$ from the center of a planet of mass $(M)$,the orbital period is given by:
$T = 2\pi \sqrt{\frac{r^3}{GM}}$
where:
$G$ is the universal gravitational constant,
$M$ is the mass of the central body,
$r$ is the orbital radius (distance from the center of the planet to the satellite).

The orbital period $T$ of a satellite revolving at a distance $r$ from the center of the Earth is given by $T = 2\pi \sqrt{\frac{r^3}{GM}}$.
For a satellite revolving very close to the Earth's surface,$r \approx R_e$ (the radius of the Earth).
Substituting $GM = gR_e^2$,we get $T = 2\pi \sqrt{\frac{R_e^3}{gR_e^2}} = 2\pi \sqrt{\frac{R_e}{g}}$.
Using $R_e \approx 6.4 \times 10^6 \ m$ and $g \approx 9.8 \ m/s^2$,we calculate:
$T = 2 \times 3.14 \times \sqrt{\frac{6.4 \times 10^6}{9.8}} \approx 6.28 \times 808 \approx 5074 \ s$.
Converting to minutes,$T \approx 84.6 \ \text{minutes}$.

(N/A) geostationary satellite is an Earth satellite that has an orbital period of $24$ hours,which is equal to the rotational period of the Earth about its own axis. It appears stationary when viewed from the Earth and revolves around the Earth in the equatorial plane in the west-to-east direction.
To achieve this,the satellite is placed at a suitable height $(h)$ from the Earth's surface.
If a satellite of mass $m$ revolves at a distance $r = R_{E} + h$ from the center of the Earth,the height $h$ from the Earth's surface is given by the formula:
$h = \left( \frac{GM_{E} T^{2}}{4 \pi^{2}} \right)^{\frac{1}{3}} - R_{E}$
For a geostationary satellite,the orbital period $T = 24$ hours. Substituting this value,we get:
$h \approx 35,800 \text{ km}$
This type of satellite is widely used for telecommunication purposes.

(N/A) The conditions necessary for a satellite to appear stationary are:
$1)$ The satellite must be in a circular orbit in the equatorial plane of the Earth.
$2)$ Its orbital time period of revolution must be exactly $24$ hours.
$3)$ The satellite must revolve in the same direction as the Earth,i.e.,from west to east.
$4)$ It must be at a height of approximately $36,000 \ km$ above the Earth's equator.
The direction of rotation of a geostationary satellite is from west to east,which is the same as the rotation of the Earth about its axis.

(N/A) polar satellite is a type of satellite that orbits the Earth in a north-south direction,passing over the poles,while the Earth rotates on its axis in an east-west direction.
These are low-altitude satellites,typically orbiting at a height of $h \approx 500$ to $800 \ km$ from the Earth's surface.
The orbital period of a polar satellite is approximately $100$ minutes,allowing it to cross any given latitude multiple times a day.
$A$ camera fixed on the satellite can view only small strips of the Earth in one orbit. Adjacent strips are viewed in subsequent orbits,effectively allowing the entire Earth to be scanned strip by strip throughout the day.
These satellites can observe polar and equatorial regions at close distances with high resolution.
Information gathered from such satellites is extremely useful for remote sensing,meteorology,and environmental studies of the Earth.

(N/A) geostationary satellite must have an orbital period $T$ equal to the Earth's rotation period,which is $T = 24 \text{ hours} = 86400 \text{ s}$.
From Kepler's third law,the square of the orbital period is proportional to the cube of the orbital radius $r$: $T^2 = \frac{4\pi^2 r^3}{GM}$.
Here,$r = R_e + h$,where $R_e$ is the Earth's radius and $h$ is the height above the surface.
Rearranging for $r$: $r = \left( \frac{GMT^2}{4\pi^2} \right)^{1/3}$.
Substituting $GM = gR_e^2$,we get $r = \left( \frac{gR_e^2 T^2}{4\pi^2} \right)^{1/3}$.
The height $h$ is given by $h = r - R_e = \left( \frac{GMT^2}{4\pi^2} \right)^{1/3} - R_e$.
Substituting standard values ($M \approx 5.97 \times 10^{24} \text{ kg}$,$G \approx 6.67 \times 10^{-11} \text{ Nm}^2/\text{kg}^2$),the calculated height $h$ is approximately $35,800 \text{ km}$ above the Earth's surface.

(N/A) Geostationary satellites are used for:
$1$. Telecommunication: They act as relay stations for television,radio,and telephone signals.
$2$. Weather forecasting: They provide continuous monitoring of weather patterns over a specific region.
Polar satellites are used for:
$1$. Remote sensing: They are used for mapping,resource exploration,and monitoring environmental changes.
$2$. Surveillance: They are used for military reconnaissance and monitoring borders.
$3$. Scientific research: They help in studying the atmosphere and Earth's magnetic field.

$(A)$ Polar satellites are low-altitude satellites that orbit the Earth from pole to pole.
These satellites typically revolve around the Earth at a height of approximately $500 \text{ km}$ to $800 \text{ km}$ above the Earth's surface.
This low altitude allows them to capture high-resolution images of the Earth's surface for weather monitoring and surveillance.
Therefore, the correct option is $A$.

(D) In an artificial satellite orbiting the Earth,the satellite and everything inside it are in a state of free fall. Because the gravitational force provides the necessary centripetal acceleration for the orbit,there is no normal force acting on objects inside the satellite to define a 'weight' or an 'upward' direction. Consequently,there is no sensation of gravity,and no physical 'up' or 'down' exists relative to the satellite's frame of reference. Therefore,there is no defined upward direction.

(A) When an astronaut is inside an orbiting satellite,both the astronaut and the satellite are in a state of free fall towards the Earth due to gravity.
Since both have the same acceleration (the acceleration due to gravity at that altitude),the relative acceleration between them is zero.
This condition is known as weightlessness.
Therefore,the astronaut appears to be floating inside the satellite.

(B) The gravitational force exerted by the Earth on the Moon provides the necessary centripetal force for the Moon to maintain its circular orbit around the Earth.
Since this gravitational force acts perpendicular to the Moon's linear velocity,it only changes the direction of the Moon's motion,not its speed.
Therefore,the Moon continues to revolve in its orbit instead of falling onto the Earth.

(B) No,a satellite does not require fuel to orbit the Earth. The gravitational force exerted by the Earth on the satellite provides the necessary centripetal force required for its circular motion. Since this force is always perpendicular to the velocity of the satellite,no work is done by the gravitational force,and the satellite continues to orbit without the need for additional fuel.

(B) No,the orbital velocity of a planet does not depend on its mass.
According to the formula for orbital velocity,$v_{0} = \sqrt{\frac{GM_{e}}{r}}$,where $G$ is the gravitational constant,$M_{e}$ is the mass of the central body (e.g.,the Sun),and $r$ is the orbital radius.
Since the mass of the orbiting planet does not appear in this expression,the orbital velocity is independent of the planet's mass.

(B) The time period $T$ of a satellite is given by the formula $T = 2 \pi \sqrt{\frac{(R_e + h)^3}{GM_e}}$,where $R_e$ is the radius of the Earth,$h$ is the height of the satellite above the Earth's surface,$G$ is the gravitational constant,and $M_e$ is the mass of the Earth.
From the formula,it is clear that $T \propto (R_e + h)^{3/2}$.
As the height $h$ decreases,the term $(R_e + h)$ decreases.
Therefore,the time period $T$ of the satellite decreases.

(B) No,the spoon will not fall to the Earth.
Since the spoon was inside the spacecraft,it shares the same orbital velocity as the spacecraft.
When thrown out,it continues to move in the same orbit with the same velocity as the spacecraft.
Therefore,it will remain in orbit around the Earth alongside the spacecraft and will not fall to the surface.

(A) satellite is defined as an object that revolves around a planet in a stable,closed orbit under the influence of gravitational force.
If a satellite were to attain the escape velocity,it would leave its orbit and move to infinity,thereby ceasing to be a satellite of that planet.
Therefore,by definition,a satellite in a stable orbit does not possess the escape velocity required to leave the gravitational influence of the central body.

(A) polar satellite orbits the Earth in a North-South direction,passing over the poles as the Earth rotates underneath it.
The Earth rotates about its own axis from West to East.

(B) polar satellite orbits the Earth in a north-south direction (passing over the poles).
Since the Earth rotates on its axis from west to east,the satellite covers a different strip of the Earth's surface with each orbit.
Over a period of time,the entire surface of the Earth is scanned by the satellite.

(N/A) The orbital period of the Moon around the Earth is approximately $27.3$ days.
Similarly,the rotational period of the Moon about its own axis is also approximately $27.3$ days.
Because these two periods are equal,the same side of the Moon always faces the Earth.

(N/A) The equatorial plane is the plane passing through the Earth's equator.
$(a)$ $A$ polar satellite orbits the Earth in a plane that passes through the North and South poles. Since this plane is perpendicular to the equatorial plane,the angle between them is $90^{\circ}$.
$(b)$ $A$ geostationary satellite orbits the Earth in the same plane as the Earth's equatorial plane to remain stationary relative to a point on the Earth's surface. Therefore,the angle between the orbital plane of a geostationary satellite and the equatorial plane is $0^{\circ}$.

(C) According to Kepler's laws of planetary motion and Newton's law of gravitation,the trajectory of a particle moving under the influence of a central gravitational force (like that of the Earth) must be a conic section (ellipse,parabola,or hyperbola) with the center of the Earth as one of its foci.
$1$. Figure $(a)$ is not a conic section.
$2$. Figure $(b)$ does not have the Earth at a focus.
$3$. Figure $(c)$ shows an elliptical path with the Earth at one of its foci,which is a valid trajectory.
$4$. Figure $(d)$ is a spiral,which is not a conic section.
$5$. Figure $(e)$ does not have the Earth at a focus.
$6$. Figure $(f)$ is a complex path that does not represent a standard conic section.
Therefore,only figure $(c)$ represents a possible trajectory for a projectile.

(N/A) Given:
Mass of the Earth,$M = 6 \times 10^{24} \ kg$
Radius of the Earth,$R = 6400 \ km = 6.4 \times 10^6 \ m$
Time period,$T = 24 \ h = 24 \times 3600 \ s = 86400 \ s$
Gravitational constant,$G = 6.67 \times 10^{-11} \ N \cdot m^2/kg^2$
$(a)$ For a geostationary satellite,the orbital period $T$ is related to the orbital radius $r = R + h$ by Kepler's Third Law:
$T^2 = \frac{4 \pi^2 r^3}{GM}$
$r^3 = \frac{T^2 GM}{4 \pi^2}$
$r = \left( \frac{T^2 GM}{4 \pi^2} \right)^{1/3}$
Substituting the values:
$r = \left( \frac{(86400)^2 \times 6.67 \times 10^{-11} \times 6 \times 10^{24}}{4 \times (3.14)^2} \right)^{1/3} \approx 4.22 \times 10^7 \ m$
Height $h = r - R = 4.22 \times 10^7 \ m - 0.64 \times 10^7 \ m = 3.58 \times 10^7 \ m = 35800 \ km$.
$(b)$ To cover the entire equator,each satellite must cover an angular width of $120^\circ$ (or $2\pi/3$ radians) to ensure global coverage with three satellites. Thus,the minimum number of satellites required is $3$.

(A) Given: $r_p = 2R$ and $r_a = 6R$.
Using the properties of an ellipse: $r_a = a(1+e) = 6R$ and $r_p = a(1-e) = 2R$.
Dividing the two equations: $\frac{1+e}{1-e} = \frac{6R}{2R} = 3 \implies 1+e = 3 - 3e \implies 4e = 2 \implies e = 0.5$.
Using conservation of angular momentum: $m v_p r_p = m v_a r_a \implies v_a = v_p \frac{r_p}{r_a} = v_p \frac{2R}{6R} = \frac{v_p}{3}$.
Using conservation of energy: $\frac{1}{2} m v_p^2 - \frac{GMm}{r_p} = \frac{1}{2} m v_a^2 - \frac{GMm}{r_a}$.
Substituting $v_a = v_p/3$: $\frac{1}{2} v_p^2 (1 - 1/9) = GM (\frac{1}{2R} - \frac{1}{6R}) = GM (\frac{2}{6R}) = \frac{GM}{3R}$.
$\frac{4}{9} v_p^2 = \frac{GM}{3R} \implies v_p^2 = \frac{3GM}{4R} \implies v_p = \sqrt{\frac{3 \times 6.67 \times 10^{-11} \times 6 \times 10^{24}}{4 \times 6400 \times 10^3}} \approx 5.59 \, km/s$.
Then $v_a = v_p/3 \approx 1.86 \, km/s$.
To transfer to a circular orbit of radius $6R$,the satellite must be at apogee $(r=6R)$ and its velocity must be increased to the orbital velocity $v_c = \sqrt{\frac{GM}{6R}} \approx 2.64 \, km/s$.

(A) Polar satellites orbit the Earth from pole to pole,providing a global view of the surface,which makes them ideal for spying and weather monitoring $(1-b, c)$.
Geostationary satellites remain fixed relative to a point on Earth's surface,making them ideal for continuous telecommunication and broadcasting services $(2-a)$.
Therefore,the correct matching is $(1-b, c)$ and $(2-a)$.

(C) The mass of the galaxy within a radius $R$ is calculated by integrating the density $\rho(r) = \frac{K}{r}$ over the volume.
$dm = \rho(r) \cdot 4\pi r^2 dr = \left(\frac{K}{r}\right) \cdot 4\pi r^2 dr = 4\pi K r dr$
$M(R) = \int_{0}^{R} 4\pi K r dr = 4\pi K \left[ \frac{r^2}{2} \right]_{0}^{R} = 2\pi K R^2$
For a star of mass $m$ in a circular orbit of radius $R$,the gravitational force provides the necessary centripetal force:
$\frac{G M(R) m}{R^2} = \frac{m v^2}{R}$
Substituting $M(R) = 2\pi K R^2$:
$\frac{G (2\pi K R^2) m}{R^2} = \frac{m v^2}{R} \Rightarrow 2\pi G K m = \frac{m v^2}{R} \Rightarrow v^2 = 2\pi G K R$
$v = \sqrt{2\pi G K R}$
The time period $T$ is given by $T = \frac{2\pi R}{v}$.
$T = \frac{2\pi R}{\sqrt{2\pi G K R}} = \sqrt{\frac{4\pi^2 R^2}{2\pi G K R}} = \sqrt{\frac{2\pi R}{G K}} \propto \sqrt{R}$
Squaring both sides,we get $T^2 \propto R$.

(B) The initial orbital speed of the satellite is $V_0 = \sqrt{\frac{GM}{R_e}}$.
After firing the rockets,the new speed is $V = \sqrt{\frac{3}{2}} V_0 = \sqrt{\frac{3GM}{2R_e}}$.
Using the conservation of angular momentum at the perigee $(R_e)$ and apogee $(R_{max} = R)$:
$L_{initial} = L_{final} \implies m V R_e = m V' R$
$V' = \frac{V R_e}{R} = \frac{R_e}{R} \sqrt{\frac{3GM}{2R_e}}$
Using the conservation of mechanical energy:
$-\frac{GMm}{R_e} + \frac{1}{2} m V^2 = -\frac{GMm}{R} + \frac{1}{2} m V'^2$
$-\frac{GM}{R_e} + \frac{1}{2} \left(\frac{3GM}{2R_e}\right) = -\frac{GM}{R} + \frac{1}{2} \left(\frac{R_e^2}{R^2} \cdot \frac{3GM}{2R_e}\right)$
$-\frac{1}{R_e} + \frac{3}{4R_e} = -\frac{1}{R} + \frac{3R_e}{4R^2}$
$-\frac{1}{4R_e} = \frac{-4R + 3R_e}{4R^2}$
$-R^2 = -4R R_e + 3R_e^2 \implies R^2 - 4R R_e + 3R_e^2 = 0$
$(R - 3R_e)(R - R_e) = 0$
Since $R > R_e$,we get $R = 3R_e$.

(C) The orbital speed of a body in a circular orbit at a distance $R$ from the center of a planet of mass $M$ is given by $V_{\text{orbit}} = \sqrt{\frac{GM}{R}}$.
The escape velocity from the surface of the planet is given by $V_{\text{escape}} = \sqrt{\frac{2GM}{R}}$.
Taking the ratio of the orbital speed to the escape velocity,we get:
$\frac{V_{\text{orbit}}}{V_{\text{escape}}} = \frac{\sqrt{\frac{GM}{R}}}{\sqrt{\frac{2GM}{R}}} = \frac{1}{\sqrt{2}}$.

(A) According to the law of conservation of angular momentum,the angular momentum of the satellite remains constant throughout its orbit.
At the closest point (periapsis) and farthest point (apoapsis),the velocity vector is perpendicular to the position vector.
Therefore,$L = m r_{\min} V_{\max} = m r_{\max} V_{\min}$.
This simplifies to $r_{\min} V_{\max} = r_{\max} V_{\min} \quad \dots(i)$.
Given that the velocity at the farthest point is $6$ times less than the velocity at the closest point,we have $V_{\min} = \frac{V_{\max}}{6}$,or $\frac{V_{\min}}{V_{\max}} = \frac{1}{6}$.
Substituting this into equation $(i)$,we get $\frac{r_{\min}}{r_{\max}} = \frac{V_{\min}}{V_{\max}} = \frac{1}{6}$.
Thus,the ratio of the distances is $1:6$.

(C) The orbital velocity $(v_o)$ of a satellite orbiting the Earth at a distance $(r)$ from the center is given by the formula:
$v_o = \sqrt{\frac{GM}{r}}$
where $G$ is the gravitational constant and $M$ is the mass of the Earth.
From this expression, it is clear that $v_o$ is inversely proportional to the square root of the distance $r$:
$v_o \propto \frac{1}{\sqrt{r}}$
Therefore, the correct relationship is $v_o \propto 1/\sqrt{r}$.

(D) According to Kepler's third law of planetary motion,the square of the time period is proportional to the cube of the orbital radius: $\frac{T_{A}^{2}}{T_{B}^{2}} = \frac{r_{A}^{3}}{r_{B}^{3}}$.
Given $T_{A} = 1\, h$,$T_{B} = 8\, h$,and $r_{A} = 10^{4}\, km$.
Substituting the values: $\frac{1^{2}}{8^{2}} = \frac{(10^{4})^{3}}{r_{B}^{3}} \Rightarrow \frac{1}{64} = \frac{(10^{4})^{3}}{r_{B}^{3}}$.
Thus,$r_{B}^{3} = 64 \times (10^{4})^{3}$,which gives $r_{B} = 4 \times 10^{4}\, km$.
The orbital speed of a satellite is given by $v = \frac{2 \pi r}{T}$.
For satellite $A$: $v_{A} = \frac{2 \pi \times 10^{4}}{1} = 2 \pi \times 10^{4}\, km/h$.
For satellite $B$: $v_{B} = \frac{2 \pi \times 4 \times 10^{4}}{8} = \pi \times 10^{4}\, km/h$.
Since both satellites revolve in the same direction in coplanar orbits,their relative speed when they are closest is the difference of their orbital speeds: $v_{rel} = v_{A} - v_{B} = 2 \pi \times 10^{4} - \pi \times 10^{4} = \pi \times 10^{4}\, km/h$.

(D) The gravitational force provides the necessary centripetal force for circular motion.
Given,$F \propto r^{-3/2}$,so $F = \frac{k}{r^{3/2}}$ where $k$ is a constant.
The centripetal force is given by $F = m \omega^2 r = m \left(\frac{2\pi}{T}\right)^2 r = \frac{4\pi^2 m r}{T^2}$.
Equating the two expressions for force:
$\frac{k}{r^{3/2}} = \frac{4\pi^2 m r}{T^2}$.
Rearranging for $T^2$:
$T^2 = \frac{4\pi^2 m}{k} \cdot r \cdot r^{3/2} = \frac{4\pi^2 m}{k} \cdot r^{5/2}$.
Since $\frac{4\pi^2 m}{k}$ is a constant,we have $T^2 \propto r^{5/2}$.

(C) According to the law of conservation of angular momentum,the angular momentum of the comet remains constant at all points in its orbit.
Let $r_1$ and $v_1$ be the distance and speed at the nearest point (perihelion),and $r_2$ and $v_2$ be the distance and speed at the farthest point (aphelion).
Given:
$r_1 = 8.0 \times 10^{10} \ m$
$v_1 = 6 \times 10^{4} \ m/s$
$r_2 = 1.6 \times 10^{12} \ m$
Using the conservation of angular momentum:
$m v_1 r_1 = m v_2 r_2$
$v_2 = \frac{v_1 r_1}{r_2}$
Substituting the values:
$v_2 = \frac{(6 \times 10^{4}) \times (8.0 \times 10^{10})}{1.6 \times 10^{12}}$
$v_2 = \frac{48 \times 10^{14}}{1.6 \times 10^{12}}$
$v_2 = 30 \times 10^{2} \ m/s = 3 \times 10^{3} \ m/s$
Thus,the speed at the farthest point is $3 \times 10^{3} \ m/s$.

(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 $r$ $(T^2 \propto r^3)$,which implies $T \propto r^{3/2}$.
The orbital radius $r$ is the sum of the planet's radius $R$ and the height $h$ above the surface $(r = R + h)$.
For the first satellite: $r_1 = R + 11R = 12R$ and $T_1 = 24\, \text{hours}$.
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: $\frac{24}{T_2} = \left(\frac{12R}{3R}\right)^{3/2} = (4)^{3/2} = (2^2)^{3/2} = 2^3 = 8$.
Therefore,$T_2 = \frac{24}{8} = 3\, \text{hours}$.

(D) Consider a small displacement $ds$ of the planet along its elliptical orbit in a small time interval $dt$.
The area $dA$ swept by the position vector $\vec{r}$ in this time is given by the area of a triangle with sides $r$ and $ds$:
$dA = \frac{1}{2} |\vec{r} \times d\vec{s}| = \frac{1}{2} r ds \sin \theta$
where $\theta$ is the angle between the position vector $\vec{r}$ and the displacement vector $d\vec{s}$.
The areal velocity is the rate of change of area with respect to time:
$\text{Areal velocity} = \frac{dA}{dt} = \frac{1}{2} r \sin \theta \frac{ds}{dt}$
Since the velocity of the planet is $v = \frac{ds}{dt}$,we have:
$\frac{dA}{dt} = \frac{1}{2} r v \sin \theta$
Multiplying and dividing by the mass $M$ of the planet:
$\frac{dA}{dt} = \frac{1}{2M} (M v r \sin \theta)$
Since the magnitude of angular momentum is $L = Mvr \sin \theta$,we get:
$\frac{dA}{dt} = \frac{L}{2M}$

(B) The gravitational force between the two stars provides the necessary centripetal force for their circular motion about the common centre of mass (c.o.m.).
The distance of mass $m$ from the c.o.m. is $r_1 = \frac{2m}{m+2m} d = \frac{2d}{3}$.
The gravitational force is $F = \frac{G(m)(2m)}{d^2} = \frac{2Gm^2}{d^2}$.
For the star of mass $m$,the centripetal force is $F = m \omega^2 r_1 = m \omega^2 (\frac{2d}{3})$.
Equating the forces: $\frac{2Gm^2}{d^2} = m \omega^2 (\frac{2d}{3})$.
Simplifying: $\frac{Gm}{d^2} = \omega^2 \frac{d}{3} \implies \omega^2 = \frac{3Gm}{d^3}$.
Thus,$\omega = \sqrt{\frac{3Gm}{d^3}}$.
The period of revolution $T$ is given by $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{d^3}{3Gm}}$.

(C) The period of revolution $T$ is related to the angular velocity $\omega$ by the formula $T = \frac{2\pi}{\omega}$.
Given the periods $T_{1} = 1\, hr$ and $T_{2} = 8\, hr$.
The ratio of the periods is $\frac{T_{1}}{T_{2}} = \frac{1}{8}$.
Substituting the formula for $T$ in terms of $\omega$:
$\frac{2\pi / \omega_{1}}{2\pi / \omega_{2}} = \frac{1}{8}$.
This simplifies to $\frac{\omega_{2}}{\omega_{1}} = \frac{1}{8}$.
Therefore,the ratio of the angular velocity of $S_{1}$ to $S_{2}$ is $\frac{\omega_{1}}{\omega_{2}} = \frac{8}{1}$ or $8:1$.

(A) The time period of a satellite at a height $h$ from the Earth's surface is given by $T = 2 \pi \sqrt{\frac{(R+h)^{3}}{GM}}$.
For satellite $A$: $h_{A} = 600 \, km = 0.6 \times 10^{6} \, m$,$R = 6.4 \times 10^{6} \, m$. Orbital radius $r_{A} = R + h_{A} = 7.0 \times 10^{6} \, m$.
$T_{A} = 2 \pi \sqrt{\frac{(7.0 \times 10^{6})^{3}}{6.67 \times 10^{-11} \times 6 \times 10^{24}}} \approx 5800 \, s$.
For satellite $B$: $h_{B} = 1600 \, km = 1.6 \times 10^{6} \, m$. Orbital radius $r_{B} = R + h_{B} = 8.0 \times 10^{6} \, m$.
$T_{B} = 2 \pi \sqrt{\frac{(8.0 \times 10^{6})^{3}}{6.67 \times 10^{-11} \times 6 \times 10^{24}}} \approx 7133 \, s$.
$T_{B} - T_{A} = 7133 - 5800 = 1333 \, s = 1.33 \times 10^{3} \, s$.

(A) Given:
$T_1 = 1\, h$,$T_2 = 8\, h$
$R_1 = 2 \times 10^3\, km$
Angular velocities are $\omega_1 = \frac{2\pi}{T_1} = 2\pi\, rad/h$ and $\omega_2 = \frac{2\pi}{T_2} = \frac{\pi}{4}\, rad/h$.
Using Kepler's third law,$\frac{R_2^3}{R_1^3} = \frac{T_2^2}{T_1^2} \Rightarrow \frac{R_2}{R_1} = (\frac{8}{1})^{2/3} = 4$.
So,$R_2 = 4 \times R_1 = 8 \times 10^3\, km$.
The linear velocities are $v_1 = \omega_1 R_1 = 2\pi \times 2 \times 10^3 = 4\pi \times 10^3\, km/h$ and $v_2 = \omega_2 R_2 = \frac{\pi}{4} \times 8 \times 10^3 = 2\pi \times 10^3\, km/h$.
When the satellites are closest,the relative angular velocity $\omega_{rel}$ as observed from the nearer satellite is given by $\omega_{rel} = \frac{v_1 - v_2}{R_2 - R_1}$.
$\omega_{rel} = \frac{4\pi \times 10^3 - 2\pi \times 10^3}{8 \times 10^3 - 2 \times 10^3} = \frac{2\pi \times 10^3}{6 \times 10^3} = \frac{\pi}{3}\, rad/h$.
Comparing this with $\frac{\pi}{x}$,we get $x = 3$.