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

(B) 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.
In a circular orbit,the gravitational force provides the necessary centripetal force to keep the satellite moving in a circle.
If the gravitational force suddenly disappears,there is no longer any centripetal force to change the direction of the satellite's velocity.
Therefore,due to the inertia of direction,the satellite will continue to move in a straight line along the direction of its velocity at that instant.
Since the velocity vector is always tangent to the circular path,the satellite will move with velocity $v$ tangentially to the original orbit.

(B) The Earth revolves around the Sun due to the gravitational pull of the Sun.
Due to this gravitational attraction between these celestial bodies,a centripetal force is generated,which keeps the Earth in its orbit.
According to Newton's laws of motion,a body moving in a circular path must be acted upon by a centripetal force directed towards the center of the circle.
Therefore,the revolution of the Earth around the Sun is the evidence that there must be a force acting on the Earth and directed towards the Sun.

(D) Given: Mass of the earth $m = 6 \times 10^{24} \ kg$,angular velocity $\omega = 2 \times 10^{-7} \ rad/s$,and radius of orbit $R = 1.5 \times 10^8 \ km = 1.5 \times 10^{11} \ m$.
The centripetal force required for the circular motion of the earth is provided by the gravitational force exerted by the sun.
The formula for centripetal force is $F = m \omega^2 R$.
Substituting the given values into the formula:
$F = (6 \times 10^{24}) \times (2 \times 10^{-7})^2 \times (1.5 \times 10^{11})$
$F = (6 \times 10^{24}) \times (4 \times 10^{-14}) \times (1.5 \times 10^{11})$
$F = (6 \times 4 \times 1.5) \times (10^{24} \times 10^{-14} \times 10^{11})$
$F = 36 \times 10^{21} \ N$.

(B) The gravitational force exerted by the Earth on the satellite acts as the required centripetal force for its circular orbit.
Since the centripetal force is not an additional force but is provided by the gravitational force itself,the gravitational force is the net force acting on the satellite.
Therefore,the net force on the satellite is equal to the gravitational force,which is $F$.

(C) The reason for weightlessness is the $zero$ reaction force exerted by the satellite surface.
Weight is defined as the normal reaction force exerted by a surface on a body.
In a satellite, both the satellite and the body inside it are in a state of free fall towards the Earth's center with the same acceleration (equal to the acceleration due to gravity at that altitude).
Because they are falling together, the surface of the satellite does not exert any normal reaction force on the body.
Therefore, the body experiences weightlessness.

(D) When a spaceship orbits the earth,both the spaceship and the objects inside it are in a state of free fall towards the earth due to the earth's gravitational pull. Since both are falling with the same acceleration (the acceleration due to gravity),there is no normal force exerted by the floor of the spaceship on the objects inside. This lack of a contact force results in the sensation of weightlessness. Therefore,the correct option is $(d)$.

(D) An astronaut inside an orbiting satellite is in a state of continuous free fall towards the Earth.
Although the Earth's gravitational force acts on the astronaut,the satellite and the astronaut are both accelerating towards the center of the Earth at the same rate (the acceleration due to gravity $g$ at that altitude).
Because both are falling together,the astronaut experiences no normal reaction force from the floor of the satellite,which results in the sensation of weightlessness.
Therefore,weightlessness is a situation of free fall.
Thus,the correct option is $D$.

(D) For a satellite of mass $m$ to revolve in a circular orbit of radius $r$ around a planet of mass $M$,the gravitational force provides the necessary centripetal force.
The gravitational force is given by $F_g = \frac{GMm}{r^2}$.
The centripetal force required for circular motion is $F_c = \frac{mv^2}{r}$.
Equating the two forces:
$\frac{mv^2}{r} = \frac{GMm}{r^2}$.
Canceling $m$ from both sides and multiplying by $r$:
$v^2 = \frac{GM}{r}$.
Thus,the correct expression is $v^2 = \frac{GM}{r}$.

(D) The orbital velocity $v_0$ of a satellite revolving around a planet of mass $M$ at a distance $r$ from its center is given by the formula: $v_0 = \sqrt{\frac{GM}{r}}$.
From this expression,it is clear that $v_0 \propto \frac{1}{\sqrt{r}}$.
Therefore,the orbital velocity is inversely proportional to the square root of the radius of the orbit.
Thus,option $D$ is the correct statement.

(D) The orbital velocity $v_0$ of a satellite at a distance $r$ from the center of the Earth is given by $v_0 = \sqrt{\frac{GM}{r}}$.
Here,the distance from the center of the Earth is $r = R + h$.
We know that the acceleration due to gravity at the surface of the Earth is $g = \frac{GM}{R^2}$,which implies $GM = gR^2$.
Substituting these values into the orbital velocity formula:
$v_0 = \sqrt{\frac{gR^2}{R + h}}$.

(B) satellite orbiting the Earth is under the influence of the Earth's gravitational force,which acts as a centripetal force.
Since there is a force acting on the satellite,it must experience acceleration,known as centripetal acceleration $(a_c = v^2/r)$.
Therefore,the statement that it suffers no acceleration is incorrect.
Option $(b)$ is the wrong statement.

(B) The orbital speed $v$ of a satellite revolving around a planet of mass $M$ at a distance $r$ from its center is given by the formula $v = \sqrt{\frac{GM}{r}}$.
From this expression,it is clear that the orbital speed $v$ is independent of the mass of the satellite ($m_1$ or $m_2$).
However,the orbital speed $v$ is inversely proportional to the square root of the orbital radius $r$ $(v \propto \frac{1}{\sqrt{r}})$.
Given that $r_1 > r_2$,it follows that $\frac{1}{\sqrt{r_1}} < \frac{1}{\sqrt{r_2}}$.
Therefore,$v_1 < v_2$.

(C) When an object is dropped from a spaceship orbiting the Earth,it possesses the same orbital velocity as the spaceship at that instant. Since there is no external force (like air resistance) to change its state of motion in the vacuum of space,the spoon will continue to move in the same circular orbit alongside the spaceship. Therefore,the spoon will move along with the spaceship.

(C) The orbital period $T$ of a satellite is given by the ratio of the circumference of the orbit to the orbital velocity:
$T = \frac{2 \pi r}{v_o}$
Since the orbital velocity $v_o = \sqrt{\frac{GM}{r}}$,we substitute this into the equation:
$T = \frac{2 \pi r}{\sqrt{\frac{GM}{r}}} = 2 \pi \sqrt{\frac{r^3}{GM}}$
Here,$M$ is the mass of the planet,$r$ is the radius of the orbit,and $G$ is the gravitational constant.
From the formula,it is clear that $T$ depends on the radius of the orbit $(r)$ and the mass of the planet $(M)$,but it does not depend on the mass of the satellite $(m)$.
Therefore,the period is independent of the mass of the satellite.

(B) The orbital velocity $v_0$ of a satellite orbiting at a height $h$ above the Earth's surface is given by $v_0 = \sqrt{\frac{GM}{R+h}}$.
For a satellite orbiting very close to the Earth's surface,the height $h$ is negligible compared to the radius of the Earth $R$ (i.e.,$h \approx 0$).
Thus,the formula becomes $v_0 = \sqrt{\frac{GM}{R}}$.
Since $g = \frac{GM}{R^2}$,we can write $GM = gR^2$.
Substituting this into the velocity formula: $v_0 = \sqrt{\frac{gR^2}{R}} = \sqrt{gR}$.
Therefore,the orbital velocity of a satellite close to the Earth's surface depends on the acceleration due to gravity $g$ and the radius of the Earth $R$.

(D) relay satellite used for $T.V.$ transmission is a geostationary satellite.
For a satellite to be geostationary,it must appear stationary relative to a point on the Earth's surface.
This requires the satellite's orbital period to be exactly equal to the Earth's rotational period about its axis,which is $24$ hours.
Therefore,the correct option is $D$.

(B) The orbital speed of a satellite at a distance $r$ from the center of a planet is given by $v = \sqrt{\frac{GM}{r}}$.
Since $v \propto \frac{1}{\sqrt{r}}$,we can write the ratio of the speeds of satellites $A$ and $B$ as:
$\frac{v_A}{v_B} = \sqrt{\frac{r_B}{r_A}}$
Given $r_A = 4R$ and $r_B = R$,we have:
$\frac{v_A}{v_B} = \sqrt{\frac{R}{4R}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$
Given the speed of satellite $A$ is $v_A = 3V$,we substitute this into the ratio:
$\frac{3V}{v_B} = \frac{1}{2}$
Therefore,$v_B = 3V \times 2 = 6V$.

(A) geostationary satellite is a satellite that orbits the Earth in the equatorial plane with an orbital period equal to the Earth's rotational period ($24$ hours).
Since the Earth rotates about its polar axis,a geostationary satellite must also revolve about the same polar axis to remain fixed relative to a point on the Earth's surface.
Therefore,the correct statement is that it revolves about the polar axis.

(A) The orbital velocity $(v_{0})$ of a satellite revolving near the Earth's surface is given by the formula:
$v_{0} = \sqrt{g R_{e}}$
Where $g$ is the acceleration due to gravity $(9.8 \ m/s^{2})$ and $R_{e}$ is the radius of the Earth $(6.4 \times 10^{6} \ m)$.
Substituting the values:
$v_{0} = \sqrt{9.8 \times 6.4 \times 10^{6}}$
$v_{0} = \sqrt{62.72 \times 10^{6}}$
$v_{0} \approx 7.92 \times 10^{3} \ m/s$
Converting to $km/s$:
$v_{0} \approx 7.92 \ km/s \approx 8 \ km/s$.
Thus,the correct option is $A$.

(B) According to Kepler's second law of planetary motion,the areal velocity of a satellite remains constant.
When a satellite is in an elliptical orbit,it moves faster when it is closer to the earth (at perigee) and slower when it is farther away (at apogee) to conserve angular momentum.
Angular momentum $L = mvr \sin(\theta)$ is conserved.
Since $L$ is constant,$v$ is inversely proportional to $r$ when the satellite is at the closest or farthest points.
Therefore,the speed is greatest when the distance $r$ from the earth is minimum.

(C) The orbital velocity of a satellite at an altitude $h$ is given by $v = \sqrt{\frac{GM}{R+h}}$.
For the first satellite,$h = 0$,so the velocity is $v_1 = \sqrt{\frac{GM}{R}} = v$.
For the second satellite,the altitude is $h = \frac{R}{2}$.
Substituting this into the formula,we get $v_2 = \sqrt{\frac{GM}{R + \frac{R}{2}}} = \sqrt{\frac{GM}{\frac{3R}{2}}} = \sqrt{\frac{2GM}{3R}}$.
We can rewrite this as $v_2 = \sqrt{\frac{2}{3}} \sqrt{\frac{GM}{R}}$.
Since $v = \sqrt{\frac{GM}{R}}$,we have $v_2 = \sqrt{\frac{2}{3}}v$.

(D) The orbital velocity of a satellite is given by $v = \sqrt{\frac{GM}{r}}$,where $G$ is the gravitational constant,$M$ is the mass of the planet,and $r$ is the orbital radius.
Kinetic energy $(K.E.)$ is given by $K.E. = \frac{1}{2}mv^2$,which implies $K.E. \propto v^2$.
Substituting the expression for $v$,we get $K.E. \propto \frac{1}{r}$.
According to Kepler's third law of planetary motion,the square of the time period is proportional to the cube of the orbital radius,i.e.,$T^2 \propto r^3$,which means $r \propto T^{2/3}$.
Substituting $r \propto T^{2/3}$ into the relation $K.E. \propto \frac{1}{r}$,we get $K.E. \propto \frac{1}{T^{2/3}}$ or $K.E. \propto T^{-2/3}$.

(D) The orbital velocity of a satellite at a height $h$ from the surface of the Earth is given by the formula:
$v_{0} = \sqrt{\frac{GM}{R+h}}$
Since $g = \frac{GM}{R^2}$,we can write $GM = gR^2$.
Substituting this into the orbital velocity formula:
$v_{0} = \sqrt{\frac{gR^2}{R+h}}$
Given that the height $h$ is negligible compared to the radius of the Earth $R$ $(R >> h)$,we can approximate $R+h \approx R$.
Therefore,the orbital velocity becomes:
$v_{0} = \sqrt{\frac{gR^2}{R}} = \sqrt{gR}$

(D) The orbital period $T$ of a satellite is given by the formula: $T = 2\pi \sqrt{\frac{r^3}{GM}}$
Squaring both sides,we get: $T^2 = 4\pi^2 \frac{r^3}{GM}$
Rearranging the equation to solve for the orbital radius $r$: $r^3 = \frac{GMT^2}{4\pi^2}$
Therefore,$r = \left( \frac{GMT^2}{4\pi^2} \right)^{1/3}$
From this expression,it is clear that the radius $r$ depends only on the mass of the Earth $(M)$,the gravitational constant $(G)$,and the time period of the satellite $(T)$. It is independent of the mass of the satellite $(m)$.

(B) The orbital speed of a satellite is given by $v = \sqrt{\frac{GM}{r}}$,which implies $v \propto r^{-1/2}$.
Taking the natural logarithm on both sides,we get $\ln v = \ln(GM)^{1/2} - \frac{1}{2} \ln r$.
Differentiating both sides,we get $\frac{dv}{v} = -\frac{1}{2} \frac{dr}{r}$.
Given that the radius is decreased by $1\%$,we have $\frac{dr}{r} = -0.01$.
Substituting this value,$\frac{dv}{v} = -\frac{1}{2} (-0.01) = +0.005$.
Therefore,the speed will increase by $0.5\%$.

(B) The orbital velocity $v$ of a satellite revolving around a planet is given by the formula: $v = \sqrt{\frac{GM}{r}}$
Where:
$G$ is the universal gravitational constant,
$M$ is the mass of the central body (Earth),
$r$ is the orbital radius of the satellite.
From the formula,it is clear that the orbital velocity $v$ depends on the mass of the Earth $(M)$ and the distance from the center of the Earth $(r)$.
It does not depend on the mass of the satellite $(m)$.
Therefore,the correct option is $B$.

(A) geostationary satellite is an earth-orbiting satellite,placed at an altitude of approximately $35,800 \text{ km}$ directly over the equator,that revolves in the same direction the earth rotates (west to east).
At this altitude,one orbit takes $24 \text{ hours}$,which is the same length of time as the earth requires to rotate once on its axis.

(A) The formula for the orbital velocity 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 orbital radius.
From this relation,we can see that $v \propto \frac{1}{\sqrt{r}}$.
Let $v_1 = 7 \, km/s$ be the velocity at the surface (where $r_1 = R_e$) and $v_2$ be the velocity at an orbit where $r_2 = 4R_e$.
Using the ratio: $\frac{v_2}{v_1} = \sqrt{\frac{r_1}{r_2}} = \sqrt{\frac{R_e}{4R_e}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Therefore,$v_2 = \frac{v_1}{2} = \frac{7}{2} = 3.5 \, km/s$.

(D) The orbital radii of the satellites are $r_1 = R + R = 2R$ and $r_2 = R + 7R = 8R$.
For a satellite of mass $m$ orbiting Earth of mass $M$,the potential energy is $U = -\frac{GMm}{r}$,kinetic energy is $K = \frac{GMm}{2r}$,and total energy is $E = -\frac{GMm}{2r}$.
Calculating the ratios:
$\frac{U_1}{U_2} = \frac{r_2}{r_1} = \frac{8R}{2R} = 4$.
$\frac{K_1}{K_2} = \frac{r_2}{r_1} = \frac{8R}{2R} = 4$.
$\frac{E_1}{E_2} = \frac{r_2}{r_1} = \frac{8R}{2R} = 4$.
Since all these ratios are equal to $4$,the statement in option $(D)$ is incorrect.

(D) For a geostationary satellite,the orbital period $T$ is equal to the rotational period of the earth,which is $T = 2\pi/\omega$.
According to Kepler's third law,the square of the orbital period is proportional to the cube of the orbital radius $r$,given by $T^2 = (4\pi^2/GM)r^3$.
We know that the acceleration due to gravity at the surface is $g = GM/R^2$,which implies $GM = gR^2$.
Substituting $GM$ into the Kepler's law equation: $(2\pi/\omega)^2 = (4\pi^2 / gR^2)r^3$.
Simplifying this: $4\pi^2/\omega^2 = (4\pi^2 / gR^2)r^3$.
Canceling $4\pi^2$ from both sides: $1/\omega^2 = r^3 / (gR^2)$.
Therefore,$r^3 = gR^2/\omega^2$.

(A) The orbital velocity $v_o$ of a satellite is given by the formula $v_o = \sqrt{\frac{GM}{r}}$,where $G$ is the gravitational constant,$M$ is the mass of the earth,and $r$ is the orbital radius.
As seen from the formula,the orbital velocity depends only on the mass of the earth and the radius of the orbit; it is independent of the mass of the satellite.
Therefore,the statement that the orbital velocity depends on the mass of the satellite is incorrect.

(A) When an object is dropped from a spacecraft orbiting the Earth,it possesses the same orbital velocity as the spacecraft at the moment of release.
According to Newton's first law of motion (Inertia),the object will continue to move with the same velocity $v$ along the original orbit of the spacecraft.
Since there is no external force acting on the ball to change its state of motion,it will not fall to the Earth nor move away into space; it will simply follow the same path as the spacecraft.

(B) For a satellite revolving in a circular orbit of radius $r$ around the Earth of mass $M_e$,the gravitational force provides the necessary centripetal force.
$\frac{GM_e M}{r^2} = \frac{M v^2}{r}$
Since $v = \frac{2\pi r}{T}$,we substitute this into the equation:
$\frac{GM_e}{r^2} = \frac{(2\pi r / T)^2}{r}$
$\frac{GM_e}{r^2} = \frac{4\pi^2 r^2}{T^2 r}$
$T^2 = \frac{4\pi^2 r^3}{GM_e}$
$T = 2\pi \sqrt{\frac{r^3}{GM_e}}$
Thus,$T \propto \sqrt{\frac{r^3}{GM_e}}$. Comparing this with the given options,the correct proportionality is $T \propto \sqrt{\frac{r^3}{GM}}$.

(D) satellite that remains stationary with respect to the Earth's surface is called a geostationary satellite.
Its time period of revolution $(T)$ is equal to the time period of rotation of the Earth,which is $24 \text{ hours}$.
The formula for the orbital radius $(r)$ is given by $T = 2\pi \sqrt{\frac{r^3}{GM}}$.
Using $T = 86400 \text{ s}$,$G = 6.67 \times 10^{-11} \text{ Nm}^2/\text{kg}^2$,and $M = 5.97 \times 10^{24} \text{ kg}$,we find the orbital radius $r \approx 42200 \text{ km}$.
The height $(h)$ above the Earth's surface is $h = r - R_e$,where $R_e \approx 6400 \text{ km}$.
Therefore,$h = 42200 \text{ km} - 6400 \text{ km} = 35800 \text{ km}$,which is approximately $36000 \text{ km}$.

(A) The weight of an object is given by the formula $W = mg$,where $m$ is the mass of the object and $g$ is the acceleration due to gravity.
In an artificial satellite revolving around the Earth,the satellite and everything inside it are in a state of free fall towards the Earth.
Because the satellite is in free fall,the effective acceleration due to gravity $(g_{eff})$ experienced by the astronaut inside the satellite is $0$.
Therefore,the weight of the astronaut is $W = m \times 0 = 0$.

(D) communication satellite is typically a geostationary satellite.
Geostationary satellites orbit the Earth at the same angular velocity as the Earth's rotation.
Therefore,the time period of a geostationary satellite is equal to the time taken by the Earth to complete one rotation about its axis,which is $24\, hours$.

(C) The orbital speed $v$ of a satellite at a distance $r$ from the center of the Earth is given by $v = \sqrt{\frac{GM}{r}}$.
For a satellite very close to the surface,the distance from the center is $r_1 = R$,so $V_o = \sqrt{\frac{GM}{R}}$.
For a satellite at a height $h = 3R$,the distance from the center is $r_2 = R + h = R + 3R = 4R$.
The new orbital speed $v'$ is given by $v' = \sqrt{\frac{GM}{4R}} = \frac{1}{2} \sqrt{\frac{GM}{R}}$.
Substituting $V_o$ into the equation,we get $v' = \frac{1}{2} V_o = 0.5\,V_o$.

(D) geostationary satellite orbits the Earth in the equatorial plane with a time period of $24 \text{ hours}$.
To calculate its height $h$,we use the formula for the orbital radius $r = R_e + h$,where $R_e$ is the radius of the Earth.
Using Kepler's Third Law,$T^2 = \frac{4\pi^2 r^3}{GM}$,where $M$ is the mass of the Earth.
Substituting $T = 86400 \text{ s}$,$G = 6.67 \times 10^{-11} \text{ Nm}^2/\text{kg}^2$,and $M = 5.97 \times 10^{24} \text{ kg}$,we find $r \approx 42,200 \text{ km}$.
Since $R_e \approx 6,400 \text{ km}$,the height $h = r - R_e \approx 42,200 - 6,400 = 35,800 \text{ km}$.
This height is approximately $5.6$ to $6$ times the radius of the Earth $(6 \times 6,400 = 38,400 \text{ km})$.
Therefore,statement $(d)$ is the correct description.

(B) geostationary satellite orbits the Earth at an altitude of approximately $36,000 \ km$ above the Earth's surface.
Given the radius of the Earth $R = 6400 \ km$.
The distance from the centre of the Earth is $r = R + h$.
Substituting the values: $r = 6400 \ km + 36000 \ km = 42400 \ km$.
Now,expressing this in terms of $R$: $r = \frac{42400}{6400} R \approx 6.625 R$.
Rounding to the nearest integer,the distance is approximately $7 R$ from the centre of the Earth.

(D) The orbital radius $r$ of a satellite is determined by the balance of gravitational force and centripetal force: $\frac{GMm}{r^2} = \frac{mv^2}{r}$.
For a circular orbit,the angular momentum $L = mvr$ is conserved if no external torque acts on the system.
Since the gravitational force is a central force,it exerts no torque on the satellite.
Therefore,the angular momentum $L$ of the satellite remains unchanged even if the gravitational constant $G$ changes over time.

(B) The formula for the time period $T$ of a satellite at a 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 can write $GM = gR^2$.
Substituting $h = R$ and $GM = gR^2$ into the formula:
$T = 2\pi \sqrt{\frac{(R+R)^3}{gR^2}}$
$T = 2\pi \sqrt{\frac{(2R)^3}{gR^2}}$
$T = 2\pi \sqrt{\frac{8R^3}{gR^2}}$
$T = 2\pi \sqrt{\frac{8R}{g}}$
$T = 2\pi \cdot 2\sqrt{2} \sqrt{\frac{R}{g}}$
$T = 4\sqrt{2}\pi \sqrt{\frac{R}{g}}$.

(C) The orbital period $T$ of a satellite at a distance $r$ from the center of the Earth is given by Kepler's Third Law: $T = 2\pi \sqrt{\frac{r^3}{GM}}$.
Here,$r = R + h$,where $R$ is the radius of the Earth and $h$ is the height of the satellite above the surface.
Squaring both sides,we get: $T^2 = \frac{4\pi^2 r^3}{GM} = \frac{4\pi^2 (R+h)^3}{GM}$.
Rearranging for $(R+h)^3$: $(R+h)^3 = \frac{GMT^2}{4\pi^2}$.
Taking the cube root of both sides: $R+h = \left( \frac{GMT^2}{4\pi^2} \right)^{1/3}$.
Finally,solving for $h$: $h = \left( \frac{GMT^2}{4\pi^2} \right)^{1/3} - R$.

(D) The orbital radius $r$ is the distance from the center of the earth,given by $r = R + h$.
For the geostationary satellite,$h_1 = 6R$,so $r_1 = R + 6R = 7R$. The time period $T_1 = 24 \text{ hr}$.
For the second satellite,$h_2 = 2.5R$,so $r_2 = R + 2.5R = 3.5R$.
According to Kepler's Third Law,$T^2 \propto r^3$,which implies $\frac{T_2}{T_1} = \left( \frac{r_2}{r_1} \right)^{3/2}$.
Substituting the values: $\frac{T_2}{24} = \left( \frac{3.5R}{7R} \right)^{3/2} = \left( \frac{1}{2} \right)^{3/2}$.
$T_2 = 24 \times \frac{1}{2\sqrt{2}} = \frac{12}{\sqrt{2}} = 6\sqrt{2} \text{ hr}$.

(A) The orbital speed $v$ of a satellite (Moon) at a distance $r$ from the centre of the Earth is given by the formula: $v = \sqrt{\frac{GM}{r}}$.
Given:
$G = 6.67 \times 10^{-11} \, N m^2/kg^2$
$M = 6 \times 10^{24} \, kg$
$r = 384000 \, km = 384000 \times 10^3 \, m = 3.84 \times 10^8 \, m$
Substituting the values:
$v = \sqrt{\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{3.84 \times 10^8}}$
$v = \sqrt{\frac{40.02 \times 10^{13}}{3.84 \times 10^8}}$
$v = \sqrt{10.42 \times 10^5} \approx \sqrt{1.042 \times 10^6}$
$v \approx 1020 \, m/s$
Converting to $km/s$:
$v \approx 1.02 \, km/s \approx 1 \, km/s$.

(A) geostationary satellite must orbit the Earth in the equatorial plane.
This is because the satellite must appear stationary with respect to an observer on the Earth's surface.
For the satellite to remain fixed relative to the Earth's rotation,its orbital period must be exactly $24$ hours,and it must rotate in the same direction as the Earth (from west to east).
Therefore,it must be positioned directly above the equator.

(C) geostationary satellite orbits at an altitude $h$ above the Earth's surface such that its orbital period is $24 \ hours$.
Using the formula for orbital radius $r = \sqrt[3]{\frac{GM T^2}{4\pi^2}}$, the distance from the center of the Earth is approximately $42400 \ km$.
The altitude $h$ from the surface is $h = r - R_e = 42400 \ km - 6400 \ km = 36000 \ km$.
The question asks for the distance from the surface in terms of $R_e$.
Distance $h = 36000 \ km$.
To express this in terms of $R_e$, we write $h = n \times R_e$.
$n = \frac{36000}{6400} = 5.625$.
However, standard approximations often use $r \approx 6.6 R_e$, leading to $h \approx 5.6 R_e$. Given the options provided, the closest value representing the distance from the surface is approximately $5.6 R_e$, but since the question asks for the distance from the surface and the options provided are numerical constants, we evaluate the ratio $h/R_e \approx 5.6$. Re-evaluating the provided options, $5.6$ is not listed, but $6.56$ is often cited in textbooks as the orbital radius $r/R_e$. Assuming the question implies the distance from the center, $r/R_e = 42400/6400 = 6.625 \approx 6.6$.

(C) The orbital speed $v_0$ of a satellite is given by the formula $v_0 = \sqrt{\frac{GM}{r}}$.
Using the relation $GM = gR^2$,where $g = 10\, m/s^2$ and $R = 6400\, km = 6.4 \times 10^6\, m$,we get:
$v_0 = \sqrt{\frac{gR^2}{r}}$
Given $r = 8000\, km = 8 \times 10^6\, m$:
$v_0 = \sqrt{\frac{10 \times (6.4 \times 10^6)^2}{8 \times 10^6}}$
$v_0 = \sqrt{\frac{10 \times 40.96 \times 10^{12}}{8 \times 10^6}}$
$v_0 = \sqrt{51.2 \times 10^6} \approx 7155\, m/s$
$v_0 \approx 7.15\, km/s$.

(B) The orbital velocity $v$ of a satellite revolving at a distance $r$ from the center of the Earth is given by the formula $v = \sqrt{\frac{GM}{r}}$.
For a satellite revolving close to the Earth's surface,the orbital radius $r \approx R$,where $R$ is the radius of the Earth.
Thus,$v = \sqrt{\frac{GM}{R}}$.
We know that the acceleration due to gravity at the Earth's surface is $g = \frac{GM}{R^2}$,which implies $GM = gR^2$.
Substituting $GM = gR^2$ into the orbital velocity formula:
$v = \sqrt{\frac{gR^2}{R}} = \sqrt{gR}$.

(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,$v \propto R^0$.