Kepler’s laws of Planetary Motion Questions in English

(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 $(r)$: $T^2 \propto r^3$,which implies $T \propto r^{3/2}$.
Given that the initial time period $T_1 = 24$ hours (for a geostationary orbit) and the new radius $r_2 = 2r_1$.
The new time period $T_2$ is given by: $T_2 = T_1 \times (r_2/r_1)^{3/2}$.
Substituting the values: $T_2 = 24 \times (2)^{3/2} = 24 \times 2\sqrt{2} = 48\sqrt{2}$ hours.

(B) According to Kepler's third law of planetary motion, the square of the time period $T$ is proportional to the cube of the orbital radius $r$, i.e., $T^2 \propto r^3$.
Given that the orbital radius of satellite $S$ is $r_s = 4r_c$, where $r_c$ is the orbital radius of the communication satellite $C$.
The time period of a communication satellite is $T_c = 1 \text{ day}$.
Using the relation $\frac{T_s}{T_c} = \left(\frac{r_s}{r_c}\right)^{3/2}$, we substitute the given values:
$\frac{T_s}{1} = (4)^{3/2} = (2^2)^{3/2} = 2^3 = 8$.
Therefore, the time period of satellite $S$ is $8 \text{ days}$.

(D) 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 orbital radius $(R)$.
Mathematically,this is expressed as $T^2 \propto R^3$.
Therefore,the ratio $\frac{T^2}{R^3} = \text{constant}$.
Since this ratio is a constant value,it does not depend on the orbital radius $(R)$ of the satellite.

(D) 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 $T \propto R^{3/2}$.
Given the initial time period $T_1 = 5$ hours.
Let the initial separation be $R_1$ and the new separation be $R_2 = 4R_1$.
The new time period $T_2$ is given by the ratio: $\frac{T_2}{T_1} = \left( \frac{R_2}{R_1} \right)^{3/2}$.
Substituting the values: $\frac{T_2}{5} = \left( \frac{4R_1}{R_1} \right)^{3/2} = (4)^{3/2}$.
Calculating the exponent: $(4)^{3/2} = (2^2)^{3/2} = 2^3 = 8$.
Therefore,$T_2 = 5 \times 8 = 40$ hours.

(B) 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 $(r)$: $T^2 \propto r^3$.
Given for the first satellite: $T_1 = 1 \text{ day}$,$r_1 = R$.
Given for the second satellite: $T_2 = 8 \text{ days}$,$r_2 = ?$.
Using the ratio: $\left( \frac{T_2}{T_1} \right)^2 = \left( \frac{r_2}{r_1} \right)^3$.
Substituting the values: $\left( \frac{8}{1} \right)^2 = \left( \frac{r_2}{R} \right)^3$.
$64 = \left( \frac{r_2}{R} \right)^3$.
Taking the cube root on both sides: $\sqrt[3]{64} = \frac{r_2}{R}$.
$4 = \frac{r_2}{R}$,which implies $r_2 = 4R$.

(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.,$T^2 \propto r^3$.
Let $T_s$ and $r_s$ be the time period and orbital radius of the satellite,and $T_m$ and $r_m$ be the time period and orbital radius of the moon.
Given: $r_s = \frac{1}{2} r_m$ and $T_m = 1$ lunar month.
Using the ratio: $\frac{T_s}{T_m} = \left( \frac{r_s}{r_m} \right)^{3/2}$.
Substituting the values: $\frac{T_s}{1} = \left( \frac{1/2 r_m}{r_m} \right)^{3/2} = \left( \frac{1}{2} \right)^{3/2} = 2^{-3/2}$.
Therefore,the satellite completes one revolution in $2^{-3/2}$ lunar month.

(A) 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: $T^2 \propto R^3$.
Given:
$R_1 = 10^{13} \ m$ (Neptune)
$R_2 = 10^{12} \ m$ (Saturn)
The ratio of the time periods is given by:
$\frac{T_1}{T_2} = \left( \frac{R_1}{R_2} \right)^{3/2}$
Substituting the values:
$\frac{T_1}{T_2} = \left( \frac{10^{13}}{10^{12}} \right)^{3/2}$
$\frac{T_1}{T_2} = (10)^{3/2}$
$\frac{T_1}{T_2} = 10^1 \cdot 10^{1/2} = 10\sqrt{10}$

(C) According to Kepler's second law of planetary motion,the areal velocity of a planet revolving around the sun is constant.
This means that the radius vector sweeps out equal areas in equal intervals of time.
Given that the shaded areas $A$ and $B$ are equal,the time taken to sweep these areas must also be equal.
Therefore,if ${t_1}$ is the time taken to move from $a$ to $b$ (sweeping area $A$) and ${t_2}$ is the time taken to move from $d$ to $c$ (sweeping area $B$),then ${t_1} = {t_2}$.

(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.,$T^2 \propto R^3$.
Given for the first satellite: $T_1 = T$ and $R_1 = R$.
For the second satellite: $R_2 = 4R$ and we need to find $T_2$.
Using the ratio formula:
$\frac{T_2}{T_1} = \left( \frac{R_2}{R_1} \right)^{3/2}$
Substituting the values:
$\frac{T_2}{T} = \left( \frac{4R}{R} \right)^{3/2}$
$\frac{T_2}{T} = (4)^{3/2}$
$\frac{T_2}{T} = (2^2)^{3/2} = 2^3 = 8$
Therefore,$T_2 = 8T$.

(B) According to Kepler's first law of planetary motion,also known as the Law of Orbits,every planet moves around the Sun (or a star) in an elliptical orbit with the Sun at one of the two foci. Therefore,the correct shape of the orbit is an ellipse.

(C) In an inverse square field,the force $F$ acting on a body of mass $m$ moving in a circular orbit of radius $r$ is given by $F = \frac{k}{r^2}$,where $k$ is a constant.
For circular motion,this force provides the necessary centripetal force: $\frac{k}{r^2} = \frac{mv^2}{r}$.
Simplifying this,we get $v^2 = \frac{k}{mr}$,which implies $v = \sqrt{\frac{k}{mr}}$.
The time period $T$ for one revolution is $T = \frac{2\pi r}{v}$.
Substituting the expression for $v$: $T = \frac{2\pi r}{\sqrt{\frac{k}{mr}}} = 2\pi \sqrt{\frac{m}{k}} \cdot r^{3/2}$.
Squaring both sides,we get $T^2 = 4\pi^2 \frac{m}{k} r^3$.
Since $4, \pi, m,$ and $k$ are constants,we have $T^2 \propto r^3$.

(B) According to Kepler's third law of planetary motion,the square of the time period $(T)$ of a planet is directly proportional to the cube of its semi-major axis $(r)$,i.e.,$T^2 \propto r^3$.
Given that the new distance $r' = \frac{1}{4}r$.
Using the ratio formula: $\left( \frac{T'}{T} \right)^2 = \left( \frac{r'}{r} \right)^3$.
Substituting the value of $r'$: $\left( \frac{T'}{T} \right)^2 = \left( \frac{1}{4} \right)^3 = \frac{1}{64}$.
Taking the square root on both sides: $\frac{T'}{T} = \sqrt{\frac{1}{64}} = \frac{1}{8}$.
Therefore,the new duration of the year $T' = \frac{1}{8}T$.

(B) According to Kepler's second law of planetary motion,the gravitational force exerted by the Sun on the Earth is a central force.
Since the torque exerted by a central force about the center of the Sun is zero $(\tau = r \times F = 0)$,the angular momentum $(L)$ of the Earth remains constant throughout its orbit.
However,the distance between the Earth and the Sun changes in an elliptical orbit,which causes both the kinetic energy and the potential energy of the Earth to vary with its position.

(C) According to Kepler's second law of planetary motion,the angular momentum of a planet revolving around the sun remains constant.
The angular momentum $L$ is given by $L = mvr \sin(\theta)$. At the points of closest approach (perihelion) and farthest distance (aphelion),the velocity vector is perpendicular to the position vector,so $\theta = 90^\circ$ and $\sin(90^\circ) = 1$.
Therefore,the conservation of angular momentum gives:
$m v_1 d_1 = m v_2 d_2$
Canceling the mass $m$ from both sides:
$v_1 d_1 = v_2 d_2$
Solving for $v_2$:
$v_2 = \frac{v_1 d_1}{d_2}$

(D) According to Kepler's third law of planetary motion,the square of the time period of revolution of a planet is directly proportional to the cube of the semi-major axis of its orbit.
Mathematically,$T^2 \propto r^3$.
For two planets,we can write the ratio as:
$\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}$
Taking the square root on both sides:
$\frac{T_1}{T_2} = \left( \frac{r_1}{r_2} \right)^{3/2}$

(B) Kepler's second law states that the areal velocity $(dA/dt)$ of a planet revolving around the Sun is constant.
Since the gravitational force exerted by the Sun on the planet is a central force,the torque $(\tau)$ acting on the planet about the Sun is zero.
According to the relation $\tau = dL/dt$,if $\tau = 0$,then the angular momentum $(L)$ of the planet remains constant.
The areal velocity is given by the expression $\frac{dA}{dt} = \frac{L}{2m}$,where $m$ is the mass of the planet.
Since $L$ and $m$ are constants,the areal velocity is constant.
Therefore,Kepler's second law is a direct consequence of the law of conservation of angular momentum.

(B) Kepler's third law of planetary motion, also known as the law of periods, states that the square of the time period of revolution $(T)$ of a planet is directly proportional to the cube of its semi-major axis $(r)$ of its orbit around the sun.
Mathematically, this is expressed as $T^2 \propto r^3$.
This can be rewritten as $\frac{T^2}{r^3} = \text{constant}$.
By using the laws of exponents, this is equivalent to $T^2 r^{-3} = \text{constant}$.
Therefore, the correct option is $B$.

(D) According to Kepler's third law of planetary motion, the square of the time period of revolution $(T)$ is proportional to the cube of the semi-major axis $(r)$ of the orbit: $T^2 \propto r^3$.
Given that the mean distance of the planet $(r_p)$ is $1.588$ times the mean distance of the earth $(r_e)$, we have $r_p = 1.588 \times r_e$.
The ratio of the time periods is given by: $\frac{T_p}{T_e} = \left( \frac{r_p}{r_e} \right)^{3/2}$.
Substituting the given values: $\frac{T_p}{T_e} = (1.588)^{3/2}$.
Since $1.588 \approx (2)^{2/3}$, we have $(1.588)^{3/2} \approx ((2)^{2/3})^{3/2} = 2^1 = 2$.
Therefore, $T_p = 2 \times T_e$. Since $T_e = 1 \text{ year}$, the revolution time of the planet is $2 \text{ years}$.

(D) 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 its orbital radius $r$,i.e.,$T^2 \propto r^3$.
This relationship is independent of the mass of the satellite.
Given for satellite $A$: $r_A = r$.
Given for satellite $B$: $r_B = 2r$.
The ratio of their time periods is given by:
$\frac{T_A}{T_B} = \left( \frac{r_A}{r_B} \right)^{3/2} = \left( \frac{r}{2r} \right)^{3/2} = \left( \frac{1}{2} \right)^{3/2} = \frac{1}{2^{3/2}} = \frac{1}{2 \cdot 2^{1/2}} = \frac{1}{2\sqrt{2}}$.
Thus,the ratio is $1:2\sqrt{2}$.

(C) According to Kepler's Third Law of Planetary Motion,the square of the time period $(T)$ is directly proportional to the cube of the semi-major axis $(r)$ of the orbit: $T^2 \propto r^3$.
Let the initial radius be $r_1$ and the initial time period be $T_1$. Let the new radius be $r_2 = \frac{1}{4} r_1$ and the new time period be $T_2$.
Using the proportionality: $\frac{T_2^2}{T_1^2} = \left( \frac{r_2}{r_1} \right)^3$.
Substituting the values: $\frac{T_2^2}{T_1^2} = \left( \frac{1}{4} \right)^3 = \frac{1}{64}$.
Taking the square root on both sides: $\frac{T_2}{T_1} = \sqrt{\frac{1}{64}} = \frac{1}{8}$.
Therefore,the new duration of the year will be $\frac{1}{8}$ times the original duration.

(B) According to Kepler's third law of planetary motion,the square of the time period $(T)$ is directly proportional to the cube of the semi-major axis $(r)$ of the orbit: $T^2 \propto r^3$.
Given the initial state: $T_1 = 1$ year and $r_1 = r$.
For the new state: $r_2 = 2r$.
Using the ratio: $\frac{T_2}{T_1} = \left( \frac{r_2}{r_1} \right)^{3/2}$.
Substituting the values: $\frac{T_2}{1} = \left( \frac{2r}{r} \right)^{3/2} = (2)^{3/2} = 2\sqrt{2}$.
Therefore,the new period of revolution $T_2 = 2\sqrt{2}$ years.

(C) Johannes Kepler formulated three fundamental laws that describe the motion of planets around the Sun. These are known as Kepler's laws of planetary motion.
$1$. The Law of Orbits: All planets move in elliptical orbits with the Sun at one of the foci.
$2$. The Law of Areas: $A$ line that connects a planet to the Sun sweeps out equal areas in equal times.
$3$. The Law of Periods: The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.
Therefore,the correct option is $C$.

(B) According to Kepler's third law,the square of the orbital period $T$ is proportional to the cube of the semi-major axis $r$: $T^2 \propto r^3$.
Since angular velocity $\omega = 2\pi / T$,we have $T = 2\pi / \omega$,which implies $T^2 \propto 1/\omega^2$.
Substituting this into Kepler's law: $1/\omega^2 \propto r^3$,which gives $\omega^2 \propto 1/r^3$ or $\omega \propto r^{-3/2}$.
Rearranging for $r$,we get $r \propto \omega^{-2/3}$.
Given that the body revolves $27$ times faster than the earth,$\omega_{\text{body}} = 27 \omega_{\text{earth}}$.
The ratio of the radii is $\frac{r_{\text{body}}}{r_{\text{earth}}} = \left( \frac{\omega_{\text{earth}}}{\omega_{\text{body}}} \right)^{2/3}$.
Substituting the values: $\frac{r_{\text{body}}}{r_{\text{earth}}} = \left( \frac{1}{27} \right)^{2/3} = \left( (3^3)^{-1} \right)^{2/3} = (3^{-3})^{2/3} = 3^{-2} = \frac{1}{9}$.

(B) According to Kepler's Third Law of Planetary Motion,the square of the time period $T$ is proportional to the cube of the mean distance $d$ from the sun: $T^2 \propto d^3$.
Since frequency $n$ is the reciprocal of the time period $(n = 1/T)$,we have $T = 1/n$.
Substituting this into the law: $(1/n)^2 \propto d^3$,which implies $1/n^2 \propto d^3$.
Rearranging this gives $n^2 d^3 = \text{constant}$.
Therefore,for two planets,we have $n_1^2 d_1^3 = n_2^2 d_2^3$.

(A) 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: $T^2 \propto R^3$ or $T \propto R^{3/2}$.
Given that the distance of the planet from the sun is $R_p = 5 R_e$,where $R_e$ is the distance between the earth and the sun.
The time period of the earth is $T_e = 1$ year.
Using the ratio: $\frac{T_p}{T_e} = \left( \frac{R_p}{R_e} \right)^{3/2}$.
Substituting the values: $\frac{T_p}{1} = (5)^{3/2}$.
Therefore,the time period of the planet is $T_p = 5^{3/2}$ years.

(A) According to Kepler's second law of planetary motion,the radius vector joining the sun and the planet sweeps out equal areas in equal intervals of time.
In the given figure,the sun $S$ is at one of the foci of the elliptical orbit.
The area swept by the planet in path $DAB$ is less than the area swept in path $BCD$ because the sun is closer to the arc $DAB$.
Since the area swept in path $DAB$ is smaller than the area swept in path $BCD$,the time taken to travel $DAB$ must be less than the time taken to travel $BCD$ to satisfy the condition of equal area in equal time.
Alternatively,according to the law of conservation of angular momentum,the orbital velocity of the planet is higher when it is closer to the sun (near $A$) and lower when it is farther away (near $C$).
Thus,the planet travels faster along the path $DAB$ and slower along the path $BCD$,confirming that the time taken for $DAB$ is less than that for $BCD$.

(C) According to Kepler's second law of planetary motion,the areal velocity of a planet remains constant.
This implies that the orbital velocity $v$ of a planet is inversely proportional to its distance $r$ from the sun $(v \propto 1/r)$.
Therefore,the orbital velocity is minimum when the distance $r$ from the sun is maximum.
Looking at the provided elliptical path,point $C$ is at the maximum distance from the sun $(S)$.
Thus,the orbital velocity of the planet is minimum at point $C$.

(B) According to Kepler's third law of planetary motion,the square of the time period $(T)$ is directly proportional to the cube of the semi-major axis $(R)$ of the orbit: $T^2 \propto R^3$.
Let $T_1$ and $R_1$ be the time period and orbital radius of the Earth,and $T_2$ and $R_2$ be the time period and orbital radius of the planet.
Given: $T_1 = 1 \text{ year}$,$R_1 = R$,and $R_2 = 2R$.
Using the ratio formula: $\frac{T_2}{T_1} = \left( \frac{R_2}{R_1} \right)^{3/2}$.
Substituting the values: $T_2 = 1 \times \left( \frac{2R}{R} \right)^{3/2} = 2^{3/2}$.
Calculating the value: $2^{3/2} = \sqrt{2^3} = \sqrt{8} \approx 2.828 \text{ years}$.
Rounding to the nearest option,the time period is $2.8 \text{ years}$.

(A) According to Kepler's third law of planetary motion,also known as the law of periods,the square of the orbital period $(T)$ of a planet (or satellite) is directly proportional to the cube of the semi-major axis $(R)$ of its orbit.
Mathematically,this is expressed as $T^2 \propto R^3$.
Therefore,the correct relationship is $T^2 \propto R^3$.

(C) The areal velocity of a planet is given by the rate of change of area swept by the position vector,which is $\frac{dA}{dt} = \frac{1}{2} r^2 \omega$.
Since the angular momentum $L = mvr = mr^2\omega$ is constant for a central force,the areal velocity is $\frac{dA}{dt} = \frac{L}{2m}$.
From the expression $\frac{dA}{dt} = \frac{1}{2} r^2 \omega$,it is clear that $\frac{dA}{dt} \propto \omega r^2$.
Therefore,the correct relation is $\frac{dA}{dt} \propto \omega r^2$.

(B) According to Kepler's third law of planetary motion,the square of the time period of revolution $(T)$ is directly proportional to the cube of the semi-major axis $(a)$ of the orbit: $T^2 \propto a^3$.
Given the ratio of distances (semi-major axes) of two planets from the sun is $\frac{a_1}{a_2} = 1.38$.
We need to find the ratio of their periods of revolution $\frac{T_1}{T_2}$.
From Kepler's third law: $\left(\frac{T_1}{T_2}\right)^2 = \left(\frac{a_1}{a_2}\right)^3$.
Taking the square root on both sides: $\frac{T_1}{T_2} = \left(\frac{a_1}{a_2}\right)^{3/2}$.
Substituting the given value: $\frac{T_1}{T_2} = (1.38)^{3/2}$.

(D) According to Kepler's second law of planetary motion,a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.
This implies that the areal velocity $(dA/dt)$ of the planet remains constant throughout its elliptical orbit.
While the tangential,angular,and radial velocities change as the distance from the Sun varies,the areal velocity is conserved due to the conservation of angular momentum in a central force field.

(A) According to Kepler's third law of planetary motion,the square of the time period $T$ is proportional to the cube of the orbital radius $R$,i.e.,$T^2 \propto R^3$.
Let $T_E$ and $R_E$ be the time period and orbital radius of the Earth,and $T_P$ and $R_P$ be those of the new planet.
Given $R_P = 2R_E$ and $T_E = 365$ days.
Using the ratio formula:
$\left( \frac{T_P}{T_E} \right)^2 = \left( \frac{R_P}{R_E} \right)^3 = \left( \frac{2R_E}{R_E} \right)^3 = 2^3 = 8$.
Taking the square root on both sides:
$\frac{T_P}{T_E} = \sqrt{8} = 2\sqrt{2}$.
Substituting the value of $T_E$:
$T_P = 2\sqrt{2} \times 365 \approx 2.8284 \times 365 \approx 1032.37$ days.
Rounding to the nearest whole number,the time period is $1032$ days.

(A) Kepler's second law (law of areas) is a direct consequence of the conservation of angular momentum,which holds for any central force. Since the force remains a central force,the torque $\tau = r \times F$ is zero,meaning angular momentum $L$ is conserved. Therefore,Kepler's law of areas still holds.
Kepler's third law (law of periods) states that $T^2 \propto r^3$. This is derived specifically for an inverse-square law $(F \propto 1/r^2)$. If the force follows an inverse-cube law $(F \propto 1/r^3)$,the centripetal force requirement $mv^2/r = k/r^3$ implies $v^2 \propto 1/r^2$,or $v \propto 1/r$. Since $v = 2\pi r/T$,we get $1/r \propto r/T$,which leads to $T \propto r^2$,or $T^2 \propto r^4$. Thus,the law of periods does not hold.

(B) According to Kepler's second law,the angular momentum of a planet remains constant. The speed of a planet in an elliptical orbit is given by $v = \frac{L}{mr}$,where $L$ is angular momentum,$m$ is mass,and $r$ is the distance from the Sun.
At perihelion (closest approach),$r_{\min} = a(1-e)$,so $v_{\max} = \frac{L}{m a(1-e)}$.
At aphelion (farthest distance),$r_{\max} = a(1+e)$,so $v_{\min} = \frac{L}{m a(1+e)}$.
The ratio of maximum speed to minimum speed is $\frac{v_{\max}}{v_{\min}} = \frac{1+e}{1-e}$.
Given $e = 0.0167$,we have $\frac{v_{\max}}{v_{\min}} = \frac{1 + 0.0167}{1 - 0.0167} = \frac{1.0167}{0.9833} \approx 1.034$.

(C) 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 $(r)$: $T^2 \propto r^3$ or $T \propto r^{3/2}$.
Given that the new distance $r_2 = \frac{1}{4} r_1$.
Substituting this into the ratio formula:
$\frac{T_2}{T_1} = \left( \frac{r_2}{r_1} \right)^{3/2} = \left( \frac{1}{4} \right)^{3/2} = \frac{1}{8}$.
Therefore,the length of the year (orbital period) is reduced to $\frac{1}{8}$ of its original value.

(B) According to Kepler's third law,the square of the orbital period $(T)$ is proportional to the cube of the semi-major axis $(r)$,i.e.,$T^2 \propto r^3$.
Let $T_1 = 365$ days and $r_1$ be the present distance.
Given $r_2 = \frac{1}{2} r_1$.
Then,$\left( \frac{T_2}{T_1} \right)^2 = \left( \frac{r_2}{r_1} \right)^3 = \left( \frac{1}{2} \right)^3 = \frac{1}{8}$.
Taking the square root on both sides,$\frac{T_2}{T_1} = \sqrt{\frac{1}{8}} = \frac{1}{2\sqrt{2}}$.
Therefore,$T_2 = \frac{T_1}{2\sqrt{2}} = \frac{365}{2 \times 1.414} \approx \frac{365}{2.828} \approx 129$ days.

(C) According to Kepler's third law of planetary motion,the square of the time period $(T)$ of a planet is directly proportional to the cube of the semi-major axis $(R)$ of its orbit.
Mathematically,this is expressed as $T^2 \propto R^3$,or $T^2 = kR^3$,where $k$ is a constant.
This equation represents a straight line passing through the origin when $T^2$ is plotted on the y-axis and $R^3$ is plotted on the x-axis.
Therefore,the graph that shows a straight line passing through the origin represents this relationship,which corresponds to Graph $C$.

(B) According to Kepler's third law of planetary motion,the square of the orbital period $T$ is directly proportional to the cube of the semi-major axis $r$ of its orbit,expressed as $T^2 \propto r^3$.
This implies that as the distance $r$ from the Sun increases,the period of revolution $T$ also increases.
The average distances of the planets from the Sun are: Mercury,Venus,Earth,Mars,Jupiter,Saturn,Uranus,Neptune,and Pluto.
Comparing the given options,the order of increasing distance from the Sun is Mars,Saturn,and Pluto.
Therefore,the period of revolution increases in the order of Mars,Saturn,and Pluto.

(C) The gravitational force exerted by the Sun on the planets is a central force. $A$ central force has zero torque about the center of force (the Sun). Since the net external torque acting on the planet is zero,the angular momentum of the planet remains constant. Therefore,the motion of planets in the solar system is an example of the conservation of angular momentum.

(A) According to Kepler's Third Law,the orbital period $T$ is related to the orbital radius $R$ as $T \propto R^{3/2}$.
Also,the orbital speed $v$ is given by $v = \frac{2\pi R}{T}$.
Substituting $T \propto R^{3/2}$ into the speed formula,we get $v \propto \frac{R}{R^{3/2}} = R^{-1/2}$,which means $v \propto \frac{1}{\sqrt{R}}$.
Let $v_1$ and $R_1$ be the speed and radius of the Earth,and $v_2$ and $R_2$ be the speed and radius of the planet.
Given $v_2 = 2v_1$,we have $\frac{v_1}{v_2} = \sqrt{\frac{R_2}{R_1}}$.
Substituting the values: $\frac{v_1}{2v_1} = \sqrt{\frac{R_2}{R}}$.
$\frac{1}{2} = \sqrt{\frac{R_2}{R}}$.
Squaring both sides: $\frac{1}{4} = \frac{R_2}{R}$.
Therefore,$R_2 = \frac{R}{4}$.

(B) According to Kepler's second law,the angular momentum of a planet revolving around the sun is conserved.
$L = mvr \sin(\theta) = \text{constant}$.
At the perihelion and aphelion,the velocity vector is perpendicular to the radius vector,so $L = mvr$.
Since the mass $m$ of the earth is constant,we have $vr = \text{constant}$,which implies $v \propto (1/r)$.
This means the orbital speed $v$ is inversely proportional to the distance $r$ from the sun.
The speed of the earth will be maximum when its distance from the sun is minimum.
Looking at the elliptical orbit,the distance is minimum at point $A$ (the perihelion).
Therefore,the speed of the earth is maximum at point $A$.

(C) The areal velocity of a planet is given by the formula $\frac{dA}{dt} = \frac{L}{2m}$,where $L$ is the angular momentum and $m$ is the mass of the planet.
Since angular momentum $L = mvr$ and linear velocity $v = r\omega$,we can substitute these into the equation.
$\frac{dA}{dt} = \frac{m(r\omega)r}{2m}$.
Simplifying the expression,we get $\frac{dA}{dt} = \frac{1}{2} \omega r^2$.
Therefore,the areal velocity is proportional to $\omega r^2$.

(A) According to Kepler's second law of planetary motion,the areal velocity of a planet revolving around the Sun is constant. This means the planet sweeps out equal areas in equal intervals of time.
In the given figure,the Sun $S$ is at one of the foci of the elliptical orbit.
The path $DAB$ is closer to the Sun compared to the path $BCD$.
Since the area swept by the radius vector in path $DAB$ is smaller than the area swept in path $BCD$,the time taken to travel $DAB$ is less than the time taken to travel $BCD$.

(D) According to Kepler's third law,the time period $T$ of a satellite is related to the orbital radius $r$ by the relation $T \propto r^{3/2}$.
Taking the logarithmic derivative of both sides,we get $\frac{\Delta T}{T} = \frac{3}{2} \frac{\Delta r}{r}$.
Given that the radius increases from $R$ to $1.02 R$,the change in radius is $\Delta r = 1.02 R - R = 0.02 R$.
Therefore,the fractional change in radius is $\frac{\Delta r}{r} = \frac{0.02 R}{R} = 0.02$ or $2\%$.
Substituting this into the derivative equation: $\frac{\Delta T}{T} = \frac{3}{2} \times 2\% = 3\%$.
Thus,the time period increases by $3\%$.

(B) The areal velocity of a radius vector is given by $\frac{dA}{dt} = \frac{1}{2} r^2 \frac{d\theta}{dt} = \frac{1}{2} r^2 \omega$.
The angular momentum of the planet is $J = I \omega = mr^2 \omega$,where $m$ is the mass of the planet,$r$ is the distance from the Sun,and $\omega$ is the angular velocity.
From this,we can express $r^2 \omega$ as $\frac{J}{m}$.
Substituting this into the areal velocity formula: $\frac{dA}{dt} = \frac{1}{2} \left( \frac{J}{m} \right) = \frac{J}{2m}$.

(B) According to Kepler's Second Law of Planetary Motion,a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.
This means the areal velocity $\frac{dA}{dt}$ is constant.
Therefore,the time taken to sweep an area is directly proportional to the area swept.
Given that the area $SCD = 2 \times \text{Area } SAB$.
Since $t_1$ is the time taken to sweep area $SCD$ and $t_2$ is the time taken to sweep area $SAB$,we have:
$\frac{t_1}{t_2} = \frac{\text{Area } SCD}{\text{Area } SAB} = \frac{2 \times \text{Area } SAB}{\text{Area } SAB} = 2$.
Thus,$t_1 = 2t_2$.

(B) The gravitational force of attraction between the sun and the planet provides the necessary centripetal force for the planet's orbit.
$\therefore \frac{GMm}{r^2} = \frac{mv^2}{r} \implies v^2 = \frac{GM}{r} \quad \dots(i)$
The time period $(T)$ of the planet is given by $T = \frac{2\pi r}{v}$.
Squaring both sides,we get $T^2 = \frac{4\pi^2 r^2}{v^2}$.
Substituting the value of $v^2$ from equation $(i)$:
$T^2 = \frac{4\pi^2 r^2}{(\frac{GM}{r})} = \frac{4\pi^2 r^3}{GM} \quad \dots(ii)$
According to the problem,Kepler's third law is given as:
$T^2 = Kr^3 \quad \dots(iii)$
Comparing equations $(ii)$ and $(iii)$,we get:
$K = \frac{4\pi^2}{GM} \implies GMK = 4\pi^2$.

(B) According to Kepler's second law of planetary motion,the areal velocity of a planet is constant. This implies that the planet moves faster when it is closer to the Sun and slower when it is farther away.
Point $A$ is the perihelion (closest to the Sun),and point $C$ is the aphelion (farthest from the Sun). Point $B$ is at an intermediate distance.
Therefore,the orbital speeds at these positions follow the relation: $v_A > v_B > v_C$.
Since the kinetic energy $K$ is given by $K = \frac{1}{2}mv^2$,it is directly proportional to the square of the speed $(K \propto v^2)$.
Thus,the kinetic energies at these positions follow the relation: $K_A > K_B > K_C$.