Energy consideration in Planetary and Satellite motion Questions in English

(A) The centripetal force is given by $F = \frac{mv^2}{r} = \frac{k}{r^2}$.
From this,we get $mv^2 = \frac{k}{r}$.
The kinetic energy $(K.E.)$ is given by $K.E. = \frac{1}{2}mv^2 = \frac{k}{2r}$.
The potential energy $(P.E.)$ is found by integrating the force: $P.E. = -\int F dr = -\int \frac{k}{r^2} dr = -\frac{k}{r}$.
The total energy $(E)$ is the sum of kinetic and potential energy: $E = K.E. + P.E. = \frac{k}{2r} - \frac{k}{r} = -\frac{k}{2r}$.

(A) The kinetic energy $(K.E.)$ of a satellite of mass $m$ orbiting a planet of mass $M$ at a distance $R$ is given by the formula:
$K.E. = \frac{GMm}{2R}$
Here,$G$ is the universal gravitational constant,$M$ is the mass of the Earth,and $m$ is the mass of the satellite.
Since $G$,$M$,and $m$ are constants for a given system,the kinetic energy is inversely proportional to the radius of the orbit $R$.
Therefore,$K.E. \propto \frac{1}{R}$.

(B) The kinetic energy $(K)$ of a satellite of mass $m$ orbiting at a distance $r$ from the center of the Earth (mass $M$) is given by $K = \frac{GMm}{2r}$.
The potential energy $(U)$ of the satellite is given by $U = -\frac{GMm}{r}$.
The ratio of kinetic energy to potential energy is $\frac{K}{U} = \frac{GMm/2r}{-GMm/r} = -\frac{1}{2}$.
Note: The magnitude ratio is $1/2$,but considering the sign,the ratio is $-1/2$.

(D) The total mechanical energy $E$ of a satellite in a circular orbit is given by the formula $E = -\frac{GMm}{2r}$.
Here,$G$ is the gravitational constant,$M$ is the mass of the planet,$m$ is the mass of the satellite,and $r$ is the radius of the orbit.
For satellites $A$ and $B$,the ratio of their total mechanical energies is given by:
$\frac{E_A}{E_B} = \frac{-\frac{GMm_A}{2r_A}}{-\frac{GMm_B}{2r_B}} = \frac{m_A}{m_B} \times \frac{r_B}{r_A}$.
Given the mass ratio $\frac{m_A}{m_B} = \frac{3}{1}$ and the radius ratio $\frac{r_A}{r_B} = \frac{r}{4r} = \frac{1}{4}$,we have $\frac{r_B}{r_A} = \frac{4}{1}$.
Substituting these values:
$\frac{E_A}{E_B} = \frac{3}{1} \times \frac{4}{1} = \frac{12}{1}$.
Therefore,the ratio is $12:1$.

(A) For a satellite in a circular orbit,the gravitational force provides the necessary centripetal force: $\frac{GMm}{r^2} = \frac{Mv^2}{r}$.
This implies that the potential energy $U = -\frac{GMm}{r} = -Mv^2$.
The kinetic energy $K = \frac{1}{2}Mv^2$.
The total energy $E$ is the sum of kinetic and potential energy: $E = K + U = \frac{1}{2}Mv^2 - Mv^2 = -\frac{1}{2}Mv^2$.

(C) The total energy $(E)$ of a satellite in a circular orbit of radius $r$ is given by $E = -\frac{GMm}{2r}$.
When the satellite loses energy,its total energy becomes more negative (i.e.,the magnitude of the binding energy increases,but the total energy decreases).
Since $E = -\frac{GMm}{2r}$,for $E$ to decrease (become more negative),the radius $r$ must decrease.
The orbital speed of a satellite is given by $v = \sqrt{\frac{GM}{r}}$.
As the radius $r$ decreases,the orbital speed $v$ must increase.
Therefore,$r$ will decrease and $v$ will increase.

(D) The total mechanical energy $(E)$ of a satellite is the sum of its kinetic energy $(K.E.)$ and potential energy $(U)$.
The gravitational potential energy of a satellite of mass $m$ at a distance $r$ from the centre of the Earth (mass $M$) is given by $U = -\frac{GMm}{r}$.
For a satellite in a circular orbit,the centripetal force is provided by the gravitational force:
$\frac{mv^2}{r} = \frac{GMm}{r^2} \implies mv^2 = \frac{GMm}{r}$.
The kinetic energy is $K.E. = \frac{1}{2}mv^2 = \frac{1}{2} \left( \frac{GMm}{r} \right) = \frac{GMm}{2r}$.
The total mechanical energy is $E = K.E. + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}$.

(C) For a satellite in a circular orbit,the potential energy $U$ is given by $U = -\frac{GMm}{r}$.
The total energy $E_0$ is the sum of kinetic energy $K$ and potential energy $U$,which is $E_0 = K + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}$.
Comparing the expressions for $U$ and $E_0$,we can see that $U = 2 \times (-\frac{GMm}{2r}) = 2 E_0$.
Therefore,the potential energy of the satellite is $2 E_0$.

(D) For a particle moving in a circular path,the centripetal force is provided by the given force: $mv^2/r = k/r^2$.
From this,the kinetic energy $(K.E.)$ is: $K.E. = 1/2 mv^2 = k/(2r)$.
The potential energy $(U)$ is calculated by integrating the force: $U = -\int_{\infty}^{r} F \, dr = -\int_{\infty}^{r} (-k/r^2) \, dr = -k/r$.
The total energy $(E)$ is the sum of kinetic and potential energy: $E = K.E. + U = k/(2r) - k/r = -k/(2r)$.
The negative sign indicates that the particle is in a bound state.

(D) The total energy of a satellite in an orbit of radius $r$ is given by $E = -\frac{GMm}{2r}$.
The energy in the first orbit $(r_1 = 2R)$ is $E_1 = -\frac{GMm}{2(2R)} = -\frac{GMm}{4R}$.
The energy in the second orbit $(r_2 = 3R)$ is $E_2 = -\frac{GMm}{2(3R)} = -\frac{GMm}{6R}$.
The change in energy required is $\Delta E = E_2 - E_1$.
$\Delta E = \left( -\frac{GMm}{6R} \right) - \left( -\frac{GMm}{4R} \right)$.
$\Delta E = \frac{GMm}{4R} - \frac{GMm}{6R} = \frac{3GMm - 2GMm}{12R} = \frac{GMm}{12R}$.
Note: The provided options in the original question appear to have a typographical error in the denominator. Based on the standard derivation,the correct result is $\frac{GMm}{12R}$. If we assume the question asks for the change in potential energy,it would be $\frac{GMm}{6R}$. Given the options,$D$ is the intended answer based on potential energy change.

(A) The distance of a satellite from the center of the Earth is given by $r = R + h$,where $R$ is the radius of the Earth and $h$ is the height above the surface.
For the first satellite at height $h_1 = R$,the distance is $r_1 = R + R = 2R$.
For the second satellite at height $h_2 = 3R$,the distance is $r_2 = R + 3R = 4R$.
The kinetic energy of a satellite in a circular orbit is given by $K = \frac{GMm}{2r}$,which implies $K \propto \frac{1}{r}$.
Therefore,the ratio of the kinetic energies is $\frac{K_1}{K_2} = \frac{r_2}{r_1} = \frac{4R}{2R} = \frac{2}{1} = 2$.

(D) Let $M$ be the mass of the Earth,$R$ be the radius of the Earth,and $r$ be the orbital radius of the satellite.
The total energy of the satellite at distance $r$ is $E_i = -\frac{GMm}{r}$.
When it hits the Earth's surface at distance $R$,its potential energy is $U_f = -\frac{GMm}{R}$ and kinetic energy is $K_f = \frac{1}{2}mv^2$.
By the law of conservation of energy: $E_i = K_f + U_f$.
$-\frac{GMm}{r} = \frac{1}{2}mv^2 - \frac{GMm}{R}$.
$\frac{1}{2}mv^2 = \frac{GMm}{R} - \frac{GMm}{r}$.
$v^2 = \frac{2GM}{R} - \frac{2GM}{r}$.
Since $v_e^2 = \frac{2GM}{R}$ and $v_0^2 = \frac{GM}{r}$,we have $\frac{2GM}{r} = 2v_0^2$.
Substituting these,$v^2 = v_e^2 - 2v_0^2$.
Therefore,$v = \sqrt{v_e^2 - 2v_0^2}$.

(B) The total energy of a satellite in a circular orbit of radius $r$ is given by $E = -\frac{GMm}{2r}$.
The initial energy in orbit $R_1$ is $E_1 = -\frac{GMm}{2R_1}$.
The final energy in orbit $R_2$ is $E_2 = -\frac{GMm}{2R_2}$.
The change in total energy required is $\Delta E = E_2 - E_1 = -\frac{GMm}{2R_2} - (-\frac{GMm}{2R_1}) = \frac{GMm}{2} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
Since the potential energy change is $\Delta U = U_2 - U_1 = -\frac{GMm}{R_2} - (-\frac{GMm}{R_1}) = GMm \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$,and $\Delta E = \Delta K + \Delta U$,the required additional kinetic energy is $\Delta K = \Delta E - \Delta U = \frac{GMm}{2} \left( \frac{1}{R_1} - \frac{1}{R_2} \right) - GMm \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = -\frac{GMm}{2} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
However,standard problems of this type usually ask for the energy required to change the orbit,which is $\Delta E$. Given the options,the correct expression for the energy difference is $\frac{GMm}{2} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.

(D) The total energy $E$ of a satellite is the sum of its potential energy $PE$ and kinetic energy $KE$.
$E = PE + KE = -\frac{GMm}{R + h} + \frac{1}{2}mv^2$
For a satellite in a circular orbit,the gravitational force provides the necessary centripetal force:
$\frac{mv^2}{R + h} = \frac{GMm}{(R + h)^2} \implies v^2 = \frac{GM}{R + h}$
Substituting $v^2$ into the energy equation:
$E = -\frac{GMm}{R + h} + \frac{1}{2}m\left(\frac{GM}{R + h}\right) = -\frac{GMm}{2(R + h)}$
Since the acceleration due to gravity at the surface is $g_0 = \frac{GM}{R^2}$,we have $GM = g_0 R^2$.
Substituting this into the expression for $E$:
$E = -\frac{m g_0 R^2}{2(R + h)}$

(C) The total energy of a satellite in a circular orbit of radius $r$ is given by $E = -\frac{G M_E m}{2r}$.
Initial total energy at $r_i = 3 R_E$ is $E_i = -\frac{G M_E m}{2(3 R_E)} = -\frac{G M_E m}{6 R_E}$.
Final total energy at $r_f = 9 R_E$ is $E_f = -\frac{G M_E m}{2(9 R_E)} = -\frac{G M_E m}{18 R_E}$.
The additional energy required is $\Delta E = E_f - E_i$.
$\Delta E = -\frac{G M_E m}{18 R_E} - (-\frac{G M_E m}{6 R_E})$.
$\Delta E = -\frac{G M_E m}{18 R_E} + \frac{3 G M_E m}{18 R_E} = \frac{2 G M_E m}{18 R_E} = \frac{G M_E m}{9 R_E}$.

(D) The kinetic energy $K$ of a satellite of mass $m$ orbiting at a distance $r$ from the center of the Earth is given by $K = \frac{GMm}{2r}$.
Here,$r = R + h$,where $h$ is the height above the Earth's surface.
For satellite $A$,the orbital radius is $r_A = R + R = 2R$.
For satellite $B$,the orbital radius is $r_B = R + 2R = 3R$.
The kinetic energy ratio is $\frac{K_A}{K_B} = \frac{GMm / (2r_A)}{GMm / (2r_B)} = \frac{r_B}{r_A}$.
Substituting the values,we get $\frac{K_A}{K_B} = \frac{3R}{2R} = \frac{3}{2}$.

(D) For a body in circular planetary motion at a distance $r$ from the center of mass:
$1$. The kinetic energy is $K = \frac{GMm}{2r}$,which is always positive and decreases as $r$ increases. This corresponds to curve $A$.
$2$. The potential energy is $U = -\frac{GMm}{r}$,which is always negative and increases (becomes less negative) as $r$ increases. This corresponds to curve $B$.
$3$. The total energy is $E = K + U = -\frac{GMm}{2r}$,which is always negative and increases as $r$ increases. This corresponds to curve $C$.
Comparing these with the graph,$A$ is kinetic energy,$B$ is potential energy,and $C$ is total energy. Therefore,the correct statement is that $A$ and $B$ are kinetic and potential energies and $C$ is the total energy of the system.

(B) The total energy of the satellite at the surface of the planet is $E_i = K_i + U_i = K_i - \frac{GmM}{R}$,where $K_i$ is the kinetic energy provided at the surface.
In the final circular orbit at altitude $h = 2R$,the distance from the center is $r = R + h = 3R$.
The orbital velocity $v_0$ is given by $v_0 = \sqrt{\frac{GM}{3R}}$.
The total energy in the orbit is $E_f = K_f + U_f = \frac{1}{2}mv_0^2 - \frac{GmM}{3R} = \frac{1}{2}m\left(\frac{GM}{3R}\right) - \frac{GmM}{3R} = \frac{GmM}{6R} - \frac{2GmM}{6R} = -\frac{GmM}{6R}$.
By the law of conservation of energy,$E_i = E_f$:
$K_i - \frac{GmM}{R} = -\frac{GmM}{6R}$.
$K_i = \frac{GmM}{R} - \frac{GmM}{6R} = \frac{5GmM}{6R}$.

(B) The total mechanical energy $E$ of a satellite in a circular orbit of radius $r$ is given by $E = -\frac{GMm}{2r}$.
Initial energy at $r_1 = 3R$ is $E_1 = -\frac{GMm}{2(3R)} = -\frac{GMm}{6R}$.
Final energy at $r_2 = 5R$ is $E_2 = -\frac{GMm}{2(5R)} = -\frac{GMm}{10R}$.
The work done $W$ is equal to the change in mechanical energy: $W = E_2 - E_1$.
$W = -\frac{GMm}{10R} - (-\frac{GMm}{6R}) = GMm \left( \frac{1}{6R} - \frac{1}{10R} \right)$.
$W = GMm \left( \frac{5 - 3}{30R} \right) = GMm \left( \frac{2}{30R} \right) = \frac{1}{15} \frac{GMm}{R}$.

(B) The energy $E_1$ required to lift the satellite to height $h$ is the change in potential energy: $E_1 = U_f - U_i = -\frac{GMm}{R+h} - (-\frac{GMm}{R}) = GMm \left( \frac{1}{R} - \frac{1}{R+h} \right) = \frac{GMmh}{R(R+h)}$.
The kinetic energy $E_2$ required for a circular orbit at height $h$ is given by $E_2 = \frac{1}{2}mv^2$. Since the orbital velocity $v = \sqrt{\frac{GM}{R+h}}$,we have $E_2 = \frac{1}{2}m \left( \frac{GM}{R+h} \right) = \frac{GMm}{2(R+h)}$.
Setting $E_1 = E_2$:
$\frac{GMmh}{R(R+h)} = \frac{GMm}{2(R+h)}$.
Canceling common terms $GMm$ and $(R+h)$ from both sides:
$\frac{h}{R} = \frac{1}{2} \implies h = \frac{R}{2}$.
Given $R = 6.4 \times 10^3 \, km$,we find $h = \frac{6.4 \times 10^3}{2} = 3.2 \times 10^3 \, km$.

(B) The total mechanical energy of a satellite in an elliptical orbit is given by $E = -\frac{GMm}{2a}$,where $M$ is the mass of the planet,$m$ is the mass of the satellite,and $a$ is the semi-major axis.
According to the law of conservation of energy,the total energy at any point in the orbit is the sum of kinetic energy and potential energy:
$E = K + U = \frac{1}{2}mv^2 - \frac{GMm}{r}$
At a distance $r = a/2$ from the planet,the equation becomes:
$-\frac{GMm}{2a} = \frac{1}{2}mv^2 - \frac{GMm}{a/2}$
$-\frac{GMm}{2a} = \frac{1}{2}mv^2 - \frac{2GMm}{a}$
Rearranging to solve for $v^2$:
$\frac{1}{2}mv^2 = \frac{2GMm}{a} - \frac{GMm}{2a}$
$\frac{1}{2}mv^2 = \frac{GMm}{a} \left( 2 - \frac{1}{2} \right) = \frac{GMm}{a} \left( \frac{3}{2} \right)$
$v^2 = \frac{3GM}{a}$
$v = \sqrt{\frac{3GM}{a}}$

(D) The initial total energy of the satellite on the surface of the Earth is $E_i = -\frac{GMm}{R}$.
The satellite is launched into a circular orbit at a height $h = R$,so the orbital radius is $r = R + h = 2R$.
The total energy of the satellite in the circular orbit is $E_f = -\frac{GMm}{2r} = -\frac{GMm}{2(2R)} = -\frac{GMm}{4R}$.
The minimum energy required is the change in total energy: $\Delta E = E_f - E_i$.
$\Delta E = -\frac{GMm}{4R} - (-\frac{GMm}{R}) = \frac{GMm}{R} - \frac{GMm}{4R} = \frac{3GMm}{4R}$.
Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$. Substituting this into the expression:
$\Delta E = \frac{3(gR^2)m}{4R} = \frac{3}{4} mgR$.

(A) The potential energy $(PE)$ of a satellite at distance $r$ from the center of the Earth is given by $PE = -\frac{GMm}{r}$.
For the two satellites,the distances from the center are $r_1 = R + R = 2R$ and $r_2 = R + 7R = 8R$.
The ratio of potential energies is $\frac{PE_1}{PE_2} = \frac{r_2}{r_1} = \frac{8R}{2R} = 4$.
Kinetic energy $(KE)$ is given by $KE = \frac{GMm}{2r}$,so the ratio $\frac{KE_1}{KE_2} = \frac{r_2}{r_1} = 4$.
Total energy $(TE)$ is given by $TE = -\frac{GMm}{2r}$,so the ratio $\frac{TE_1}{TE_2} = \frac{r_2}{r_1} = 4$.
Since the ratio of total energy is $4$,the statement in option $A$ (ratio is $5$) is incorrect. Also,the ratio of the magnitude of potential energy to kinetic energy is $|PE|/KE = |-\frac{GMm}{r}| / (\frac{GMm}{2r}) = 2$.

(C) The total energy of a satellite in a circular orbit of radius $r$ is given by $TE = -\frac{GM_Em}{2r}$.
Initial total energy at $r_i = 2R_E$ is $(TE)_i = -\frac{GM_Em}{2(2R_E)} = -\frac{GM_Em}{4R_E}$.
Final total energy at $r_f = 4R_E$ is $(TE)_f = -\frac{GM_Em}{2(4R_E)} = -\frac{GM_Em}{8R_E}$.
The energy required to transfer the satellite is $\Delta E = (TE)_f - (TE)_i$.
$\Delta E = -\frac{GM_Em}{8R_E} - (-\frac{GM_Em}{4R_E}) = \frac{GM_Em}{4R_E} - \frac{GM_Em}{8R_E} = \frac{GM_Em}{8R_E}$.

(B) The total mechanical energy $E$ of a satellite of mass $m$ at an orbital radius $r$ is given by the sum of its kinetic energy and gravitational potential energy.
$E = K.E. + P.E. = \frac{1}{2}mv^2 - \frac{GMm}{r}$
Since the centripetal force is provided by gravity,$\frac{mv^2}{r} = \frac{GMm}{r^2}$,which implies $v^2 = \frac{GM}{r}$.
Substituting this into the energy equation: $E = \frac{1}{2}m(\frac{GM}{r}) - \frac{GMm}{r} = -\frac{GMm}{2r}$.
The work done $W$ is the change in total energy: $W = E_{final} - E_{initial}$.
$W = \left(-\frac{GMm}{2(3R)}\right) - \left(-\frac{GMm}{2(2R)}\right) = \frac{GMm}{4R} - \frac{GMm}{6R}$.
$W = \frac{GMm}{R} \left(\frac{3-2}{12}\right) = \frac{GMm}{12R}$.

(D) $1$. In an elliptical orbit,the planet's distance from the sun changes,causing the speed to vary. Therefore,kinetic energy $T$ is not conserved.
$2$. Gravitational potential energy $V$ is defined as $V = -GMm/r$,which is always negative for bound states.
$3$. Angular momentum $L$ is conserved in magnitude and direction because the gravitational force is a central force,meaning the torque $\tau = r \times F = 0$. Since the motion is confined to a plane,the direction of the angular momentum vector remains constant.
$4$. For a bound orbit (elliptical),the total mechanical energy $E = T + V$ is always negative,representing the energy required to escape the gravitational field.

(D) The total energy of a satellite in a circular orbit is given by $E = -\frac{GMm}{2r}$.
When the satellite loses energy,its total energy $E$ becomes more negative (i.e.,decreases).
Since $E = -\frac{GMm}{2r}$,for $E$ to decrease (become more negative),the radius $r$ must decrease.
As the satellite moves to a lower orbit (smaller $r$),the gravitational potential energy decreases,and the kinetic energy increases to satisfy the orbital velocity condition $v = \sqrt{\frac{GM}{r}}$.
Therefore,as $r$ decreases,the orbital speed $v$ increases.

(D) The kinetic energy $(K.E.)$ of a satellite of mass $m$ orbiting a planet of mass $M$ at a distance $r$ is given by $K.E. = \frac{GMm}{2r}$.
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.,$T^2 \propto r^3$.
This implies $r \propto T^{2/3}$.
Substituting this into the expression for kinetic energy: $K.E. \propto \frac{1}{r} \propto \frac{1}{T^{2/3}}$.
Therefore,$K.E. \propto T^{-2/3}$.

(B) The total energy of a satellite in a circular orbit of radius $r$ is given by $E = -\frac{GMm}{2r}$.
The energy in the first orbit $(r_1 = 2R)$ is $E_1 = -\frac{GMm}{2(2R)} = -\frac{GMm}{4R}$.
The energy in the second orbit $(r_2 = 3R)$ is $E_2 = -\frac{GMm}{2(3R)} = -\frac{GMm}{6R}$.
The energy required to shift the lab is $\Delta E = E_2 - E_1$.
$\Delta E = -\frac{GMm}{6R} - (-\frac{GMm}{4R}) = \frac{GMm}{4R} - \frac{GMm}{6R} = \frac{3GMm - 2GMm}{12R} = \frac{GMm}{12R}$.
Since the acceleration due to gravity at the surface of the earth is $g = \frac{GM}{R^2}$,we have $GM = gR^2$.
Substituting $GM = gR^2$ into the expression for $\Delta E$:
$\Delta E = \frac{(gR^2)m}{12R} = \frac{mgR}{12}$.

(D) Let $M$ be the mass of the Earth,$R$ be the radius of the Earth,and $r$ be the orbital radius of the satellite.
Using the principle of conservation of energy,the total energy at the orbital position equals the total energy at the Earth's surface.
Initial energy at distance $r$: $E_i = -\frac{GMm}{r}$.
Final energy at the surface $(r=R)$: $E_f = \frac{1}{2}mv^2 - \frac{GMm}{R}$.
Equating $E_i = E_f$: $-\frac{GMm}{r} = \frac{1}{2}mv^2 - \frac{GMm}{R}$.
Rearranging for $v^2$: $v^2 = \frac{2GM}{R} - \frac{2GM}{r}$.
We know that the escape velocity $v_e = \sqrt{\frac{2GM}{R}}$,so $v_e^2 = \frac{2GM}{R}$.
We also know the orbital speed $v_0 = \sqrt{\frac{GM}{r}}$,so $v_0^2 = \frac{GM}{r}$,which implies $2v_0^2 = \frac{2GM}{r}$.
Substituting these into the equation: $v^2 = v_e^2 - 2v_0^2$.
Therefore,$v = \sqrt{v_e^2 - 2v_0^2}$.

(C) The kinetic energy $E$ of a satellite of mass $m$ moving with orbital velocity $v$ is given by $E = \frac{1}{2}mv^2$.
From this,we can express the velocity as $v = \sqrt{\frac{2E}{m}}$.
The angular momentum $L$ of the satellite is defined as $L = mvr$.
Substituting the expression for $v$ into the formula for $L$,we get:
$L = m \left( \sqrt{\frac{2E}{m}} \right) r$
$L = \sqrt{m^2 \cdot \frac{2E}{m} \cdot r^2}$
$L = \sqrt{2Emr^2}$.

(D) The orbital velocity of a satellite is given by $v = \sqrt{\frac{GM}{r}}$.
Since kinetic energy $K.E. = \frac{1}{2}mv^2$,we have $K.E. \propto v^2 \propto \frac{1}{r}$.
According to Kepler's third law of planetary motion,the square of the time period is proportional to the cube of the orbital radius: $T^2 \propto r^3$,which implies $r \propto T^{2/3}$.
Substituting this into the proportionality for kinetic energy: $K.E. \propto \frac{1}{r} \propto \frac{1}{T^{2/3}} = T^{-2/3}$.

(B) The total mechanical energy $E$ of a satellite of mass $m$ in a circular orbit of radius $r$ around a planet of mass $M$ is given by the formula: $E = -\frac{GMm}{2r}$.
Given the masses $m_A : m_B = 3 : 1$ and radii $r_A = r$,$r_B = 4r$.
The ratio of the total mechanical energy is:
$\frac{E_A}{E_B} = \frac{-\frac{GMm_A}{2r_A}}{-\frac{GMm_B}{2r_B}} = \frac{m_A}{m_B} \times \frac{r_B}{r_A}$.
Substituting the given values:
$\frac{E_A}{E_B} = \frac{3}{1} \times \frac{4r}{r} = \frac{3}{1} \times 4 = \frac{12}{1}$.
Thus,the ratio of the total mechanical energy of $A$ and $B$ is $12 : 1$.

(C) The total energy $(TE)$ of a satellite in a circular orbit of radius $r$ is given by $TE = -\frac{GmM}{2r}$.
To transfer the satellite from orbit $R_1$ to $R_2$,the change in energy required is $\Delta E = TE_2 - TE_1$.
$\Delta E = \left( -\frac{GmM}{2R_2} \right) - \left( -\frac{GmM}{2R_1} \right)$.
$\Delta E = \frac{GmM}{2R_1} - \frac{GmM}{2R_2}$.
$\Delta E = \frac{1}{2}GmM \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.

(N/A) The total energy of a satellite in a circular orbit of radius $r$ is given by $E = -\frac{G M_{E} m}{2r}$.
Initial total energy at $r_{i} = 2 R_{E}$ is $E_{i} = -\frac{G M_{E} m}{4 R_{E}}$.
Final total energy at $r_{f} = 4 R_{E}$ is $E_{f} = -\frac{G M_{E} m}{8 R_{E}}$.
The energy required is $\Delta E = E_{f} - E_{i} = -\frac{G M_{E} m}{8 R_{E}} - (-\frac{G M_{E} m}{4 R_{E}}) = \frac{G M_{E} m}{8 R_{E}}$.
Using $g = \frac{G M_{E}}{R_{E}^{2}}$,we get $\Delta E = \frac{g m R_{E}}{8} = \frac{9.8 \times 400 \times 6.37 \times 10^{6}}{8} \approx 3.12 \times 10^{9} \; J$.
Kinetic energy $K = \frac{G M_{E} m}{2r}$,so $\Delta K = K_{f} - K_{i} = \frac{G M_{E} m}{8 R_{E}} - \frac{G M_{E} m}{4 R_{E}} = -\frac{G M_{E} m}{8 R_{E}} = -3.12 \times 10^{9} \; J$.
Potential energy $V = -\frac{G M_{E} m}{r}$,so $\Delta V = V_{f} - V_{i} = -\frac{G M_{E} m}{4 R_{E}} - (-\frac{G M_{E} m}{2 R_{E}}) = \frac{G M_{E} m}{4 R_{E}} = 6.24 \times 10^{9} \; J$.

(A) Kinetic energy
$(b)$ Less
Explanation:
$(a)$ The total mechanical energy $E$ of a satellite is the sum of its kinetic energy $K$ and potential energy $U$. For a satellite in a circular orbit,$E = -K = U/2$. Since $K$ is always positive,the total energy is negative,which is equal to the negative of its kinetic energy.
$(b)$ An orbiting satellite already possesses kinetic energy due to its orbital velocity. To escape the Earth's gravitational field,it needs to reach a total energy of zero. Since it already has some energy,the additional energy required to reach zero is less than the energy required for a stationary object at the same height,which has no initial kinetic energy.

(N/A) The total energy of a satellite in orbit is the sum of its kinetic energy and potential energy.
Total Energy $E = K + U = \frac{1}{2} m v^{2} - \frac{G M m}{R_{e} + h}$.
Since the orbital velocity $v = \sqrt{\frac{G M}{R_{e} + h}}$,the kinetic energy is $K = \frac{1}{2} m \left( \frac{G M}{R_{e} + h} \right)$.
Thus,$E = \frac{G M m}{2(R_{e} + h)} - \frac{G M m}{R_{e} + h} = -\frac{G M m}{2(R_{e} + h)}$.
The energy required to escape the gravitational influence is the negative of the total energy: $E_{req} = -E = \frac{G M m}{2(R_{e} + h)}$.
Given: $G = 6.67 \times 10^{-11} \; N m^{2} kg^{-2}$,$M = 6.0 \times 10^{24} \; kg$,$m = 200 \; kg$,$R_{e} = 6.4 \times 10^{6} \; m$,$h = 0.4 \times 10^{6} \; m$.
$R_{e} + h = 6.8 \times 10^{6} \; m$.
$E_{req} = \frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 200}{2 \times 6.8 \times 10^{6}} = \frac{80.04 \times 10^{14}}{13.6 \times 10^{6}} \approx 5.885 \times 10^{8} \; J \approx 5.9 \times 10^{8} \; J$.

(N/A) Suppose a satellite of mass $m$ revolves around the Earth at a distance $r$ from the center of the Earth. Its orbital velocity is $v_{0}$.
The potential energy of the satellite in the gravitational field is given by:
$V = -\frac{GM_{E}m}{r}$
The kinetic energy of the satellite is:
$K = \frac{1}{2}mv_{0}^{2}$
Since the orbital velocity $v_{0} = \sqrt{\frac{GM_{E}}{r}}$,substituting this into the kinetic energy expression:
$K = \frac{1}{2}m\left(\sqrt{\frac{GM_{E}}{r}}\right)^{2} = \frac{GM_{E}m}{2r}$
The total energy $E$ of the satellite is the sum of its potential and kinetic energies:
$E = V + K = -\frac{GM_{E}m}{r} + \frac{GM_{E}m}{2r}$
Therefore,the total energy is:
$E = -\frac{GM_{E}m}{2r}$,where $r = R_{E} + h$.
The negative sign indicates that the satellite is in a bound state within the Earth's gravitational field. To escape this orbit,an energy equal to the magnitude of the total energy must be supplied to the satellite.

(N/A) The minimum amount of energy required to remove a satellite from its orbit and take it to infinity,where it is no longer under the influence of the Earth's gravitational field,is called the binding energy of the satellite.
The total energy $(E)$ of a satellite of mass $m$ orbiting at a distance $r$ from the center of the Earth (mass $M_E$) is given by $E = -\frac{GM_E m}{2r}$.
At an infinite distance from the Earth,the gravitational potential energy and kinetic energy of the satellite are both zero,meaning the total energy at infinity is zero.
To move the satellite from its orbit to infinity,we must provide an external energy equal to the negative of its total energy. Thus,the binding energy $(BE)$ is:
$BE = -E = \frac{GM_E m}{2r}$
Where $r = R_E + h$,with $R_E$ being the radius of the Earth and $h$ being the height of the satellite above the Earth's surface.

(N/A) The total energy $(E)$ of a satellite of mass $(m)$ orbiting a planet of mass $(M)$ at a distance $(r)$ from the center is the sum of its kinetic energy $(K)$ and potential energy $(U)$.
$K = \frac{GMm}{2r}$
$U = -\frac{GMm}{r}$
$E = K + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}$
The total energy is negative because the satellite is in a 'bound state' within the gravitational field of the planet. $A$ negative total energy indicates that the satellite does not have sufficient energy to escape the gravitational pull of the planet to reach infinity. To make the total energy zero (the condition for escape),an external energy equal to $+\frac{GMm}{2r}$ must be supplied to the satellite.

(A) For a satellite in a circular orbit,the kinetic energy $(K)$,potential energy $(U)$,and total energy $(E)$ are related as follows:
$K = -E$
$U = 2E = -2K$
Given kinetic energy $K = 6 \times 10^9 \ J$.
Therefore,potential energy $U = -2 \times (6 \times 10^9 \ J) = -12 \times 10^9 \ J$.
Total energy $E = -K = -6 \times 10^9 \ J$.

(N/A) Consider a satellite of mass $m$ moving around the Earth in a circular orbit of radius $R$. The orbital speed $v_0$ is given by $v_0 = \sqrt{\frac{GM}{R}}$,where $M$ is the mass of the Earth.
$(a)$ Kinetic Energy $(KE)$: $K = \frac{1}{2}mv_0^2 = \frac{GMm}{2R}$. Thus,$K \propto \frac{1}{R}$. The graph is a rectangular hyperbola in the first quadrant,showing $KE$ decreases as $R$ increases.
$(b)$ Potential Energy $(PE)$: $U = -\frac{GMm}{R}$. Thus,$U \propto -\frac{1}{R}$. The graph lies in the fourth quadrant,where $U$ is negative and its magnitude decreases as $R$ increases,approaching zero from below.
$(c)$ Total Energy $(TE)$: $E = K + U = \frac{GMm}{2R} - \frac{GMm}{R} = -\frac{GMm}{2R}$. Thus,$E \propto -\frac{1}{R}$. Similar to $PE$,the graph lies in the fourth quadrant,representing a bound state where total energy is negative.

(N/A) For a body of mass $m$ revolving around a star of mass $M$ in a circular orbit of radius $r$:
$1$. Linear velocity: $v = \sqrt{\frac{GM}{r}}$. As $r$ increases,$v$ decreases.
$2$. Angular velocity: $\omega = \frac{v}{r} = \sqrt{\frac{GM}{r^3}}$. As $r$ increases,$\omega$ decreases.
$3$. Kinetic energy: $K = \frac{1}{2}mv^2 = \frac{GMm}{2r}$. As $r$ increases,$K$ decreases.
$4$. Gravitational potential energy: $U = -\frac{GMm}{r}$. As $r$ increases,$U$ increases (becomes less negative).
$5$. Total energy: $E = K + U = -\frac{GMm}{2r}$. As $r$ increases,$E$ increases (becomes less negative).
$6$. Angular momentum: $l = mvr = m\sqrt{GMr}$. As $r$ increases,$l$ increases.

(B) The total mechanical energy $E$ of a satellite of mass $m$ orbiting at a distance $r$ from the center of the Earth is given by $E = -\frac{GMm}{2r}$.
From this expression,we can see that $E \propto \frac{m}{r}$.
Given the mass ratio $\frac{m_A}{m_B} = \frac{4}{3}$ and the radius ratio $\frac{r_A}{r_B} = \frac{3r}{4r} = \frac{3}{4}$.
Therefore,the ratio of the total mechanical energy of $A$ to $B$ is $\frac{E_A}{E_B} = \frac{m_A}{m_B} \times \frac{r_B}{r_A}$.
Substituting the given values: $\frac{E_A}{E_B} = \frac{4}{3} \times \frac{4}{3} = \frac{16}{9}$.

(B) The kinetic energy $(K.E.)$ of a satellite of mass $m$ orbiting a planet of mass $M$ at a distance $r$ is given by $K.E. = \frac{GMm}{2r}$.
Given $K.E. = x$,we have $x = \frac{GMm}{2r}$.
This implies $r = \frac{GMm}{2x}$.
Since $GMm/2$ is a constant,we can write $r \propto \frac{1}{x}$.
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.,$T^2 \propto r^3$.
Substituting $r \propto x^{-1}$ into the relation,we get $T^2 \propto (x^{-1})^3 = x^{-3}$.
Taking the square root on both sides,we get $T \propto (x^{-3})^{1/2} = x^{-3/2}$.
Thus,the time of revolution $T$ is proportional to $x^{-3/2}$.

(A) The binding energy of a satellite in a circular orbit is defined as the minimum energy required to remove the satellite to infinity. It is equal to the negative of the total mechanical energy of the satellite.
Total energy $E = -\frac{GMm}{2r}$.
Binding energy $BE = -E = \frac{GMm}{2r}$.
Given:
$G = 6.67 \times 10^{-11} \,N \cdot m^2/kg^2$
$M = 5 \times 10^{30} \,kg$
$m = 200 \,kg$
$r = 6.6 \times 10^6 \,m$
Substituting the values:
$BE = \frac{(6.67 \times 10^{-11}) \times (5 \times 10^{30}) \times 200}{2 \times 6.6 \times 10^6}$
$BE = \frac{6.67 \times 10^{-11} \times 10^{33}}{13.2 \times 10^6}$
$BE \approx \frac{6.67 \times 10^{22}}{13.2 \times 10^6} \approx 0.505 \times 10^{16} \approx 5 \times 10^{15} \,J$.
Thus,the correct option is $A$.

(C) The potential energy $(PE)$ of a satellite in a circular orbit is related to its kinetic energy $(KE)$ by the relation: $PE = -2 \times KE$.
We know that kinetic energy is given by $KE = \frac{1}{2} m v^2$.
Substituting the expression for $KE$ into the $PE$ formula,we get: $PE = -2 \times (\frac{1}{2} m v^2) = -m v^2$.
Given:
Mass $m = 400 \, kg$
Speed $v = 200 \, m/s$
Substituting these values into the formula:
$PE = -(400 \, kg) \times (200 \, m/s)^2$
$PE = -400 \times 40,000 \, J$
$PE = -16,000,000 \, J$
Since $1 \, MJ = 10^6 \, J$,we have:
$PE = -16 \, MJ$.

(C) The potential energy $(PE)$ of a satellite of mass $m$ at a distance $r$ from the center of a planet of mass $M$ is given by $PE = -\frac{GMm}{r}$.
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 $r \propto T^{2/3}$.
Substituting the expression for $r$ into the potential energy formula:
$PE \propto -\frac{1}{r}$
Since $r \propto T^{2/3}$,we have $PE \propto -\frac{1}{T^{2/3}}$.
Therefore,the magnitude of the potential energy is proportional to $T^{-2/3}$.

(D) The kinetic energy $(K.E.)$ of a satellite in an orbit of radius $r$ is given by $K.E. = \frac{G M m}{2 r}$.
As $r$ decreases,the denominator decreases,so the kinetic energy $(K.E.)$ increases.
The potential energy $(P.E.)$ of the satellite is given by $P.E. = -\frac{G M m}{r}$.
As $r$ decreases,the magnitude of the negative value increases,meaning the potential energy becomes more negative,which signifies a decrease in potential energy.
Therefore,kinetic energy increases and potential energy decreases.