Energy consideration in Planetary and Satellite motion Questions in English

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

(B) The total energy $E$ of a satellite in an orbit of radius $R$ is given by $E = -\frac{GMm}{2R}$.
The potential energy $U$ of the satellite is given by $U = -\frac{GMm}{R}$.
Comparing these,we see that $U = 2E$. Therefore,Statement $I$ is incorrect.
The kinetic energy $K$ of the satellite is given by $K = \frac{GMm}{2R}$.
Comparing this with the total energy $E$,we see that $K = -E$,which means the magnitude of kinetic energy $|K|$ is equal to the magnitude of total energy $|E|$.
Thus,Statement $II$ is also incorrect.

(A) The initial total mechanical energy of the ball on the Earth's surface is $TE_i = -\frac{GM_e m}{R_e}$.
Given the altitude $h = 318.5 \ km$ and $R_e = 6370 \ km$,we find $h = \frac{R_e}{20}$.
The final total mechanical energy of the ball in a circular orbit at altitude $h$ is $TE_f = -\frac{GM_e m}{2(R_e + h)}$.
Substituting $h = \frac{R_e}{20}$,we get $TE_f = -\frac{GM_e m}{2(R_e + R_e/20)} = -\frac{GM_e m}{2(21R_e/20)} = -\frac{10 GM_e m}{21 R_e}$.
The change in total mechanical energy is $\Delta TE = TE_f - TE_i = -\frac{10 GM_e m}{21 R_e} - (-\frac{GM_e m}{R_e})$.
$\Delta TE = \frac{GM_e m}{R_e} (1 - \frac{10}{21}) = \frac{11 GM_e m}{21 R_e}$.
Comparing this with $x \frac{GM_e m}{21 R_e}$,we get $x = 11$.

(D) The total energy of a satellite in a circular orbit of radius $r$ is given by $E = -\frac{GMm}{2r}$.
Initially,the satellite is in an orbit of radius $r_1 = 2R$. Its initial energy is $E_1 = -\frac{GMm}{2(2R)} = -\frac{GMm}{4R}$.
Given that an energy $\Delta E = \frac{10^4 R}{6} \text{ J}$ is supplied to the satellite,the new total energy $E_2$ will be $E_1 + \Delta E$.
$E_2 = -\frac{GMm}{4R} + \frac{10^4 R}{6}$.
Let the new radius be $r_2$. Then $E_2 = -\frac{GMm}{2r_2}$.
Equating the two expressions for $E_2$:
$-\frac{GMm}{4R} + \frac{10^4 R}{6} = -\frac{GMm}{2r_2}$.
Using $g = \frac{GM}{R^2}$,we have $GM = gR^2$. Substituting this into the equation:
$-\frac{mgR^2}{4R} + \frac{10^4 R}{6} = -\frac{mgR^2}{2r_2}$.
Given $m = 10^3 \text{ kg}$ and $g = 10 \text{ m/s}^2$,so $mg = 10^4 \text{ N}$.
$-\frac{10^4 R}{4} + \frac{10^4 R}{6} = -\frac{10^4 R^2}{2r_2}$.
Dividing by $10^4 R$:
$-\frac{1}{4} + \frac{1}{6} = -\frac{R}{2r_2}$.
$-\frac{3-2}{12} = -\frac{R}{2r_2} \implies -\frac{1}{12} = -\frac{R}{2r_2}$.
$\frac{1}{12} = \frac{R}{2r_2} \implies 2r_2 = 12R \implies r_2 = 6R$.

(D) The total energy of the satellite at the surface of the Earth is $E_i = K_i + U_i = K_i - \frac{G M m}{R}$.
At an altitude of $2R$ from the surface,the distance from the center of the Earth is $r = R + 2R = 3R$.
The orbital velocity $v$ at distance $r = 3R$ is given by $v = \sqrt{\frac{G M}{3R}}$.
The total energy at the orbit is $E_f = K_f + U_f = \frac{1}{2} m v^2 - \frac{G M m}{3R} = \frac{1}{2} m \left(\frac{G M}{3R}\right) - \frac{G M m}{3R} = \frac{G M m}{6R} - \frac{G M m}{3R} = -\frac{G M m}{6R}$.
By the law of conservation of energy,$E_i = E_f$:
$K_i - \frac{G M m}{R} = -\frac{G M m}{6R}$.
$K_i = \frac{G M m}{R} - \frac{G M m}{6R} = \frac{5 G M m}{6R}$.

(B) The total energy $E$ of a satellite orbiting at a height $h$ is given by $E = -\frac{GMm}{2(R+h)}$.
We know that the acceleration due to gravity on the Earth's surface is $g = \frac{GM}{R^2}$,which implies $GM = gR^2$.
Given that the height $h = R$,the orbital radius is $r = R + h = R + R = 2R$.
Substituting $GM = gR^2$ and $r = 2R$ into the energy formula:
$E = -\frac{(gR^2)m}{2(2R)}$
$E = -\frac{gR^2m}{4R}$
$E = -\frac{mgR}{4}$.

(D) The kinetic energy $(K.E.)$ of a satellite of mass $m$ orbiting at a distance $r$ from the center of the Earth (mass $M$) is given by $K.E. = \frac{GMm}{2r}$.
The potential energy $(P.E.)$ of the satellite at the same distance $r$ is given by $P.E. = -\frac{GMm}{r}$.
We are asked to find the ratio of potential energy to kinetic energy. Note that the magnitude of potential energy is $|P.E.| = \frac{GMm}{r}$.
The ratio is $\frac{P.E.}{K.E.} = \frac{-GMm/r}{GMm/2r} = -2$.
Since the question asks for the ratio of potential energy to kinetic energy,and considering the sign,the value is $-2$. However,if the question implies the ratio of magnitudes,the answer is $2$. Given standard physics convention,the potential energy is negative,so the ratio is $-2$. If the options provided are positive,we select the magnitude $2$.

(B) The binding energy $(BE)$ of a satellite is defined as the minimum energy required to remove the satellite from its orbit to infinity.
For a satellite of mass $m$ at a distance $r = R + h$ from the center of the Earth,the potential energy is $U = -\frac{GMm}{r}$.
The total energy of the satellite is $E = K + U = -\frac{GMm}{2r}$.
The binding energy is the negative of the total energy: $BE = -E = \frac{GMm}{2r}$.
Comparing $BE$ and $U$,we see that $U = -2 \times BE$.
Given $BE = 3.5 \times 10^8 \ J$,the potential energy is $U = -2 \times (3.5 \times 10^8 \ J) = -7 \times 10^8 \ J$.

(B) The binding energy $(BE)$ of a satellite is defined as the negative of the total mechanical energy $(E)$.
$BE = -E = - (K.E. + P.E.) = - (\frac{GMm}{2r} - \frac{GMm}{r}) = \frac{GMm}{2r}$.
The potential energy $(P.E.)$ of a satellite is given by $P.E. = -\frac{GMm}{r}$.
Comparing the two expressions, we see that $P.E. = -2 \times BE$.
Given $BE = 3.5 \times 10^{8} \,J$.
Therefore, $P.E. = -2 \times (3.5 \times 10^{8} \,J) = -7.0 \times 10^{8} \,J$.

(C) The binding energy of a satellite at rest on the earth's surface is given by $E_1 = \frac{GMm}{2R}$.
The binding energy of a satellite of mass $m$ revolving in a circular orbit at a height $h$ above the earth's surface is $E_2 = \frac{GMm}{2(R+h)}$.
Taking the ratio of the two energies:
$\frac{E_1}{E_2} = \frac{\frac{GMm}{2R}}{\frac{GMm}{2(R+h)}} = \frac{R+h}{R}$.

(B) For a satellite of mass $m$ orbiting a planet of mass $M$ at a distance $r$,the potential energy is $U = -\frac{GMm}{r}$.
The kinetic energy required for a circular orbit is $K = \frac{GMm}{2r}$.
The total energy $E$ is the sum of kinetic and potential energy: $E = K + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}$.
Comparing $E$ with $K$: $E = -K$,which means the total energy is the negative of the kinetic energy.
Comparing $E$ with $U$: $E = \frac{1}{2} U$,which means the total energy is half the potential energy.

(A) The total energy of a satellite in a circular orbit at a distance $r = R + h$ from the center of the planet is given by $E_0 = -\frac{GMm}{2r}$.
Given the altitude $h = 2R$,the orbital radius is $r = R + 2R = 3R$.
Thus,the orbital energy is $E_0 = -\frac{GMm}{2(3R)} = -\frac{GMm}{6R}$.
The potential energy of the satellite on the surface of the planet is $E_i = -\frac{GMm}{R}$.
The minimum energy required to launch the satellite is the difference between the final orbital energy and the initial surface energy:
$\Delta E = E_0 - E_i = -\frac{GMm}{6R} - (-\frac{GMm}{R}) = \frac{GMm}{R} - \frac{GMm}{6R} = \frac{5GMm}{6R}$.

(A) The energy required to raise a satellite of mass $m$ to a height $h$ above the earth's surface is the change in potential energy:
$E_1 = U_h - U_R = -\frac{GMm}{R+h} - (-\frac{GMm}{R}) = GMm \left( \frac{1}{R} - \frac{1}{R+h} \right) = \frac{GMmh}{R(R+h)}$.
Using $GM = gR^2$,we get $E_1 = \frac{mgR^2h}{R(R+h)} = \frac{mghR}{R+h}$.
The total energy of a satellite in a circular orbit at height $h$ is $E_{orbit} = -\frac{GMm}{2(R+h)}$.
The energy required to put it into orbit at that height is the difference between the energy in orbit and the energy at the surface:
$E_2 = E_{orbit} - U_R = -\frac{GMm}{2(R+h)} + \frac{GMm}{R} = GMm \left( \frac{1}{R} - \frac{1}{2(R+h)} \right) = GMm \left( \frac{2R+2h-R}{2R(R+h)} \right) = \frac{GMm(R+2h)}{2R(R+h)}$.
Using $GM = gR^2$,we get $E_2 = \frac{mgR^2(R+2h)}{2R(R+h)} = \frac{mgR(R+2h)}{2(R+h)}$.
The ratio is $\frac{E_1}{E_2} = \frac{mghR}{R+h} \times \frac{2(R+h)}{mgR(R+2h)} = \frac{2h}{R+2h}$.
For $h \ll R$,$R+2h \approx R$,so the ratio is $\frac{2h}{R}$.

(D) The energy required to raise a satellite of mass '$m$' to a height '$h$' above the Earth's surface is equal to the change in its gravitational potential energy:
$W = U_f - U_i = -\frac{GMm}{R+h} - (-\frac{GMm}{R}) = GMm \left( \frac{1}{R} - \frac{1}{R+h} \right) = \frac{GMmh}{R(R+h)}$.
The total energy of a satellite of mass '$m$' in a circular orbit at height '$h$' (distance $r = R+h$ from the center) is given by:
$E = \frac{GMm}{2(R+h)}$.
Now,the ratio of the energy required to raise the satellite to the energy required to put it into orbit is:
$\frac{W}{E} = \frac{\frac{GMmh}{R(R+h)}}{\frac{GMm}{2(R+h)}} = \frac{GMmh}{R(R+h)} \times \frac{2(R+h)}{GMm} = \frac{2h}{R}$.

(D) The orbital velocity of a satellite is given by $v = \sqrt{\frac{GM}{r}}$.
Since kinetic energy $KE = \frac{1}{2}mv^2$,we have $KE \propto v^2 \propto \frac{1}{r}$.
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: $KE \propto \frac{1}{r} \propto \frac{1}{T^{2/3}} = T^{-2/3}$.

(C) The kinetic energy of the satellite is given by $E = \frac{1}{2}mv^{2}$.
From this,we can express the velocity $v$ as $v = \sqrt{\frac{2E}{m}}$.
The angular momentum $L$ of a particle moving in a circular orbit of radius $r$ is defined as $L = mvr$.
Substituting the expression for $v$ into the formula for $L$:
$L = m \cdot \sqrt{\frac{2E}{m}} \cdot r$
$L = \sqrt{m^{2} \cdot \frac{2E}{m} \cdot r^{2}}$
$L = \sqrt{2mEr^{2}}$
$L = (2mEr^{2})^{\frac{1}{2}}$.

(A) The total energy $(E)$ of a satellite in a circular orbit is given by the formula:
$E = -\frac{GMm}{2r}$
Where:
- $G$ is the universal gravitational constant.
- $M$ is the mass of the Earth.
- $m$ is the mass of the satellite.
- $r = R + h$ is the distance from the center of the Earth.
Substituting $r = R + h$ into the energy equation,we get:
$E = -\frac{GMm}{2(R+h)}$
Since $G, M, m,$ and $2$ are constants,the total energy $E$ is proportional to $-\frac{1}{R+h}$.
Thus,the total energy varies as $-\frac{1}{R+h}$.

(D) The potential energy of a satellite at a distance $h$ from the centre of the Earth is given by $U = -\frac{GMm}{h}$.
The total energy of a satellite in a circular orbit is given by $E_0 = -\frac{GMm}{2h}$.
Comparing these two expressions,we can see that $U = 2 \times (-\frac{GMm}{2h})$.
Therefore,the potential energy $U = 2 E_0$.

(D) The total energy of a satellite in an orbit of radius $r$ is given by $E = -\frac{GMm}{2r}$.
Initial energy $E_1 = -\frac{GMm}{2r}$.
Final energy in an orbit of radius $r' = \frac{3r}{2}$ is $E_2 = -\frac{GMm}{2(3r/2)} = -\frac{GMm}{3r}$.
The change in energy is $\Delta E = E_2 - E_1 = -\frac{GMm}{3r} - (-\frac{GMm}{2r}) = \frac{GMm}{2r} - \frac{GMm}{3r} = \frac{GMm}{6r}$.
The percentage increase is given by $\frac{\Delta E}{|E_1|} \times 100$.
Percentage increase $= \frac{GMm/6r}{GMm/2r} \times 100 = \frac{2}{6} \times 100 = \frac{1}{3} \times 100 = 33.33 \%$.

(D) The time period $T$ of a satellite revolving in a circular orbit of radius $R$ is given by Kepler's Third Law: $T = 2\pi \sqrt{\frac{R^3}{GM}}$.
Squaring both sides,we get $T^2 = \frac{4\pi^2 R^3}{GM}$,which implies $R^3 \propto T^2$,or $R \propto T^{2/3}$.
The kinetic energy $K$ of a satellite in a circular orbit is given by $K = \frac{GMm}{2R}$.
Since $G$,$M$,and $m$ are constants,$K \propto \frac{1}{R}$.
Substituting the relation $R \propto T^{2/3}$,we get $K \propto \frac{1}{T^{2/3}}$.
Therefore,$K \propto T^{-2/3}$.

(B) For a satellite in a circular orbit of radius '$r$',the kinetic energy is given by $K = \frac{GMm}{2r}$ and the potential energy is $U = -\frac{GMm}{r}$.
Thus,the total mechanical energy is $E_{total} = K + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r} = -K = -E$.
To escape the Earth's gravitational field,the final total energy of the satellite must be at least $0$.
Let the energy to be added be $\Delta E$.
$E_{final} = E_{initial} + \Delta E = 0$.
$-E + \Delta E = 0$.
$\Delta E = E$.

(A) The total energy $E$ of a satellite of mass $m$ orbiting the Earth of mass $M_e$ at a distance $r$ is given by:
$E = -\frac{G M_e m}{2r}$
This shows that $E \propto \frac{1}{r}$.
Let the initial radius be $r_1$ and the initial energy be $E_1 = -\frac{G M_e m}{2r_1}$.
When the radius is reduced to half,$r_2 = \frac{r_1}{2}$.
The new energy is $E_2 = -\frac{G M_e m}{2r_2} = -\frac{G M_e m}{2(r_1/2)} = -\frac{G M_e m}{r_1} = 2 E_1$.
The change in total energy is $\Delta E = E_2 - E_1 = 2E_1 - E_1 = E_1$.
The percentage change is given by $\frac{|\Delta E|}{|E_1|} \times 100\% = \frac{|E_1|}{|E_1|} \times 100\% = 100\%$.

(A) The kinetic energy $(KE)$ of a satellite of mass $m$ revolving in a circular orbit of radius $R$ around a planet of mass $M$ is given by the formula:
$KE = \frac{G M m}{2 R}$
Here,$G$ is the universal gravitational constant,$M$ is the mass of the planet,and $m$ is the mass of the satellite.
Since $G$,$M$,and $m$ are constants for a given system,we can observe that:
$KE \propto \frac{1}{R}$
Therefore,the kinetic energy of the satellite is inversely proportional to the radius of the orbit $R$.

(A) The potential energy $(U)$ of a satellite at a distance $r$ from the center of the Earth is given by $U = -\frac{GMm}{r}$. Since the gravitational force is attractive,the potential energy is always negative.
The kinetic energy $(K)$ of a satellite is given by $K = \frac{GMm}{2r}$. Since $G, M, m,$ and $r$ are all positive,the kinetic energy is always positive.
The total energy $(TE)$ of the satellite is the sum of its kinetic and potential energies:
$TE = K + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}$.
Thus,the total energy is always negative.
The gravitational potential energy of a satellite at infinity $(r = \infty)$ is:
$U = -\frac{GMm}{\infty} = 0$.
Comparing these results with the given columns:
$A$ (Potential energy) $\rightarrow II$ (Negative)
$B$ (Total energy) $\rightarrow II$ (Negative)
$C$ (Kinetic energy) $\rightarrow I$ (Positive)
$D$ (Potential energy at infinity) $\rightarrow III$ (Zero)
The correct matching is $A-II, B-II, C-I, D-III$.

(B) The total energy of the satellite on the surface of the earth is the sum of its potential energy and kinetic energy. Since it is at rest on the surface,its kinetic energy is $0$. Thus,$E_1 = -\frac{GMm}{R}$.
When the satellite is in a circular orbit of radius $r = 7R$,its total energy is given by $E_2 = -\frac{GMm}{2r}$.
Substituting $r = 7R$,we get $E_2 = -\frac{GMm}{2(7R)} = -\frac{GMm}{14R}$.
The minimum energy required to be spent by the launching vehicle is the difference between the final energy and the initial energy: $\Delta E = E_2 - E_1$.
$\Delta E = -\frac{GMm}{14R} - (-\frac{GMm}{R}) = -\frac{GMm}{14R} + \frac{GMm}{R}$.
$\Delta E = \frac{-GMm + 14GMm}{14R} = \frac{13GMm}{14R}$.

(B) The moment of inertia of a satellite of mass $m$ at a distance $R$ from the center of the Earth is $I = m R^2$.
The kinetic energy $(K)$ of the satellite in terms of angular momentum $(J)$ is given by $K = \frac{J^2}{2 I}$.
Substituting $I = m R^2$,we get $K = \frac{J^2}{2 m R^2}$.
For a satellite in a circular orbit,the total energy $(E)$ is related to the kinetic energy $(K)$ by the relation $E = -K$.
Therefore,$E = -\frac{J^2}{2 m R^2}$.
Thus,the correct option is $B$.

(A) For a satellite of mass $m$ orbiting a planet of mass $M$ at a distance $a$ (orbital radius),the energies are given by:
Kinetic energy,$K = \frac{GMm}{2a}$
Potential energy,$V = -\frac{GMm}{a}$
Total energy,$E = K + V = \frac{GMm}{2a} - \frac{GMm}{a} = -\frac{GMm}{2a}$
Comparing these expressions:
$V = -\frac{GMm}{a} = -2 \left( \frac{GMm}{2a} \right) = -2K$
Therefore,$K = -V / 2$.

(B) For a satellite in a circular orbit of radius $r$ around the Earth of mass $M$,the potential energy $U$ is given by $U = -\frac{G M m}{r}$.
The kinetic energy $K$ of the satellite is given by $K = \frac{G M m}{2 r}$.
The total energy $E$ is the sum of potential and kinetic energy:
$E = U + K = -\frac{G M m}{r} + \frac{G M m}{2 r} = -\frac{G M m}{2 r}$.
Comparing the expressions for $U$ and $E$:
$U = -\frac{G M m}{r}$
$E = -\frac{G M m}{2 r}$
Therefore,$U = 2 \times (- \frac{G M m}{2 r}) = 2 E$.
Wait,let us re-evaluate: $U = -\frac{G M m}{r}$ and $E = -\frac{G M m}{2 r}$.
Thus,$U = 2 E$ is incorrect. The correct relation is $U = 2 E$ if $E$ were defined differently,but here $U = 2 E$ is not correct. Let's re-check: $U = -\frac{G M m}{r}$ and $E = -\frac{G M m}{2 r}$.
So,$U = 2 \times (-\frac{G M m}{2 r}) = 2 E$.
Actually,$U = 2 E$ is correct because $2 \times (- \frac{G M m}{2 r}) = - \frac{G M m}{r} = U$.

(D) The energy of a satellite in a circular orbit is given by $E = -\frac{GM_E m}{2r}$,where $r$ is the radius of the circular orbit.
The energy to be supplied is $\Delta E = E_f - E_i$.
$\Delta E = \left( -\frac{GM_E m}{2(3R_E)} \right) - \left( -\frac{GM_E m}{2(1.5R_E)} \right)$
$\Delta E = \frac{GM_E m}{2R_E} \left( \frac{1}{1.5} - \frac{1}{3} \right) = \frac{GM_E m}{2R_E} \left( \frac{2}{3} - \frac{1}{3} \right) = \frac{GM_E m}{6R_E}$.
Using the relation $g = \frac{GM_E}{R_E^2}$,we have $GM_E = gR_E^2$.
Substituting this into the expression for $\Delta E$:
$\Delta E = \frac{gR_E^2 m}{6R_E} = \frac{1}{6} mgR_E$.
Given $m = 100 \ kg$,$g = 10 \ m/s^2$,and $R_E = 6 \times 10^6 \ m$:
$\Delta E = \frac{1}{6} \times 100 \times 10 \times 6 \times 10^6 = 1000 \times 10^6 \ J$.
Comparing this with $\alpha \times 10^6 \ J$,we get $\alpha = 1000$.