Escape Velocity and Escape Energy Questions in English

(C) The orbital velocity of a satellite close to the Earth's surface is given by $v_{\text{orbit}} = \sqrt{gR}$.
Substituting the values: $v_{\text{orbit}} = \sqrt{10 \times 6.4 \times 10^6} = \sqrt{64 \times 10^6} = 8000 \, m/s = 8 \, km/s$.
The escape velocity from the Earth's surface is given by $v_{\text{escape}} = \sqrt{2gR} = \sqrt{2} \times v_{\text{orbit}}$.
Substituting the value: $v_{\text{escape}} = 8\sqrt{2} \, km/s$.
The additional velocity required to overcome the gravitational pull is $\Delta v = v_{\text{escape}} - v_{\text{orbit}}$.
$\Delta v = 8\sqrt{2} - 8 = 8(\sqrt{2}-1) \, km/s$.

(B) The formula for escape velocity is $V_e = \sqrt{\frac{2GM}{R}}$.
Let $M_E$ and $R_E$ be the mass and radius of the Earth,and $M_P$ and $R_P$ be the mass and radius of the planet.
Given: $M_P = 16 M_E$ and $R_P = 4 R_E$.
The escape velocity of the planet is $V_P = \sqrt{\frac{2GM_P}{R_P}} = \sqrt{\frac{2G(16M_E)}{4R_E}} = \sqrt{4 \times \frac{2GM_E}{R_E}} = 2 \sqrt{\frac{2GM_E}{R_E}}$.
Since $V_E = \sqrt{\frac{2GM_E}{R_E}}$,we have $V_P = 2 V_E$.
Therefore,the ratio $\frac{V_P}{V_E} = \frac{2}{1}$,which is $2:1$.

(B) The formula for escape velocity is $V = \sqrt{\frac{2GM}{R}}$.
For Earth,$V_e = \sqrt{\frac{2GM_e}{R_e}} = 11.2 \, km/s$.
For the planet,$V_p = \sqrt{\frac{2G(9M_e)}{4R_e}} = \sqrt{\frac{9}{4}} \times \sqrt{\frac{2GM_e}{R_e}}$.
$V_p = \frac{3}{2} \times V_e$.
Substituting the value of $V_e = 11.2 \, km/s$:
$V_p = 1.5 \times 11.2 \, km/s = 16.8 \, km/s$.

(B) The formula for escape velocity $V_e$ is given by $V_e = \sqrt{\frac{2GM}{R}}$,where $G$ is the gravitational constant,$M$ is the mass of the planet,and $R$ is its radius.
From the formula,we can see that $V_e \propto \sqrt{\frac{M}{R}}$.
Therefore,if the ratio $\frac{M}{R}$ increases,the escape velocity $V_e$ must also increase. Thus,Statement $I$ is correct.
Furthermore,since $V_e = \sqrt{\frac{2GM}{R}}$,it is clear that $V_e$ depends on the radius $R$ of the planet $(V_e \propto \frac{1}{\sqrt{R}})$. Therefore,Statement $II$ is incorrect.

(D) The formula for escape velocity is $V_e = \sqrt{\frac{2GM}{R}}$.
Given for Earth: $V_E = 11.2 \,km/s$, $M_E$, and $R_E$.
For the planet: $M_P = \frac{M_E}{6}$ and $R_P = \frac{R_E}{3}$.
The escape velocity for the planet is $V_P = \sqrt{\frac{2GM_P}{R_P}}$.
Substituting the values: $V_P = \sqrt{\frac{2G(M_E/6)}{(R_E/3)}} = \sqrt{\frac{2GM_E}{R_E} \times \frac{3}{6}} = \sqrt{\frac{2GM_E}{R_E} \times \frac{1}{2}}$.
Since $V_E = \sqrt{\frac{2GM_E}{R_E}} = 11.2 \,km/s$, we have $V_P = \frac{V_E}{\sqrt{2}}$.
$V_P = \frac{11.2}{1.414} \approx 7.9 \,km/s$.

(A) The formula for escape velocity is $V_e = \sqrt{\frac{2GM}{R}}$.
Let $M_p$ and $R_p$ be the mass and radius of the planet,and $M_m$ and $R_m$ be the mass and radius of the moon.
Given: $M_m = \frac{M_p}{144}$ and $D_m = \frac{D_p}{16}$,which implies $R_m = \frac{R_p}{16}$.
Escape velocity on the planet is $V_p = \sqrt{\frac{2GM_p}{R_p}} = v$.
Escape velocity on the moon is $V_m = \sqrt{\frac{2GM_m}{R_m}} = \sqrt{\frac{2G(M_p/144)}{(R_p/16)}} = \sqrt{\frac{2GM_p}{R_p} \times \frac{16}{144}} = \sqrt{\frac{2GM_p}{R_p} \times \frac{1}{9}}$.
Substituting $V_p = v$,we get $V_m = \sqrt{\frac{v^2}{9}} = \frac{v}{3}$.

(B) The escape velocity $v_e$ is given by the formula $v_e = \sqrt{\frac{2GM}{R_E}}$.
To project a body to infinity,the kinetic energy $K$ must be equal to the magnitude of the gravitational potential energy at the surface of the Earth.
$K = \frac{GMm}{R_E}$.
We know that the acceleration due to gravity on the surface of the Earth is $g = \frac{GM}{R_E^2}$,which implies $GM = gR_E^2$.
Substituting $GM$ into the kinetic energy expression:
$K = \frac{(gR_E^2)m}{R_E} = mgR_E$.
Therefore,the required kinetic energy is $mgR_E$.

(B) Let $M$ be the mass of the Earth and $R$ be its radius. The mass of the Sun is $M_s = 3 \times 10^5 M$ and the distance of the Sun from the Earth is $d = 2.5 \times 10^4 R$.
The total energy of the rocket at the surface of the Earth is $E = \frac{1}{2}mv_s^2 - \frac{GMm}{R} - \frac{GM_sm}{d}$.
For the rocket to escape the system,the final energy must be at least $0$.
$\frac{1}{2}mv_s^2 = \frac{GMm}{R} + \frac{G(3 \times 10^5 M)m}{2.5 \times 10^4 R}$.
Since $v_e^2 = \frac{2GM}{R}$,we have $\frac{GM}{R} = \frac{v_e^2}{2}$.
Substituting this into the energy equation: $\frac{1}{2}v_s^2 = \frac{v_e^2}{2} + \frac{3 \times 10^5}{2.5 \times 10^4} \times \frac{v_e^2}{2}$.
$v_s^2 = v_e^2 + 12 v_e^2 = 13 v_e^2$.
$v_s = v_e \sqrt{13} = 11.2 \times 3.605 \approx 40.38 \text{ km s}^{-1}$.
The closest value is $42 \text{ km s}^{-1}$.

(B) The gravitational acceleration is given by $g = \frac{4}{3} \pi G \rho R$.
Thus,$\frac{g'}{g} = \frac{\rho' R'}{\rho R}$.
Given $\frac{g'}{g} = \frac{\sqrt{6}}{11}$ and $\frac{\rho'}{\rho} = \frac{2}{3}$,we have $\frac{\sqrt{6}}{11} = \frac{2}{3} \cdot \frac{R'}{R}$,which implies $\frac{R'}{R} = \frac{3\sqrt{6}}{22}$.
The escape velocity is $v_e = \sqrt{2gR} = \sqrt{2 (\frac{4}{3} \pi G \rho R) R} = R \sqrt{\frac{8}{3} \pi G \rho}$.
Therefore,$\frac{v_e'}{v_e} = \frac{R'}{R} \sqrt{\frac{\rho'}{\rho}} = \left( \frac{3\sqrt{6}}{22} \right) \sqrt{\frac{2}{3}} = \frac{3\sqrt{6}}{22} \cdot \frac{\sqrt{2}}{\sqrt{3}} = \frac{3\sqrt{2}\sqrt{3}}{22} \cdot \frac{\sqrt{2}}{\sqrt{3}} = \frac{3 \cdot 2}{22} = \frac{6}{22} = \frac{3}{11}$.
Given $v_e = 11 \ km/s$,we get $v_e' = 11 \cdot \frac{3}{11} = 3 \ km/s$.

(B) Let the radius of the circular orbit be $r$. The orbital speed of the satellite is given by $V = \sqrt{\frac{GM}{r}}$,which implies $V^2 = \frac{GM}{r}$.
For an object to just escape the gravitational pull of the Earth,its total mechanical energy at the point of ejection must be equal to its total mechanical energy at infinity,which is $0$.
Let $K$ be the kinetic energy of the object of mass $m$ at the time of ejection.
The potential energy of the object at distance $r$ from the center of the Earth is $U = -\frac{GMm}{r}$.
According to the law of conservation of energy: $K + U = 0$.
$K - \frac{GMm}{r} = 0$.
$K = \frac{GMm}{r}$.
Substituting $\frac{GM}{r} = V^2$,we get $K = m V^2$.

(A) The escape velocity $V_e$ is given by $V_e = \sqrt{\frac{2GM}{R}}$. Since $M = \rho \cdot \frac{4}{3}\pi R^3$,we have $V_e = \sqrt{\frac{2G}{R} \cdot \rho \cdot \frac{4}{3}\pi R^3} = \sqrt{\frac{8\pi G \rho}{3}} R$. Thus,$V_e \propto R$.
Given surface area $A_P = A = 4\pi R_P^2$ and $A_Q = 4A = 4\pi R_Q^2$. This implies $R_Q^2 = 4R_P^2$,so $R_Q = 2R_P$.
For planet $R$,$M_R = M_P + M_Q$. Since density $\rho$ is uniform,$\frac{4}{3}\pi R_R^3 \rho = \frac{4}{3}\pi R_P^3 \rho + \frac{4}{3}\pi R_Q^3 \rho$.
Thus,$R_R^3 = R_P^3 + R_Q^3 = R_P^3 + (2R_P)^3 = R_P^3 + 8R_P^3 = 9R_P^3$.
So,$R_R = 9^{1/3} R_P$.
Comparing radii: $R_R = 9^{1/3} R_P \approx 2.08 R_P$,$R_Q = 2 R_P$,and $R_P = R_P$.
Therefore,$R_R > R_Q > R_P$,which implies $V_R > V_Q > V_P$. This confirms statement $(B)$ is correct.
For statement $(D)$,$\frac{V_P}{V_Q} = \frac{R_P}{R_Q} = \frac{R_P}{2R_P} = \frac{1}{2}$. This confirms statement $(D)$ is correct.

(B) Let $R$ be the radius of the planet and $M$ be its mass. The acceleration due to gravity at the surface is $g = GM/R^2$.
At a height $h$ above the surface,the acceleration due to gravity is $g' = GM/(R+h)^2$.
Given $g' = g/4$,we have $GM/(R+h)^2 = (1/4) \cdot (GM/R^2)$,which implies $(R+h)^2 = 4R^2$,so $R+h = 2R$,or $h = R$.
Using the conservation of energy at the surface and at the maximum height $h=R$:
$-(GMm/R) + (1/2)mv^2 = -(GMm/(R+R))$
$-(GMm/R) + (1/2)mv^2 = -(GMm/2R)$
$(1/2)mv^2 = GMm/2R \implies v^2 = GM/R$.
The escape velocity is defined as $v_{esc} = \sqrt{2GM/R}$.
Substituting $GM/R = v^2$,we get $v_{esc} = \sqrt{2v^2} = v\sqrt{2}$.
Comparing this with $v_{esc} = v\sqrt{N}$,we find $N = 2$.

(A) Initial state: $R_A = R_B = R$,$M_A = M$,$M_B = 2M = 2M_A$. Density $\rho_A = \frac{M_A}{\frac{4}{3}\pi R^3}$.
After interaction,star $A$ has radius $R_A' = R/2$. Since density $\rho_A$ is constant,the new mass $M_A' = \rho_A \cdot \frac{4}{3}\pi (R/2)^3 = M_A/8$.
The escape velocity $v_A = \sqrt{\frac{2GM_A'}{R_A'}} = \sqrt{\frac{2G(M_A/8)}{R/2}} = \sqrt{\frac{GM_A}{2R}}$.
Mass lost by $A$ is $\Delta M = M_A - M_A/8 = \frac{7}{8}M_A$. This mass is added to $B$ as a shell of density $\rho_A$. Let the new radius of $B$ be $R_B'$.
Volume of shell = $\frac{4}{3}\pi(R_B'^3 - R^3) = \frac{\Delta M}{\rho_A} = \frac{7/8 M_A}{\rho_A} = \frac{7}{8} \cdot \frac{4}{3}\pi R^3$.
$R_B'^3 - R^3 = \frac{7}{8}R^3 \Rightarrow R_B'^3 = \frac{15}{8}R^3 \Rightarrow R_B' = R \left(\frac{15}{8}\right)^{1/3}$.
The new mass of $B$ is $M_B' = M_B + \Delta M = 2M_A + \frac{7}{8}M_A = \frac{23}{8}M_A$.
The escape velocity $v_B = \sqrt{\frac{2GM_B'}{R_B'}} = \sqrt{\frac{2G(23/8)M_A}{R(15/8)^{1/3}}} = \sqrt{\frac{23GM_A}{4R(15/8)^{1/3}}} = \sqrt{\frac{23GM_A}{2R(15^{1/3})}}$.
Ratio $\frac{v_B}{v_A} = \sqrt{\frac{23GM_A}{2R(15^{1/3})} \cdot \frac{2R}{GM_A}} = \sqrt{\frac{23}{15^{1/3}}} = \sqrt{\frac{10 \times 2.3}{15^{1/3}}}$.
Comparing with $\sqrt{\frac{10n}{15^{1/3}}}$,we get $n = 2.30$.

(B) The formula for escape velocity is $V_e = \sqrt{\frac{2GM}{R}}$.
Given: $M_E = 8M_P$ and $R_E = 2R_P$,where $E$ denotes Earth and $P$ denotes the planet.
Therefore,the ratio of escape velocities is:
$\frac{V_P}{V_E} = \sqrt{\frac{M_P}{M_E} \times \frac{R_E}{R_P}}$
Substituting the given values:
$\frac{V_P}{V_E} = \sqrt{\frac{M_P}{8M_P} \times \frac{2R_P}{R_P}} = \sqrt{\frac{1}{8} \times 2} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Thus,$V_P = \frac{1}{2} V_E = \frac{1}{2} \times 11.2 \ km/s = 5.6 \ km/s$.

(A) The escape velocity of a body from the Earth's surface is given by $v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}$.
The kinetic energy required to project the body to infinity is $KE = \frac{1}{2}mv_e^2 = \frac{1}{2}m(2gR) = mgR$.
Since the assertion states the energy is $\frac{1}{2}mgR$,the Assertion $A$ is false.
The potential energy of a body at infinity is defined as zero,which is the maximum value of potential energy for a body in the Earth's gravitational field. Thus,Reason $R$ is true.

(A) To ensure the object does not return to Earth,its total mechanical energy at infinity must be at least zero.
Let $m$ be the mass of the object.
The distance of the object from the center of the Earth is $r = R + 3R = 4R$.
The potential energy at this point is $U = -\frac{GMm}{4R}$.
Let $v$ be the projection speed. The kinetic energy is $K = \frac{1}{2}mv^2$.
By the law of conservation of energy:
$K_i + U_i = K_f + U_f$
$\frac{1}{2}mv^2 - \frac{GMm}{4R} = 0 + 0$
$\frac{1}{2}mv^2 = \frac{GMm}{4R}$
$v^2 = \frac{GM}{2R}$
$v = \sqrt{\frac{GM}{2R}}$

(B) Apply the principle of conservation of mechanical energy.
Total energy at the surface of the earth = Total energy at the maximum height $h$.
$\frac{-GMm}{R} + \frac{1}{2}m\left(\frac{v_e}{2}\right)^2 = \frac{-GMm}{R+h} + 0$
Since the escape velocity $v_e = \sqrt{\frac{2GM}{R}}$,we have $v_e^2 = \frac{2GM}{R}$.
Substituting this into the energy equation:
$\frac{-GMm}{R} + \frac{1}{2}m \left(\frac{1}{4} \cdot \frac{2GM}{R}\right) = \frac{-GMm}{R+h}$
$\frac{-GMm}{R} + \frac{GMm}{4R} = \frac{-GMm}{R+h}$
$-\frac{3GMm}{4R} = \frac{-GMm}{R+h}$
$\frac{3}{4R} = \frac{1}{R+h}$
$3(R+h) = 4R$
$3R + 3h = 4R$
$3h = R$
$h = \frac{R}{3}$

(B) The formula for the escape velocity $(v_e)$ of an object from the surface of the Earth is given by:
$v_e = \sqrt{\frac{2GM}{R}}$
where:
$G$ is the gravitational constant,
$M$ is the mass of the Earth,
$R$ is the radius of the Earth.
From this formula,it is clear that the escape velocity depends only on the mass of the Earth $(M)$,the radius of the Earth $(R)$,and the gravitational constant $(G)$.
It does not depend on the mass of the object $(m)$ being projected.
Therefore,the correct option is $B$.

(B) The escape velocity from the earth's surface is given by $v_e = \sqrt{\frac{2GM}{R}}$.
The initial velocity of projection is $v = \frac{1}{3} v_e = \frac{1}{3} \sqrt{\frac{2GM}{R}}$.
Using the law of conservation of energy between the surface of the earth and the maximum height $h$:
Total energy at surface = Total energy at maximum height $h$.
$-\frac{GMm}{R} + \frac{1}{2}mv^2 = -\frac{GMm}{R+h} + 0$.
Substituting $v^2 = \frac{1}{9} \left(\frac{2GM}{R}\right) = \frac{2GM}{9R}$:
$-\frac{GMm}{R} + \frac{1}{2}m \left(\frac{2GM}{9R}\right) = -\frac{GMm}{R+h}$.
$-\frac{GMm}{R} + \frac{GMm}{9R} = -\frac{GMm}{R+h}$.
Dividing by $-GMm$: $\frac{1}{R} - \frac{1}{9R} = \frac{1}{R+h}$.
$\frac{8}{9R} = \frac{1}{R+h}$.
$8(R+h) = 9R \implies 8R + 8h = 9R$.
$8h = R \implies h = \frac{R}{8}$.

(D) The gravitational potential energy $U$ of the body of mass $M$ at a distance $r/3$ from the Earth and $2r/3$ from the moon is given by:
$U = -\frac{G M_1 M}{r/3} - \frac{G M_2 M}{2r/3} = -\frac{3 G M_1 M}{r} - \frac{3 G M_2 M}{2r} = -\frac{3 G M}{r} \left( M_1 + \frac{M_2}{2} \right)$.
To escape to infinity,the total energy must be at least zero. If $v$ is the required escape speed,the kinetic energy is $\frac{1}{2} M v^2$.
By conservation of energy: $\frac{1}{2} M v^2 + U = 0$.
$\frac{1}{2} M v^2 = \frac{3 G M}{r} \left( M_1 + \frac{M_2}{2} \right)$.
Solving for $v$:
$v^2 = \frac{6 G}{r} \left( M_1 + \frac{M_2}{2} \right)$.
$v = \left[ \frac{6 G}{r} \left( M_1 + \frac{M_2}{2} \right) \right]^{1/2}$.

(A) The escape velocity $v_e$ is given by the formula: $v_e = \sqrt{\frac{2GM}{R}}$.
Since the mass $M$ of a planet can be expressed in terms of its density $\rho$ and radius $R$ as $M = \frac{4}{3} \pi R^3 \rho$, we substitute this into the formula:
$v_e = \sqrt{\frac{2G}{R} \cdot \frac{4}{3} \pi R^3 \rho} = \sqrt{\frac{8}{3} G \pi \rho R^2} = R \sqrt{\frac{8}{3} G \pi \rho}$.
Given that the density $\rho$ is the same for both the Earth and the planet, we have $v_e \propto R$.
Let $v_e$ be the escape velocity of Earth and $v_e'$ be the escape velocity of the planet.
Given $R' = 2R$, we have:
$\frac{v_e'}{v_e} = \frac{R'}{R} = \frac{2R}{R} = 2$.
Therefore, $v_e' = 2 \times v_e = 2 \times 11 \,km/s = 22 \,km/s$.

(C) The escape velocity $v_e$ of a planet is given by the formula: $v_e = \sqrt{\frac{2GM}{R}}$.
Since the mass $M$ of a planet can be expressed in terms of its density $\rho$ and radius $R$ as $M = \frac{4}{3}\pi R^3 \rho$,we substitute this into the escape velocity formula:
$v_e = \sqrt{\frac{2G}{R} \cdot \frac{4}{3}\pi R^3 \rho} = \sqrt{\frac{8}{3}G\pi\rho R^2} = R \sqrt{\frac{8}{3}G\pi\rho}$.
Given that the densities $\rho$ are the same for both the planet and the earth,the escape velocity is directly proportional to the radius: $v_e \propto R$.
Therefore,the ratio of the escape velocities is: $\frac{V_p}{V_E} = \frac{R_p}{R_E}$.
Given that the radius of the planet is double that of the earth $(R_p = 2R_E)$,we have:
$\frac{V_p}{V_E} = \frac{2R_E}{R_E} = 2$.
Thus,the value of $x$ is $2$.

(B) The escape velocity $v_e$ is given by the formula $v_e = \sqrt{\frac{2GM}{R}}$.
Substituting $M = \frac{4}{3} \pi R^3 \rho$,we get $v_e = \sqrt{\frac{2G}{R} \cdot \frac{4}{3} \pi R^3 \rho} = R \sqrt{\frac{8}{3} \pi G \rho}$.
This implies $v_e \propto R \sqrt{\rho}$.
Given $R_p = 4R_E$ and $\rho_p = 4\rho_E$,the ratio of escape velocity on Earth $(v_E)$ to that on the planet $(v_p)$ is:
$\frac{v_E}{v_p} = \frac{R_E \sqrt{\rho_E}}{R_p \sqrt{\rho_p}} = \frac{R_E \sqrt{\rho_E}}{(4R_E) \sqrt{4\rho_E}} = \frac{1}{4 \times 2} = \frac{1}{8}$.
Thus,the ratio is $1: 8$.

(A) According to the law of conservation of energy:
$(K.E + P.E)_{\text{surface}} = (K.E + P.E)_{\text{max height}}$
Since the velocity at maximum height is $0$,we have:
$-\frac{GMm}{R} + \frac{1}{2} m(n V_e)^2 = -\frac{GMm}{R + h} + 0$
We know that the escape velocity $V_e = \sqrt{\frac{2GM}{R}}$,so $V_e^2 = \frac{2GM}{R}$.
Substituting this into the equation:
$\frac{1}{2} m n^2 \left( \frac{2GM}{R} \right) = GMm \left( \frac{1}{R} - \frac{1}{R + h} \right)$
$\frac{n^2 GMm}{R} = GMm \left( \frac{R + h - R}{R(R + h)} \right)$
$\frac{n^2}{R} = \frac{h}{R(R + h)}$
$n^2 = \frac{h}{R + h}$
$n^2(R + h) = h$
$n^2 R + n^2 h = h$
$n^2 R = h(1 - n^2)$
$h = \frac{n^2 R}{1 - n^2}$

(C) According to the law of conservation of energy,the total energy at the surface equals the total energy at the maximum height $h$.
At the surface: $E_i = K + U = \frac{1}{2}mv^2 - \frac{GMm}{R}$.
Given $v = \frac{v_e}{2}$,where $v_e = \sqrt{\frac{2GM}{R}}$,so $v^2 = \frac{v_e^2}{4} = \frac{2GM}{4R} = \frac{GM}{2R}$.
$E_i = \frac{1}{2}m(\frac{GM}{2R}) - \frac{GMm}{R} = \frac{GMm}{4R} - \frac{GMm}{R} = -\frac{3GMm}{4R}$.
At maximum height $h$,velocity is $0$,so $E_f = 0 - \frac{GMm}{R+h}$.
Equating $E_i = E_f$: $-\frac{3GMm}{4R} = -\frac{GMm}{R+h}$.
$\frac{3}{4R} = \frac{1}{R+h} \implies 3R + 3h = 4R \implies 3h = R \implies h = \frac{R}{3}$.

(C) The formula for escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$.
Since the mass $M$ of a planet can be expressed in terms of its density $\rho$ and radius $R$ as $M = \frac{4}{3} \pi R^3 \rho$,we substitute this into the formula:
$v_e = \sqrt{\frac{2G(\frac{4}{3} \pi R^3 \rho)}{R}} = \sqrt{\frac{8}{3} G \pi R^2 \rho} = R \sqrt{\frac{8}{3} G \pi \rho}$.
From this expression,we see that $v_e \propto R$ when the density $\rho$ is constant.
Given that the planet has the same density as the earth but twice the radius $(R' = 2R_e)$,the new escape velocity $v_e'$ is:
$v_e' = 2 \times v_e = 2 \times 11.2 \,km/s = 22.4 \,km/s$.

(C) The escape velocity $V_{e}$ from the surface of a planet of mass $M$ and radius $R$ is given by the formula: $V_{e} = \sqrt{\frac{2GM}{R}}$.
For Earth,$V_{e} = \sqrt{\frac{2GM}{R}}$.
For the given planet,the mass $M' = 6M$ and the radius $R' = 2R$.
The escape velocity $V_{e}'$ for this planet is:
$V_{e}' = \sqrt{\frac{2G(6M)}{2R}}$
$V_{e}' = \sqrt{3 \times \frac{2GM}{R}}$
$V_{e}' = \sqrt{3} \times \sqrt{\frac{2GM}{R}}$
$V_{e}' = \sqrt{3} V_{e}$.

(A) The gravitational potential energy $U$ of a particle of mass $m_0$ at the midpoint between the earth and the moon is given by the sum of potentials due to both bodies:
$U = -\frac{G M m_0}{d/2} - \frac{G m m_0}{d/2} = -\frac{2 G m_0}{d} (M + m)$
To reach infinity,the total energy of the particle must be at least zero. Let $V$ be the minimum projection velocity.
Applying the principle of conservation of energy:
$K_i + U_i = K_f + U_f$
$\frac{1}{2} m_0 V^2 - \frac{2 G m_0}{d} (M + m) = 0 + 0$
$\frac{1}{2} m_0 V^2 = \frac{2 G m_0}{d} (M + m)$
$V^2 = \frac{4 G}{d} (M + m)$
$V = 2 \sqrt{\frac{G(M+m)}{d}}$

(A) The escape velocity $V_e$ is given by $V_e = \sqrt{\frac{2GM}{R}}$.
Applying the law of conservation of energy between the Earth's surface and infinity:
Total energy at surface = Total energy at infinity
$-\frac{GMm}{R} + \frac{1}{2}m(2V_e)^2 = 0 + \frac{1}{2}mV^2$
Here,$V$ is the final velocity at infinity.
Substituting $V_e^2 = \frac{2GM}{R}$,we get:
$-\frac{GM}{R} + \frac{1}{2}(4 \cdot \frac{2GM}{R}) = \frac{1}{2}V^2$
$-\frac{GM}{R} + \frac{4GM}{R} = \frac{1}{2}V^2$
$\frac{3GM}{R} = \frac{1}{2}V^2$
$V^2 = \frac{6GM}{R} = 3 \left( \frac{2GM}{R} \right) = 3V_e^2$
$V = \sqrt{3}V_e$

(A) Let the mass $m$ be at a distance $r_1 = \frac{2d}{3}$ from the centre of $M_1$. Then,its distance from the centre of $M_2$ is $r_2 = d - \frac{2d}{3} = \frac{d}{3}$.
To escape the gravitational influence of the system,the total energy of the body must be at least zero at infinity.
By the law of conservation of energy,the kinetic energy provided at the initial position must equal the magnitude of the gravitational potential energy at that position:
$\frac{1}{2} m v_e^2 = \frac{G M_1 m}{r_1} + \frac{G M_2 m}{r_2}$
Substituting the values of $r_1$ and $r_2$:
$\frac{1}{2} m v_e^2 = \frac{G M_1 m}{(2d/3)} + \frac{G M_2 m}{(d/3)}$
$\frac{1}{2} m v_e^2 = \frac{3 G M_1 m}{2d} + \frac{3 G M_2 m}{d}$
$\frac{1}{2} m v_e^2 = \frac{3 G m}{2d} (M_1 + 2 M_2)$
$v_e^2 = \frac{3 G (M_1 + 2 M_2)}{d}$
$v_e = \left[\frac{3 G (M_1 + 2 M_2)}{d}\right]^{\frac{1}{2}}$

(B) According to the law of conservation of mechanical energy,the total energy at the height $h = \frac{R}{2}$ is equal to the total energy at the Earth's surface.
Total energy at height $h = \frac{R}{2}$:
$E_i = K_i + U_i = 0 + \left( -\frac{GMm}{R + \frac{R}{2}} \right) = -\frac{GMm}{\frac{3R}{2}} = -\frac{2GMm}{3R}$
Total energy at the Earth's surface:
$E_f = K_f + U_f = \frac{1}{2}mv^2 + \left( -\frac{GMm}{R} \right)$
Equating $E_i = E_f$:
$-\frac{2GMm}{3R} = \frac{1}{2}mv^2 - \frac{GMm}{R}$
$\frac{1}{2}mv^2 = \frac{GMm}{R} - \frac{2GMm}{3R} = \frac{GMm}{3R}$
$v^2 = \frac{2GM}{3R}$
Since the escape velocity $v_e = \sqrt{\frac{2GM}{R}}$,we have $v_e^2 = \frac{2GM}{R}$.
Substituting this into the expression for $v^2$:
$v^2 = \frac{v_e^2}{3}$
$v = \frac{v_e}{\sqrt{3}}$

(C) The energy required to escape the earth's gravitational field is $E_{esc} = \frac{1}{2} m V_{e}^2$.
The initial kinetic energy given to the body is $K_i = \frac{1}{2} m (3 V_{e})^2 = \frac{9}{2} m V_{e}^2$.
According to the law of conservation of energy, the final kinetic energy $K_f$ after escaping the gravitational pull is the difference between the initial energy and the energy required to escape:
$K_f = K_i - E_{esc}$
$\frac{1}{2} m V^2 = \frac{9}{2} m V_{e}^2 - \frac{1}{2} m V_{e}^2$
$\frac{1}{2} m V^2 = 4 m V_{e}^2$
$V^2 = 8 V_{e}^2$
$V = \sqrt{8} V_{e} = 2 \sqrt{2} V_{e}$.

(A) To ensure the particle does not return,its total mechanical energy at infinity must be at least zero.
Let $v$ be the projection speed at height $h = 3R$ from the surface.
The distance from the center of the Earth is $r = R + 3R = 4R$.
Using the law of conservation of energy: $K_i + U_i = K_f + U_f$.
Here,$K_i = \frac{1}{2}mv^2$,$U_i = -\frac{GMm}{4R}$,$K_f = 0$,and $U_f = 0$.
$\frac{1}{2}mv^2 - \frac{GMm}{4R} = 0$.
$\frac{1}{2}v^2 = \frac{GM}{4R}$.
$v^2 = \frac{GM}{2R}$.
$v = [\frac{GM}{2R}]^{1/2}$.

(A) The formula for escape velocity is $V = \sqrt{\frac{2GM}{R}}$.
For Earth,the escape velocity is $V_e = \sqrt{\frac{2GM}{R}}$.
For the planet,the mass is $M_p = 6M$ and the radius is $R_p = 2R$.
Thus,the escape velocity from the planet is $V_p = \sqrt{\frac{2G(6M)}{2R}} = \sqrt{3 \times \frac{2GM}{R}}$.
Substituting $V_e$ into the equation,we get $V_p = \sqrt{3} V_e$.

(A) The escape velocity $v_e$ from the surface of a planet is given by the formula $v_e = \sqrt{2 g R}$,where $g$ is the acceleration due to gravity and $R$ is the radius of the planet.
Given the ratio of acceleration due to gravity $\frac{g_1}{g_2} = K_1$ and the ratio of radii $\frac{R_1}{R_2} = K_2$.
The ratio of escape velocities is $\frac{v_{e1}}{v_{e2}} = \sqrt{\frac{2 g_1 R_1}{2 g_2 R_2}}$.
Substituting the given ratios,we get $\frac{v_{e1}}{v_{e2}} = \sqrt{\frac{g_1}{g_2} \cdot \frac{R_1}{R_2}} = \sqrt{K_1 K_2}$.

(C) The formula for escape velocity is $v = \sqrt{\frac{2GM}{R}}$.
Since mass $M = \text{Volume} \times \text{density} = \frac{4}{3} \pi R^3 \rho$,we substitute this into the formula:
$v = \sqrt{\frac{2G}{R} \cdot \frac{4}{3} \pi R^3 \rho} = \sqrt{\frac{8}{3} \pi G \rho R^2} = R \sqrt{\frac{8 \pi G \rho}{3}}$.
Thus,$v \propto R \sqrt{\rho}$.
Given that the average mass densities $\rho$ are equal,the escape velocity is directly proportional to the radius: $v \propto R$.
Therefore,$\frac{V_P}{V_E} = \frac{R_P}{R_E}$.
Given $R_P = 2 R_E$,we get $\frac{V_P}{V_E} = 2$,which implies $V_P = 2 V_E$.

(B) According to the law of conservation of energy,the total energy at the surface equals the total energy at infinity.
Initial energy at the surface: $E_i = K_i + U_i = \frac{1}{2} m(2 v_e)^2 - \frac{G M m}{R} = 2 m v_e^2 - m v_e^2 = m v_e^2$ (since $v_e^2 = \frac{2 G M}{R}$).
Final energy at infinity: $E_f = K_f + U_f = \frac{1}{2} m v^2 + 0$.
Equating $E_i = E_f$:
$m v_e^2 = \frac{1}{2} m v^2$
$v^2 = 2 v_e^2$
$v = \sqrt{2} v_e$ is incorrect based on the standard energy balance approach. Let's re-evaluate: The energy provided is $K_i = \frac{1}{2} m (2 v_e)^2 = 2 m v_e^2$. The energy required to escape is $K_{req} = \frac{1}{2} m v_e^2$. The remaining kinetic energy at infinity is $K_f = K_i - K_{req} = 2 m v_e^2 - 0.5 m v_e^2 = 1.5 m v_e^2$.
Thus,$\frac{1}{2} m v^2 = \frac{3}{2} m v_e^2$,which gives $v^2 = 3 v_e^2$,so $v = \sqrt{3} v_e$.

(A) The formula for escape velocity from the surface of a planet of mass $M$ and radius $R$ is given by $V_{e} = \sqrt{\frac{2GM}{R}}$.
For the Earth,$V_{e} = \sqrt{\frac{2GM}{R}}$.
For the new planet,the mass $M^{\prime} = 3M$ and the radius $R^{\prime} = 3R$.
The escape velocity $V_{e}^{\prime}$ for this planet is $V_{e}^{\prime} = \sqrt{\frac{2G(3M)}{3R}}$.
Simplifying the expression,we get $V_{e}^{\prime} = \sqrt{\frac{2GM}{R}} = V_{e}$.
Therefore,the escape velocity remains the same.

(A) The formula for escape velocity is $V_{e} = \sqrt{\frac{2GM}{R}}$.
For the Earth,$V_{e} = \sqrt{\frac{2GM}{R}}$.
For the given planet,the mass $M' = 6M$ and the radius $R' = 2R$.
The escape velocity $V_{e}'$ for this planet is:
$V_{e}' = \sqrt{\frac{2G(6M)}{2R}}$
$V_{e}' = \sqrt{3 \times \frac{2GM}{R}}$
$V_{e}' = \sqrt{3} V_{e}$.

(C) The formula for escape velocity is $v_{e} = \sqrt{\frac{2GM}{R}}$.
Since the mass $M$ of a planet can be expressed in terms of its density $\rho$ and radius $R$ as $M = \text{Volume} \times \text{density} = \frac{4}{3} \pi R^{3} \rho$.
Substituting this into the escape velocity formula:
$v_{e} = \sqrt{\frac{2G}{R} \times \frac{4}{3} \pi R^{3} \rho} = \sqrt{\frac{8}{3} G \pi R^{2} \rho}$.
Simplifying the expression,we get $v_{e} = R \sqrt{\frac{8}{3} G \pi \rho}$.
Since $\frac{8}{3}$,$G$,and $\pi$ are constants,we find that $v_{e} \propto R \sqrt{\rho}$.

(B) The escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$.
Given that the projection velocity $v = \frac{1}{3} v_e = \frac{1}{3} \sqrt{\frac{2GM}{R}}$.
Using the law of conservation of energy,the total energy at the surface equals the total energy at the maximum height $h$:
$TE_{\text{surface}} = TE_{\text{height } h}$
$-\frac{GMm}{R} + \frac{1}{2}mv^2 = -\frac{GMm}{R+h} + 0$
Substituting $v^2 = \frac{1}{9} \times \frac{2GM}{R}$:
$-\frac{GMm}{R} + \frac{1}{2}m \left( \frac{2GM}{9R} \right) = -\frac{GMm}{R+h}$
$-\frac{GMm}{R} + \frac{GMm}{9R} = -\frac{GMm}{R+h}$
$-\frac{8GMm}{9R} = -\frac{GMm}{R+h}$
$\frac{8}{9R} = \frac{1}{R+h}$
$8(R+h) = 9R$
$8R + 8h = 9R$
$8h = R$
$h = \frac{R}{8}$

(B) Let the mass of the body be $m$ and the mass of the earth be $M$. The escape velocity is $v_e = \sqrt{\frac{2GM}{R}}$.
The initial velocity of the body is $v = \frac{v_e}{2} = \frac{1}{2} \sqrt{\frac{2GM}{R}}$.
Using the law of conservation of mechanical energy between the surface of the earth and the maximum height $h$ (where final velocity is $0$):
$K_i + U_i = K_f + U_f$
$\frac{1}{2}mv^2 - \frac{GMm}{R} = 0 - \frac{GMm}{R+h}$
Substituting $v^2 = \frac{1}{4} \cdot \frac{2GM}{R} = \frac{GM}{2R}$:
$\frac{1}{2}m \left( \frac{GM}{2R} \right) - \frac{GMm}{R} = - \frac{GMm}{R+h}$
$\frac{GMm}{4R} - \frac{GMm}{R} = - \frac{GMm}{R+h}$
$-\frac{3GMm}{4R} = - \frac{GMm}{R+h}$
$\frac{3}{4R} = \frac{1}{R+h}$
$3(R+h) = 4R$
$3R + 3h = 4R$
$3h = R$
$h = R/3$

(C) The formula for escape velocity is $v_{e} = \sqrt{\frac{2GM}{R}}$.
Let $M_{e}$ and $R_{e}$ be the mass and radius of the Earth,and $M_{m}$ and $R_{m}$ be the mass and radius of the Moon.
Given: $M_{e} = 81 M_{m}$ and $R_{e} = 3.5 R_{m}$.
The ratio of escape velocity on Earth $(v_{e})$ to that on the Moon $(v_{m})$ is:
$\frac{v_{e}}{v_{m}} = \sqrt{\frac{M_{e}}{R_{e}} \times \frac{R_{m}}{M_{m}}}$
Substituting the given values:
$\frac{v_{e}}{v_{m}} = \sqrt{\frac{81 M_{m}}{3.5 R_{m}} \times \frac{R_{m}}{M_{m}}}$
$\frac{v_{e}}{v_{m}} = \sqrt{\frac{81}{3.5}} = \sqrt{23.14} \approx 4.81$.

(C) The escape velocity $v$ at a distance $r$ from the center of the Earth is given by the formula $v = \sqrt{\frac{2GM}{r}}$.
For the surface of the Earth,the distance $r = R$,so the escape velocity is $V_{E} = \sqrt{\frac{2GM}{R}}$.
For a space station at a height $h = R$ above the surface,the distance from the center of the Earth is $r = R + h = R + R = 2R$.
The escape velocity $v_{s}$ at the space station is $v_{s} = \sqrt{\frac{2GM}{2R}} = \sqrt{\frac{GM}{R}}$.
Comparing $v_{s}$ with $V_{E}$:
$v_{s} = \frac{1}{\sqrt{2}} \sqrt{\frac{2GM}{R}} = \frac{1}{\sqrt{2}} V_{E}$.
Thus,the escape velocity on the space station is $\frac{1}{\sqrt{2}}$ times $V_{E}$.

(B) The orbital velocity of a satellite orbiting close to the Earth is given by $v_0 = \sqrt{\frac{GM}{R}}$.
The kinetic energy of the satellite in this orbit is $K = \frac{1}{2}mv_0^2 = \frac{GMm}{2R}$.
To escape the gravitational pull of the Earth,the satellite must reach the escape velocity,which is $v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2}v_0$.
The required kinetic energy to escape is $K_e = \frac{1}{2}mv_e^2 = \frac{1}{2}m(2v_0^2) = 2(\frac{1}{2}mv_0^2) = 2K$.
The extra kinetic energy required is $\Delta K = K_e - K = 2K - K = K$.

(A) Let the body be at a distance $r$ from the centre of the earth with velocity $v$. By the law of conservation of mechanical energy:
$(TE)_{\text{at surface}} = (TE)_{\text{at distance } r}$
$\frac{1}{2} m v_e^2 - \frac{GMm}{R} = \frac{1}{2} m v^2 - \frac{GMm}{r}$
Since $v_e = \sqrt{\frac{2GM}{R}}$,we have $\frac{1}{2} v_e^2 = \frac{GM}{R} = gR$.
Substituting this,$\frac{1}{2} v^2 = gR - gR + \frac{GM}{r} = \frac{GM}{r} = \frac{gR^2}{r}$.
Thus,$v = \frac{dr}{dt} = R \sqrt{\frac{2g}{r}}$.
Rearranging and integrating from $r = R$ to $r = R + h = 4R$:
$\int_0^t dt = \int_R^{4R} \sqrt{\frac{r}{2gR^2}} dr = \frac{1}{R\sqrt{2g}} \int_R^{4R} r^{1/2} dr$
$t = \frac{1}{R\sqrt{2g}} \left[ \frac{2}{3} r^{3/2} \right]_R^{4R} = \frac{2}{3R\sqrt{2g}} \left[ (4R)^{3/2} - R^{3/2} \right]$
$t = \frac{2}{3R\sqrt{2g}} R^{3/2} [8 - 1] = \frac{2}{3} \sqrt{\frac{R}{2g}} \times 7 = \frac{7}{3} \sqrt{\frac{2R}{g}}$
Substituting $R = 6.4 \times 10^6 \ m$ and $g = 9.8 \ m/s^2$:
$t = \frac{7}{3} \sqrt{\frac{2 \times 6.4 \times 10^6}{9.8}} = \frac{7}{3} \sqrt{1.306 \times 10^6} \approx \frac{7}{3} \times 1142.8 \approx 2666.5 \ s$
$t = \frac{2666.5}{60} \approx 44.44 \ min$.

(D) The formula for escape velocity $v_e$ from the surface of a sphere of mass $M$ and radius $R$ is given by:
$v_e = \sqrt{\frac{2GM}{R}}$
Squaring both sides,we get:
$v_e^2 = \frac{2GM}{R}$
Rearranging to solve for $R$:
$R = \frac{2GM}{v_e^2}$
Given values:
$M = 6 \times 10^{24} \,kg$
$v_e = 3 \times 10^4 \,ms^{-1}$
$G = 6.66 \times 10^{-11} \,N \,m^2 \,kg^{-2}$
Substituting the values:
$R = \frac{2 \times 6.66 \times 10^{-11} \times 6 \times 10^{24}}{(3 \times 10^4)^2}$
$R = \frac{79.92 \times 10^{13}}{9 \times 10^8}$
$R = 8.88 \times 10^5 \,m$
Converting to kilometers:
$R = 888 \,km$
Thus,the correct option is $D$.

(A) The formula for escape velocity $v_e$ is given by $v_e = \sqrt{\frac{2GM}{R}}$.
Given for the first planet: $v_{e1} = \sqrt{\frac{2GM}{R}} = 14 \,km \,s^{-1}$.
For the second planet,the mass is $M$ and the diameter is $8R$,so the radius $R' = \frac{8R}{2} = 4R$.
The escape velocity for the second planet is $v_{e2} = \sqrt{\frac{2GM}{R'}} = \sqrt{\frac{2GM}{4R}}$.
This can be written as $v_{e2} = \frac{1}{\sqrt{4}} \sqrt{\frac{2GM}{R}} = \frac{1}{2} v_{e1}$.
Substituting the value of $v_{e1}$: $v_{e2} = \frac{14}{2} = 7 \,km \,s^{-1}$.

(C) The initial speed of the rocket is $v_i = 0.5 v_e = \frac{1}{2} \sqrt{\frac{2GM}{R}}$.
At maximum height $h$,the final velocity $v_f = 0$.
Applying the law of conservation of mechanical energy:
$K_i + U_i = K_f + U_f$
$\frac{1}{2} m v_i^2 - \frac{GMm}{R} = 0 - \frac{GMm}{R+h}$
Substituting $v_i^2 = \frac{1}{4} \left( \frac{2GM}{R} \right) = \frac{GM}{2R}$:
$\frac{1}{2} m \left( \frac{GM}{2R} \right) - \frac{GMm}{R} = - \frac{GMm}{R+h}$
$\frac{GMm}{4R} - \frac{GMm}{R} = - \frac{GMm}{R+h}$
$-\frac{3GMm}{4R} = - \frac{GMm}{R+h}$
$\frac{3}{4R} = \frac{1}{R+h}$
$3(R+h) = 4R$
$3R + 3h = 4R$
$3h = R$
$h = \frac{R}{3}$