Escape Velocity and Escape Energy Questions in English

(A) The formula for escape velocity is $v = \sqrt{2gR}$.
Given the ratio of radii $\frac{R_1}{R_2} = K$ and the ratio of acceleration due to gravity $\frac{g_1}{g_2} = g$.
The ratio of their escape velocities is $\frac{v_1}{v_2} = \sqrt{\frac{g_1}{g_2} \times \frac{R_1}{R_2}}$.
Substituting the given values,we get $\frac{v_1}{v_2} = \sqrt{g \times K} = (Kg)^{1/2}$.

(B) The escape velocity of a body from the surface of the Earth is given by the formula: ${v_e} = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}$.
The orbital velocity of a satellite in a circular orbit of radius $R$ is given by the formula: ${v_o} = \sqrt{\frac{GM}{R}} = \sqrt{gR}$.
By comparing the two expressions,we can see that:
${v_e} = \sqrt{2} \times \sqrt{gR}$
${v_e} = \sqrt{2} {v_o}$.
Therefore,the correct relation is $\sqrt{2} {v_o} = {v_e}$.

(A) The escape velocity of a body from the surface of a planet is given by the formula $v_e = \sqrt{\frac{2GM}{R}}$.
This expression depends only on the mass of the planet $(M)$ and the radius of the planet $(R)$.
It is independent of the mass of the object being launched and the angle of projection.
Therefore,regardless of the angle at which the satellite is launched,the escape velocity remains constant at $11 \ km/s$.

(A) For a satellite in a circular orbit of radius $r$ around the Earth of mass $M$,the gravitational potential energy is $U = -\frac{GMm}{r}$ and the kinetic energy is $E_k = \frac{GMm}{2r}$.
The total mechanical energy of the satellite is $E = E_k + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r} = -E_k$.
To escape into outer space,the total energy of the satellite must become at least $0$.
Let the additional kinetic energy required be $\Delta E$. Then,$E + \Delta E = 0$.
Substituting the value of $E$,we get $-E_k + \Delta E = 0$,which implies $\Delta E = E_k$.
Therefore,an additional kinetic energy equal to $E_k$ must be provided to the satellite.

(C) The escape velocity $v_e$ from the surface of a spherical mass $m$ of radius $r$ is given by the formula: $v_e = \sqrt{\frac{2Gm}{r}}$.
For an object to be a black hole,not even light can escape its gravitational pull.
Therefore,the escape velocity must be greater than or equal to the speed of light $c$.
Thus,the condition is $\sqrt{\frac{2Gm}{r}} \ge c$ or $(2Gm/r)^{1/2} \ge c$.

(C) According to the law of conservation of energy,the total energy at the surface of the Earth is equal to the total energy at the maximum height $h$ from the surface.
Total energy at surface = $-\frac{GMm}{R} + \frac{1}{2}m(kv_e)^2$
Total energy at maximum height = $-\frac{GMm}{R+h} + 0$
Since $v_e = \sqrt{\frac{2GM}{R}}$,we have $v_e^2 = \frac{2GM}{R}$.
Equating the energies: $-\frac{GMm}{R} + \frac{1}{2}m k^2 \left(\frac{2GM}{R}\right) = -\frac{GMm}{R+h}$
Dividing by $GMm$: $-\frac{1}{R} + \frac{k^2}{R} = -\frac{1}{R+h}$
$\frac{k^2 - 1}{R} = -\frac{1}{R+h} \implies \frac{1 - k^2}{R} = \frac{1}{R+h}$
$R+h = \frac{R}{1 - k^2}$
Thus,the distance from the center of the Earth is $r = R+h = \frac{R}{1 - k^2}$.

(B) The atmosphere of a planet or moon is retained due to its gravitational pull. For a gas molecule to be retained,its root mean square velocity $(v_{rms})$ must be significantly less than the escape velocity $(v_e)$ of the celestial body. On the moon,the acceleration due to gravity is very low,which results in a very low escape velocity (approximately $2.38 \ km/s$). The thermal velocity of gas molecules at the moon's surface temperature is higher than this escape velocity. Consequently,gas molecules escape into space,preventing the formation of an atmosphere. Therefore,the correct option is $B$.

(C) The velocity $v$ required to jump to a height $h$ is given by $v = \sqrt{2gh}$.
To escape the gravitational pull,the boy's jump velocity must be equal to the escape velocity $v_e$ of the planet.
The escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$.
Substituting the mass $M = \frac{4}{3}\pi R^3 d$,we get $v_e = \sqrt{\frac{2G}{R} \cdot \frac{4}{3}\pi R^3 d} = \sqrt{\frac{8}{3}\pi G R^2 d} = R \sqrt{\frac{8}{3}\pi Gd}$.
Equating $v = v_e$,we have $\sqrt{2gh} = R \sqrt{\frac{8}{3}\pi Gd}$.
Squaring both sides: $2gh = R^2 (\frac{8}{3}\pi Gd)$.
Solving for $R^2$: $R^2 = \frac{2gh \cdot 3}{8\pi Gd} = \frac{3gh}{4\pi Gd}$.
Therefore,$R = [\frac{3}{4\pi} \frac{gh}{Gd}]^{1/2}$.

(A) The escape velocity of a body from the surface of the Earth is given by the formula $v_e = \sqrt{\frac{2GM}{R}}$.
This expression shows that the escape velocity depends only on the mass of the planet $(M)$ and the radius of the planet $(R)$.
It is independent of the angle of projection or the mass of the body being projected.
Therefore,even if the body is projected at an angle of $60^\circ$ with the vertical,the escape velocity remains the same as the vertical escape velocity.
Thus,the escape velocity is $11 \, km/s$.

(A) The formula for escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$.
Since $M = \rho V = \rho (\frac{4}{3}\pi R^3)$,we substitute this into the formula:
$v_e = \sqrt{\frac{2G(\frac{4}{3}\pi \rho R^3)}{R}} = \sqrt{\frac{8}{3}\pi \rho G R^2} = R \sqrt{\frac{8}{3}\pi \rho G}$.
Given that the density $\rho$ is the same,we have $v_e \propto R$.
Let $v_{e1} = 11 \ km/s$ and $R_1 = R$. For the new planet,$R_2 = 2R$.
Therefore,$\frac{v_{e2}}{v_{e1}} = \frac{R_2}{R_1} = \frac{2R}{R} = 2$.
$v_{e2} = 2 \times v_{e1} = 2 \times 11 \ km/s = 22 \ km/s$.

(C) According to the law of conservation of energy,the total energy at the surface equals the total energy at the maximum height $h$ from the surface.
Initial energy at surface: $E_i = -\frac{GMm}{R} + \frac{1}{2}m(k{v_e})^2$
Final energy at maximum height: $E_f = -\frac{GMm}{R+h}$
Since ${v_e} = \sqrt{\frac{2GM}{R}}$,we have ${v_e}^2 = \frac{2GM}{R}$.
Setting $E_i = E_f$:
$-\frac{GMm}{R} + \frac{1}{2}m k^2 \left(\frac{2GM}{R}\right) = -\frac{GMm}{R+h}$
$-\frac{GMm}{R} + \frac{GMm k^2}{R} = -\frac{GMm}{R+h}$
$-\frac{1}{R}(1 - k^2) = -\frac{1}{R+h}$
$R+h = \frac{R}{1-k^2}$
Thus,the maximum distance from the center of the earth is $r = R+h = \frac{R}{1-k^2}$.

(C) The gravitational potential energy $U$ of an object of mass $m$ at a height $h$ above the Earth's surface is given by $U = -\frac{GMm}{R_e + h}$.
Since $g = \frac{GM}{R_e^2}$,we have $GM = gR_e^2$.
Substituting this into the potential energy formula: $U = -\frac{gR_e^2 m}{R_e + h} = -\frac{mgR_e}{1 + h/R_e}$.
Given $h = 4R_e$,the potential energy is $U = -\frac{mgR_e}{1 + 4} = -\frac{mgR_e}{5}$.
The escape energy is the energy required to bring the object to infinity (where potential energy is $0$),which is $E = 0 - U = \frac{mgR_e}{5}$.

(B) The gravitational potential at the surface of the planet is $V_s = -\frac{GM}{R}$.
The gravitational potential at the center of the planet is $V_c = -\frac{3GM}{2R}$.
Using the law of conservation of energy,the kinetic energy gained by the body equals the loss in potential energy: $\frac{1}{2}mv^2 = m(V_s - V_c)$.
Substituting the values: $\frac{1}{2}mv^2 = m\left(-\frac{GM}{R} - (-\frac{3GM}{2R})\right)$.
$\frac{1}{2}v^2 = \frac{3GM}{2R} - \frac{GM}{R} = \frac{GM}{2R}$.
$v^2 = \frac{GM}{R}$.
Since the escape velocity is $v_e = \sqrt{\frac{2GM}{R}}$,we have $v_e^2 = \frac{2GM}{R}$,which implies $\frac{GM}{R} = \frac{v_e^2}{2}$.
Therefore,$v^2 = \frac{v_e^2}{2}$,which gives $v = \frac{v_e}{\sqrt{2}}$.

(C) The escape velocity from the surface of the Earth is given by $v = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}$.
For a body at a height $h = R$ from the surface,the distance from the center of the Earth is $r = R + h = 2R$.
The total energy required to escape from this distance must be zero:
$-\frac{GMm}{2R} + \frac{1}{2}m(fv)^2 = 0$.
Solving for the escape velocity at height $R$ $(v')$:
$\frac{1}{2}m(v')^2 = \frac{GMm}{2R} \Rightarrow v' = \sqrt{\frac{GM}{R}}$.
Since $v = \sqrt{\frac{2GM}{R}}$,we can write $v' = \frac{1}{\sqrt{2}} \sqrt{\frac{2GM}{R}} = \frac{v}{\sqrt{2}}$.
Given that the escape velocity from the platform is $fv$,we have $fv = \frac{v}{\sqrt{2}}$.
Therefore,$f = \frac{1}{\sqrt{2}}$.

(C) To ensure the particle does not return to Earth, its total mechanical energy must be at least zero at infinity.
According to the law of conservation of mechanical energy:
$E_{initial} = E_{final}$
$\frac{1}{2}mu^2 - \frac{GMm}{R} = 0 + 0$
$\frac{1}{2}mu^2 = \frac{GMm}{R}$
$u^2 = \frac{2GM}{R}$
$u = \sqrt{\frac{2GM}{R}}$
Since $g = \frac{GM}{R^2}$, we can also write $u = \sqrt{2gR}$.

(D) The escape velocity of an object from the earth's surface is given by the formula:
$v_e = \sqrt{\frac{2GM}{R}} \quad ...(i)$
Where $M$ is the mass of the earth and $R$ is the radius of the earth.
The orbital velocity of a satellite orbiting close to the earth's surface is given by:
$v_0 = \sqrt{\frac{GM}{R}} \quad ...(ii)$
Comparing equation $(i)$ and equation $(ii)$,we can substitute $\sqrt{\frac{GM}{R}}$ with $v_0$ in the expression for $v_e$:
$v_e = \sqrt{2} \times \sqrt{\frac{GM}{R}}$
$v_e = \sqrt{2} v_0$

(C) For an object to become a black hole,its escape velocity must be equal to the speed of light $(c)$.
The formula for escape velocity is $v_{esc} = \sqrt{\frac{2GM}{R}}$.
Setting $v_{esc} = c$,we get $c = \sqrt{\frac{2GM}{R}}$,which implies $R = \frac{2GM}{c^2}$.
Given $G = 6.67 \times 10^{-11} \, N \cdot m^2/kg^2$,$M = 5.98 \times 10^{24} \, kg$,and $c = 3 \times 10^8 \, m/s$:
$R = \frac{2 \times 6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{(3 \times 10^8)^2}$
$R = \frac{79.77 \times 10^{13}}{9 \times 10^{16}}$
$R \approx 8.86 \times 10^{-3} \, m \approx 10^{-2} \, m$.

(A) The formula for escape velocity is $v = \sqrt{\frac{2GM}{R}}$.
Since mass $M = \rho \times V = \rho \times \frac{4}{3}\pi R^3$,we substitute this into the formula:
$v = \sqrt{\frac{2G}{R} \times \frac{4}{3}\pi R^3 \rho} = R \sqrt{\frac{8\pi G \rho}{3}}$.
Thus,the escape velocity is proportional to $R\sqrt{\rho}$,i.e.,$v \propto R\sqrt{\rho}$.
Given for the planet: $R_p = 2R_e$ and $\rho_p = 2\rho_e$.
Therefore,the ratio is:
$\frac{v_e}{v_p} = \frac{R_e}{R_p} \times \sqrt{\frac{\rho_e}{\rho_p}} = \frac{R_e}{2R_e} \times \sqrt{\frac{\rho_e}{2\rho_e}} = \frac{1}{2} \times \frac{1}{\sqrt{2}} = \frac{1}{2\sqrt{2}}$.
Hence,the ratio is $1 : 2\sqrt{2}$.

(D) The atmosphere of a planet or satellite is retained if the escape velocity $(v_e)$ is significantly greater than the root mean square velocity $(v_{rms})$ of the gas molecules present in its vicinity.
On the moon,the acceleration due to gravity is very low,which results in a low escape velocity $(v_e \approx 2.38 \ km/s)$.
At the surface temperature of the moon,the root mean square velocity $(v_{rms})$ of gas molecules (like $O_2$,$N_2$,etc.) is higher than this escape velocity.
Consequently,the gas molecules easily overcome the gravitational pull of the moon and escape into space.
Therefore,the correct reason is that the escape velocity of gas molecules is lesser than their root mean square velocity on the moon.

(C) The formula for escape velocity is $v_e = \sqrt{2gR}$.
Therefore,the ratio of escape velocity from the Earth $(v_e)$ to that from the Moon $(v_m)$ is given by:
$\frac{v_e}{v_m} = \sqrt{\frac{g_e}{g_m} \times \frac{R_e}{R_m}}$
Given that $\frac{R_e}{R_m} = 10$ and $\frac{g_e}{g_m} = 6$,we substitute these values:
$\frac{v_e}{v_m} = \sqrt{6 \times 10} = \sqrt{60}$
Since $\sqrt{60} \approx 7.74$,the ratio is nearly $8$.

(B) The formula for escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$,where $G$ is the gravitational constant,$M$ is the mass of the planet/moon,and $R$ is its radius.
Since the mass of the moon $(M_m \approx 7.35 \times 10^{22} \ kg)$ is much smaller than the mass of the earth $(M_e \approx 5.97 \times 10^{24} \ kg)$ and its radius $(R_m \approx 1.74 \times 10^6 \ m)$ is also smaller than the earth's radius $(R_e \approx 6.37 \times 10^6 \ m)$,the ratio $\frac{M}{R}$ for the moon is significantly smaller than that for the earth.
Consequently,the escape velocity from the moon's surface is approximately $2.38 \ km/s$,which is much less than the earth's escape velocity of approximately $11.2 \ km/s$.
Therefore,the correct reason is that the moon's mass and radius result in a smaller gravitational potential at its surface compared to the earth.

(C) The formula for escape velocity is given by $v_e = \sqrt{2gR}$,where $g$ is the acceleration due to gravity and $R$ is the radius of the planet.
Given the ratio of radii $\frac{R_A}{R_B} = r$ and the ratio of acceleration due to gravity $\frac{g_A}{g_B} = x$.
The ratio of escape velocities is $\frac{v_A}{v_B} = \sqrt{\frac{2g_A R_A}{2g_B R_B}} = \sqrt{\frac{g_A}{g_B} \times \frac{R_A}{R_B}}$.
Substituting the given ratios,we get $\frac{v_A}{v_B} = \sqrt{x \times r} = \sqrt{rx}$.

(A) The acceleration due to gravity at the equator $(g_e)$ is related to the acceleration due to gravity at the poles $(g_p)$ by the formula: $g_e = g_p - R\omega^2$.
Given that $g_e = g_p / 2$,we have: $g_p / 2 = g_p - R\omega^2$.
This simplifies to: $R\omega^2 = g_p / 2$.
Since the velocity of a point on the equator is $V = R\omega$,we have $V^2 = R^2\omega^2 = R(R\omega^2) = R(g_p / 2)$.
Thus,$V^2 = g_p R / 2$,which implies $g_p R = 2V^2$.
The escape velocity $V_e$ is given by the formula: $V_e = \sqrt{2g_p R}$.
Substituting $g_p R = 2V^2$ into the escape velocity formula:
$V_e = \sqrt{2(2V^2)} = \sqrt{4V^2} = 2V$.

(A) The escape velocity $V_e$ of a planet is given by the formula $V_e = \sqrt{\frac{2GM}{R}}$.
Since the density $\rho$ is the same for both planets,we can express the mass $M$ as $M = \rho \times \text{Volume} = \rho \times \frac{4}{3} \pi R^3$.
Substituting this into the escape velocity formula:
$V_e = \sqrt{\frac{2G}{R} \times \frac{4}{3} \pi R^3 \rho} = \sqrt{\frac{8}{3} \pi G \rho} \times R$.
This shows that $V_e \propto R$ when the density $\rho$ is constant.
Therefore,the ratio of the escape velocities is $\frac{V_A}{V_B} = \frac{R_A}{R_B}$.
Given that $R_A = 2R_B$,we have $\frac{V_A}{V_B} = \frac{2R_B}{R_B} = 2$.

(D) For ball $A$ falling from infinity,the velocity $v_A$ upon reaching the earth's surface is the escape velocity:
$v_A = \sqrt{\frac{2GM}{R}}$
For ball $B$ falling from a height $h = 6R$,we use the conservation of energy:
$\frac{1}{2}mv_B^2 - \frac{GMm}{R+h} = 0 - \frac{GMm}{R}$
$\frac{1}{2}v_B^2 = GM \left( \frac{1}{R} - \frac{1}{7R} \right) = GM \left( \frac{6}{7R} \right)$
$v_B = \sqrt{\frac{12GM}{7R}}$
Now,the ratio of the velocities of $A$ and $B$ is:
$\frac{v_A}{v_B} = \frac{\sqrt{\frac{2GM}{R}}}{\sqrt{\frac{12GM}{7R}}} = \sqrt{\frac{2GM}{R} \times \frac{7R}{12GM}} = \sqrt{\frac{14}{12}} = \sqrt{\frac{7}{6}}$
Thus,the ratio of the velocity of $A$ to $B$ is $\sqrt{\frac{7}{6}}$.

(C) Let $R$ be the radius of the earth and $M$ be its mass. The escape velocity at the surface is $v_e = \sqrt{\frac{2GM}{R}} = 11.2 \ km/s$.
When the rocket is at an altitude $h = \frac{R}{4}$,its distance from the center of the earth is $r = R + h = R + \frac{R}{4} = \frac{5R}{4}$.
To escape the earth's gravitational pull from this point,the rocket must have a velocity $v'$ such that its total energy is at least zero.
Using the conservation of energy: $\frac{1}{2}mv'^2 - \frac{GMm}{r} = 0$.
$v' = \sqrt{\frac{2GM}{r}} = \sqrt{\frac{2GM}{5R/4}} = \sqrt{\frac{8GM}{5R}} = \sqrt{\frac{4}{5}} \times \sqrt{\frac{2GM}{R}}$.
Substituting $v_e = 11.2 \ km/s$:
$v' = \sqrt{0.8} \times 11.2 \approx 0.8944 \times 11.2 \approx 10.02 \ km/s$.
Thus,the minimum velocity is approximately $10 \ km/s$.

(B) The escape velocity of a planet is given by $v_e = \sqrt{\frac{2GM}{R}}$.
Using the principle of conservation of energy,the total energy at the surface equals the total energy at the center.
The gravitational potential at the surface is $V_S = -\frac{GM}{R}$.
The gravitational potential at the center is $V_C = -\frac{3GM}{2R}$.
Loss in potential energy = Gain in kinetic energy:
$m(V_S - V_C) = \frac{1}{2}mv^2$
$m\left(-\frac{GM}{R} - (-\frac{3GM}{2R})\right) = \frac{1}{2}mv^2$
$m\left(\frac{3GM}{2R} - \frac{2GM}{2R}\right) = \frac{1}{2}mv^2$
$\frac{GM}{2R} = \frac{1}{2}v^2$
$v^2 = \frac{GM}{R}$
Since $v_e^2 = \frac{2GM}{R}$,we have $\frac{GM}{R} = \frac{v_e^2}{2}$.
Therefore,$v^2 = \frac{v_e^2}{2}$,which gives $v = \frac{v_e}{\sqrt{2}}$.

(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 $M_p$ and $R_p$ be the mass and radius of the planet.
Given: $M_p = 10 M_e$ and $R_p = \frac{R_e}{10}$.
The ratio of escape velocities is given by:
$\frac{(v_e)_p}{(v_e)_e} = \sqrt{\frac{M_p}{M_e} \times \frac{R_e}{R_p}}$
Substituting the given values:
$\frac{(v_e)_p}{(v_e)_e} = \sqrt{\frac{10 M_e}{M_e} \times \frac{R_e}{R_e/10}} = \sqrt{10 \times 10} = \sqrt{100} = 10$.
Therefore,$(v_e)_p = 10 \times (v_e)_e = 10 \times 11 \ km/s = 110 \ km/s$.

(A) The energy required to launch an object of mass $m$ from the earth's surface to infinity (free space) is equal to the gravitational potential energy at the surface,which is given by the formula: $E = \frac{GMm}{R}$.
Since the acceleration due to gravity at the surface is $g = \frac{GM}{R^2}$,we can write $GM = gR^2$.
Substituting this into the energy equation: $E = \frac{(gR^2)m}{R} = mgR$.
Given values: $m = 1000 \ kg$,$g = 10 \ m/s^2$,and $R = 6400 \ km = 6.4 \times 10^6 \ m$.
Calculating the energy: $E = 1000 \times 10 \times 6.4 \times 10^6 = 6.4 \times 10^{10} \ J$.

(D) Let the mass of the projected particle be $m_0$. The distance of the particle from each fixed mass $m$ is $l/2$.
The gravitational potential energy $U$ of the particle $m_0$ at the midpoint is given by:
$U = -\frac{Gmm_0}{l/2} - \frac{Gmm_0}{l/2} = -\frac{2Gmm_0}{l} - \frac{2Gmm_0}{l} = -\frac{4Gmm_0}{l}$.
For the particle to escape to infinity,its total mechanical energy must be at least zero.
$E_{total} = KE + PE = 0$
$\frac{1}{2}m_0v^2 + U = 0$
$\frac{1}{2}m_0v^2 - \frac{4Gmm_0}{l} = 0$
$\frac{1}{2}m_0v^2 = \frac{4Gmm_0}{l}$
$v^2 = \frac{8Gm}{l}$
$v = \sqrt{\frac{8Gm}{l}} = 2\sqrt{\frac{2Gm}{l}}$.

(B) The potential energy $(PE)$ of a particle of mass $m$ placed at a point midway between the Earth and the Moon is given by:
$U = -\left(\frac{GM_1 m}{d/2} + \frac{GM_2 m}{d/2}\right) = -\frac{2Gm(M_1 + M_2)}{d}$
The particle will escape to infinity if its total energy becomes at least zero.
By the law of conservation of energy,$U + K = 0$,where $K$ is the kinetic energy $\frac{1}{2}mv^2$.
$-\frac{2Gm(M_1 + M_2)}{d} + \frac{1}{2}mv^2 = 0$
$\frac{1}{2}mv^2 = \frac{2Gm(M_1 + M_2)}{d}$
$v^2 = \frac{4G(M_1 + M_2)}{d}$
$v = \sqrt{\frac{4G(M_1 + M_2)}{d}}$

(B) The gravitational acceleration on the surface of a planet is given by $g = \frac{GM}{R^2} = \frac{G}{R^2} \cdot \frac{4}{3} \pi R^3 \rho = \frac{4}{3} \pi G R \rho$.
Thus,the ratio of gravitational acceleration is $\frac{g_p}{g_e} = \frac{R_p \rho_p}{R_e \rho_e}$.
Given $\frac{g_p}{g_e} = \frac{\sqrt{6}}{11}$ and $\frac{\rho_p}{\rho_e} = \frac{2}{3}$,we have $\frac{R_p}{R_e} = \frac{g_p}{g_e} \cdot \frac{\rho_e}{\rho_p} = \frac{\sqrt{6}}{11} \cdot \frac{3}{2} = \frac{3\sqrt{6}}{22}$.
The escape velocity is $v_e = \sqrt{2gR}$.
Therefore,$\frac{v_p}{v_e} = \sqrt{\frac{g_p R_p}{g_e R_e}} = \sqrt{\frac{g_p}{g_e} \cdot \frac{R_p}{R_e}} = \sqrt{\frac{\sqrt{6}}{11} \cdot \frac{3\sqrt{6}}{22}} = \sqrt{\frac{3 \cdot 6}{11 \cdot 22}} = \sqrt{\frac{18}{242}} = \sqrt{\frac{9}{121}} = \frac{3}{11}$.
Given $v_e = 11 \ km/s$,we get $v_p = \frac{3}{11} \times 11 = 3 \ km/s$.

(B) The forces acting on the spaceship are the gravitational pull due to the Earth and the gravitational pull due to the Moon.
Since the mass of the Earth $(M_E)$ is significantly greater than the mass of the Moon $(M_M)$,the gravitational potential energy required to escape the Earth's gravitational field is much higher than that required for the Moon.
Specifically,the escape velocity from the Earth is approximately $11.2 \ km/s$,whereas from the Moon it is only about $2.38 \ km/s$.
Therefore,the spaceship requires the greatest amount of energy to overcome the gravitational pull during take-off from the Earth's surface.

(A) The gravitational potential energy $U$ of a mass $m$ placed at the midpoint between $M_1$ and $M_2$ is the sum of the potentials due to both masses:
$U = -\frac{G M_1 m}{d/2} - \frac{G M_2 m}{d/2} = -\frac{2 G m}{d} (M_1 + M_2)$
To escape the system,the total energy must be at least zero. By the law of conservation of energy,the kinetic energy provided must equal the magnitude of the potential energy:
$\frac{1}{2} m v_e^2 = |U| = \frac{2 G m}{d} (M_1 + M_2)$
Solving for the escape velocity $v_e$:
$v_e^2 = \frac{4 G (M_1 + M_2)}{d}$
$v_e = \sqrt{\frac{4 G (M_1 + M_2)}{d}}$

(B) For a satellite in a circular orbit of radius $r$,the orbital speed $v$ is given by $v = \sqrt{\frac{GM_e}{r}}$,which implies $v^2 = \frac{GM_e}{r}$.
The potential energy of the object of mass $m$ at distance $r$ is $U = -\frac{GM_em}{r}$.
For the object to just escape the gravitational pull of the Earth,its total mechanical energy must be zero at infinity,meaning its total energy at the point of ejection must also be zero.
Let $K$ be the kinetic energy of the object at the time of ejection.
Total energy $E = K + U = 0$.
$K - \frac{GM_em}{r} = 0$.
$K = \frac{GM_em}{r}$.
Substituting $\frac{GM_e}{r} = v^2$,we get $K = mv^2$.

(D) Let the mass of the meteorite be $m$. The distance of each star from the center of mass $O$ is $r = d/2 = (2\times10^{11} \ m)/2 = 1\times10^{11} \ m$.
The gravitational potential energy of the meteorite at point $O$ due to the two stars is $U = -\frac{GMm}{r} - \frac{GMm}{r} = -\frac{2GMm}{r}$.
To escape the gravitational field,the total mechanical energy of the meteorite at $O$ must be at least zero. By the law of conservation of energy:
$\frac{1}{2}mv^2 + U = 0$
$\frac{1}{2}mv^2 - \frac{2GMm}{r} = 0$
Solving for $v$:
$v^2 = \frac{4GM}{r}$
$v = \sqrt{\frac{4 \times 6.67\times10^{-11} \times 3\times10^{31}}{1\times10^{11}}}$
$v = \sqrt{4 \times 6.67 \times 3 \times 10^9}$
$v = \sqrt{80.04 \times 10^9} = \sqrt{8.004 \times 10^{10}}$
$v \approx 2.83 \times 10^5 \ m/s$.

(B) The minimum energy required to escape the gravitational field of a planet is equal to the magnitude of the gravitational potential energy of the object at the surface: $E = \frac{GMm}{R}$.
Given that the densities $\rho$ of the Earth and the Moon are equal,we have $M = \rho V = \rho \cdot \frac{4}{3}\pi R^3$.
Since $V_e = 64 V_m$,we have $R_e^3 = 64 R_m^3$,which implies $R_e = 4 R_m$.
Also,$M_e = \rho V_e = 64 \rho V_m = 64 M_m$.
The energy required is $E = \frac{GMm}{R}$.
Therefore,$\frac{E_m}{E_e} = \frac{M_m}{M_e} \cdot \frac{R_e}{R_m} = \frac{1}{64} \cdot \frac{4}{1} = \frac{1}{16}$.
Thus,$E_m = \frac{E}{16}$.

(C) The escape velocity $v_e$ at a distance $r$ from the center of the Earth is given by the formula $v_e = \sqrt{\frac{2GM}{r}}$.
Here,the particle is at a distance $R$ from the surface of the Earth,so its distance from the center of the Earth is $r = R + R = 2R$.
Substituting $r = 2R$ into the formula,we get:
$v_e = \sqrt{\frac{2GM}{2R}}$
$v_e = \sqrt{\frac{GM}{R}}$
Thus,the minimum speed required is $\sqrt{\frac{GM}{R}}$.

(B) The forces acting on the spaceship are the gravitational pull due to the Earth and the gravitational pull due to the Moon.
Since the mass of the Earth $(M_E)$ is significantly greater than the mass of the Moon $(M_M)$,the gravitational potential energy and the gravitational force at the Earth's surface are much higher than at the Moon's surface.
To escape the Earth's gravitational influence,the spaceship must overcome a much larger gravitational potential barrier compared to the Moon.
Therefore,the greatest energy is required to perform a take-off from the Earth's gravitational field.

(C) The total energy of a body at the earth's surface is the sum of its kinetic energy $(K.E.)$ and gravitational potential energy $(U)$.
$U = -\frac{GM_E m}{R}$
To project the body to infinity,the final total energy must be at least zero.
By the law of conservation of energy: $K.E. + U = 0$
$K.E. - \frac{GM_E m}{R} = 0$
$K.E. = \frac{GM_E m}{R}$
Since the acceleration due to gravity at the surface is $g = \frac{GM_E}{R^2}$,we can write $GM_E = gR^2$.
Substituting this into the expression for kinetic energy:
$K.E. = \frac{(gR^2)m}{R} = mgR$

(D) The escape velocity $v_e$ from the surface of the Earth is given by the formula:
$v_e = \sqrt{\frac{2GM}{R}}$
The orbital velocity $v_0$ of a satellite in a circular orbit close to the Earth's surface is given by the formula:
$v_0 = \sqrt{\frac{GM}{R}}$
By comparing the two expressions,we can see that:
$v_e = \sqrt{2} \times \sqrt{\frac{GM}{R}}$
$v_e = \sqrt{2} v_0$
Thus,the correct relation is $v_e = \sqrt{2} v_0$.

(A) The escape velocity from the surface of the earth is given by $v_e = \sqrt{\frac{2GM}{R}}$.
At a height $h = R$ from the surface,the distance from the center of the earth is $r = R + h = R + R = 2R$.
The escape velocity $v'$ at this height is given by $v' = \sqrt{\frac{2GM}{r}} = \sqrt{\frac{2GM}{2R}}$.
Simplifying this,we get $v' = \sqrt{\frac{1}{2} \cdot \frac{2GM}{R}} = \frac{1}{\sqrt{2}} \sqrt{\frac{2GM}{R}}$.
Substituting $v_e$ into the equation,we get $v' = \frac{v_e}{\sqrt{2}}$.

(B) The distance of the body from the center of the Earth is $r = R + 2R = 3R$.
The gravitational potential energy $(PE)$ of the body at this distance is given by $PE = -\frac{GMm}{r} = -\frac{GMm}{3R}$.
Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$.
Substituting this into the potential energy expression: $PE = -\frac{(gR^2)m}{3R} = -\frac{mgR}{3}$.
The energy required to escape the gravitational field is equal to the magnitude of the potential energy,which is the binding energy: $E_{escape} = -PE = \frac{mgR}{3}$.

(C) The gravitational potential energy $U$ of a body of mass $m$ at a height $h$ above the Earth's surface is given by $U = -\frac{GMm}{R_e + h}$.
Given $h = 4R_e$,the potential energy is $U = -\frac{GMm}{R_e + 4R_e} = -\frac{GMm}{5R_e}$.
Since $g = \frac{GM}{R_e^2}$,we have $GM = gR_e^2$.
Substituting this into the expression for $U$,we get $U = -\frac{(gR_e^2)m}{5R_e} = -\frac{mgR_e}{5}$.
To escape the Earth's gravitational field,the total energy of the body must be at least zero. Therefore,the minimum energy required is $E = -U = \frac{mgR_e}{5}$.

(D) The gravitational potential $V$ at a distance $r = \frac{R}{2}$ inside a uniform sphere of mass $M$ and radius $R$ is given by:
$V = -\frac{GM}{2R^3} (3R^2 - r^2)$
Substituting $r = \frac{R}{2}$:
$V = -\frac{GM}{2R^3} (3R^2 - \frac{R^2}{4}) = -\frac{GM}{2R^3} (\frac{11R^2}{4}) = -\frac{11GM}{8R}$
The total energy of the particle of mass $m_0$ at this position is $E_i = K_i + U_i = \frac{1}{2}m_0 v_e^2 + m_0 V$.
For the particle to escape,the final total energy $E_f$ at infinity must be at least $0$.
$E_i = E_f \implies \frac{1}{2}m_0 v_e^2 - \frac{11GMm_0}{8R} = 0$
$\frac{1}{2} v_e^2 = \frac{11GM}{8R}$
$v_e^2 = \frac{11GM}{4R}$
$v_e = \sqrt{\frac{11GM}{4R}}$

(A) Let $h$ be the maximum height reached by the object. According to the law of conservation of energy,the total energy at the surface of the earth is equal to the total energy at the maximum height $h$.
Total energy at the surface: $E_i = KE_i + PE_i = \frac{1}{2} m v_p^2 - \frac{GMm}{R}$,where $v_p$ is the projection velocity.
Given that the projection velocity is half the escape velocity: $v_p = \frac{1}{2} v_e = \frac{1}{2} \sqrt{\frac{2GM}{R}}$.
Thus,$KE_i = \frac{1}{2} m (\frac{1}{2} \sqrt{\frac{2GM}{R}})^2 = \frac{1}{2} m (\frac{1}{4} \cdot \frac{2GM}{R}) = \frac{GMm}{4R}$.
Total energy at the surface: $E_i = \frac{GMm}{4R} - \frac{GMm}{R} = -\frac{3GMm}{4R}$.
At maximum height $h$,the velocity is zero,so $KE_f = 0$. The potential energy is $PE_f = -\frac{GMm}{R+h}$.
Equating initial and final energies: $-\frac{3GMm}{4R} = -\frac{GMm}{R+h}$.
$\frac{3}{4R} = \frac{1}{R+h} \implies 3(R+h) = 4R \implies 3R + 3h = 4R \implies 3h = R \implies h = R/3$.

(D) The formula for escape velocity $V_e$ is given by $V_e = \sqrt{\frac{2GM}{R}}$.
Since mass $M = \text{density} (\rho) \times \text{volume} (V) = \rho \times \frac{4}{3} \pi R^3$,we substitute $M$ into the formula:
$V_e = \sqrt{\frac{2G(\rho \cdot \frac{4}{3} \pi R^3)}{R}} = \sqrt{\frac{8}{3} \pi G \rho R^2} = R \sqrt{\frac{8}{3} \pi G \rho}$.
This shows that $V_e \propto R$ when density $\rho$ is constant.
Given $R' = 2R$ and $\rho' = \rho$,the new escape velocity $V_e'$ is:
$\frac{V_e'}{V_e} = \frac{R'}{R} = \frac{2R}{R} = 2$.
Therefore,$V_e' = 2 \times V_e = 2 \times 11.2 \, km/s = 22.4 \, km/s$.

(C) The escape velocity from the surface of the Earth is given by $v_e = \sqrt{2gR}$.
Given that the initial speed of the object is $v = 50\% \text{ of } v_e = \frac{v_e}{2} = \frac{\sqrt{2gR}}{2}$.
Using the principle of conservation of energy,the total energy at the surface equals the total energy at the maximum height $h$:
$\frac{1}{2}mv^2 - \frac{GMm}{R} = 0 - \frac{GMm}{R+h}$.
Substituting $v = \frac{\sqrt{2gR}}{2}$ and $GM = gR^2$:
$\frac{1}{2}m \left(\frac{2gR}{4}\right) - \frac{mgR^2}{R} = - \frac{mgR^2}{R+h}$.
$\frac{mgR}{4} - mgR = - \frac{mgR^2}{R+h}$.
$-\frac{3}{4}mgR = - \frac{mgR^2}{R+h}$.
$\frac{3}{4} = \frac{R}{R+h}$.
$3(R+h) = 4R \Rightarrow 3R + 3h = 4R \Rightarrow 3h = R \Rightarrow h = \frac{R}{3}$.

(B) The orbital velocity of a satellite is given by $v_{0}$.
The escape velocity required to leave the gravitational field is $v_{e} = \sqrt{2} v_{0} \approx 1.414 v_{0}$.
To find the required percentage increase in velocity,we calculate the fractional change:
$\frac{\Delta v}{v_{0}} = \frac{v_{e} - v_{0}}{v_{0}} = \frac{1.414 v_{0} - v_{0}}{v_{0}} = 0.414$.
Therefore,the percentage increase required is $0.414 \times 100 \% = 41.4 \%$.

(D) According to the law of conservation of mechanical energy,the total energy at the surface of the Earth is equal to the total energy at the maximum height $r$ reached by the rocket.
At the surface: $K_1 + U_1 = \frac{1}{2}m(kv_e)^2 - \frac{GMm}{R_e}$
At maximum height $r$: $K_2 + U_2 = 0 - \frac{GMm}{r}$ (since velocity at maximum height is zero).
We know that the escape velocity $v_e = \sqrt{\frac{2GM}{R_e}}$,so $v_e^2 = \frac{2GM}{R_e}$.
Substituting this into the energy equation:
$\frac{1}{2}m k^2 (\frac{2GM}{R_e}) - \frac{GMm}{R_e} = - \frac{GMm}{r}$
Dividing by $GMm$:
$\frac{k^2}{R_e} - \frac{1}{R_e} = - \frac{1}{r}$
$\frac{k^2 - 1}{R_e} = - \frac{1}{r}$
$\frac{1 - k^2}{R_e} = \frac{1}{r}$
Therefore,$r = \frac{R_e}{1 - k^2}$.