Escape Velocity and Escape Energy Questions in English

(A) Earth is surrounded by an atmosphere of gases (air).
The reason is that in the Earth's atmosphere,the average thermal velocity of even the lightest molecules at the maximum possible temperature is small compared to the escape velocity,which in turn depends upon gravity.
The escape velocity is given by $v_{e} = \sqrt{2gR_{e}}$.
Since the thermal velocity of gas molecules is much less than the escape velocity,the molecules cannot escape the Earth's gravitational pull. Hence,an atmosphere exists around the Earth.

(A) The gravitational potential $V$ at the midpoint between the centers of the Earth and the Moon is given by the sum of potentials due to both masses:
$V = -\frac{GM_1}{d/2} - \frac{GM_2}{d/2} = -\frac{2G}{d}(M_1 + M_2)$
The potential energy $PE$ of a particle of mass $m$ at this point is:
$PE = m \times V = -\frac{2G m}{d}(M_1 + M_2)$
To escape to infinity,the total energy of the particle must be at least zero. Therefore,the kinetic energy $KE$ provided must equal the magnitude of the potential energy:
$KE = |PE|$
$\frac{1}{2}mv^2 = \frac{2G m}{d}(M_1 + M_2)$
Solving for the velocity $v$:
$v^2 = \frac{4G}{d}(M_1 + M_2)$
$v = 2\sqrt{\frac{G}{d}(M_1 + M_2)}$

(C) According to the law of conservation of energy,the total energy at the surface of the Earth equals the total energy at the maximum height $h$.
At the surface: $E_i = \frac{1}{2}mv^2 - \frac{GMm}{R}$.
At maximum height $h$: $E_f = 0 - \frac{GMm}{R+h}$.
Equating $E_i = E_f$: $\frac{1}{2}mv^2 - \frac{GMm}{R} = - \frac{GMm}{R+h}$.
Using $V_e = \sqrt{\frac{2GM}{R}}$,we have $GM = \frac{V_e^2 R}{2}$.
Substituting this: $\frac{v^2}{2} - \frac{V_e^2}{2} = - \frac{V_e^2 R}{2(R+h)}$.
Rearranging gives: $\frac{1}{R+h} = \frac{1}{R} \left(1 - \frac{v^2}{V_e^2}\right)$.
Given $v = 10 \ km/s$ and $V_e = 11.2 \ km/s$:
$h = \frac{R}{\left(\frac{V_e^2}{v^2} - 1\right)} = \frac{R}{\left(\left(\frac{11.2}{10}\right)^2 - 1\right)} = \frac{R}{(1.2544 - 1)} = \frac{R}{0.2544} \approx 3.93R$.
Thus,the maximum height attained is approximately $4R$.

(A) The escape velocity $v_e$ is given by the formula $v_e = \sqrt{\frac{2GM}{R}}$.
Given $v_e = 100\, m/s$,we have $\sqrt{\frac{2GM}{R}} = 100$.
Squaring both sides,we get $\frac{2GM}{R} = 10000$,which implies $\frac{GM}{R} = 5000$.
The gravitational potential energy $U$ of a body of mass $m$ on the surface of a planet is given by $U = -\frac{GMm}{R}$.
Substituting the values $m = 1\, kg$ and $\frac{GM}{R} = 5000$,we get $U = -(5000) \times 1 = -5000\, J$.

(A) According to the law of conservation of mechanical energy,the total energy at the surface is equal to the total energy at the maximum height $h$.
At the surface: $E_i = K_i + U_i = \frac{1}{2}mv^2 - \frac{GMm}{R}$
Given $v = \frac{v_e}{2}$,where $v_e = \sqrt{\frac{2GM}{R}}$. Thus,$v^2 = \frac{v_e^2}{4} = \frac{2GM}{4R} = \frac{GM}{2R}$.
$E_i = \frac{1}{2}m(\frac{GM}{2R}) - \frac{GMm}{R} = \frac{GMm}{4R} - \frac{GMm}{R} = -\frac{3GMm}{4R}$.
At maximum height $h$,the velocity is zero,so $E_f = K_f + U_f = 0 - \frac{GMm}{R+h}$.
Equating $E_i = E_f$: $-\frac{3GMm}{4R} = -\frac{GMm}{R+h}$.
$\frac{3}{4R} = \frac{1}{R+h} \implies 3(R+h) = 4R \implies 3R + 3h = 4R \implies 3h = R \implies h = \frac{R}{3}$.

(C) The escape velocity $v_e$ of a body from the surface of the earth is given by the formula $v_e = \sqrt{2gR}$.
To project a body of mass $m$ to infinity,the required kinetic energy is equal to the work done against the gravitational pull of the earth,which is equal to the kinetic energy at the escape velocity.
Kinetic Energy $(K)$ = $\frac{1}{2}mv_e^2$.
Substituting the value of $v_e$:
$K = \frac{1}{2}m(\sqrt{2gR})^2$.
$K = \frac{1}{2}m(2gR)$.
$K = mgR$.

(B) The velocity of a particle falling from infinity to the Earth's surface is equal to the escape velocity of the Earth.
By the principle of conservation of energy,the total energy at infinity is zero (kinetic energy + potential energy = $0 + 0 = 0$).
At the surface of the Earth,the total energy is $\frac{1}{2}mv^2 - \frac{GMm}{R} = 0$.
Solving for $v$,we get $v = \sqrt{\frac{2GM}{R}}$.
Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$.
Substituting this,$v = \sqrt{\frac{2gR^2}{R}} = \sqrt{2gR}$.

(C) The escape velocity $(v_e)$ of an object from the surface of a planet is given by the formula:
$v_e = \sqrt{\frac{2GM}{R}}$
Where:
$G$ is the universal gravitational constant,
$M$ is the mass of the planet,
$R$ is the radius of the planet.
From the formula,it is clear that the escape velocity depends only on the mass $(M)$ and the radius $(R)$ of the planet.
It does not depend on the mass of the escaping particle or the temperature of the planet.
Therefore,the correct factors are $I$ and $IV$.

(B) The formula for escape velocity is $v = \sqrt{\frac{2GM}{R}}$.
Since mass $M = \text{density} (\rho) \times \text{volume} = \rho \times \frac{4}{3}\pi R^3$,we substitute $M$ into the formula:
$v = \sqrt{\frac{2G(\rho \cdot \frac{4}{3}\pi R^3)}{R}} = \sqrt{\frac{8}{3}\pi G \rho R^2} = R \sqrt{\frac{8}{3}\pi G \rho}$.
Given that the mean density $\rho$ is constant,we find that $v \propto R$.
For the Earth $(e)$ and the planet $(p)$,we have $\frac{v_e}{v_p} = \frac{R_e}{R_p}$.
Given $R_p = 2R_e$,we get $\frac{v_e}{v_p} = \frac{R_e}{2R_e} = \frac{1}{2}$.
Therefore,$v_e = \frac{v_p}{2}$.

(A) The escape velocity $v_e$ of an object from the surface of the Earth is the minimum velocity required for the object to escape the Earth's gravitational field.
It is derived by equating the kinetic energy of the object to the magnitude of its gravitational potential energy at the Earth's surface: $\frac{1}{2}mv_e^2 = \frac{GM_em}{R_e}$.
Solving for $v_e$,we get $v_e = \sqrt{\frac{2GM_e}{R_e}}$.
Note that the mass of the object $m$ cancels out,meaning the escape velocity is independent of the mass of the projectile.

(C) The formula for escape velocity is $v_e = \sqrt{2gR}$.
Given,for the planet: $g_p = 2g_e$ and $R_p = 2R_e$ (since diameter is twice,radius is also twice).
Taking the ratio of escape velocities:
$\frac{v_p}{v_e} = \sqrt{\frac{g_p}{g_e} \times \frac{R_p}{R_e}}$
$\frac{v_p}{v_e} = \sqrt{2 \times 2} = \sqrt{4} = 2$
Therefore,$v_p = 2 \times v_e = 2 \times 11.2 \ km/s = 22.4 \ km/s$.

(A) The formula for escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$.
Since mass $M = \text{density} (\rho) \times \text{volume} = \rho \times \frac{4}{3}\pi R^3$,we can substitute $M$ into the formula:
$v_e = \sqrt{\frac{2G(\rho \cdot \frac{4}{3}\pi R^3)}{R}} = \sqrt{\frac{8}{3}\pi G \rho} \cdot R$.
Given that the mean density $\rho$ is constant,we find that $v_e \propto R$.
If the radius of the planet is twice that of the earth $(R' = 2R)$,then the new escape velocity $v_e'$ will be:
$v_e' = 2 \times v_e = 2 \times 11 \, km/s = 22 \, km/s$.

(B) To escape the earth's gravitational field,the total mechanical energy of the body must be at least zero.
$K.E. + P.E. = 0$
$\frac{1}{2}mv^2 - \frac{GMm}{r} = 0$
$\frac{1}{2}mv^2 = \frac{GMm}{r}$
$v = \sqrt{\frac{2GM}{r}}$
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 velocity equation:
$v = \sqrt{\frac{2(gr^2)}{r}}$
$v = \sqrt{2gr}$

(C) The formula for escape velocity is $V_{e} = \sqrt{2gR}$.
Given that the acceleration due to gravity $g \approx 9.8 \ m/s^2$ (or $10 \ m/s^2$ for approximation) and the radius of the earth $R \approx 6400 \ km = 6.4 \times 10^6 \ m$.
Substituting the values:
$V_{e} = \sqrt{2 \times 9.8 \times 6.4 \times 10^6} \approx 11.2 \times 10^3 \ m/s$.
Therefore,the escape velocity is $11.2 \ km/s$.

(C) The formula for the escape velocity $(v_e)$ of a body from the surface of a planet of mass $(M)$ and radius $(R)$ is given by $v_e = \sqrt{\frac{2GM}{R}}$.
Here,$G$ is the universal gravitational constant.
It is evident from the formula that the escape velocity depends only on the mass and radius of the planet (or celestial body) from which the object is being launched.
It does not depend on the mass $(m)$ of the body (projectile) being launched.
Therefore,the escape velocity is proportional to $m^0$.

(B) The orbital velocity of a satellite is given by $v_0 = \sqrt{\frac{GM}{r}}$.
The escape velocity from the same distance $r$ is given by $v_e = \sqrt{\frac{2GM}{r}}$.
Comparing the two expressions,we get $v_e = \sqrt{2} \, v_0$.
Therefore,for the moon to cease to be a satellite and escape the earth's gravitational field,its orbital velocity must be increased by a factor of $\sqrt{2}$.

(A) The escape velocity $v$ is given by the energy conservation principle: $\frac{-GMm}{R} + \frac{1}{2}mv^2 = 0$.
Solving for $v$,we get $v = \sqrt{\frac{2GM}{R}}$.
The mass of the Earth $M$ in terms of its mean density $\rho$ and radius $R$ is $M = \rho \times \frac{4}{3}\pi R^3$.
Substituting this value of $M$ into the escape velocity formula:
$v = \sqrt{\frac{2G}{R} \times \rho \times \frac{4}{3}\pi R^3}$.
Simplifying the expression:
$v = \sqrt{\frac{8}{3}G\pi R^2 \rho} = R\sqrt{\frac{8\pi}{3}G\rho}$.

(B) The formula for escape velocity is $v_e = \sqrt{\frac{2GM}{R}}$.
Since $G$ is the universal gravitational constant,if the radius $R$ remains constant,then $v_e \propto \sqrt{M}$.
Let the initial mass be $M$ and the new mass be $M' = 4M$.
The new escape velocity $v_e'$ will be $v_e' = \sqrt{\frac{2G(4M)}{R}} = 2 \sqrt{\frac{2GM}{R}} = 2v_e$.
Therefore,the escape velocity becomes $2$ times the original value.

(A) The formula for escape velocity is given by $V_e = \sqrt{\frac{2GM}{R}}$.
From this expression,we can see that $V_e \propto \sqrt{\frac{M}{R}}$.
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: $M_p = 3M_e$ and $R_p = 3R_e$.
The escape velocity of the planet $V_p$ is given by:
$V_p = \sqrt{\frac{2GM_p}{R_p}} = \sqrt{\frac{2G(3M_e)}{3R_e}} = \sqrt{\frac{2GM_e}{R_e}} = V_e$.
Therefore,the escape velocity remains the same.

(C) The gravitational potential energy of a body of mass $m$ at the surface of the Earth is given by $PE = -\frac{GMm}{R}$.
Since $g = \frac{GM}{R^2}$,we can write $GM = gR^2$.
Substituting this into the potential energy formula: $PE = -\frac{(gR^2)m}{R} = -mgR$.
Given $m = 500 \, kg$,$g = 9.8 \, m/s^2$,and $R = 6.4 \times 10^6 \, m$:
$PE = -(500) \times (9.8) \times (6.4 \times 10^6) = -3.136 \times 10^{10} \, J$.
To make the body escape,we must provide energy equal to the magnitude of its potential energy,which is $3.1 \times 10^{10} \, J$.

(D) The formula for escape velocity is $v = \sqrt{\frac{2GM}{R}}$.
Given that the escape velocity of the earth is $v_e = 11.2 \ km/s$.
For another planet,the mass $M_p = 100 M_e$ and the radius $R_p = 4 R_e$.
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: $\frac{v_p}{v_e} = \sqrt{\frac{100 M_e}{M_e} \times \frac{R_e}{4 R_e}} = \sqrt{100 \times \frac{1}{4}} = \sqrt{25} = 5$.
Therefore,$v_p = 5 \times v_e = 5 \times 11.2 \ km/s = 56.0 \ km/s$.

(A) 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 $V_e$ be the escape velocity of the Earth.
For the given planet,$M_p = 6M_e$ and $R_p = 2R_e$.
The escape velocity of the planet $V_p$ is given by $V_p = \sqrt{\frac{2GM_p}{R_p}}$.
Taking the ratio: $\frac{V_p}{V_e} = \sqrt{\frac{M_p}{M_e} \times \frac{R_e}{R_p}} = \sqrt{6 \times \frac{1}{2}} = \sqrt{3}$.
Therefore,$V_p = \sqrt{3} \, V_e$.

(A) The formula for escape velocity is $v_e = \sqrt{2gR}$.
Given,$g_p = 9g_e$ and $R_p = 4R_e$,where $g_p$ and $R_p$ are the acceleration due to gravity and radius of the planet,and $g_e$ and $R_e$ are those of Earth.
The ratio of escape velocities is $\frac{v_p}{v_e} = \sqrt{\frac{g_p}{g_e} \times \frac{R_p}{R_e}}$.
Substituting the given values: $\frac{v_p}{v_e} = \sqrt{9 \times 4} = \sqrt{36} = 6$.
Since the escape velocity on Earth is approximately $v_e = 11.2 \ km/s$,the escape velocity on the planet is $v_p = 6 \times 11.2 \ km/s = 67.2 \ km/s$.

(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: $M_p = 8M_e$ and $R_p = 2R_e$.
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 values: $\frac{v_p}{v_e} = \sqrt{8 \times \frac{1}{2}} = \sqrt{4} = 2$.
Therefore,$v_p = 2 \times v_e = 2 \times 11.2 \, km/s = 22.4 \, km/s$.

(C) The formula for escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$.
From this formula,we can see that $v_e \propto \sqrt{\frac{M}{R}}$.
Let the initial mass be $M$ and radius be $R$. The initial escape velocity is $v_{e1} = 11.2 \, km/s$.
According to the problem,the new mass $M' = 2M$ and the new radius $R' = \frac{R}{2}$.
The new escape velocity $v_{e2}$ is given by:
$v_{e2} = \sqrt{\frac{2G(2M)}{R/2}} = \sqrt{4 \cdot \frac{2GM}{R}} = 2 \cdot \sqrt{\frac{2GM}{R}} = 2 \cdot v_{e1}$.
Substituting the value of $v_{e1}$:
$v_{e2} = 2 \cdot 11.2 \, km/s = 22.4 \, km/s$.

(C) The formula for escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$.
For the Earth,$v_e = \sqrt{\frac{2GM_e}{R_e}} = 11.2 \, km/s$.
For the Moon,the mass $M_m = \frac{M_e}{81}$ and the radius $R_m = \frac{R_e}{4}$.
The escape velocity on the Moon is $v_m = \sqrt{\frac{2GM_m}{R_m}} = \sqrt{\frac{2G(M_e/81)}{(R_e/4)}} = \sqrt{\frac{2GM_e}{R_e} \times \frac{4}{81}}$.
Substituting the values,$v_m = \sqrt{\frac{4}{81}} \times \sqrt{\frac{2GM_e}{R_e}} = \frac{2}{9} \times 11.2 \, km/s$.
$v_m = 0.222 \times 11.2 \approx 2.488 \, km/s$,which is approximately $2.5 \, km/s$.

(C) For matter to escape from the equator of a rotating star,the centrifugal force acting on a particle of mass $m$ at the equator must be equal to the gravitational force acting on it.
At the equator,the condition for the particle to just lift off is $m\omega^2 R = \frac{GMm}{R^2}$.
However,the question asks for the angular velocity at which matter escapes,which relates to the condition where the effective gravity becomes zero.
Equating the centripetal force required for circular motion at the surface to the gravitational force: $m\omega^2 R = \frac{GMm}{R^2}$.
Solving for $\omega$: $\omega^2 = \frac{GM}{R^3}$.
Thus,$\omega = \sqrt{\frac{GM}{R^3}}$.
Note: If the question implies the velocity equivalent to escape velocity $v_e = \sqrt{\frac{2GM}{R}}$,then $\omega = \frac{v_e}{R} = \sqrt{\frac{2GM}{R^3}}$.
Given the options provided,the correct expression is $\sqrt{\frac{2GM}{R^3}}$.

(D) The minimum velocity required for an object to escape the gravitational pull of a planet is known as the escape velocity $(v_e)$.
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: $g = 9.8 \ m/s^2$ and $R = 6.4 \times 10^6 \ m$.
Substituting these values into the formula:
$v_e = \sqrt{2 \times 9.8 \times 6.4 \times 10^6}$
$v_e = \sqrt{19.6 \times 6.4 \times 10^6}$
$v_e = \sqrt{125.44 \times 10^6}$
$v_e = 11.2 \times 10^3 \ m/s$.
Thus,the correct option is $D$.

(A) The formula for escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$.
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: $M_p = 1000 M_e$ and $R_p = 10 R_e$.
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: $\frac{v_p}{v_e} = \sqrt{1000 \times \frac{1}{10}} = \sqrt{100} = 10$.
Therefore,$v_p = 10 \times v_e = 10 \times 11.2 \, km/s = 112 \, km/s$.

(B) The escape velocity $v_e$ from the surface of a planet of mass $M$ and radius $R$ is given by the formula:
$v_e = \sqrt{\frac{2GM}{R}}$
Assuming the planet is a sphere of uniform density $\rho$,its mass $M$ can be expressed in terms of its volume and density:
$M = \text{Volume} \times \text{Density} = \left( \frac{4}{3} \pi R^3 \right) \rho$
Substituting the expression for $M$ into the escape velocity formula:
$v_e = \sqrt{\frac{2G}{R} \left( \frac{4}{3} \pi R^3 \rho \right)}$
Simplifying the expression:
$v_e = \sqrt{\frac{8}{3} G \pi R^2 \rho}$
$v_e = R \sqrt{\frac{8}{3} G \pi \rho}$
Since $G$,$\pi$,and the constant factor are constants,we have:
$v_e \propto R \sqrt{\rho}$
Therefore,the correct option is $B$.

(D) The escape velocity $v_e$ of a body from a height $h$ above the surface of a planet of mass $M$ and radius $R$ is given by the formula:
$v_e = \sqrt{\frac{2GM}{R+h}}$
From this expression,it is clear that the escape velocity depends on the distance from the center of the planet,which is $(R+h)$.
Therefore,the escape velocity depends upon the height $h$ from which the body is projected.
Thus,option $D$ is correct.

(A) The formula for escape velocity is given by $v_e = \sqrt{2gR}$.
Let $g_e$ and $R_e$ be the acceleration due to gravity and the radius of the Earth,respectively.
For the planet,we are given $g_p = 2g_e$ and $R_p = 2R_e$.
The escape velocity from the planet is $v_p = \sqrt{2g_p R_p}$.
Substituting the given values,we get $v_p = \sqrt{2(2g_e)(2R_e)} = \sqrt{4(2g_e R_e)} = 2\sqrt{2g_e R_e}$.
Since $v_e = \sqrt{2g_e R_e}$,it follows that $v_p = 2v_e$.

(A) The formula for escape velocity $(v_e)$ is given by $v_e = \sqrt{\frac{2GM}{R}}$,where $G$ is the universal gravitational constant,$M$ is the mass of the Earth,and $R$ is the radius of the Earth.
From this expression,it is clear that the escape velocity depends only on the mass and radius of the planet from which the object is launched.
It is independent of the mass of the object (rocket) being launched.
Therefore,option $A$ is correct.

(B) The escape velocity $v$ from the surface of a planet is given by the formula $v = \sqrt{2gR}$,where $g$ is the acceleration due to gravity and $R$ is the radius of the planet.
Given that the ratio of the radii is $\frac{R_A}{R_B} = k_1$ and the ratio of acceleration due to gravity is $\frac{g_A}{g_B} = k_2$.
Therefore,the ratio of the escape velocities $v_A$ and $v_B$ is:
$\frac{v_A}{v_B} = \frac{\sqrt{2g_A R_A}}{\sqrt{2g_B R_B}} = \sqrt{\frac{g_A}{g_B} \times \frac{R_A}{R_B}}$
Substituting the given ratios:
$\frac{v_A}{v_B} = \sqrt{k_2 \times k_1} = \sqrt{k_1 k_2}$.

(D) The formula for escape velocity is $v_e = \sqrt{\frac{2GM}{R}}$.
Given values are $M = 6 \times 10^{24} \ kg$,$v_e = 3 \times 10^8 \ m/s$,and $G = 6.67 \times 10^{-11} \ N \cdot m^2/kg^2$.
Squaring both sides,we get $v_e^2 = \frac{2GM}{R}$.
Rearranging for $R$,we get $R = \frac{2GM}{v_e^2}$.
Substituting the values: $R = \frac{2 \times 6.67 \times 10^{-11} \times 6 \times 10^{24}}{(3 \times 10^8)^2}$.
$R = \frac{80.04 \times 10^{13}}{9 \times 10^{16}} = 8.893 \times 10^{-3} \ m \approx 9 \times 10^{-3} \ m$.
Since $1 \ mm = 10^{-3} \ m$,the radius is $9 \ mm$.

(A) The formula for escape velocity is $v = \sqrt{\frac{2GM}{R}}$.
For Earth,$v_e = \sqrt{\frac{2GM_e}{R_e}}$.
For the imaginary planet,$M_p = 2M_e$ and $R_p = 3R_e$.
Thus,the escape velocity on the planet $v_p$ is:
$v_p = \sqrt{\frac{2G(2M_e)}{3R_e}} = \sqrt{\frac{2}{3}} \sqrt{\frac{2GM_e}{R_e}}$.
Substituting $v_e$ into the equation:
$v_p = \sqrt{\frac{2}{3}} v_e$.

(C) The formula for escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$.
Given that the mass of the planet $M_p = M_e$ (mass of Earth) and the radius $R_p = \frac{R_e}{4}$ (where $R_e$ is the radius of Earth).
The escape velocity for the planet is $v_p = \sqrt{\frac{2GM_p}{R_p}} = \sqrt{\frac{2GM_e}{R_e/4}}$.
This simplifies to $v_p = \sqrt{4 \times \frac{2GM_e}{R_e}} = 2 \times \sqrt{\frac{2GM_e}{R_e}}$.
Since the escape velocity of Earth is $v_e = 11.2 \ km/s$,the escape velocity for the planet is $v_p = 2 \times 11.2 \ km/s = 22.4 \ km/s$.

(B) The escape velocity is the minimum initial velocity required for an object to escape the gravitational field of a planet and never return.
For an object of mass $m$ to escape from the surface of the Earth (mass $M_e$,radius $R_e$),its initial kinetic energy must be equal to the magnitude of its gravitational potential energy.
$\frac{1}{2} m v_e^2 = \frac{G M_e m}{R_e}$
Solving for $v_e$,we get:
$v_e = \sqrt{\frac{2 G M_e}{R_e}}$
From this formula,it is clear that the escape velocity $v_e$ depends only on the gravitational constant $G$,the mass of the Earth $M_e$,and the radius of the Earth $R_e$.
It is independent of the mass of the projectile $m$. Therefore,the correct option is $B$.

(A) The formula for escape velocity is $v = \sqrt{2gR}$.
Let $g_e$ and $R_e$ be the acceleration due to gravity and radius of the earth,and $g_p$ and $R_p$ be the acceleration due to gravity and radius of the planet.
Given: $R_p = \frac{1}{4} R_e$ and $g_p = 2g_e$.
The ratio of escape velocity on the planet $(v_p)$ to the escape velocity on earth $(v_e)$ is:
$\frac{v_p}{v_e} = \sqrt{\frac{g_p}{g_e} \times \frac{R_p}{R_e}}$
Substituting the given values:
$\frac{v_p}{v_e} = \sqrt{2 \times \frac{1}{4}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$
Therefore,the escape velocity on the planet is $\frac{1}{\sqrt{2}}$ times the escape velocity on earth.

(B) The formula for escape velocity is $v = \sqrt{\frac{2GM}{R}}$.
Since mass $M = \text{density} (\rho) \times \text{volume} (V) = \rho \times \frac{4}{3}\pi R^3$,we can substitute this into the formula:
$v = \sqrt{\frac{2G(\rho \cdot \frac{4}{3}\pi R^3)}{R}} = \sqrt{\frac{8}{3}\pi G \rho R^2} = R \sqrt{\frac{8}{3}\pi G \rho}$.
Thus,$v \propto R\sqrt{\rho}$.
Given for the planet: $R_p = 4R_e$ and $\rho_p = 9\rho_e$.
Therefore,the ratio of escape velocities is:
$\frac{v_p}{v_e} = \frac{R_p}{R_e} \times \sqrt{\frac{\rho_p}{\rho_e}} = 4 \times \sqrt{9} = 4 \times 3 = 12$.
Hence,$v_p = 12v_e$.

(D) The escape velocity of a body from the surface of the Earth is given by the formula $v_e = \sqrt{2gR_e}$,where $g$ is the acceleration due to gravity and $R_e$ is the radius of the Earth.
This expression shows that the escape velocity depends only on the mass and radius of the planet (or the gravitational potential at the point of projection).
It is independent of the direction or the angle at which the body is projected.
Therefore,if the body is projected at an angle of $45^o$ with the vertical,the escape velocity remains the same,which is $11.2 \ km/s$.

(D) To escape the surface of the earth,the kinetic energy of an object must be equal to the work done against gravity to move it from the surface to infinity.
The energy balance equation is given by: $\frac{1}{2}mv^2 = \frac{GMm}{R}$,where $M$ is the mass of the earth and $R$ is the radius of the earth.
Solving for $v$,we get: $v = \sqrt{\frac{2GM}{R}}$.
We know that the acceleration due to gravity at the surface is $g = \frac{GM}{R^2}$,which implies $GM = gR^2$.
Substituting $GM = gR^2$ into the expression for $v$:
$v = \sqrt{\frac{2(gR^2)}{R}} = \sqrt{2gR}$.
Thus,the correct equation is $V = \sqrt{2gR}$.

(B) The formula for escape velocity $(v_e)$ is given by $v_e = \sqrt{\frac{2GM}{R}}$,where $G$ is the universal gravitational constant,$M$ is the mass of the Earth,and $R$ is the radius of the Earth.
As seen from the formula,the escape velocity depends only on the mass and radius of the planet (or celestial body) from which the object is being launched.
It is independent of the mass of the object being launched.
Therefore,for a body of mass $100\, kg$,the escape velocity remains the same as that for a body of mass $1\, kg$,which is $11.2\, km/s$.

(B) The formula for escape velocity is given by $v = \sqrt{2gR}$.
Given that the acceleration due to gravity on the planet is the same as on Earth $(g_p = g_e)$ and the radius of the planet is four times that of Earth $(R_p = 4R_e)$.
Taking the ratio of escape velocities:
$\frac{v_p}{v_e} = \sqrt{\frac{g_p}{g_e} \times \frac{R_p}{R_e}}$
Substituting the given values:
$\frac{v_p}{v_e} = \sqrt{1 \times 4} = \sqrt{4} = 2$
Therefore,the escape velocity on the planet is $v_p = 2v_e$.

(C) The formula for escape velocity is given by $v = \sqrt{2gR}$.
Given that the value of $g$ is the same for both the planet and the Earth,we have $g_p = g_e$.
The radius of the planet is four times that of the Earth,so $R_p = 4R_e$.
Taking the ratio of the escape velocities:
$\frac{v_p}{v_e} = \sqrt{\frac{g_p}{g_e} \times \frac{R_p}{R_e}} = \sqrt{1 \times 4} = 2$.
Therefore,the escape velocity on the planet is $v_p = 2 \times v_e = 2 \times 11.2 \; km/s = 22.4 \; km/s$.

(B) 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 Earth.
Given that the new radius $R' = 2R$ and the new acceleration due to gravity $g' = 2g$.
Substituting these values into the formula,the new escape velocity $v_e'$ is:
$v_e' = \sqrt{2g'R'} = \sqrt{2(2g)(2R)} = \sqrt{4(2gR)} = 2\sqrt{2gR}$.
Since the initial escape velocity $v_e = 11.2 \, km/s$,the new escape velocity is $v_e' = 2 \times 11.2 \, km/s = 22.4 \, km/s$.

(C) The formula for escape velocity $v_e$ in terms of radius $R$ and mean density $\rho$ is given by $v_e = R \sqrt{\frac{8}{3} \pi \rho G}$.
Given for the planet: $R_p = 2R_e$ and $\rho_p = \frac{1}{4} \rho_e$.
The ratio of escape velocity from the planet $(v_p)$ to that from Earth $(v_e)$ is:
$\frac{v_p}{v_e} = \frac{R_p}{R_e} \sqrt{\frac{\rho_p}{\rho_e}} = 2 \sqrt{\frac{1}{4}} = 2 \times \frac{1}{2} = 1$.
Therefore,the ratio of escape velocity from Earth to that from the planet is $1:1$.

(C) According to the law of conservation of energy,the total energy of the body at the surface of the Earth must be equal to the total energy in interplanetary space.
Let $v$ be the projection velocity and $v'$ be the velocity in interplanetary space.
Total energy at surface = $\frac{1}{2}mv^2 - \frac{GMm}{R} = \frac{1}{2}mv^2 - \frac{1}{2}mv_{es}^2$.
Total energy in interplanetary space = $\frac{1}{2}m(v')^2$.
Equating the two: $\frac{1}{2}m(v')^2 = \frac{1}{2}mv^2 - \frac{1}{2}mv_{es}^2$.
$v' = \sqrt{v^2 - v_{es}^2}$.
Given $v = 2v_{es}$,we have $v' = \sqrt{(2v_{es})^2 - v_{es}^2} = \sqrt{4v_{es}^2 - v_{es}^2} = \sqrt{3v_{es}^2}$.
Therefore,$v' = \sqrt{3}v_{es}$.