Escape Velocity and Escape Energy Questions in English

(C) The relationship between escape velocity $(v_e)$ and orbital velocity $(v_o)$ for a satellite orbiting close to the surface of a planet is given by the formula: $v_e = \sqrt{2} \cdot v_o$.
Rearranging this for orbital velocity,we get: $v_o = \frac{v_e}{\sqrt{2}}$.
Given that the escape velocity $v_e = 2 \, km/s$,we substitute this value into the equation:
$v_o = \frac{2}{\sqrt{2}} = \sqrt{2} \, km/s$.

(B) The escape velocity of a body from the surface of the Earth is given by the formula $v_e = \sqrt{\frac{2GM}{R_e}}$,where $G$ is the gravitational constant,$M$ is the mass of the Earth,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 body from which the object is being projected).
It is independent of the angle of projection.
Therefore,if the body is projected at an angle of $45^o$ with the vertical,the escape velocity remains the same,i.e.,$11 \ km/s$.

(B) The escape speed from the surface of the Earth is given by the formula $v_e = \sqrt{\frac{2GM}{R}}$. This formula depends only on the gravitational constant $G$,the mass of the Earth $M$,and the radius of the Earth $R$. It does not depend on the direction of projection. Thus,the $Assertion$ is correct.
The $Reason$ states that it is easier to attain escape speed if fired in the direction of Earth's rotation. This is true because the rotation of the Earth provides an initial tangential velocity to the projectile. However,this initial velocity effectively reduces the amount of additional velocity required to reach the escape speed relative to the Earth's surface,but it does not change the value of the escape speed itself. Therefore,the $Reason$ is a correct statement,but it does not explain why the escape speed (a fixed physical constant for a given planet) is independent of direction. Thus,the $Reason$ is not the correct explanation of the $Assertion$.

(A) The formula for escape velocity is $v_{e} = \sqrt{\frac{2GM}{R}}$.
For Planet $A$,the escape velocity is $v_{A} = \sqrt{\frac{2GM}{R}}$.
For Planet $B$,the mass is $M' = \frac{M}{2}$ and the radius is $R' = \frac{R}{2}$.
Thus,the escape velocity for Planet $B$ is $v_{B} = \sqrt{\frac{2G(M/2)}{R/2}} = \sqrt{\frac{2GM}{R}}$.
Comparing the two,we get $\frac{v_{A}}{v_{B}} = \frac{\sqrt{2GM/R}}{\sqrt{2GM/R}} = 1$.
Given that $\frac{v_{A}}{v_{B}} = \frac{n}{4}$,we have $1 = \frac{n}{4}$,which implies $n = 4$.

(D) No.
$(b)$ No.
$(c)$ No.
$(d)$ Yes.
The escape velocity $v_{esc}$ of a body from the Earth is given by the formula:
$v_{esc} = \sqrt{\frac{2GM}{R+h}}$
where $G$ is the gravitational constant,$M$ is the mass of the Earth,$R$ is the radius of the Earth,and $h$ is the height of the location from the surface.
$1$. It is clear from the formula that $v_{esc}$ is independent of the mass of the body $(m)$.
$2$. It is independent of the direction of projection as long as it is not directed into the Earth.
$3$. It depends on the location $(R+h)$ from where the body is launched. As the height $h$ increases,the escape velocity decreases.

(B) The escape velocity of a projectile from the Earth is $v_{esc} = 11.2 \; km/s$.
The projection velocity of the projectile is $v_p = 3 v_{esc}$.
Let the mass of the projectile be $m$.
According to the law of conservation of energy,the total energy at the Earth's surface equals the total energy far away from the Earth (where gravitational potential energy is zero).
$\frac{1}{2} m v_p^2 - \frac{G M m}{R} = \frac{1}{2} m v_f^2 + 0$
Since the escape velocity is defined as $v_{esc} = \sqrt{\frac{2GM}{R}}$,we have $\frac{GM}{R} = \frac{v_{esc}^2}{2}$.
Substituting this into the energy equation: $\frac{1}{2} m v_p^2 - \frac{1}{2} m v_{esc}^2 = \frac{1}{2} m v_f^2$.
$v_f = \sqrt{v_p^2 - v_{esc}^2} = \sqrt{(3 v_{esc})^2 - v_{esc}^2} = \sqrt{8 v_{esc}^2} = v_{esc} \sqrt{8}$.
$v_f = 11.2 \times 2.828 = 31.68 \; km/s$.

(D) Mass of the spaceship,$m_{s} = 1000 \; kg$.
Mass of the Sun,$M = 2 \times 10^{30} \; kg$.
Mass of Mars,$m_{m} = 6.4 \times 10^{23} \; kg$.
Orbital radius of Mars,$R = 2.28 \times 10^{8} \; km = 2.28 \times 10^{11} \; m$.
Radius of Mars,$r = 3395 \; km = 3.395 \times 10^{6} \; m$.
Gravitational constant,$G = 6.67 \times 10^{-11} \; N m^{2} kg^{-2}$.
Total potential energy of the spaceship is the sum of potential energy due to the Sun and Mars: $U = -\frac{GMm_{s}}{R} - \frac{Gm_{m}m_{s}}{r}$.
Since the spaceship is stationary on Mars,its kinetic energy is zero.
Total energy $E = -Gm_{s} \left( \frac{M}{R} + \frac{m_{m}}{r} \right)$.
Energy required to escape the solar system is $E_{req} = -E = Gm_{s} \left( \frac{M}{R} + \frac{m_{m}}{r} \right)$.
$E_{req} = 6.67 \times 10^{-11} \times 1000 \times \left( \frac{2 \times 10^{30}}{2.28 \times 10^{11}} + \frac{6.4 \times 10^{23}}{3.395 \times 10^{6}} \right)$.
$E_{req} = 6.67 \times 10^{-8} \times (8.772 \times 10^{18} + 1.885 \times 10^{17}) \approx 6.67 \times 10^{-8} \times 8.96 \times 10^{18} \approx 5.97 \times 10^{11} \; J \approx 6 \times 10^{11} \; J$.

(N/A) If we throw a stone upwards, it reaches a certain height and then falls back towards the Earth due to gravity.
If it is thrown with a higher initial speed, it reaches a greater height.
If a stone is thrown with a specific initial speed such that it reaches an infinite distance from the Earth's gravitational field, it will never return. In this state, there is no gravitational attraction acting on it from the Earth.
Consider a body at a distance $r$ from the center of the Earth. If the body is stationary, its total energy $E_i$ is:
$E_i = \text{Kinetic Energy} + \text{Potential Energy} = 0 + \left(-\frac{GM_E m}{r}\right) = -\frac{GM_E m}{R_E + h}$
At an infinite distance, the total energy of the body is considered to be zero. If energy equal to $+\frac{GM_E m}{R_E + h}$ is supplied to the body, its total energy becomes zero, and it escapes the Earth's gravitational field. This required energy is known as escape energy.
To provide this energy as kinetic energy, we must give the body an initial speed $v_e$, known as the escape speed:
$\frac{1}{2} m v_e^2 = \frac{GM_E m}{R_E + h}$
Thus, the escape speed is:
$v_e = \sqrt{\frac{2GM_E}{R_E + h}}$
If the body is on the surface of the Earth, $h = 0$, so:
$v_e = \sqrt{\frac{2GM_E}{R_E}} = \sqrt{2gR_E}$

(N/A) The escape velocity of a body on the surface of a planet is given by $v_{e} = \sqrt{\frac{2GM}{R}}$.
For the Moon,the escape velocity is $(v_{e})_{\text{moon}} = \sqrt{\frac{2GM_{m}}{R_{m}}}$.
Given the mass of the Moon $M_{m} \approx 7.36 \times 10^{22} \text{ kg}$ and its radius $R_{m} \approx 1.74 \times 10^{6} \text{ m}$,the escape velocity is calculated as:
$(v_{e})_{\text{moon}} \approx 2.38 \text{ km/s}$.
The root mean square velocity of gas molecules at the surface temperature of the Moon is higher than this escape velocity. Because the thermal velocity of the gas molecules exceeds the escape velocity of the Moon,the gas molecules easily escape the Moon's gravitational pull. Consequently,the Moon cannot retain an atmosphere.

(N/A) Escape energy is defined as the minimum amount of energy required to be provided to an object of mass $m$ on the surface of a planet so that it can escape the gravitational field of the planet and reach infinity.
Mathematically,it is given by $E_e = \frac{GMm}{R}$,where $G$ is the universal gravitational constant,$M$ is the mass of the planet,$m$ is the mass of the object,and $R$ is the radius of the planet.
Since energy is a form of work,its $SI$ unit is the Joule $(J)$.
The dimensional formula for energy is $[ML^2T^{-2}]$.

(N/A) The escape energy of a body is defined as the minimum energy required to move the body from the surface of the Earth to infinity,such that its final kinetic energy at infinity is zero.
The gravitational potential energy $U$ of a body of mass $m$ at the surface of the Earth (at distance $R_E$ from the center) is given by:
$U = -\frac{GM_Em}{R_E}$
To move the body to infinity,where the potential energy is zero,we must provide an amount of energy equal to the magnitude of this potential energy.
Therefore,the escape energy $E_e$ is:
$E_e = -U = -\left(-\frac{GM_Em}{R_E}\right) = +\frac{GM_Em}{R_E}$
This energy is typically provided in the form of kinetic energy to the body to allow it to escape the Earth's gravitational pull.

(N/A) Escape speed is defined as the minimum speed required for an object to be projected vertically upwards from the surface of a planet or moon so that it escapes the gravitational influence of that body and never returns to the surface.
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 planet,and $R$ is the radius of the planet.
Escape velocity does not depend on the following factors:
$1$. The mass of the object being projected $(m)$.
$2$. The direction of projection (as long as it is not directed into the planet).
$3$. The location on the planet's surface from where it is projected.

(A) The escape velocity $v_e$ of a body from the surface of a planet is given by the formula $v_e = \sqrt{\frac{2GM}{R}}$,where $G$ is the gravitational constant,$M$ is the mass of the planet,and $R$ is its radius.
For Earth,the mass $M_e \approx 5.97 \times 10^{24} \text{ kg}$ and radius $R_e \approx 6.37 \times 10^6 \text{ m}$. Substituting these values,we get $v_e \approx 11.2 \text{ km/s}$.
For the Moon,the mass $M_m \approx 7.35 \times 10^{22} \text{ kg}$ and radius $R_m \approx 1.74 \times 10^6 \text{ m}$. Substituting these values,we get $v_e \approx 2.38 \text{ km/s}$.

(N/A) This statement is incorrect.
The correct statement is: "The value of the escape velocity $v_e$ for a stationary object on the surface of a planet is directly proportional to the square root of the ratio of the planet's mass to its radius."
Explanation: The formula for escape velocity is $v_e = \sqrt{\frac{2GM}{R}}$,where $G$ is the gravitational constant,$M$ is the mass of the planet,and $R$ is the radius of the planet. Thus,$v_e \propto \sqrt{\frac{M}{R}}$.

(B) The escape velocity $v_e$ of a body 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,and $R$ is the radius of the planet.
From this formula,it is clear that the escape velocity $v_e$ is independent of the mass of the body being projected.
Therefore,the escape velocity remains the same regardless of the mass of the object.
Since the escape velocity for a $1\,kg$ mass is $11.2\,km/s$,the escape velocity for a $10\,kg$ mass will also be $11.2\,km/s$.

(A, 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 Earth,and $R$ is the radius of the Earth.
From this formula,it is clear that the escape velocity depends only on the mass and radius of the planet (or celestial body) from which the object is projected.
It does not depend on the mass of the object being projected $(m)$ or the angle of projection $( \theta)$.

(B) No,it is not necessary to provide an initial speed of $11.2 \, km/s$ to a rocket.
Escape velocity is the speed required for an object to escape Earth's gravitational pull without any further propulsion.
$A$ rocket,however,is a self-propelled vehicle.
It carries fuel that burns to produce thrust,which continuously increases its speed and allows it to overcome gravity gradually,even if it starts with a much lower initial speed.

(N/A) The two factors responsible for the presence of an atmosphere on a planet are:
$(i)$ The acceleration due to gravity on the planet.
$(ii)$ The temperature on the surface of the planet.
This is because the speed of gas molecules depends on the temperature of the planet.

(B) The orbital velocity of a satellite orbiting close to the Earth's surface is given by $v_{0} = \sqrt{\frac{GM_{e}}{R_{e}}}$.
The escape velocity from the Earth's surface is given by $v_{e} = \sqrt{\frac{2GM_{e}}{R_{e}}}$.
Dividing the expression for escape velocity by the expression for orbital velocity:
$\frac{v_{e}}{v_{0}} = \frac{\sqrt{\frac{2GM_{e}}{R_{e}}}}{\sqrt{\frac{GM_{e}}{R_{e}}}} = \sqrt{2}$.
Therefore,the relationship is $v_{e} = \sqrt{2} v_{0}$.

(A) The kinetic energy required to project an object to infinity is equal to the magnitude of its gravitational potential energy at the surface of the Earth.
The gravitational potential energy at the surface of the Earth is $U = -\frac{GMm}{R}$.
Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$.
Substituting this into the potential energy expression: $U = -\frac{(gR^2)m}{R} = -mgR$.
Therefore,the required kinetic energy $K = |U| = mgR$.

(C) The escape velocity $(v_{e})$ from the surface of a celestial body is given by the formula $v_{e} = \sqrt{\frac{2GM}{R}}$.
For a black hole,the gravitational pull is so intense that not even light can escape its surface.
By definition,the escape velocity at the event horizon of a black hole must be equal to or greater than the speed of light $(c)$.
Therefore,for a black hole,$v_{e} \geq c$.

(A) The potential energy $(U)$ of a satellite in orbit is given as $-8 \times 10^9 \ J$.
Binding energy $(BE)$ is defined as the minimum energy required to remove the satellite from its orbit to infinity,where the total energy becomes zero.
$BE = -U$
$BE = -(-8 \times 10^9 \ J)$
$BE = 8 \times 10^9 \ J$
Therefore,the binding energy is $8 \times 10^9 \ J$.

(A) The formula for escape velocity is $v_e = \sqrt{\frac{2GM}{R}}$.
Here,$G$ is the gravitational constant,$M$ is the mass of the Earth,and $R$ is the radius of the Earth.
From the formula,it is clear that the escape velocity depends only on the mass and radius of the planet (Earth).
It does not depend on the mass of the object being projected $(m)$ or the angle of projection $(\theta)$.

(D) Escape energy is defined as the minimum energy required by a passive particle (like a projectile) to escape the gravitational field of a planet.
$A$ rocket is not a passive particle; it is a self-propelled system.
As the rocket burns fuel,it continuously gains momentum and kinetic energy through the ejection of gases.
Therefore,it does not rely on an initial 'escape energy' to overcome gravity; it can escape the gravitational field even with a lower initial velocity by continuously burning fuel.

(D) The escape velocity $v_e$ is given by the formula $v_e = \sqrt{\frac{2GM}{R}}$.
For the Earth,the escape velocity is approximately $11.2 \, km \, s^{-1}$. Thus,$(1)$ matches with $(c)$.
For the Moon,the escape velocity is approximately $2.38 \, km \, s^{-1}$. Thus,$(2)$ matches with $(a)$.
Therefore,the correct matching is $(1-c), (2-a)$.

(A) The formula for escape velocity is $V_e = \sqrt{\frac{2GM}{R}}$.
Let the new radius be $R'$ such that the new escape velocity $V_e' = 10V_e$.
Thus,$10V_e = \sqrt{\frac{2GM}{R'}}$.
Dividing the two equations,we get $10 = \sqrt{\frac{R}{R'}}$.
Squaring both sides,$100 = \frac{R}{R'}$.
Therefore,$R' = \frac{R}{100}$.
Given $R = 6400 \ km$,we have $R' = \frac{6400}{100} = 64 \ km$.

(B) The escape velocity $V_{e}$ of a planet is given by the formula $V_{e} = \sqrt{\frac{2GM}{R}}$,where $G$ is the gravitational constant,$M$ is the mass of the planet,and $R$ is its radius.
If the escape velocities of two planets $A$ and $B$ are equal,then $\sqrt{\frac{2GM_{1}}{R_{1}}} = \sqrt{\frac{2GM_{2}}{R_{2}}}$.
Squaring both sides,we get $\frac{M_{1}}{R_{1}} = \frac{M_{2}}{R_{2}}$,which implies $\frac{M_{1}}{M_{2}} = \frac{R_{1}}{R_{2}}$.
This condition allows for different masses $(M_{1} \neq M_{2})$ as long as their radii are also different in the same proportion. Thus,Assertion $A$ is correct.
Reason $R$ states that the product $M_{1}R_{1} = M_{2}R_{2}$ must be the same,which is mathematically incorrect for the given condition. Therefore,$R$ is incorrect.

(C) Using the law of conservation of energy between the surface of the earth and the maximum height $h = 10 R$:
Total energy at surface = Total energy at height $h$.
$-\frac{GMm}{R} + \frac{1}{2}mv_{i}^{2} = -\frac{GMm}{R+h} + 0$.
Since $h = 10R$,the total distance from the center is $R + 10R = 11R$.
$-\frac{GMm}{R} + \frac{1}{2}mv_{i}^{2} = -\frac{GMm}{11R}$.
$\frac{1}{2}mv_{i}^{2} = GMm \left( \frac{1}{R} - \frac{1}{11R} \right) = GMm \left( \frac{10}{11R} \right)$.
$v_{i}^{2} = \frac{20GM}{11R}$.
We know the escape velocity $v_{e} = \sqrt{\frac{2GM}{R}}$,so $v_{e}^{2} = \frac{2GM}{R}$.
Substituting this into the equation for $v_{i}^{2}$:
$v_{i}^{2} = \frac{10}{11} \times \left( \frac{2GM}{R} \right) = \frac{10}{11} v_{e}^{2}$.
Comparing this with $v_{i} = \sqrt{\frac{x}{y}} v_{e}$,we get $\frac{x}{y} = \frac{10}{11}$.
Thus,$x = 10$.

(B) To escape the gravitational influence of both masses,the particle must reach a point where the total gravitational potential energy is zero,which is at infinity. By the law of conservation of energy,the total energy at the midpoint must be equal to the total energy at infinity (which is $0$).
At the midpoint,the distance from each mass is $r/2$.
The total energy at the midpoint is the sum of kinetic energy and gravitational potential energy:
$E_i = \frac{1}{2}mV^2 - \frac{GM_1m}{r/2} - \frac{GM_2m}{r/2}$
Setting the total energy to zero:
$\frac{1}{2}mV^2 - \frac{2GM_1m}{r} - \frac{2GM_2m}{r} = 0$
$\frac{1}{2}mV^2 = \frac{2Gm}{r}(M_1 + M_2)$
$V^2 = \frac{4G(M_1 + M_2)}{r}$
$V = \sqrt{\frac{4G(M_1 + M_2)}{r}}$

(D) The formula for escape velocity is $v_{e} = \sqrt{\frac{2GM}{R}}$.
Since mass $M = \text{Volume} \times \text{Density} = \frac{4}{3} \pi R^{3} \rho$,we substitute this into the formula:
$v_{e} = \sqrt{\frac{2G}{R} \times \frac{4}{3} \pi R^{3} \rho} = \sqrt{\frac{8 \pi G \rho}{3} R^{2}} = R \sqrt{\frac{8 \pi G \rho}{3}}$.
This shows that $v_{e} \propto R$ when the density $\rho$ is constant.
Given the radius of the new planet is $R' = 4R$,the new escape velocity $v'$ will be:
$v' = 4 \times v_{e} = 4v$.

(D) By the law of conservation of energy, the total energy at the surface equals the total energy at the maximum height $r$ where velocity is zero.
$-\frac{GMm}{R} + \frac{1}{2} m v^2 = -\frac{GMm}{r}$
Given $v = k V_{e}$ and $V_{e} = \sqrt{\frac{2GM}{R}}$, we have $v^2 = k^2 \frac{2GM}{R}$.
Substituting this into the energy equation:
$-\frac{GMm}{R} + \frac{1}{2} m \left( k^2 \frac{2GM}{R} \right) = -\frac{GMm}{r}$
Dividing by $GMm$:
$-\frac{1}{R} + \frac{k^2}{R} = -\frac{1}{r}$
$\frac{1}{r} = \frac{1}{R} - \frac{k^2}{R} = \frac{1-k^2}{R}$
$r = \frac{R}{1-k^2}$
Since $r = R + h$, the maximum height $h$ is:
$h = r - R = \frac{R}{1-k^2} - R = R \left( \frac{1 - (1-k^2)}{1-k^2} \right) = \frac{R k^2}{1-k^2}$.

(A) The body is projected with escape velocity $v_e = \sqrt{\frac{2GM}{R_e}}$.
By conservation of energy at the surface and at a distance $r$ from the center of the Earth:
$\frac{1}{2}mv^2 - \frac{GMm}{r} = 0$ (since total energy is zero for escape velocity).
$\frac{1}{2}mv^2 = \frac{GMm}{r} \Rightarrow v = \sqrt{\frac{2GM}{r}} = \frac{dr}{dt}$.
Separating variables: $dt = \frac{dr}{\sqrt{2GM}} \cdot \sqrt{r}$.
Integrating from $t=0$ at $r=R_e$ to time $t$ at $r=R_e+h$:
$t = \frac{1}{\sqrt{2GM}} \int_{R_e}^{R_e+h} r^{1/2} dr = \frac{1}{\sqrt{2GM}} \cdot \frac{2}{3} [r^{3/2}]_{R_e}^{R_e+h}$.
$t = \frac{2}{3\sqrt{2GM}} [ (R_e+h)^{3/2} - R_e^{3/2} ] = \frac{2}{3\sqrt{2GM}} R_e^{3/2} [ (1 + \frac{h}{R_e})^{3/2} - 1 ]$.
Since $GM = gR_e^2$,we have $\sqrt{GM} = \sqrt{g}R_e$.
Substituting this: $t = \frac{2 R_e^{3/2}}{3 \sqrt{2} \sqrt{g} R_e} [ (1 + \frac{h}{R_e})^{3/2} - 1 ] = \frac{1}{3} \sqrt{\frac{2R_e}{g}} [ (1 + \frac{h}{R_e})^{3/2} - 1 ]$.

(A) The formula for escape velocity is $V_e = \sqrt{\frac{2GM}{R}}$.
Since mass $M = \rho \times \text{Volume} = \rho \times \frac{4}{3} \pi R^3$,we can write $V_e = \sqrt{\frac{2G \rho \frac{4}{3} \pi R^3}{R}} = \sqrt{\frac{8}{3} G \pi \rho R^2}$.
This implies $V_e \propto R \sqrt{\rho}$.
Given for planet $B$: $\rho_B = 4\rho_A$ and $R_B = \frac{1}{2}R_A$.
Therefore,$\frac{V_{eB}}{V_{eA}} = \frac{R_B}{R_A} \sqrt{\frac{\rho_B}{\rho_A}} = \left(\frac{1}{2}\right) \sqrt{4} = \frac{1}{2} \times 2 = 1$.
Thus,$V_{eB} = V_{eA} = 12 \, km/s$.

(A) Let $v_e$ be the escape velocity,given by $v_e = \sqrt{\frac{2GM}{R}}$.
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: $E_i = -\frac{GMm}{R} + \frac{1}{2}m v^2$,where $v = \frac{v_e}{3}$.
At maximum height $h$: $E_f = -\frac{GMm}{R+h}$.
Equating $E_i = E_f$:
$-\frac{GMm}{R} + \frac{1}{2}m \left(\frac{v_e}{3}\right)^2 = -\frac{GMm}{R+h}$
$-\frac{GMm}{R} + \frac{1}{2}m \left(\frac{2GM}{9R}\right) = -\frac{GMm}{R+h}$
$-\frac{GMm}{R} + \frac{GMm}{9R} = -\frac{GMm}{R+h}$
$-\frac{8GMm}{9R} = -\frac{GMm}{R+h}$
$\frac{8}{9R} = \frac{1}{R+h}$
$8(R+h) = 9R$
$8R + 8h = 9R$
$8h = R$
$h = \frac{R}{8} = \frac{6400 \ km}{8} = 800 \ km$.

(B) 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 center of the earth.
Total energy at surface = Total energy at height $h$
$-\frac{GMm}{R} + \frac{1}{2}m(\lambda v_{e})^{2} = -\frac{GMm}{h} + 0$
Since the escape velocity $v_{e} = \sqrt{\frac{2GM}{R}}$,we have $v_{e}^{2} = \frac{2GM}{R}$.
Substituting this into the equation:
$-\frac{GMm}{R} + \frac{1}{2}m\lambda^{2}(\frac{2GM}{R}) = -\frac{GMm}{h}$
$-\frac{GMm}{R} + \frac{\lambda^{2}GMm}{R} = -\frac{GMm}{h}$
Dividing both sides by $-GMm$:
$\frac{1}{R} - \frac{\lambda^{2}}{R} = \frac{1}{h}$
$\frac{1-\lambda^{2}}{R} = \frac{1}{h}$
$h = \frac{R}{1-\lambda^{2}}$

(A) The total energy of the rocket at the instant of launch must be less than or equal to zero for it to remain bound to the solar system.
The orbital speed of the asteroid is given by $v_0 = \sqrt{\frac{GM}{r_0}}$,which implies $v_0^2 = \frac{GM}{r_0}$.
The total energy $E$ of the rocket relative to the sun is the sum of its gravitational potential energy and its kinetic energy:
$E = -\frac{GMm}{r_0} + \frac{1}{2}mv^2$
Given that the speed of the rocket relative to the sun is $v = \alpha v_0$,we substitute this into the energy equation:
$E = -\frac{GMm}{r_0} + \frac{1}{2}m(\alpha v_0)^2$
For the rocket to remain bound,$E \leq 0$:
$-\frac{GMm}{r_0} + \frac{1}{2}m \alpha^2 v_0^2 \leq 0$
Substituting $v_0^2 = \frac{GM}{r_0}$:
$-\frac{GMm}{r_0} + \frac{1}{2}m \alpha^2 \left(\frac{GM}{r_0}\right) \leq 0$
Dividing by $\frac{GMm}{r_0}$:
$-1 + \frac{1}{2} \alpha^2 \leq 0$
$\frac{1}{2} \alpha^2 \leq 1$
$\alpha^2 \leq 2$
$\alpha \leq \sqrt{2}$
Wait,re-evaluating the launch condition: If the rocket is launched *from* the asteroid,its initial velocity relative to the sun is the vector sum of the asteroid's orbital velocity and the launch velocity. However,the problem states $v = \alpha v_0$ is the speed relative to the sun. Thus,the condition for being bound is $E \leq 0$,leading to $\alpha \leq \sqrt{2}$. The highest value is $\sqrt{2}$.

(D) Given: Escape velocity $= u$.
The initial velocity of the projectile is $200 \%$ more than the escape velocity.
$V_{\text{initial}} = u + (200/100)u = u + 2u = 3u$.
According to the Law of Conservation of Energy:
$\frac{1}{2} m V_{\text{initial}}^2 + (P.E.)_{\text{initial}} = \frac{1}{2} m V_{\text{final}}^2 + (P.E.)_{\text{final}}$.
At the surface of the planet,the potential energy is $P.E. = -\frac{GMm}{R} = -m u^2$ (since $u = \sqrt{\frac{2GM}{R}}$,so $u^2 = \frac{2GM}{R}$,which implies $\frac{GM}{R} = \frac{u^2}{2}$,thus $P.E. = -\frac{m u^2}{2}$).
In interstellar space,the potential energy is $0$.
$\frac{1}{2} m (3u)^2 - \frac{1}{2} m u^2 = \frac{1}{2} m V_{\text{final}}^2$.
$\frac{9}{2} u^2 - \frac{1}{2} u^2 = \frac{1}{2} V_{\text{final}}^2$.
$8 u^2 = V_{\text{final}}^2$.
$V_{\text{final}} = \sqrt{8} u = 2\sqrt{2} u$.

(C) The escape velocity $V_e$ at a height $h$ from the Earth's surface is given by the formula: $V_e = \sqrt{\frac{2GM}{R+h}}$.
We know that the escape velocity at the Earth's surface $(h=0)$ is $V_0 = \sqrt{\frac{2GM}{R}} \approx 11.2 \, km/s$.
Therefore,the escape velocity at height $h$ is $V_h = V_0 \sqrt{\frac{R}{R+h}}$.
Given $R = 6400 \, km$ and $h = 1000 \, km$,we have $R+h = 7400 \, km$.
Substituting the values: $V_h = 11.2 \times \sqrt{\frac{6400}{7400}}$.
$V_h = 11.2 \times \sqrt{\frac{64}{74}} = 11.2 \times \sqrt{0.8648} \approx 11.2 \times 0.93 = 10.416 \, km/s$.
Rounding to the nearest option,we get $10.4 \, km/s$.

(B) Let $m$ be the mass of the satellite and $M$ be the mass of the Earth. The radius of the Earth is $R$.
$1$. Kinetic energy required for escape $(E_1)$: The escape velocity is $v_e = \sqrt{\frac{2GM}{R}}$. Thus,$E_1 = \frac{1}{2} m v_e^2 = \frac{1}{2} m \left(\frac{2GM}{R}\right) = \frac{GMm}{R}$.
$2$. Kinetic energy required for orbital motion just above the surface $(E_2)$: The orbital velocity is $v_0 = \sqrt{\frac{GM}{R}}$. Thus,$E_2 = \frac{1}{2} m v_0^2 = \frac{1}{2} m \left(\frac{GM}{R}\right) = \frac{GMm}{2R}$.
$3$. The ratio is $\frac{E_1}{E_2} = \frac{GMm/R}{GMm/2R} = 2:1$.

(A) The gravitational potential energy $(U)$ of an object of mass $m$ at the surface of the Earth (mass $M$,radius $R$) is given by $U = -\frac{G M m}{R}$.
The potential energy per unit mass is defined as $\frac{U}{m} = -\frac{G M}{R}$.
The magnitude of this value is given as $E = \frac{G M}{R}$.
The formula for escape velocity $(v_e)$ is $v_e = \sqrt{\frac{2 G M}{R}}$.
Substituting $E = \frac{G M}{R}$ into the escape velocity formula,we get $v_e = \sqrt{2 E}$.

(B) Using the conservation of mechanical energy between the Earth's surface $(r = R_e)$ and the maximum height $(r = r_{max})$:
$U_i + K_i = U_f + K_f$
$-\frac{G M m}{R_e} + \frac{1}{2} m v^2 = -\frac{G M m}{r_{max}} + 0$
Given $v^2 = \frac{4 g R_e}{3}$ and $g = \frac{G M}{R_e^2}$,so $v^2 = \frac{4 G M}{3 R_e}$.
$-\frac{G M m}{R_e} + \frac{1}{2} m (\frac{4 G M}{3 R_e}) = -\frac{G M m}{r_{max}}$
$-\frac{G M m}{R_e} + \frac{2 G M m}{3 R_e} = -\frac{G M m}{r_{max}}$
$-\frac{1}{3} \frac{G M m}{R_e} = -\frac{G M m}{r_{max}} \Rightarrow r_{max} = 3 R_e$.
The maximum height $h_{max} = r_{max} - R_e = 2 R_e$.
We need the velocity $v'$ at height $h = \frac{h_{max}}{2} = R_e$,which corresponds to $r = 2 R_e$.
Applying conservation of energy again:
$-\frac{G M m}{R_e} + \frac{1}{2} m v^2 = -\frac{G M m}{2 R_e} + \frac{1}{2} m (v')^2$
$-\frac{G M m}{R_e} + \frac{2 G M m}{3 R_e} = -\frac{G M m}{2 R_e} + \frac{1}{2} m (v')^2$
$-\frac{1}{3} \frac{G M m}{R_e} + \frac{1}{2} \frac{G M m}{R_e} = \frac{1}{2} m (v')^2$
$\frac{1}{6} \frac{G M m}{R_e} = \frac{1}{2} m (v')^2$
$(v')^2 = \frac{G M}{3 R_e} = \frac{g R_e}{3} \Rightarrow v' = \sqrt{\frac{g R_e}{3}}$.

(D) The total energy $E$ of a satellite-planet system is given by $E = K + U$,where $K$ is the kinetic energy and $U$ is the potential energy.
If $E < 0$,the satellite is in a bound state (circular or elliptical orbit).
If $E = 0$,the satellite is at the threshold of escaping the gravitational field (it moves with escape velocity $v_e$).
If $E > 0$,the kinetic energy is greater than the magnitude of the potential energy,meaning the satellite has excess energy.
Therefore,the satellite will escape the gravitational field of the planet with a speed greater than the escape velocity.

(D) The escape speed of Earth is given by $v_e = \sqrt{\frac{2GM}{R}}$.
Given that the projection speed $v = \frac{1}{2} v_e = \frac{1}{2} \sqrt{\frac{2GM}{R}}$.
Using the law of conservation of mechanical energy between the surface of the Earth and the maximum height $H$ (where the final velocity is zero):
$E_i = E_f$
$-\frac{GMm}{R} + \frac{1}{2}mv^2 = -\frac{GMm}{R+H} + 0$
Substituting $v^2 = \frac{1}{4} \left(\frac{2GM}{R}\right) = \frac{GM}{2R}$:
$-\frac{GMm}{R} + \frac{1}{2}m \left(\frac{GM}{2R}\right) = -\frac{GMm}{R+H}$
$-\frac{GMm}{R} + \frac{GMm}{4R} = -\frac{GMm}{R+H}$
$-\frac{3GMm}{4R} = -\frac{GMm}{R+H}$
$\frac{3}{4R} = \frac{1}{R+H}$
$3(R+H) = 4R$
$3R + 3H = 4R$
$3H = R$
$H = \frac{R}{3}$

(A) According to the principle of conservation of mechanical energy,the total energy at the surface of the Earth must equal the total energy at a large distance (infinity) from the Earth.
Let $m$ be the mass of the body,$M$ be the mass of the Earth,and $R$ be the radius of the Earth.
At the surface of the Earth,the velocity is $v = n v_e$,where $v_e = \sqrt{\frac{2GM}{R}}$.
Total energy at the surface: $E_i = \frac{1}{2} m (n v_e)^2 - \frac{GMm}{R}$.
At a large distance,the potential energy is $0$. Let the final velocity be $v_f$.
Total energy at a large distance: $E_f = \frac{1}{2} m v_f^2$.
Equating $E_i = E_f$:
$\frac{1}{2} m n^2 v_e^2 - \frac{GMm}{R} = \frac{1}{2} m v_f^2$.
Since $v_e^2 = \frac{2GM}{R}$,we have $\frac{GM}{R} = \frac{v_e^2}{2}$.
Substituting this into the equation:
$\frac{1}{2} m n^2 v_e^2 - m \left(\frac{v_e^2}{2}\right) = \frac{1}{2} m v_f^2$.
Dividing by $\frac{1}{2} m$:
$n^2 v_e^2 - v_e^2 = v_f^2$.
$v_f^2 = v_e^2 (n^2 - 1)$.
$v_f = v_e \sqrt{n^2 - 1}$.

(B) The formula for escape velocity is $V_e = \sqrt{\frac{2GM}{R}}$.
Given that the mass of the Earth $(M_e)$ is $81$ times the mass of the Moon $(M_m)$,so $M_e = 81 M_m$.
The radius of the Earth $(R_e)$ is $4$ times the radius of the Moon $(R_m)$,so $R_e = 4 R_m$.
The ratio of escape velocities is given by:
$\frac{V_{e,e}}{V_{e,m}} = \sqrt{\frac{M_e}{M_m} \times \frac{R_m}{R_e}} = \sqrt{81 \times \frac{1}{4}} = \sqrt{\frac{81}{4}} = \frac{9}{2} = 4.5$.
Therefore,the escape velocity from the Moon is $V_{e,m} = \frac{V_{e,e}}{4.5} = \frac{11.2}{4.5} \approx 2.48 \, km/s$.

(C) The escape velocity $v_e$ of an object from the surface of a planet is given by the formula $v_e = \sqrt{\frac{2 G M}{R}}$.
$A$ black hole is defined as a celestial object from which even light cannot escape due to its intense gravitational field.
For light to be unable to escape,the escape velocity of the planet must be greater than or equal to the speed of light $c$.
Therefore,the condition is $v_e \geq c$.
Substituting the expression for $v_e$,we get $\sqrt{\frac{2 G M}{R}} \geq c$.

(B) For a satellite in a circular orbit near the Earth's surface,the orbital speed is given by $v_0 = \sqrt{\frac{GM_e}{R_e}}$.
The escape velocity required to leave the Earth's gravitational field is $v_e = \sqrt{\frac{2GM_e}{R_e}}$.
Comparing these two expressions,we find that $v_e = \sqrt{2} v_0$.
Since $\sqrt{2} \approx 1.414$,the required escape velocity is $1.414$ times the orbital velocity.
The percentage increase $x$ required is given by $x = \left( \frac{v_e - v_0}{v_0} \right) \times 100$.
Substituting the values: $x = (\sqrt{2} - 1) \times 100 = (1.414 - 1) \times 100 = 41.4 \%$.

(A) Let the mass of the particle be $m$,the mass of the planet be $M$,and its radius be $R$.
Taking the gravitational potential at a large distance (infinity) as zero,the total energy of the particle at infinity is $E_i = 0$.
The gravitational potential at the centre of the planet is $V_c = -\frac{3GM}{2R}$.
By the law of conservation of energy,the total energy at infinity equals the total energy at the centre of the planet:
$K_i + U_i = K_c + U_c$
$0 + 0 = \frac{1}{2}mv^2 + m(-\frac{3GM}{2R})$
$\frac{1}{2}mv^2 = \frac{3GMm}{2R}$
$v^2 = \frac{3GM}{R}$
$v = \sqrt{\frac{3GM}{R}}$.
We know that the escape velocity at the surface of the planet is $v_e = \sqrt{\frac{2GM}{R}}$,which implies $\frac{GM}{R} = \frac{v_e^2}{2}$.
Substituting this into the expression for $v$:
$v = \sqrt{3 \cdot \frac{v_e^2}{2}} = \sqrt{1.5} v_e$.

(A) The escape velocity of a planet is given by $v = \sqrt{\frac{2GM}{R}}$.
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 planet $P$.
Given: $M_e = 9M_p$ (so $M_p = \frac{M_e}{9}$) and $R_e = 2R_p$ (so $R_p = \frac{R_e}{2}$).
The escape velocity on Earth is $v_e = \sqrt{\frac{2GM_e}{R_e}}$.
The escape velocity on planet $P$ is $v_p = \sqrt{\frac{2GM_p}{R_p}} = \sqrt{\frac{2G(M_e/9)}{(R_e/2)}} = \sqrt{\frac{4GM_e}{9R_e}} = \frac{2}{3} \sqrt{\frac{GM_e}{R_e}}$.
Since $v_e = \sqrt{\frac{2GM_e}{R_e}}$,we have $\sqrt{\frac{GM_e}{R_e}} = \frac{v_e}{\sqrt{2}}$.
Substituting this into the expression for $v_p$: $v_p = \frac{2}{3} \left( \frac{v_e}{\sqrt{2}} \right) = \frac{\sqrt{2} v_e}{3} = \frac{v_e}{3} \sqrt{2}$.
Comparing this with $\frac{v_e}{3} \sqrt{x}$,we get $x = 2$.

(D) The escape velocity of a planet is given by $v_e = \sqrt{2gR}$.
Since the Moon has a much smaller mass and radius compared to the Earth,its escape velocity is significantly lower (approx $2.38 \ km/s$) than that of the Earth (approx $11.2 \ km/s$).
Because the escape velocity on the Moon is low,the thermal velocity (rms velocity) of gas molecules at the Moon's surface temperature exceeds the escape velocity.
Consequently,gas molecules escape from the Moon's gravitational pull,preventing the formation of an atmosphere.
Therefore,both Assertion $A$ and Reason $R$ are correct,and $R$ is the correct explanation of $A$.