Change in Gravitational Potential Energy, Energy Conservation Questions in English

(D) The gravitational potential energy at the surface of the earth is $U_i = -\frac{GMm}{R}$.
At a height $h = nR$ from the surface,the distance from the center of the earth is $r = R + h = R + nR = R(1+n)$.
The potential energy at this height is $U_f = -\frac{GMm}{R(1+n)}$.
The change in potential energy is $\Delta U = U_f - U_i = -\frac{GMm}{R(1+n)} - (-\frac{GMm}{R})$.
$\Delta U = \frac{GMm}{R} (1 - \frac{1}{1+n}) = \frac{GMm}{R} (\frac{1+n-1}{1+n}) = \frac{GMm}{R} (\frac{n}{n+1})$.
Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$.
Substituting this into the expression: $\Delta U = \frac{gR^2 m}{R} (\frac{n}{n+1}) = mgR \frac{n}{n+1}$.

(B) The gravitational potential energy $U$ at a distance $r$ from the center of the earth is given by $U = -\frac{GMm}{r}$.
At the earth's surface,$r_1 = R$,so $U_1 = -\frac{GMm}{R}$.
At a height $h = 3R$,the distance from the center is $r_2 = R + h = R + 3R = 4R$.
So,$U_2 = -\frac{GMm}{4R}$.
The change in gravitational potential energy is $\Delta U = U_2 - U_1 = -\frac{GMm}{4R} - (-\frac{GMm}{R}) = \frac{GMm}{R} - \frac{GMm}{4R} = \frac{3GMm}{4R}$.
Since the acceleration due to gravity at the surface is $g = \frac{GM}{R^2}$,we have $GM = gR^2$.
Substituting this into the expression,$\Delta U = \frac{3(gR^2)m}{4R} = \frac{3}{4}mgR$.

(B) The gravitational potential energy of a body of mass $m$ at a distance $r$ from the center of the Earth is given by $U = -\frac{GMm}{r}$.
The initial position is $2R$ above the Earth's surface,so the distance from the center is $r_i = R + 2R = 3R$.
Initial potential energy: $U_i = -\frac{GMm}{3R}$.
The final position is $R$ above the Earth's surface,so the distance from the center is $r_f = R + R = 2R$.
Final potential energy: $U_f = -\frac{GMm}{2R}$.
According to the law of conservation of energy,the loss in potential energy equals the gain in kinetic energy $(KE)$:
$KE = U_i - U_f = -\frac{GMm}{3R} - (-\frac{GMm}{2R})$
$KE = \frac{GMm}{2R} - \frac{GMm}{3R} = \frac{3GMm - 2GMm}{6R} = \frac{GMm}{6R}$.

(D) The gravitational potential energy of a body of mass $m$ at a distance $r$ from the center of a planet of mass $M$ is given by $U = -\frac{GMm}{r}$.
To move the body from an initial radius $r_1 = 2R$ to a final radius $r_2 = 3R$,the energy required is equal to the change in potential energy,$\Delta U = U_f - U_i$.
$\Delta U = \left( -\frac{GMm}{3R} \right) - \left( -\frac{GMm}{2R} \right)$.
$\Delta U = GMm \left( \frac{1}{2R} - \frac{1}{3R} \right)$.
$\Delta U = GMm \left( \frac{3 - 2}{6R} \right) = \frac{GMm}{6R}$.

(B) The total energy $E$ of a body in a gravitational field is the sum of its kinetic energy $K$ and potential energy $U$.
If a missile is launched with the escape velocity $v_e = \sqrt{2GM/R}$,its total energy is zero,allowing it to escape to infinity.
If the launch velocity $v$ is less than the escape velocity $(v < v_e)$,the body is gravitationally bound to the Earth.
For a bound system,the total energy $E = K + U$ is always negative.
This negative value indicates that the gravitational force is attractive and the object cannot escape the gravitational influence of the Earth.

(A) In the solar system,the orbit of a planet is an ellipse with the Sun at one of the two foci.
Since the distance between the Sun and the planet changes as it moves along the elliptical orbit,the velocity of the planet at every point is not the same.
Because the velocity changes,both the linear momentum $(p = mv)$ and the kinetic energy $(K.E. = \frac{1}{2}mv^2)$ are not conserved.
The angular velocity also varies according to Kepler's second law.
However,since the gravitational force is a conservative force,the total mechanical energy $(K.E. + P.E.)$ of the planet-Sun system remains conserved.

(C) By 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: $E_i = K_i + U_i = \frac{1}{2}MV^2 - \frac{GM_eM}{R}$
At maximum height $h$: $E_f = K_f + U_f = 0 - \frac{GM_eM}{R+h}$
Equating $E_i = E_f$: $\frac{1}{2}MV^2 - \frac{GM_eM}{R} = - \frac{GM_eM}{R+h}$
$\frac{1}{2}V^2 = GM_e \left( \frac{1}{R} - \frac{1}{R+h} \right)$
Using $g = \frac{GM_e}{R^2}$,we have $GM_e = gR^2$.
$\frac{V^2}{2} = gR^2 \left( \frac{R+h-R}{R(R+h)} \right) = \frac{gR^2h}{R(R+h)} = \frac{gRh}{R+h}$
$\frac{V^2}{2gR} = \frac{h}{R+h}$
$V^2(R+h) = 2gRh$
$V^2R + V^2h = 2gRh$
$V^2R = h(2gR - V^2)$
$h = \frac{V^2R}{2gR - V^2}$
Dividing numerator and denominator by $V^2$: $h = \frac{R}{\frac{2gR}{V^2} - 1}$

(C) central force is a force that is directed along the line joining the particle to the origin (center of force).
Since the torque $\vec{\tau} = \vec{r} \times \vec{F} = 0$ for a central force (because $\vec{r}$ and $\vec{F}$ are collinear),the angular momentum $\vec{L}$ remains constant.
Additionally,central forces are conservative forces,which means the work done by them is path-independent and the total mechanical energy $E = K + U$ of the system is conserved.
Therefore,both angular momentum and mechanical energy are conserved under the influence of a central force.

(B) According to the law of conservation of energy,the loss in gravitational potential energy is equal to the gain in kinetic energy.
Initial height from the center of the Earth,$r_1 = R + 2R = 3R$.
Final height from the center of the Earth,$r_2 = R + R = 2R$.
Initial potential energy,$U_1 = -\frac{GMm}{r_1} = -\frac{GMm}{3R}$.
Final potential energy,$U_2 = -\frac{GMm}{r_2} = -\frac{GMm}{2R}$.
Change in potential energy,$\Delta U = U_1 - U_2 = -\frac{GMm}{3R} - (-\frac{GMm}{2R})$.
$\Delta U = \frac{GMm}{2R} - \frac{GMm}{3R} = \frac{3GMm - 2GMm}{6R} = \frac{GMm}{6R}$.
Since the body is released from rest,the gain in kinetic energy $K = \Delta U = \frac{1}{6} \frac{GMm}{R}$.

(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: $E_i = \frac{1}{2}mV^2 - \frac{GMm}{R}$
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 $g = \frac{GM}{R^2}$,we have $GM = gR^2$. Substituting this:
$\frac{V^2}{2} = gR^2 (\frac{1}{R} - \frac{1}{R+h})$
$\frac{V^2}{2} = gR^2 (\frac{R+h-R}{R(R+h)}) = \frac{gRh}{R+h}$
$\frac{V^2}{2gR} = \frac{h}{R+h}$
Taking the reciprocal: $\frac{2gR}{V^2} = \frac{R+h}{h} = \frac{R}{h} + 1$
$\frac{R}{h} = \frac{2gR}{V^2} - 1$
$h = \frac{R}{(\frac{2gR}{V^2} - 1)}$

(A) The gravitational potential energy at a distance $r$ from the center of the earth is given by $U = -\frac{GMm}{r}$.
At the surface of the earth,the distance from the center is $r_i = R$.
Therefore,the initial potential energy is $U_i = -\frac{GMm}{R}$.
At a height $h = 2R$ from the surface,the distance from the center is $r_f = R + h = R + 2R = 3R$.
Therefore,the final potential energy is $U_f = -\frac{GMm}{3R}$.
The change in potential energy is $\Delta U = U_f - U_i$.
$\Delta U = -\frac{GMm}{3R} - (-\frac{GMm}{R}) = \frac{GMm}{R}(1 - \frac{1}{3}) = \frac{2}{3}\frac{GMm}{R}$.
Since the acceleration due to gravity at the surface is $g = \frac{GM}{R^2}$,we have $GM = gR^2$.
Substituting this,$\Delta U = \frac{2}{3} \frac{(gR^2)m}{R} = \frac{2}{3}mgR$.

(B) The change in gravitational potential energy $\Delta U$ when a body of mass $m$ is lifted to a height $h$ from the surface of the earth is given by the formula: $\Delta U = \frac{mgh}{1 + \frac{h}{R}}$.
Given that the height $h = 3R$,where $R$ is the radius of the earth.
Substituting the value of $h$ into the formula:
$\Delta U = \frac{mg(3R)}{1 + \frac{3R}{R}}$
$\Delta U = \frac{3mgR}{1 + 3}$
$\Delta U = \frac{3mgR}{4}$
Therefore,the change in potential energy is $\frac{3}{4} mgR$.

(B) According to the law of conservation of energy,the initial total energy at infinity is $0$.
At a separation $d$,the total energy is the sum of kinetic energy and potential energy: $K.E. + P.E. = 0$.
Using the reduced mass $\mu = \frac{m_1 m_2}{m_1 + m_2}$,the kinetic energy is $\frac{1}{2} \mu v_{rel}^2$.
The potential energy is $-\frac{G m_1 m_2}{d}$.
Thus,$\frac{1}{2} \left( \frac{m_1 m_2}{m_1 + m_2} \right) v_{rel}^2 - \frac{G m_1 m_2}{d} = 0$.
Solving for $v_{rel}$:
$\frac{1}{2} \left( \frac{m_1 m_2}{m_1 + m_2} \right) v_{rel}^2 = \frac{G m_1 m_2}{d}$.
$v_{rel}^2 = \frac{2G(m_1 + m_2)}{d}$.
$v_{rel} = \sqrt{\frac{2G(m_1 + m_2)}{d}}$.

(B) Let an object of mass $m$ be projected with an initial velocity $u = \sqrt{Rg}$ from the surface of the earth. Let it reach a maximum height $h$ above the surface of the earth.
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 $h$:
$K_i + U_i = K_f + U_f$
$\frac{1}{2} m u^2 - \frac{GMm}{R} = 0 - \frac{GMm}{R+h}$
Given that $u^2 = Rg$ and $g = \frac{GM}{R^2}$,we have $GM = gR^2$. Substituting $u^2 = Rg$ and $GM = gR^2$ into the equation:
$\frac{1}{2} m (Rg) - \frac{(gR^2) m}{R} = - \frac{(gR^2) m}{R+h}$
$\frac{1}{2} mgR - mgR = - \frac{mgR^2}{R+h}$
$-\frac{1}{2} mgR = - \frac{mgR^2}{R+h}$
$\frac{1}{2} = \frac{R}{R+h}$
$R + h = 2R$
$h = R$

(D) The gravitational potential energy at the surface of the Earth is $U_i = -\frac{GMm}{R}$.
At a height $h = \frac{R}{2}$,the distance from the center of the Earth is $r = R + \frac{R}{2} = \frac{3R}{2}$.
The potential energy at this height is $U_f = -\frac{GMm}{r} = -\frac{GMm}{3R/2} = -\frac{2GMm}{3R}$.
The increase in potential energy is $\Delta U = U_f - U_i$.
$\Delta U = -\frac{2GMm}{3R} - (-\frac{GMm}{R}) = -\frac{2GMm}{3R} + \frac{GMm}{R} = \frac{GMm}{3R}$.
Since the gravitational acceleration at the surface is $g = \frac{GM}{R^2}$,we have $GM = gR^2$.
Substituting this into the expression for $\Delta U$,we get $\Delta U = \frac{(gR^2)m}{3R} = \frac{mgR}{3}$.

(A) The potential energy at the surface of the planet is $U_s = -\frac{GMm}{R}$.
The potential energy at the center of the planet is $U_c = -\frac{3GMm}{2R}$.
By the law of conservation of energy,the total energy at the surface equals the total energy at the center:
$-\frac{GMm}{R} + 0 = -\frac{3GMm}{2R} + \frac{1}{2}mv^2$.
Rearranging the terms,we get $\frac{1}{2}mv^2 = \frac{3GMm}{2R} - \frac{GMm}{R} = \frac{GMm}{2R}$.
Thus,$v^2 = \frac{GM}{R}$.
Since the escape velocity is $V_e = \sqrt{\frac{2GM}{R}}$,we have $V_e^2 = \frac{2GM}{R}$,which implies $\frac{GM}{R} = \frac{V_e^2}{2}$.
Substituting this into the expression for $v^2$,we get $v^2 = \frac{V_e^2}{2}$,so $v = \frac{V_e}{\sqrt{2}}$.

(B) The initial total energy of the system at an infinite distance is $0$.
By the law of conservation of energy,the total energy at separation $d$ must also be $0$:
$\frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 - \frac{G m_1 m_2}{d} = 0$
$\frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 = \frac{G m_1 m_2}{d} \quad ... (i)$
By the law of conservation of linear momentum (as no external force acts on the system):
$m_1 v_1 - m_2 v_2 = 0 \implies v_2 = \frac{m_1}{m_2} v_1$
Substituting $v_2$ into equation $(i)$:
$\frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 \left( \frac{m_1}{m_2} v_1 \right)^2 = \frac{G m_1 m_2}{d}$
$\frac{1}{2}m_1 v_1^2 \left( 1 + \frac{m_1}{m_2} \right) = \frac{G m_1 m_2}{d}$
$\frac{1}{2} v_1^2 \left( \frac{m_1 + m_2}{m_2} \right) = \frac{G m_2}{d}$
$v_1^2 = \frac{2 G m_2^2}{d(m_1 + m_2)} \implies v_1 = m_2 \sqrt{\frac{2G}{d(m_1 + m_2)}}$
Similarly,$v_2 = m_1 \sqrt{\frac{2G}{d(m_1 + m_2)}}$.

(A) 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 $h$.
Initial energy at the surface: $E_i = K_i + U_i = \frac{1}{2}mv^2 - \frac{GMm}{R}$
At maximum height $h$,the velocity is zero,so the final energy is: $E_f = K_f + U_f = 0 - \frac{GMm}{R+h}$
Equating $E_i = E_f$:
$\frac{1}{2}mv^2 - \frac{GMm}{R} = - \frac{GMm}{R+h}$
$\frac{1}{2}v^2 = GM \left( \frac{1}{R} - \frac{1}{R+h} \right) = GM \left( \frac{R+h-R}{R(R+h)} \right) = \frac{GMh}{R(R+h)}$
Given $h = R$,substitute this into the equation:
$\frac{1}{2}v^2 = \frac{GMR}{R(R+R)} = \frac{GM}{2R}$
$v^2 = \frac{GM}{R}$
$v = \sqrt{\frac{GM}{R}} = \left( \frac{GM}{R} \right)^{1/2}$

(B) According to the law of conservation of energy,the total mechanical energy at height $R$ above the surface (distance $2R$ from the center) must equal the total mechanical energy at the surface (distance $R$ from the center).
Initial energy at height $R$ (where $r = 2R$): $E_i = -\frac{GMm}{2R} + 0$
Final energy at the surface (where $r = R$): $E_f = -\frac{GMm}{R} + \frac{1}{2}mv^2$
Equating $E_i = E_f$:
$-\frac{GMm}{2R} = -\frac{GMm}{R} + \frac{1}{2}mv^2$
$\frac{1}{2}mv^2 = \frac{GMm}{R} - \frac{GMm}{2R} = \frac{GMm}{2R}$
$v^2 = \frac{GM}{R}$
Since the acceleration due to gravity on the surface is $g = \frac{GM}{R^2}$,we have $GM = gR^2$.
Substituting $GM = gR^2$ into the equation for $v^2$:
$v^2 = \frac{gR^2}{R} = gR$
Therefore,$v = \sqrt{gR}$.

(C) By the principle of conservation of energy,the total mechanical energy at the surface of the earth must equal the total mechanical energy at the maximum height $h$.
At the surface: $E_i = K_i + U_i = \frac{1}{2}MV^2 - \frac{GM_eM}{R}$
At maximum height $h$: $E_f = K_f + U_f = 0 - \frac{GM_eM}{R+h}$
Equating $E_i = E_f$:
$\frac{1}{2}MV^2 - \frac{GM_eM}{R} = - \frac{GM_eM}{R+h}$
$\frac{1}{2}V^2 = GM_e \left( \frac{1}{R} - \frac{1}{R+h} \right)$
Using $g = \frac{GM_e}{R^2}$,we have $GM_e = gR^2$:
$\frac{V^2}{2} = gR^2 \left( \frac{R+h-R}{R(R+h)} \right) = \frac{gR^2h}{R(R+h)} = \frac{gRh}{R+h}$
$\frac{V^2}{2gR} = \frac{h}{R+h}$
$V^2(R+h) = 2gRh$
$V^2R + V^2h = 2gRh$
$V^2R = h(2gR - V^2)$
$h = \frac{V^2R}{2gR - V^2} = \frac{R}{\left( \frac{2gR}{V^2} - 1 \right)}$

(D) The gravitational potential energy of a body of mass $m$ at a distance $r$ from the center of the earth is given by $U = -\frac{GMm}{r}$.
Initial potential energy at the surface of the earth $(r = R)$: $U_i = -\frac{GMm}{R}$.
Final potential energy at a height $h = 3R$ from the surface,the distance from the center is $r = R + 3R = 4R$. Thus,$U_f = -\frac{GMm}{4R}$.
The change in potential energy is $\Delta U = U_f - U_i = -\frac{GMm}{4R} - (-\frac{GMm}{R})$.
$\Delta U = \frac{GMm}{R} - \frac{GMm}{4R} = \frac{3GMm}{4R}$.
Since the acceleration due to gravity at the surface is $g = \frac{GM}{R^2}$,we have $GM = gR^2$.
Substituting this into the expression: $\Delta U = \frac{3(gR^2)m}{4R} = \frac{3}{4}mgR$.

(B) According to the law of conservation of mechanical energy,the total energy at the initial height must equal the total energy at the surface of the Earth.
Initial height $h = R$. The distance from the center of the Earth is $r_i = R + h = 2R$.
Initial potential energy $U_i = -\frac{GMm}{2R}$ and initial kinetic energy $K_i = 0$.
Final potential energy at the surface $U_f = -\frac{GMm}{R}$ and final kinetic energy $K_f = \frac{1}{2}mv^2$.
Applying conservation of energy: $U_i + K_i = U_f + K_f$.
$-\frac{GMm}{2R} + 0 = -\frac{GMm}{R} + \frac{1}{2}mv^2$.
$\frac{1}{2}mv^2 = \frac{GMm}{R} - \frac{GMm}{2R} = \frac{GMm}{2R}$.
$v^2 = \frac{GM}{R}$.
$v = \sqrt{\frac{GM}{R}} = \left[ \frac{GM}{R} \right]^{1/2}$.

(B) According to the law of conservation of energy,the total mechanical energy at the initial point $(r = R_0)$ is equal to the total mechanical energy at the surface of the Earth $(r = R)$.
Initial Energy $(E_i)$ = Initial Kinetic Energy + Initial Potential Energy = $0 + \left( -\frac{GMm}{R_0} \right) = -\frac{GMm}{R_0}$.
Final Energy $(E_f)$ = Final Kinetic Energy + Final Potential Energy = $\frac{1}{2}mv^2 + \left( -\frac{GMm}{R} \right)$.
Equating $E_i = E_f$:
$-\frac{GMm}{R_0} = \frac{1}{2}mv^2 - \frac{GMm}{R}$.
Rearranging for $v^2$:
$\frac{1}{2}mv^2 = GMm\left( \frac{1}{R} - \frac{1}{R_0} \right)$.
$v^2 = 2GM\left( \frac{1}{R} - \frac{1}{R_0} \right)$.
$v = \sqrt{2GM\left( \frac{1}{R} - \frac{1}{R_0} \right)}$.

(C) The gravitational potential energy at the surface of the earth is $U_i = -\frac{GMm}{R}$.
The gravitational potential energy at height $h$ is $U_f = -\frac{GMm}{R+h}$.
The work done $W$ is the change in potential energy: $W = U_f - U_i = GMm \left( \frac{1}{R} - \frac{1}{R+h} \right)$.
Given $h = R$,we substitute this into the equation:
$W = GMm \left( \frac{1}{R} - \frac{1}{R+R} \right) = GMm \left( \frac{1}{R} - \frac{1}{2R} \right) = GMm \left( \frac{1}{2R} \right) = \frac{GMm}{2R}$.
Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$.
Substituting $GM = gR^2$ into the expression for $W$:
$W = \frac{(gR^2)m}{2R} = \frac{1}{2}mgR$.

(B) The gravitational potential energy $U$ at a distance $r$ from the center of the Earth is given by $U = -\frac{GMm}{r}$.
Initial distance from the center is $r_i = R$.
Final distance from the center is $r_f = R + h$.
The change in potential energy is $\Delta U = U_f - U_i$.
$\Delta U = \left( -\frac{GMm}{R+h} \right) - \left( -\frac{GMm}{R} \right)$.
$\Delta U = GMm \left( \frac{1}{R} - \frac{1}{R+h} \right)$.
$\Delta U = GMm \left( \frac{R+h-R}{R(R+h)} \right)$.
$\Delta U = \frac{GMmh}{R(R+h)}$.

(D) Using the principle of conservation of mechanical energy: $U_1 + K_1 = U_2 + K_2$
Here,$U = -\frac{GM_e m}{r}$ and $K = \frac{1}{2}mv^2$. The escape velocity is given by $v_e = \sqrt{\frac{2GM_e}{R}}$,so $\frac{GM_e}{R} = \frac{v_e^2}{2}$.
At distance $r_1 = 10R$,speed $v_1 = 12 \; km/s$. At surface $r_2 = R$,speed is $v_2$.
$-\frac{GM_e m}{10R} + \frac{1}{2}mv_1^2 = -\frac{GM_e m}{R} + \frac{1}{2}mv_2^2$
$\frac{1}{2}v_2^2 = \frac{1}{2}v_1^2 + \frac{GM_e}{R} - \frac{GM_e}{10R} = \frac{1}{2}v_1^2 + \frac{9}{10} \left( \frac{GM_e}{R} \right)$
Substituting $\frac{GM_e}{R} = \frac{v_e^2}{2}$:
$v_2^2 = v_1^2 + \frac{9}{10} v_e^2$
$v_2^2 = (12)^2 + 0.9 \times (11.2)^2 = 144 + 0.9 \times 125.44 = 144 + 112.896 = 256.896$
$v_2 = \sqrt{256.896} \approx 16.028 \; km/s$.
The nearest integer is $16 \; km/s$.

(N/A) The projectile is acted upon by two mutually opposing gravitational forces from the two spheres. The neutral point $N$ is defined as the position where the two forces cancel each other exactly. If $ON = r$,we have:
$\frac{GMm}{r^2} = \frac{4GMm}{(6R - r)^2}$
$(6R - r)^2 = 4r^2$
$6R - r = \pm 2r$
$r = 2R$ or $r = -6R$ (ignoring the negative value as it lies outside the region between the spheres).
Thus,the neutral point is at $r = 2R$ from the centre $O$.
It is sufficient to project the particle with a speed that enables it to reach $N$. Thereafter,the greater gravitational pull of the sphere of mass $4M$ will pull it towards the second sphere.
The mechanical energy at the surface of the sphere of mass $M$ is:
$E_i = \frac{1}{2}mv^2 - \frac{GMm}{R} - \frac{4GMm}{5R}$
At the neutral point $N$,the speed approaches zero. The mechanical energy at $N$ is purely potential:
$E_N = -\frac{GMm}{2R} - \frac{4GMm}{4R} = -\frac{GMm}{2R} - \frac{GMm}{R} = -\frac{3GMm}{2R}$
From the principle of conservation of mechanical energy $(E_i = E_N)$:
$\frac{1}{2}mv^2 - \frac{GMm}{R} - \frac{4GMm}{5R} = -\frac{3GMm}{2R}$
$\frac{1}{2}v^2 = \frac{GM}{R} + \frac{4GM}{5R} - \frac{3GM}{2R} = \frac{GM}{R} \left( 1 + 0.8 - 1.5 \right) = \frac{GM}{R} (0.3) = \frac{3GM}{10R}$
$v^2 = \frac{6GM}{10R} = \frac{3GM}{5R}$
$v = \sqrt{\frac{3GM}{5R}}$

(A) Let the mass of the rocket be $m$ and the height reached be $h$.
At the Earth's surface,the total energy $E_i$ is the sum of kinetic and potential energy:
$E_i = \frac{1}{2} m v^2 - \frac{G M_e m}{R_e}$
At the highest point $h$,the velocity $v = 0$,so the total energy $E_f$ is purely potential:
$E_f = -\frac{G M_e m}{R_e + h}$
By the law of conservation of energy,$E_i = E_f$:
$\frac{1}{2} m v^2 - \frac{G M_e m}{R_e} = -\frac{G M_e m}{R_e + h}$
Dividing by $m$ and rearranging:
$\frac{v^2}{2} = G M_e \left( \frac{1}{R_e} - \frac{1}{R_e + h} \right) = G M_e \left( \frac{h}{R_e(R_e + h)} \right)$
Using $g = \frac{G M_e}{R_e^2}$,we have $G M_e = g R_e^2$:
$\frac{v^2}{2} = \frac{g R_e^2 h}{R_e(R_e + h)} = \frac{g R_e h}{R_e + h}$
Solving for $h$:
$v^2(R_e + h) = 2 g R_e h \implies v^2 R_e = h(2 g R_e - v^2)$
$h = \frac{R_e v^2}{2 g R_e - v^2}$
Substituting values $v = 5 \times 10^3 \; m/s$,$R_e = 6.4 \times 10^6 \; m$,$g = 9.8 \; m/s^2$:
$h = \frac{6.4 \times 10^6 \times (5 \times 10^3)^2}{2 \times 9.8 \times 6.4 \times 10^6 - (5 \times 10^3)^2}$
$h = \frac{6.4 \times 10^6 \times 25 \times 10^6}{125.44 \times 10^6 - 25 \times 10^6} = \frac{160 \times 10^{12}}{100.44 \times 10^6} \approx 1.593 \times 10^6 \; m \approx 1.6 \times 10^6 \; m$.

(D) Mass of each star,$M = 2 \times 10^{30} \; kg$.
Radius of each star,$R = 10^{4} \; km = 10^{7} \; m$.
Initial distance between the centers of the stars,$r = 10^{9} \; km = 10^{12} \; m$.
Initial kinetic energy is $0$ as speeds are negligible.
Initial potential energy,$U_{i} = -\frac{GM^{2}}{r}$.
Total initial energy,$E_{i} = -\frac{GM^{2}}{r}$.
At the moment of collision,the distance between the centers is $2R$.
Let the speed of each star be $v$.
Total kinetic energy,$K_{f} = \frac{1}{2}Mv^{2} + \frac{1}{2}Mv^{2} = Mv^{2}$.
Total potential energy,$U_{f} = -\frac{GM^{2}}{2R}$.
Total final energy,$E_{f} = Mv^{2} - \frac{GM^{2}}{2R}$.
By the law of conservation of energy,$E_{i} = E_{f}$:
$-\frac{GM^{2}}{r} = Mv^{2} - \frac{GM^{2}}{2R}$.
$v^{2} = GM \left( \frac{1}{2R} - \frac{1}{r} \right)$.
Since $r \gg 2R$,$\frac{1}{r}$ is negligible compared to $\frac{1}{2R}$.
$v^{2} \approx \frac{GM}{2R} = \frac{6.67 \times 10^{-11} \times 2 \times 10^{30}}{2 \times 10^{7}} = 6.67 \times 10^{12} \; m^{2}/s^{2}$.
$v = \sqrt{6.67 \times 10^{12}} \approx 2.58 \times 10^{6} \; m/s$.

(A) The gravitational potential energy $U$ of an object of mass $m$ at a distance $r$ from the center of the Earth is given by $U = -\frac{GMm}{r}$,where $G$ is the gravitational constant,$M$ is the mass of the Earth,and $r$ is the distance from the center of the Earth.
$1$. When an object moves away from the Earth,the distance $r$ increases. Since $U$ is negative and inversely proportional to $r$,as $r$ increases,the value of $U$ becomes less negative,which means it increases.
$2$. When an object comes closer to the Earth,the distance $r$ decreases. As $r$ decreases,the value of $U$ becomes more negative,which means it decreases.

(N/A) The gravitational potential energy $U$ of a body of mass $m$ at a height $h$ above the surface of the Earth is given by the work done against the gravitational force.
Since the force of gravity near the Earth's surface is $F = mg$,the work done to lift the body to height $h$ is $W = F \times h = mgh$.
Therefore,the change in gravitational potential energy is $\Delta U = mgh$.

(N/A) The gravitational potential energy of an object of mass $m$ at a distance $r$ from the centre of the earth of mass $M$ is given by $U = -\frac{GMm}{r}$.
Potential energy of the object at the surface of the earth $(r = R)$: $U_i = -\frac{GMm}{R}$.
Potential energy of the object at a height equal to the radius of the earth $(r = R + R = 2R)$: $U_f = -\frac{GMm}{2R}$.
Gain in potential energy $(\Delta U)$ = $U_f - U_i$.
$\Delta U = -\frac{GMm}{2R} - \left(-\frac{GMm}{R}\right) = \frac{GMm}{R} - \frac{GMm}{2R} = \frac{GMm}{2R}$.
Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$.
Substituting this into the expression for gain in potential energy:
$\Delta U = \frac{(gR^2)m}{2R} = \frac{1}{2}mgR$.

(B) To reach the smaller planet,the body must cross the point of zero gravitational field (neutral point) between the two planets.
Let the distance of the neutral point from the center of the smaller planet be $x$.
Equating the gravitational fields from both planets at this point:
$\frac{GM}{x^2} = \frac{G(16M)}{(10a - x)^2}$
Taking the square root on both sides:
$\frac{1}{x} = \frac{4}{10a - x} \implies 10a - x = 4x \implies x = 2a$.
Now,apply the Law of Conservation of Mechanical Energy between the surface of the larger planet and the neutral point.
The potential at the surface of the larger planet (radius $2a$,mass $16M$) is $V_L = -\frac{G(16M)}{2a} - \frac{GM}{8a} = -\frac{8GM}{a} - \frac{GM}{8a} = -\frac{65GM}{8a}$.
The potential at the neutral point (distance $8a$ from larger,$2a$ from smaller) is $V_P = -\frac{G(16M)}{8a} - \frac{GM}{2a} = -\frac{2GM}{a} - \frac{GM}{2a} = -\frac{5GM}{2a}$.
Using $KE_i + PE_i = KE_f + PE_f$,where $KE_f = 0$ at the neutral point:
$\frac{1}{2}mv^2 + m V_L = 0 + m V_P$
$\frac{1}{2}v^2 = V_P - V_L = -\frac{5GM}{2a} - (-\frac{65GM}{8a}) = \frac{-20GM + 65GM}{8a} = \frac{45GM}{8a}$.
$v^2 = \frac{45GM}{4a} \implies v = \sqrt{\frac{45GM}{4a}} = \frac{3}{2} \sqrt{\frac{5GM}{a}}$.

(A) To reach the surface of the second planet,the satellite must cross the point where the net gravitational field is zero.
Let the distance of this point from the centre of the planet of mass $M$ be $x$.
$\frac{GM}{x^2} = \frac{G(9M)}{(8R-x)^2}$
$\frac{1}{x} = \frac{3}{8R-x}$
$8R-x = 3x \Rightarrow x = 2R$.
Now,apply the law of conservation of energy between the surface of the first planet and the point $x = 2R$ (where the velocity is minimum,i.e.,$v_{min} = 0$):
$E_{initial} = E_{final}$
$\frac{1}{2}mv^2 - \frac{GMm}{R} - \frac{G(9M)m}{7R} = 0 - \frac{GMm}{2R} - \frac{G(9M)m}{6R}$
$\frac{1}{2}v^2 = \frac{GM}{R} + \frac{9GM}{7R} - \frac{GM}{2R} - \frac{9GM}{6R}$
$\frac{1}{2}v^2 = \frac{GM}{R} [1 + \frac{9}{7} - \frac{1}{2} - \frac{3}{2}]$
$\frac{1}{2}v^2 = \frac{GM}{R} [1 + \frac{9}{7} - 2] = \frac{GM}{R} [\frac{9}{7} - 1] = \frac{GM}{R} [\frac{2}{7}]$
$v^2 = \frac{4}{7} \frac{GM}{R}$
$v = \sqrt{\frac{4}{7} \frac{GM}{R}}$
Comparing this with $\sqrt{\frac{a}{7} \frac{GM}{R}}$,we get $a = 4$.

(A) The gravitational potential energy at the surface of the Earth is $U_i = -\frac{GMm}{R}$.
The object is taken to a height $h = 3R$. The final distance from the center of the Earth is $r = R + h = R + 3R = 4R$.
The gravitational potential energy at this height is $U_f = -\frac{GMm}{4R}$.
The gain in potential energy is $\Delta U = U_f - U_i = -\frac{GMm}{4R} - (-\frac{GMm}{R}) = \frac{GMm}{R} - \frac{GMm}{4R} = \frac{3GMm}{4R}$.
Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$. Substituting this into the equation:
$\Delta U = \frac{3(gR^2)m}{4R} = \frac{3}{4} mgR$.
Given $m = 1 \, kg$,$g = 10 \, m/s^2$,and $R = 6400 \, km = 6.4 \times 10^6 \, m$:
$\Delta U = \frac{3}{4} \times 1 \times 10 \times 6.4 \times 10^6 = 3 \times 2.5 \times 6.4 \times 10^6 = 48 \times 10^6 \, J = 48 \, MJ$.

(A) Let $m$ be the mass of the object and $v$ be its speed when it returns to the surface of the planet.
At the maximum height,the distance from the center of the planet is $r_{max} = R + 4R = 5R$. At this point,the velocity of the object is $0$.
Using the principle of conservation of mechanical energy between the maximum height and the surface of the planet:
$E_{initial} = E_{final}$
$U_{max} + K_{max} = U_{surface} + K_{surface}$
$-\frac{G M m}{5R} + 0 = -\frac{G M m}{R} + \frac{1}{2} m v^2$
$\frac{1}{2} v^2 = \frac{G M}{R} - \frac{G M}{5R}$
$\frac{1}{2} v^2 = \frac{G M}{R} (1 - \frac{1}{5}) = \frac{G M}{R} (\frac{4}{5})$
$v^2 = \frac{8 G M}{5 R}$
$v = \sqrt{\frac{8 G M}{5 R}} = 2 \sqrt{\frac{2 G M}{5 R}}$
Thus,the speed of the object when it returns to the surface is $2 \sqrt{\frac{2 G M}{5 R}}$.

(B) By applying the principle of conservation of mechanical energy between the initial point $P$ (at distance $3R$ from the center) and the surface of the moon $S$ (at distance $R$ from the center):
Initial Mechanical Energy $(E_i)$ = Final Mechanical Energy $(E_f)$
$E_i = K_i + U_i = 0 + \left(-\frac{G M m}{3 R}\right) = -\frac{G M m}{3 R}$
$E_f = K_f + U_f = \frac{1}{2} m v^2 + \left(-\frac{G M m}{R}\right)$
Equating $E_i$ and $E_f$:
$-\frac{G M m}{3 R} = \frac{1}{2} m v^2 - \frac{G M m}{R}$
$\frac{1}{2} m v^2 = \frac{G M m}{R} - \frac{G M m}{3 R}$
$\frac{1}{2} m v^2 = \frac{G M m}{R} \left(1 - \frac{1}{3}\right) = \frac{G M m}{R} \left(\frac{2}{3}\right)$
$\frac{1}{2} v^2 = \frac{2 G M}{3 R}$
$v^2 = \frac{4 G M}{3 R}$
$v = \sqrt{\frac{4 G M}{3 R}}$
Thus,the correct option is $(B)$.

(B) By applying the law of conservation of mechanical energy between the Earth's surface and the maximum height reached by the body:
Initial Mechanical Energy $(E_i)$ = Final Mechanical Energy $(E_f)$
$P.E._i + K.E._i = P.E._f + K.E._f$
$-\frac{G M m}{R} + \frac{1}{2} m \left(\sqrt{\frac{G M}{R}}\right)^2 = -\frac{G M m}{r} + 0$
$-\frac{G M m}{R} + \frac{G M m}{2 R} = -\frac{G M m}{r}$
$-\frac{G M m}{2 R} = -\frac{G M m}{r}$
$r = 2 R$
Since $r = R + h$,where $h$ is the height above the surface:
$R + h = 2 R$
$h = R$

(B) According to the law of conservation of mechanical energy,the loss in gravitational potential energy equals the gain in kinetic energy.
Initial potential energy at height $h = R$ is $U_i = -\frac{GMm}{R+R} = -\frac{GMm}{2R}$.
Final potential energy at the earth's surface $(r = R)$ is $U_f = -\frac{GMm}{R}$.
Loss in potential energy = $U_i - U_f = -\frac{GMm}{2R} - (-\frac{GMm}{R}) = \frac{GMm}{2R}$.
Gain in kinetic energy = $\frac{1}{2}mv^2$.
Equating the two: $\frac{1}{2}mv^2 = \frac{GMm}{2R}$.
Using $GM = gR^2$,we get $\frac{1}{2}v^2 = \frac{gR^2}{2R} = \frac{gR}{2}$.
Therefore,$v^2 = gR$,which gives $v = \sqrt{gR}$.

(A) By the principle of conservation of mechanical energy,the total energy at the initial position (height $h = R$ from the surface,distance $r = 2R$ from the center) is equal to the total energy at the surface of the earth (distance $r = R$ from the center).
$U_i + K_i = U_f + K_f$
$-\frac{GMm}{2R} + 0 = -\frac{GMm}{R} + \frac{1}{2}mv^2$
$\frac{1}{2}mv^2 = \frac{GMm}{R} - \frac{GMm}{2R}$
$\frac{1}{2}mv^2 = \frac{GMm}{2R}$
$v^2 = \frac{GM}{R}$
Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$.
Substituting this into the equation for $v^2$:
$v^2 = \frac{gR^2}{R} = gR$
$v = \sqrt{gR}$

(B) Let $M$ be the mass of the earth and $R$ be its radius. The body of mass $m$ is initially at a height $R$ above the surface,so its distance from the center of the earth is $r_A = R + R = 2R$.
Using the Law of Conservation of Mechanical Energy between point $A$ (at height $R$) and point $B$ (at the surface):
$E_A = E_B$
$K_A + U_A = K_B + U_B$
Since the body is initially at rest,$K_A = 0$.
$0 - \frac{G M m}{2R} = \frac{1}{2} m v^2 - \frac{G M m}{R}$
$\frac{1}{2} m v^2 = \frac{G M m}{R} - \frac{G M m}{2R}$
$\frac{1}{2} m v^2 = \frac{G M m}{2R}$
$v^2 = \frac{G M}{R}$
Since $g = \frac{G M}{R^2}$,we have $G M = g R^2$.
Substituting this into the equation for $v^2$:
$v^2 = \frac{g R^2}{R} = g R$
$v = \sqrt{g R}$

(B) According to the law of conservation of energy,the total mechanical energy at the initial position is equal to the total mechanical energy at the surface of the earth.
Initial energy $E_i = K_i + U_i = 0 - \frac{GMm}{R_0}$.
Final energy $E_f = K_f + U_f = \frac{1}{2}mv^2 - \frac{GMm}{R}$.
Equating $E_i = E_f$:
$-\frac{GMm}{R_0} = \frac{1}{2}mv^2 - \frac{GMm}{R}$.
Rearranging the terms:
$\frac{1}{2}mv^2 = GMm\left(\frac{1}{R} - \frac{1}{R_0}\right)$.
$v^2 = 2GM\left(\frac{1}{R} - \frac{1}{R_0}\right)$.
Therefore,the velocity $v$ is:
$v = \sqrt{2GM\left(\frac{1}{R} - \frac{1}{R_0}\right)}$.

(D) The total mechanical energy is conserved. Let $r_1$ be the initial distance from the center of the Earth and $r_2$ be the final distance from the center of the Earth.
Initial distance from center: $r_1 = 3R + R = 4R$.
Final distance from center: $r_2 = R + R = 2R$.
Initial potential energy: $U_1 = -\frac{GMm}{r_1} = -\frac{GMm}{4R}$.
Final potential energy: $U_2 = -\frac{GMm}{r_2} = -\frac{GMm}{2R}$.
Since the body starts from rest,initial kinetic energy $K_1 = 0$.
By the law of conservation of energy: $K_1 + U_1 = K_2 + U_2$.
$0 + (-\frac{GMm}{4R}) = K_2 + (-\frac{GMm}{2R})$.
$K_2 = \frac{GMm}{2R} - \frac{GMm}{4R} = \frac{GMm}{4R}$.

(A) Using the law of conservation of energy,the total energy at the surface equals the total energy at the maximum height $h$.
Total Energy at surface = Total Energy at height $h$
$\frac{1}{2}mv^2 - \frac{GMm}{R} = 0 - \frac{GMm}{R+h}$
Given $v = \frac{v_e}{3}$,where $v_e = \sqrt{\frac{2GM}{R}}$ is the escape velocity.
Substituting $v^2 = \frac{v_e^2}{9} = \frac{2GM}{9R}$:
$\frac{1}{2}m\left(\frac{2GM}{9R}\right) - \frac{GMm}{R} = - \frac{GMm}{R+h}$
$\frac{GMm}{9R} - \frac{GMm}{R} = - \frac{GMm}{R+h}$
Dividing by $GMm$:
$\frac{1}{9R} - \frac{1}{R} = - \frac{1}{R+h}$
$\frac{1-9}{9R} = - \frac{1}{R+h}$
$\frac{-8}{9R} = - \frac{1}{R+h}$
$\frac{8}{9R} = \frac{1}{R+h}$
$8(R+h) = 9R$
$8R + 8h = 9R$
$8h = R$
$h = \frac{R}{8}$

(B) According to the law of conservation of energy,the total mechanical energy at the initial point (at distance $R_0$) is equal to the total mechanical energy at the surface of the Earth (at distance $R$).
Initial energy $E_i = K_i + U_i = 0 + \left( -\frac{GMm}{R_0} \right) = -\frac{GMm}{R_0}$.
Final energy $E_f = K_f + U_f = \frac{1}{2} mv^2 + \left( -\frac{GMm}{R} \right)$.
By conservation of energy,$E_i = E_f$:
$-\frac{GMm}{R_0} = \frac{1}{2} mv^2 - \frac{GMm}{R}$.
Rearranging the terms:
$\frac{1}{2} mv^2 = \frac{GMm}{R} - \frac{GMm}{R_0} = GMm \left( \frac{1}{R} - \frac{1}{R_0} \right)$.
$v^2 = 2 GM \left( \frac{1}{R} - \frac{1}{R_0} \right)$.
Therefore,the velocity $v$ is:
$v = \left[ 2 GM \left( \frac{1}{R} - \frac{1}{R_0} \right) \right]^{\frac{1}{2}}$.

(B) Using the law of conservation of energy,the total energy at the surface equals the total energy at the maximum height $h$.
Initial energy at surface: $E_i = \frac{1}{2}mV^2 - \frac{GMm}{R}$
Final energy at height $h$: $E_f = 0 - \frac{GMm}{R+h}$
Equating $E_i = E_f$:
$\frac{1}{2}mV^2 - \frac{GMm}{R} = - \frac{GMm}{R+h}$
$\frac{1}{2}V^2 = GM \left( \frac{1}{R} - \frac{1}{R+h} \right)$
Since $GM = gR^2$:
$\frac{1}{2}V^2 = gR^2 \left( \frac{R+h-R}{R(R+h)} \right) = gR \left( \frac{h}{R+h} \right)$
$\frac{V^2}{2gR} = \frac{h}{R+h}$
$V^2(R+h) = 2gRh$
$V^2R + V^2h = 2gRh$
$V^2R = h(2gR - V^2)$
$h = \frac{V^2R}{2gR - V^2}$

(B) The minimum energy required to launch a satellite from the surface of the Earth is equal to the change in gravitational potential energy as it moves from the surface to the final altitude.
Initial potential energy at the surface $(r_1 = R)$: $U_i = -\frac{G M m}{R}$
Final potential energy at altitude $h = 2R$ $(r_2 = R + 2R = 3R)$: $U_f = -\frac{G M m}{3R}$
The energy required is $\Delta U = U_f - U_i$
$\Delta U = -\frac{G M m}{3R} - (-\frac{G M m}{R})$
$\Delta U = G M m (\frac{1}{R} - \frac{1}{3R})$
$\Delta U = G M m (\frac{3-1}{3R}) = \frac{2 G M m}{3 R}$

(A) By 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 $h$.
At the surface: $E_i = -\frac{GMm}{R} + \frac{1}{2}mu^2$
At maximum height $h$: $E_f = -\frac{GMm}{R+h} + 0$
Equating $E_i = E_f$: $-\frac{GMm}{R} + \frac{1}{2}mu^2 = -\frac{GMm}{R+h}$
Dividing by $m$ and rearranging: $\frac{GM}{R+h} = \frac{GM}{R} - \frac{u^2}{2}$
Using $GM = gR^2$: $\frac{gR^2}{R+h} = gR - \frac{u^2}{2} = \frac{2gR - u^2}{2}$
$\frac{R+h}{R^2} = \frac{2g}{2gR - u^2}$
$R+h = \frac{2gR^2}{2gR - u^2}$
$h = \frac{2gR^2}{2gR - u^2} - R = \frac{2gR^2 - 2gR^2 + u^2R}{2gR - u^2} = \frac{u^2R}{2gR - u^2}$

(B) The minimum kinetic energy required to escape from the gravitational influence of the earth is given by $K_e = \frac{GMm}{R}$.
Given that the kinetic energy of projection is $K = \frac{1}{2} K_e = \frac{GMm}{2R}$.
At the highest point,the velocity of the body becomes zero,so its kinetic energy is zero.
By the law of conservation of energy,the loss in kinetic energy equals the gain in potential energy:
$K = U_f - U_i$
$\frac{GMm}{2R} = \left( -\frac{GMm}{R+h} \right) - \left( -\frac{GMm}{R} \right)$
$\frac{GMm}{2R} = \frac{GMm}{R} - \frac{GMm}{R+h}$
Dividing both sides by $GMm$:
$\frac{1}{2R} = \frac{1}{R} - \frac{1}{R+h}$
$\frac{1}{R+h} = \frac{1}{R} - \frac{1}{2R} = \frac{1}{2R}$
$R+h = 2R$
$h = R$.

(A) The gravitational potential energy on the surface of the Earth is $U_E = -\frac{G M m}{R}$,where $M$ is the mass of the Earth,$m$ is the mass of the object,$R$ is the radius of the Earth,and $G$ is the gravitational constant.
When the mass $m$ is shifted to a height equal to the radius of the Earth,the distance from the center becomes $r = R + R = 2R$.
The potential energy at this height is $U = -\frac{G M m}{2R}$.
The work done in the process is $W = U - U_E = -\frac{G M m}{2R} - (-\frac{G M m}{R}) = -\frac{G M m}{2R} + \frac{G M m}{R} = \frac{G M m}{2R}$.
Using the relation $g = \frac{G M}{R^2}$,we have $G M = g R^2$.
Substituting this into the work equation: $W = \frac{(g R^2) m}{2R} = \frac{m g R}{2}$.