Gravitational Potential and Potential Energy of system Questions in English

(B) The gravitational potential energy $U$ of a system is defined as the work done in assembling the system by bringing mass elements from infinity to their respective positions.
For a thin spherical shell of mass $M$ and radius $R$,we consider building the shell by adding infinitesimal mass elements $dm$ to the surface.
However,for a thin shell,the potential $V$ at the surface is constant and equal to $-\frac{GM}{R}$.
When we assemble the shell,the total work done is $W = \int V dm = \int_{0}^{M} -\frac{GM}{R} dm$.
Since the potential $V$ is constant for the entire shell of radius $R$,we integrate $dm$ from $0$ to $M$:
$U = -\frac{G}{R} \int_{0}^{M} M' dm$ is not correct here; rather,we consider the work to bring the mass $M$ to the surface of a shell of radius $R$ where the potential is constant.
The correct expression for the self-energy of a thin spherical shell is $U = -\frac{GM^2}{2R}$.

(C) The gravitational potential energy $U$ of a particle of mass $m$ at the surface of a sphere of mass $M$ and radius $R$ is given by $U = -\frac{GMm}{R}$.
To take the particle to infinity (where potential energy is $0$),the work done $W$ against the gravitational force is equal to the change in potential energy: $W = U_{final} - U_{initial} = 0 - (- \frac{GMm}{R}) = \frac{GMm}{R}$.
Given values:
$M = 100\, kg$
$m = 10\, g = 0.01\, kg$
$R = 10\, cm = 0.1\, m$
$G = 6.67 \times 10^{-11}\, Nm^2/kg^2$
Substituting these values into the formula:
$W = \frac{6.67 \times 10^{-11} \times 100 \times 0.01}{0.1}$
$W = \frac{6.67 \times 10^{-11} \times 1}{0.1} = 6.67 \times 10^{-10}\, J$.

(B) The gravitational potential energy $U$ of a body of mass $m$ at a distance $r$ from the center of the Earth is given by $U = -\frac{G M_e m}{r}$.
Here,the satellite is at a height $h = R_e$ above the surface of the Earth.
Therefore,the distance from the center of the Earth is $r = R_e + h = R_e + R_e = 2 R_e$.
Substituting this into the potential energy formula,we get $U = -\frac{G M_e m}{2 R_e}$.
We know that the acceleration due to gravity at the surface of the Earth is $g = \frac{G M_e}{R_e^2}$,which implies $G M_e = g R_e^2$.
Substituting $G M_e = g R_e^2$ into the expression for $U$,we get $U = -\frac{(g R_e^2) m}{2 R_e} = -\frac{m g R_e}{2}$.

(B) The gravitational potential $V$ at a point is defined as the work done per unit mass in bringing a test mass from a reference point to that point.
The potential difference between a point $A$ at distance $r$ and a reference point at infinity is given by $V_A - V_{\infty} = -\int_{\infty}^{r} \vec{E} \cdot d\vec{r} = -\frac{GM}{r}$.
Case $(i)$: When the potential at infinity is $V_{\infty} = 0$,the potential at point $A$ is $V_A = -5 \, unit$.
Thus,$-5 - 0 = -\frac{GM}{r}$,which implies $-\frac{GM}{r} = -5$.
Case $(ii)$: When the potential at infinity is $V_{\infty}' = +10 \, unit$,the potential at point $A$ becomes $V_A'$.
The potential difference remains the same because it depends only on the mass and the distance: $V_A' - V_{\infty}' = -\frac{GM}{r}$.
Substituting the values: $V_A' - 10 = -5$.
Therefore,$V_A' = -5 + 10 = +5 \, unit$.

(N/A) Consider four masses,each of mass $m$,placed at the corners of a square of side $l$. The total gravitational potential energy $U$ is the sum of the potential energies of all distinct pairs of masses.
There are $4$ pairs of masses separated by distance $l$ (the sides of the square) and $2$ pairs of masses separated by distance $\sqrt{2}l$ (the diagonals of the square).
The potential energy of a pair of masses $m_1$ and $m_2$ at distance $r$ is $U = -\frac{G m_1 m_2}{r}$.
Thus,the total potential energy is:
$U = -4 \left( \frac{G m^2}{l} \right) - 2 \left( \frac{G m^2}{\sqrt{2} l} \right)$
$U = -\frac{G m^2}{l} \left( 4 + \frac{2}{\sqrt{2}} \right) = -\frac{G m^2}{l} (4 + \sqrt{2}) \approx -5.414 \frac{G m^2}{l}$.
The gravitational potential $V$ at the centre of the square is the sum of the potentials due to each of the four masses. The distance of each mass from the centre is $r = \frac{\sqrt{2}l}{2} = \frac{l}{\sqrt{2}}$.
$V = 4 \times \left( -\frac{G m}{r} \right) = 4 \times \left( -\frac{G m}{l/\sqrt{2}} \right) = -4\sqrt{2} \frac{G m}{l}$.

(N/A) Mass of each sphere,$M = 100 \; kg$.
Separation between the centres of the spheres,$r = 1.0 \; m$.
Let $X$ be the mid-point between the centres of the spheres.
$1$. Gravitational Force at $X$:
The gravitational force exerted by each sphere on an object at $X$ is equal in magnitude but opposite in direction. Therefore,the net gravitational force at $X$ is $0 \; N$.
$2$. Gravitational Potential at $X$:
The gravitational potential $V$ at a distance $d$ from a mass $M$ is given by $V = -GM/d$. At the mid-point $X$,the distance from each sphere is $d = r/2 = 0.5 \; m$.
$V_{total} = V_1 + V_2 = -\frac{GM}{r/2} - \frac{GM}{r/2} = -\frac{4GM}{r}$
$V_{total} = -\frac{4 \times 6.67 \times 10^{-11} \times 100}{1.0} = -2.668 \times 10^{-8} \; J/kg \approx -2.67 \times 10^{-8} \; J/kg$.
$3$. Equilibrium:
Since the net force is zero,an object placed at $X$ is in equilibrium. If the object is displaced slightly towards one sphere,the gravitational force from that sphere increases while the force from the other decreases. This creates a net force in the direction of displacement,pulling the object further away from $X$. Thus,the equilibrium is unstable.

(B) Mass of the Earth,$M = 6.0 \times 10^{24} \; kg$.
Radius of the Earth,$R = 6400 \; km = 6.4 \times 10^{6} \; m$.
Height of the geostationary satellite,$h = 36000 \; km = 3.6 \times 10^{7} \; m$.
The distance of the satellite from the center of the Earth is $r = R + h = 6.4 \times 10^{6} \; m + 36.0 \times 10^{6} \; m = 42.4 \times 10^{6} \; m = 4.24 \times 10^{7} \; m$.
Gravitational potential $V$ at a distance $r$ is given by $V = -\frac{GM}{r}$.
Substituting the values: $V = -\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24}}{4.24 \times 10^{7}}$.
$V = -\frac{40.02 \times 10^{13}}{4.24 \times 10^{7}}$.
$V \approx -9.44 \times 10^{6} \; J/kg$.

(N/A) Potential energy is the energy possessed by a body by virtue of its position or configuration. If the configuration of the system changes,its potential energy changes.
Gravitational potential energy is the energy stored in an object due to its position in a gravitational field. When an object of mass $m$ is lifted to a height $h$ above the Earth's surface,the work done against the gravitational force is stored as gravitational potential energy.
The work done by an external force $F$ is given by $W = F d \cos \theta$.
Here,$F = mg$,$d = h$,and $\theta = 0^{\circ}$ (since the force and displacement are in the same upward direction).
Therefore,$W = (mg)(h) \cos 0^{\circ} = mgh$.
This work is stored as gravitational potential energy,$U = mgh$.
The dimensional formula for energy is derived from $U = mgh$:
$[M] \times [LT^{-2}] \times [L] = [M^{1} L^{2} T^{-2}]$.
The $SI$ unit of gravitational potential energy is Joule $(J)$.

(N/A) Gravitational potential energy is defined as the work done by an external agent in bringing a body of mass $m$ from an infinite distance to a specific point within the gravitational field of another body.
Mathematically,the gravitational potential energy $U$ at a distance $r$ from a mass $M$ is given by $U = -\frac{GMm}{r}$.
Zero potential energy is defined as the state where the gravitational potential energy of a system is considered to be zero. By convention,the gravitational potential energy of a system of two bodies is taken to be zero when the distance between them is infinite $(r = \infty)$,because at an infinite distance,the gravitational force exerted on the body is zero.

(N/A) Gravitational potential energy is defined as the work done in bringing a body of mass $m$ from an infinite distance to a given point in the gravitational field of another body.
Let the mass of the Earth be $M_E$ and its radius be $R_E$. We want to determine the gravitational potential energy of a mass $m$ at a point $P$ at a distance $r$ from the centre of the Earth $(O)$.
Consider the body of mass $m$ at an arbitrary point $A$ at a distance $x$ from the centre of the Earth $(OP = x)$.
The gravitational force on the body at this point is given by:
$F = \frac{G M_E m}{x^2}$
The work done to move the body by a small displacement $dx$ towards the Earth is:
$dW = F dx = \frac{G M_E m}{x^2} dx$
The total work done to bring the body from infinity to a distance $r$ is:
$W = \int_{\infty}^{r} dW = \int_{\infty}^{r} \frac{G M_E m}{x^2} dx$
$W = G M_E m \int_{\infty}^{r} x^{-2} dx$
$W = G M_E m \left[ -\frac{1}{x} \right]_{\infty}^{r}$
$W = G M_E m \left( -\frac{1}{r} - (-\frac{1}{\infty}) \right)$
Since $\frac{1}{\infty} = 0$,we get:
$W = -\frac{G M_E m}{r}$
Thus,the gravitational potential energy $U$ at distance $r$ is:
$U = -\frac{G M_E m}{r}$

(N/A) Gravitational Potential $V$: The gravitational potential at a point in a gravitational field is defined as the amount of work done by an external agent in bringing a unit mass from infinity to that point without acceleration.
Definition: The work done per unit mass in bringing a body of unit mass from an infinite distance to a given point in a gravitational field is called the gravitational potential at that point.
Formula: Gravitational potential $V = \frac{W}{m}$,where $W$ is the work done and $m$ is the mass.
Nature: It is a scalar quantity.
Units: Its $SI$ unit is $J/kg$ (joule per kilogram) and its $CGS$ unit is $erg/g$ (erg per gram).
Dimensional Formula: $M^{0} L^{2} T^{-2}$.

(N/A) The work done in bringing a body of mass $m$ from an infinite distance in the gravitational field of the Earth to a point at a distance $r$ $(r > R_{E})$ from the center of the Earth is given by the gravitational potential energy $U = -\frac{GM_{E}m}{r}$.
Gravitational potential $V$ at a point is defined as the work done per unit mass to bring a body from infinity to that point:
$V = \frac{W}{m} = \frac{U}{m}$
Substituting the expression for $U$:
$V = \frac{-GM_{E}m/r}{m} = -\frac{GM_{E}}{r}$
On the surface of the Earth,where $r = R_{E}$,the gravitational potential is:
$V_{E} = -\frac{GM_{E}}{R_{E}}$
Relation between gravitational potential and gravitational potential energy:
Since $U = -\frac{GM_{E}m}{r}$ and $V = -\frac{GM_{E}}{r}$,we can write:
$U = V \cdot m$
Therefore,Gravitational potential energy = (Gravitational potential) $\times$ (mass).

(N/A) Gravitational potential energy is the energy possessed by an object due to its position in a gravitational field. It is defined as the work done by an external agent in bringing a body from infinity to a point in the gravitational field without acceleration.
The gravitational potential $V$ at a point is given by $V = -GM/r$,where $G$ is the gravitational constant,$M$ is the mass of the source,and $r$ is the distance from the source.
The negative sign indicates that the gravitational force is attractive in nature. It means that the gravitational potential energy of a system of two masses is zero at infinity and decreases as the masses are brought closer together. Thus,work must be done by the gravitational field to bring the masses together,or conversely,an external agent must do negative work to move the body from infinity to a point $r$.

(C) The gravitational potential energy $U$ of a system of two masses $m_1$ and $m_2$ separated by a distance $r$ is given by the formula $U = -\frac{G m_1 m_2}{r}$.
As the distance $r$ approaches infinity $(r \to \infty)$,the term $\frac{1}{r}$ approaches zero.
Therefore,the potential energy at infinity distance is defined as $U = 0$.

(N/A) Gravitational potential $(V)$ at a point is defined as the work done per unit mass in bringing a test mass from infinity to that point.
Mathematically,$V = \frac{W}{m}$.
In $SI$ units,work is measured in Joules $(J)$ and mass in kilograms $(kg)$. Therefore,the $SI$ unit is $J/kg$ or $m^2/s^2$.
In $CGS$ units,work is measured in ergs $(erg)$ and mass in grams $(g)$. Therefore,the $CGS$ unit is $erg/g$ or $cm^2/s^2$.

(A) The gravitational potential $V$ at a distance $r$ from the centre of the Earth (where $r > R_E$) is defined as the work done per unit mass in bringing a test mass from infinity to that point.
The formula is given by:
$V = -\frac{GM_E}{r}$
Where:
$G$ is the universal gravitational constant,
$M_E$ is the mass of the Earth,
$r$ is the distance from the centre of the Earth.

(N/A) The gravitational potential energy $U$ of a body of mass $m$ at the surface of the Earth is given by $U = -\frac{GM_{E}m}{R_{E}}$.
The gravitational potential $V$ at the surface of the Earth is defined as the gravitational potential energy per unit mass,given by $V = \frac{U}{m} = -\frac{GM_{E}}{R_{E}}$.
Here,$G$ is the universal gravitational constant,$M_{E}$ is the mass of the Earth,and $R_{E}$ is the radius of the Earth.

(N/A) $1$. Definition: Gravitational potential energy $(U)$ is the energy possessed by a body due to its position in a gravitational field. Gravitational potential $(V)$ is the work done per unit mass to bring a test mass from infinity to a point in a gravitational field.
$2$. Formula: For a point mass $M$,gravitational potential energy at distance $r$ is $U = -GMm/r$. Gravitational potential at distance $r$ is $V = -GM/r$.
$3$. Units: The $SI$ unit of gravitational potential energy is Joule $(J)$. The $SI$ unit of gravitational potential is Joule per kilogram ($J/kg$ or $m^2/s^2$).
$4$. Dependence: Gravitational potential energy depends on the mass of the object $(m)$ placed at that point. Gravitational potential is independent of the mass of the object placed at that point; it depends only on the source mass $(M)$ and the position $(r)$.

(A) The gravitational potential energy $U$ of a system of two masses $m_1$ and $m_2$ separated by a distance $r$ is given by $U = -\frac{G m_1 m_2}{r}$.
As the distance $r$ between the masses approaches infinity $(r \to \infty)$,the potential energy $U$ approaches zero $(U \to 0)$.
Since the masses are at an infinite distance,their kinetic energy $K$ is also zero (assuming they are at rest relative to each other).
The total energy $E$ of the system is the sum of kinetic energy and potential energy: $E = K + U$.
Therefore,$E = 0 + 0 = 0$.
Thus,the total energy at an infinite distance is zero.

(B) The gravitational potential energy of a body of mass $m$ at a distance $r$ from the center of the Earth is given by $U = -\frac{GMm}{r}$.
Since $G$,$M$,$m$,and $r$ are all positive,the potential energy $U$ is negative.
$A$ negative value of potential energy indicates that the body is in a bound state within the Earth's gravitational field.
To move the body to infinity (where potential energy is zero),we must provide an amount of energy equal to $|U|$ (i.e.,positive work must be done on the body).
Therefore,negative energy signifies that the body is gravitationally bound to the Earth.

(N/A) The binding energy of a body is defined as the minimum energy required to remove the body from the gravitational influence of the Earth to infinity.
The gravitational potential energy $U$ of a body of mass $m$ at a distance $r$ from the center of the Earth (of mass $M_E$) is given by $U = -\frac{G M_E m}{r}$.
To remove the body to infinity,where the potential energy is zero,we must provide an amount of energy equal to the magnitude of the potential energy.
Therefore,the binding energy $BE = -U = -\left(-\frac{G M_E m}{r}\right) = \frac{G M_E m}{r}$.

(N/A) The gravitational potential energy $U$ of a satellite of mass $m$ orbiting a planet of mass $M$ at a distance $r$ from the center of the planet is given by the equation:
$U = -\frac{GMm}{r}$
where $G$ is the universal gravitational constant.
The potential energy is negative because the gravitational force is attractive in nature. By convention,the potential energy of a system is defined as zero at an infinite distance $(r = \infty)$ from the source mass. Since the gravitational force acts to pull the satellite towards the planet,work must be done by an external agent to move the satellite from its position $r$ to infinity. Because the system is in a bound state,the energy required to reach the zero-potential state at infinity is positive,implying that the energy at any finite distance $r$ must be less than zero.

$(A)$ The gravitational potential energy $(U)$ of a system of two masses $m_1$ and $m_2$ separated by a distance $r$ is given by $U = -G \frac{m_1 m_2}{r}$. Since $G$, $m_1$, $m_2$, and $r$ are all positive, $U$ is always negative. Thus, $(1)$ matches with $(b)$.
The potential energy of a system is negative if the force between the constituents is attractive. Since the gravitational force between galaxies is always attractive, their gravitational potential energy is negative. Thus, $(2)$ matches with $(c)$.
Therefore, the correct matching is $(1-b, 2-c)$.

(A) When an object is raised above the surface of the Earth,work must be done against the gravitational force exerted by the Earth.
Since the displacement is in the direction opposite to the gravitational force,the work done by the gravitational force is negative.
According to the work-energy theorem,this work done against the conservative gravitational force is stored in the system as gravitational potential energy.
Therefore,the potential energy of the object increases as its height above the Earth's surface increases.

(A) The gravitational potential at a point in a gravitational field is defined as the work done per unit mass in bringing a test mass from infinity to that point.
Mathematically,$\phi = -GM/R$,where $G$ is the gravitational constant,$M$ is the mass of the Earth,and $R$ is the radius of the Earth.
Since the gravitational potential $\phi$ depends only on the mass of the source $(M)$ and the distance from the center $(R)$,it is independent of the mass of the object $(m)$ placed at that point.
Therefore,for an object of mass $2m$,the gravitational potential remains $\phi$.

(A) The gravitational potential energy $U$ of a body of mass $m$ at a distance $r$ from the center of the Earth is given by $U = \frac{G M_e m}{r}$.
Here,$G$ is the universal gravitational constant,$M_e$ is the mass of the Earth,and $m$ is the mass of the body.
The weight of the body at distance $r$ is $W = mg$,where $g$ is the acceleration due to gravity at that distance.
The acceleration due to gravity at distance $r$ is $g = \frac{G M_e}{r^2}$.
Substituting $g$ into the weight formula,we get $W = m \left( \frac{G M_e}{r^2} \right) = \frac{G M_e m}{r^2}$.
We can rewrite the expression for $U$ as $U = \left( \frac{G M_e m}{r} \right)$.
Comparing the two,we see that $W = \frac{U}{r}$.
Therefore,the weight of the body at this point is $U/r$.

(B) The gravitational potential energy of an object at a distance $r$ from a mass $M$ is given by $U = -\frac{GMm}{r}$.
As the object is brought from infinity (where $U = 0$) to a distance $r$,the potential energy changes from $0$ to a negative value.
Since the value becomes negative,the gravitational potential energy decreases.

(A) The relationship between gravitational field $E_G$ and gravitational potential $V$ is given by $V(x) = -\int_{\infty}^{x} E_G \cdot dx$.
Given $E_G = \frac{Ax}{(x^2+a^2)^{3/2}}$.
Substituting this into the integral:
$V(x) = -\int_{\infty}^{x} \frac{Ax}{(x^2+a^2)^{3/2}} dx$.
Let $u = x^2 + a^2$,then $du = 2x dx$,or $x dx = \frac{du}{2}$.
As $x \to \infty$,$u \to \infty$. As $x = x$,$u = x^2 + a^2$.
$V(x) = -\int_{\infty}^{x^2+a^2} \frac{A}{u^{3/2}} \cdot \frac{du}{2} = -\frac{A}{2} \int_{\infty}^{x^2+a^2} u^{-3/2} du$.
$V(x) = -\frac{A}{2} \left[ \frac{u^{-1/2}}{-1/2} \right]_{\infty}^{x^2+a^2} = A \left[ \frac{1}{\sqrt{u}} \right]_{\infty}^{x^2+a^2}$.
$V(x) = A \left( \frac{1}{\sqrt{x^2+a^2}} - 0 \right) = \frac{A}{(x^2+a^2)^{1/2}}$.

(A) The gravitational self-energy of a solid sphere of mass $M$ and radius $R$ is given by $U_i = -\frac{3}{5} \frac{GM^2}{R}$.
To break the earth up completely and move all its mass to infinity,the final potential energy $U_f$ will be $0$.
The energy required to be supplied is $\Delta U = U_f - U_i$.
$\Delta U = 0 - (-\frac{3}{5} \frac{GM^2}{R}) = \frac{3}{5} \frac{GM^2}{R}$.
Comparing this with the given expression $\frac{x}{5} \frac{GM^2}{R}$,we get $x = 3$.

(A) The gravitational potential energy $U$ of a system of particles is given by $U = -\sum \frac{G m_i m_j}{r_{ij}}$.
For the given square configuration with side $d$,there are four sides of length $d$ and two diagonals of length $\sqrt{2}d$.
The masses at the corners are $m_1 = m$,$m_2 = M-m$,$m_3 = m$,and $m_4 = M-m$.
The potential energy is:
$U = -\frac{G}{d} [m(M-m) + (M-m)m + m(M-m) + (M-m)m] - \frac{G}{\sqrt{2}d} [m^2 + (M-m)^2]$
$U = -\frac{G}{d} [4m(M-m)] - \frac{G}{\sqrt{2}d} [m^2 + M^2 - 2Mm + m^2]$
$U = -\frac{G}{d} [4Mm - 4m^2 + \frac{1}{\sqrt{2}}(M^2 - 2Mm + 2m^2)]$
To maximize $U$ (or minimize $|U|$),we differentiate $U$ with respect to $m$ and set it to zero:
$\frac{dU}{dm} = -\frac{G}{d} [4M - 8m + \frac{1}{\sqrt{2}}(-2M + 4m)] = 0$
$4M - 8m - \sqrt{2}M + 2\sqrt{2}m = 0$
$M(4 - \sqrt{2}) = m(8 - 2\sqrt{2})$
$M(4 - \sqrt{2}) = 2m(4 - \sqrt{2})$
$\frac{M}{m} = 2$
Thus,$x = 2$.

(D) The gravitational potential $V$ at a point inside a spherical shell is the sum of the potential due to the point mass at the centre and the potential due to the shell.
$1$. Potential due to the point mass $M_1 = 50 \, \text{kg}$ at a distance $r = 25 \, \text{m}$ is $V_1 = -\frac{GM_1}{r} = -\frac{G \times 50}{25} = -2G$.
$2$. Potential due to the spherical shell of mass $M_2 = 100 \, \text{kg}$ and radius $R = 50 \, \text{m}$ at any point inside it is constant and equal to the potential at its surface: $V_2 = -\frac{GM_2}{R} = -\frac{G \times 100}{50} = -2G$.
$3$. The total gravitational potential $V$ is $V = V_1 + V_2 = -2G + (-2G) = -4G$.

(A) The total gravitational potential energy $U$ of the system is the sum of the potential energies of all pairs of spheres.
$1$. Potential energy of the four spheres of mass $m$ at the corners of the square:
There are $4$ sides of length $d$ and $2$ diagonals of length $\sqrt{2}d$.
$U_{m-m} = -\frac{G m^2}{d} \times 4 - \frac{G m^2}{\sqrt{2}d} \times 2 = -\frac{G m^2}{d} (4 + \sqrt{2})$.
$2$. Potential energy of the central sphere of mass $M$ with the four spheres of mass $m$:
The distance of each corner from the centre is $r = \frac{\sqrt{2}d}{2} = \frac{d}{\sqrt{2}}$.
$U_{M-m} = -\frac{G M m}{r} \times 4 = -\frac{G M m}{d/\sqrt{2}} \times 4 = -\frac{4\sqrt{2} G M m}{d}$.
$3$. Total potential energy $U = U_{m-m} + U_{M-m} = -\frac{G m}{d} [(4 + \sqrt{2})m + 4\sqrt{2}M]$.

(D) The work done by the gravitational force $(W_g)$ is equal to the negative change in the gravitational potential energy of the particle $D$ due to the other particles $(A, B, C)$.
$W_g = -\Delta U = -(U_{\text{final}} - U_{\text{initial}}) = U_{\text{initial}} - U_{\text{final}}$.
When the particle $D$ is at infinity,the potential energy $U_{\text{final}} = 0$.
Therefore,$W_g = U_{\text{initial}}$.
The initial gravitational potential energy of particle $D$ due to particles $A, B,$ and $C$ is:
$U_{\text{initial}} = -\frac{G m_A m_D}{r_{AD}} - \frac{G m_B m_D}{r_{BD}} - \frac{G m_C m_D}{r_{CD}}$
Given $m_A = m_B = m_C = m_D = m$,$r_{AD} = L$,$r_{CD} = L$,and $r_{BD} = \sqrt{2} L$ (diagonal of the square).
$U_{\text{initial}} = -\frac{G m^2}{L} - \frac{G m^2}{\sqrt{2} L} - \frac{G m^2}{L} = -\frac{G m^2}{L} \left( 2 + \frac{1}{\sqrt{2}} \right)$.
$U_{\text{initial}} = -\frac{G m^2}{L} \left( \frac{2 \sqrt{2} + 1}{\sqrt{2}} \right)$.
Thus,the work done by the gravitational force is $-\frac{G m^2}{L} \left( \frac{2 \sqrt{2} + 1}{\sqrt{2}} \right)$.

(B) The gravitational potential on the surface of the earth is given by $V_0 = -\frac{G M_e}{R_e}$.
The gravitational potential at a height $h$ above the surface is given by $V_h = -\frac{G M_e}{R_e + h}$.
Given that the height $h = \frac{R_e}{2}$,we substitute this into the formula:
$V_h = -\frac{G M_e}{R_e + \frac{R_e}{2}}$
Simplifying the denominator:
$V_h = -\frac{G M_e}{\frac{3 R_e}{2}}$
$V_h = -\frac{2}{3} \left( \frac{G M_e}{R_e} \right)$
Since $V_0 = -\frac{G M_e}{R_e}$,we can write:
$V_h = \frac{2}{3} V_0$.

(C) The gravitational potential energy of a system of particles is defined as the work done by an external agent in bringing the particles from infinity to their respective positions.
For a system of particles bound together by gravitational forces,such as a galaxy,the particles are in a state of attraction.
When the potential energy at infinity is taken to be zero,the potential energy of any bound system is always negative.
This is because the gravitational force is attractive,and the system is in a lower energy state compared to the state where the particles are at infinity.
Therefore,the gravitational potential energy of a galaxy is negative.

(B) Since particle $C$ is moved without any acceleration,the change in kinetic energy is zero,$\Delta K.E. = 0$.
By the work-energy theorem,the work done by an external agent plus the work done by gravitational forces is zero: $W_{\text{ext}} + W_{\text{grav}} = 0$.
Therefore,$W_{\text{ext}} = -W_{\text{grav}} = \Delta U = U_f - U_i$.
The gravitational potential energy of the system involving particle $C$ is $U = -\frac{GMm}{r_A} - \frac{GMm}{r_B}$,where $r_A$ and $r_B$ are the distances of $C$ from $A$ and $B$ respectively.
Initial position: $C$ is at the midpoint,so $r_A = r/2$ and $r_B = r/2$.
$U_i = -\frac{GMm}{r/2} - \frac{GMm}{r/2} = -\frac{2GMm}{r} - \frac{2GMm}{r} = -\frac{4GMm}{r}$.
Final position: $C$ is at a point equidistant $r$ from $A$ and $B$,so $r_A = r$ and $r_B = r$.
$U_f = -\frac{GMm}{r} - \frac{GMm}{r} = -\frac{2GMm}{r}$.
Work done $W_{\text{ext}} = U_f - U_i = -\frac{2GMm}{r} - (-\frac{4GMm}{r}) = \frac{2GMm}{r}$.

(A) Let the point $O$ be at a distance $d$ from mass $M$ and at a distance $(r-d)$ from mass $4M$.
At point $O$,the gravitational field intensity is zero,so the magnitudes of the gravitational fields due to both masses must be equal:
$\frac{G M}{d^2} = \frac{G (4 M)}{(r-d)^2}$
$\Rightarrow \frac{(r-d)^2}{d^2} = 4$
Taking the square root on both sides:
$\frac{r-d}{d} = 2$ (since $d$ must be between the masses,we take the positive root)
$r-d = 2d \Rightarrow 3d = r \Rightarrow d = \frac{r}{3}$
Thus,the distance from mass $4M$ is $r - d = r - \frac{r}{3} = \frac{2r}{3}$.
Now,calculate the gravitational potential $V$ at point $O$:
$V = V_1 + V_2 = -\frac{G M}{d} - \frac{G (4 M)}{(r-d)}$
$V = -\frac{G M}{r/3} - \frac{4 G M}{2r/3}$
$V = -\frac{3 G M}{r} - \frac{6 G M}{r} = -\frac{9 G M}{r}$

(B) The gravitational potential difference between any two points in a gravitational field is independent of the choice of reference point.
When the potential at infinity is assigned a zero value:
Potential at the surface,$V_s = -\frac{GM}{R}$
Potential at the center,$V_c = -\frac{3GM}{2R}$
The potential difference between the surface and the center is:
$V_s - V_c = -\frac{GM}{R} - (-\frac{3GM}{2R}) = -\frac{GM}{R} + \frac{3GM}{2R} = \frac{GM}{2R}$
Now,if the potential at the surface is assigned a zero value $(V_s' = 0)$,let the new potential at the center be $V_c'$.
Since the potential difference remains constant:
$V_s - V_c = V_s' - V_c'$
$\frac{GM}{2R} = 0 - V_c'$
$V_c' = -\frac{GM}{2R}$
Therefore,the potential at the center is $-\frac{GM}{2R}$.

(C) The gravitational potential energy $U$ of an object of mass $m$ at a distance $r$ from the center of the Earth is given by $U = -\frac{GMm}{r}$.
At the surface of the Earth,the distance from the center is $r_1 = R$. Therefore,the potential energy at the surface is $U_i = -\frac{GMm}{R}$.
At a height of $2R$ above the surface,the distance from the center is $r_2 = R + 2R = 3R$. Therefore,the potential energy at this height is $U_f = -\frac{GMm}{3R}$.
The increase in potential energy is $\Delta U = U_f - U_i = -\frac{GMm}{3R} - (-\frac{GMm}{R}) = \frac{GMm}{R} - \frac{GMm}{3R}$.
Taking $\frac{GMm}{R}$ as a common factor,we get $\Delta U = \frac{GMm}{R} (1 - \frac{1}{3}) = \frac{GMm}{R} (\frac{2}{3}) = \frac{2}{3} \frac{GMm}{R}$.
Since the acceleration due to gravity at the surface is $g = \frac{GM}{R^2}$,we can write $GM = gR^2$.
Substituting this into the expression for $\Delta U$,we get $\Delta U = \frac{2}{3} \frac{(gR^2)m}{R} = \frac{2}{3} mgR$.

(C) The gravitational potential energy of a body of mass $m$ at a distance $r$ from the center of the earth is given by $U = -\frac{G M m}{r}$.
At the earth's surface,the distance from the center is $r_1 = R$. Therefore,the potential energy at the surface is $U_i = -\frac{G M m}{R}$.
When the body is raised to a height $h = n R$ from the surface,the distance from the center of the earth becomes $r_2 = R + n R = (n+1) R$. Therefore,the potential energy at this height is $U_f = -\frac{G M m}{(n+1) R}$.
The change in potential energy is $\Delta U = U_f - U_i = -\frac{G M m}{(n+1) R} - \left( -\frac{G M m}{R} \right)$.
$\Delta U = \frac{G M m}{R} - \frac{G M m}{(n+1) R} = \frac{G M m}{R} \left( 1 - \frac{1}{n+1} \right)$.
$\Delta U = \frac{G M m}{R} \left( \frac{n+1-1}{n+1} \right) = \frac{G M m}{R} \left( \frac{n}{n+1} \right)$.
Since the acceleration due to gravity at the surface is $g = \frac{G M}{R^2}$,we can write $G M = g R^2$.
Substituting this into the expression for $\Delta U$:
$\Delta U = \frac{(g R^2) m}{R} \left( \frac{n}{n+1} \right) = m g R \left( \frac{n}{n+1} \right)$.

(C) The gravitational potential energy difference between any two points in a gravitational field is independent of the choice of reference point.
When the potential energy at infinity is taken as zero,the potential energy of a body of mass $m$ at a distance $r$ from the center of the Earth is given by $U = -\frac{G M m}{r}$.
Potential energy at the surface of the Earth $(r = R)$: $U_s = -\frac{G M m}{R}$.
Potential energy at height $h$ above the surface $(r = R+h)$: $U_h = -\frac{G M m}{R+h}$.
The difference in potential energy between the surface and height $h$ is:
$\Delta U = U_h - U_s = -\frac{G M m}{R+h} - \left(-\frac{G M m}{R}\right) = G M m \left( \frac{1}{R} - \frac{1}{R+h} \right)$.
Simplifying the expression:
$\Delta U = G M m \left( \frac{R+h-R}{R(R+h)} \right) = \frac{G M m h}{R(R+h)}$.
If the potential energy at the surface is taken as zero $(U_s' = 0)$,then the new potential energy at height $h$ $(U_h')$ is defined by the same difference:
$U_h' - U_s' = \Delta U$
$U_h' - 0 = \frac{G M m h}{R(R+h)}$.
Therefore,the potential energy at height $h$ is $\frac{G M m h}{R(R+h)}$.

(C) The gravitational potential energy of a body of mass $m$ at a distance $r$ from the center of the Earth is given by $U = -\frac{GM_e m}{r}$.
Initial potential energy at the surface $(r = R_e)$ is $U_i = -\frac{GM_e m}{R_e}$.
Final potential energy at height $h = 2R_e$ $(r = R_e + h = 3R_e)$ is $U_f = -\frac{GM_e m}{3R_e}$.
The increase in potential energy is $\Delta U = U_f - U_i$.
$\Delta U = -\frac{GM_e m}{3R_e} - (-\frac{GM_e m}{R_e}) = \frac{GM_e m}{R_e} (1 - \frac{1}{3}) = \frac{2}{3} \frac{GM_e m}{R_e}$.
Since $g = \frac{GM_e}{R_e^2}$,we have $GM_e = gR_e^2$.
Substituting this into the expression: $\Delta U = \frac{2}{3} \frac{(gR_e^2)m}{R_e} = \frac{2}{3} mgR_e$.

(B) The relationship between gravitational field $E$ and potential $V$ is given by $E = -\frac{dV}{dr}$.
Given $E = -\frac{K}{r^2}$,we have $-\frac{dV}{dr} = -\frac{K}{r^2}$,which implies $dV = \frac{K}{r^2} dr$.
Integrating both sides from the reference point $(r_1 = 2\,cm, V_1 = 10\,J/kg)$ to the target point $(r_2 = 3\,cm, V_2 = V)$:
$\int_{10}^{V} dV = \int_{2}^{3} \frac{K}{r^2} dr$.
$V - 10 = K \left[ -\frac{1}{r} \right]_{2}^{3} = K \left( -\frac{1}{3} - (-\frac{1}{2}) \right) = K \left( \frac{1}{2} - \frac{1}{3} \right) = K \left( \frac{1}{6} \right)$.
Substituting $K = 6\,J\cdot cm/kg$:
$V - 10 = 6 \times \frac{1}{6} = 1$.
$V = 10 + 1 = 11\,J/kg$.

(A) The gravitational potential $V$ inside a uniform solid sphere at a distance $r$ from the center is given by $V(r) = \frac{GM}{2R^3}(3R^2 - r^2)$.
At the surface,$r = R$,so $V_{surface} = \frac{GM}{2R^3}(3R^2 - R^2) = \frac{GM}{2R^3}(2R^2) = \frac{GM}{R} = V$.
At the center,$r = 0$,so $V_{center} = \frac{GM}{2R^3}(3R^2 - 0) = \frac{3GM}{2R}$.
Substituting $V = \frac{GM}{R}$,we get $V_{center} = \frac{3}{2} V$.

(D) Let the point $P$ be the neutral point where the gravitational field is zero. Let $r_1$ be the distance from mass $m$ and $r_2$ be the distance from mass $9m$.
The condition for zero gravitational field is:
$\frac{G m}{r_1^2} = \frac{G (9m)}{r_2^2}$
$\frac{1}{r_1^2} = \frac{9}{r_2^2} \implies \frac{r_2}{r_1} = 3 \implies r_2 = 3 r_1$
Since $r_1 + r_2 = R$,we have $r_1 + 3 r_1 = R \implies 4 r_1 = R \implies r_1 = \frac{R}{4}$ and $r_2 = \frac{3R}{4}$.
The gravitational potential $V$ at point $P$ is the sum of potentials due to both masses:
$V = -\frac{G m}{r_1} - \frac{G (9m)}{r_2}$
$V = -\frac{G m}{R/4} - \frac{9 G m}{3R/4}$
$V = -\frac{4 G m}{R} - \frac{36 G m}{3R} = -\frac{4 G m}{R} - \frac{12 G m}{R}$
$V = -\frac{16 G m}{R}$

(A) The surface mass density $\sigma$ of the annular disc is $\sigma = \frac{M}{\pi(4R)^2 - \pi(3R)^2} = \frac{M}{7\pi R^2}$.
Consider a thin ring element of radius $r$ and width $dr$. The mass of this element is $dm = \sigma (2\pi r dr)$.
The gravitational potential $V_P$ at point $P$ (at distance $h=4R$ from the center) due to this ring is $dV_P = -\frac{G dm}{\sqrt{r^2 + h^2}} = -\frac{G \sigma 2\pi r dr}{\sqrt{r^2 + (4R)^2}}$.
Integrating from $r=3R$ to $r=4R$:
$V_P = -\int_{3R}^{4R} \frac{G \sigma 2\pi r dr}{\sqrt{r^2 + 16R^2}} = -2\pi G \sigma [\sqrt{r^2 + 16R^2}]_{3R}^{4R}$.
$V_P = -2\pi G \left(\frac{M}{7\pi R^2}\right) [\sqrt{16R^2 + 16R^2} - \sqrt{9R^2 + 16R^2}] = -\frac{2GM}{7R^2} [4R\sqrt{2} - 5R] = -\frac{2GM}{7R}(4\sqrt{2}-5)$.
The work required to move a unit mass from $P$ to infinity is $W = U_{\infty} - U_P = 1 \cdot V_{\infty} - 1 \cdot V_P = 0 - V_P = -V_P$.
Therefore,$W = \frac{2GM}{7R}(4\sqrt{2}-5)$.

(C) The gravitational potential $V$ at a point due to a mass element $dm$ at a distance $r$ is given by $dV = -\frac{G dm}{r}$.
For an arc of mass $M$ and radius $R$,every mass element $dm$ of the arc is at the same distance $R$ from the center $O$.
Therefore,the gravitational potential at the center $O$ is the integral of the potential due to all mass elements: $V_O = \int dV = \int -\frac{G dm}{R} = -\frac{G}{R} \int dm = -\frac{GM}{R}$.
Since the potential depends only on the total mass $M$ and the distance $R$,and both are identical for all three arcs,the gravitational potential at the center is the same for all three cases.
Thus,$(V_O)_1 = (V_O)_2 = (V_O)_3$.