Gravitational Potential and Potential Energy of system Questions in English

(D) According to the work-energy theorem,the work done by all forces is equal to the change in kinetic energy: $W_{ext} + W_{grav} = \Delta K.$
Here,$W_{ext} = -5.5 \ J$ is the work done by the person.
The work done by the gravitational force is $W_{grav} = -\Delta U = -(U_A - U_{\infty}) = -(m V_A - 0) = -m V_A.$
The change in kinetic energy is $\Delta K = K_f - K_i = \frac{1}{2} m v^2 - 0 = \frac{1}{2} \times 1 \times (3)^2 = 4.5 \ J.$
Substituting these into the work-energy theorem: $-5.5 + (-1 \times V_A) = 4.5.$
$-V_A = 4.5 + 5.5 = 10 \ J/kg.$
Therefore,$V_A = -10 \ J/kg.$

(C) The gravitational potential energy $U$ of a body of mass $m$ at a distance $r$ from the centre of the earth is given by $U = -\frac{GMm}{r}$.
The magnitude of this energy at $r = R$ is $E = \frac{GMm}{R}$,which implies $GMm = ER$.
Weight $W$ of the body at a distance $r$ is the gravitational force $F = \frac{GMm}{r^2}$.
At $r = 1.5 R = \frac{3}{2} R$,the weight is $W = \frac{GMm}{(1.5 R)^2} = \frac{GMm}{2.25 R^2} = \frac{GMm}{\frac{9}{4} R^2} = \frac{4 GMm}{9 R^2}$.
Substituting $GMm = ER$ into the expression for $W$:
$W = \frac{4 (ER)}{9 R^2} = \frac{4 E}{9 R}$.

(C) The gravitational potential energy of a body of mass $m$ at the surface of the Earth is $U_i = -\frac{GMm}{R}$.
At a height $h$ above the surface,the potential energy is $U_f = -\frac{GMm}{R+h}$.
The increase in potential energy is $\Delta U = U_f - U_i = GMm \left( \frac{1}{R} - \frac{1}{R+h} \right) = GMm \left( \frac{h}{R(R+h)} \right)$.
Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$.
Substituting this,$\Delta U = mgR^2 \left( \frac{h}{R(R+h)} \right) = mgR \left( \frac{h}{R+h} \right)$.
Given $\Delta U = \frac{mgR}{5}$,we equate: $\frac{mgR}{5} = mgR \left( \frac{h}{R+h} \right)$.
$\frac{1}{5} = \frac{h}{R+h} \implies R+h = 5h \implies 4h = R \implies h = \frac{R}{4}$.

(B) The gravitational potential energy $U$ of the body of mass $M$ at a distance $r/3$ from the Earth's centre is given by the sum of potentials due to Earth and Moon:
$U = -\frac{G M_{1} M}{r/3} - \frac{G M_{2} M}{2r/3} = -\frac{3 G M_{1} M}{r} - \frac{3 G M_{2} M}{2r} = -\frac{3 G M}{r} \left( M_{1} + \frac{M_{2}}{2} \right)$.
To escape to infinity,the total energy must be at least zero. Thus,the kinetic energy $K$ required is equal to the magnitude of the potential energy:
$K = \frac{1}{2} M V^{2} = |U| = \frac{3 G M}{r} \left( M_{1} + \frac{M_{2}}{2} \right)$.
Solving for $V$:
$V^{2} = \frac{6 G}{r} \left( M_{1} + \frac{M_{2}}{2} \right)$,
$V = \left[ \frac{6 G}{r} \left( M_{1} + \frac{M_{2}}{2} \right) \right]^{\frac{1}{2}}$.
Therefore,the correct option is $B$.

(C) The gravitational field is a conservative force field.
In a conservative force field,the work done in moving a particle from one point to another is independent of the path taken.
It only depends on the initial and final positions of the particle.
Since all three paths $(1, 2, 3)$ start at point $A$ and end at point $B$,the work done along each path must be equal.
Therefore,$W_1 = W_2 = W_3$.

(C) For the gravitational field to be zero,the intensities produced by both particles must be equal in magnitude and opposite in direction.
$\frac{G m}{r_1^2} = \frac{G(9 m)}{r_2^2}$
where $r_1$ is the distance from mass $m$ and $r_2$ is the distance from mass $9m$.
Since $r_1 + r_2 = r$,we have $\frac{r_2}{r_1} = \sqrt{9} = 3 \Rightarrow r_2 = 3 r_1$.
Therefore,$r_1 + 3 r_1 = r \Rightarrow 4 r_1 = r \Rightarrow r_1 = \frac{r}{4}$ and $r_2 = \frac{3r}{4}$.
The gravitational potential $V$ is given by $V = -\frac{G m}{r_1} - \frac{G(9 m)}{r_2}$.
Substituting the values: $V = -\frac{G m}{r/4} - \frac{9 G m}{3r/4} = -\frac{4 G m}{r} - \frac{36 G m}{3r} = -\frac{4 G m}{r} - \frac{12 G m}{r} = -\frac{16 G m}{r}$.

(C) The gravitational potential energy of an object of mass $m$ at the surface of the earth is given by $U_{1} = -\frac{GMm}{R}$.
At a height $h = R$ from the surface,the distance from the center of the earth is $r = R + h = R + R = 2R$.
The potential energy at this height is $U_{2} = -\frac{GMm}{2R}$.
The gain in potential energy is $\Delta U = U_{2} - U_{1} = -\frac{GMm}{2R} - (-\frac{GMm}{R}) = \frac{GMm}{R} - \frac{GMm}{2R} = \frac{GMm}{2R}$.
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 for $\Delta U$:
$\Delta U = \frac{1}{2} \frac{(gR^{2})m}{R} = \frac{1}{2} mgR$.

(D) The gravitational potential $V$ at a point due to a mass $m$ at distance $r$ is given by $V = -\frac{Gm}{r}$.
Here,$m = 1 \ kg$. The objects are placed at $x = \pm 1, \pm 2, \pm 4, \pm 8, \ldots \ m$.
Since there are two objects at each distance $r$ (one at $+r$ and one at $-r$),the total potential $V_{total}$ at $x=0$ is:
$V_{total} = \sum -\frac{Gm}{r_i} = -G(1) \left[ \frac{1}{1} + \frac{1}{1} + \frac{1}{2} + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \ldots \right]$
$V_{total} = -G \left[ 2 \left( 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots \right) \right]$
The term in the bracket is a geometric progression with first term $a=1$ and common ratio $r=1/2$. The sum is $S = \frac{a}{1-r} = \frac{1}{1 - 1/2} = 2$.
Therefore,$V_{total} = -G \times 2 \times 2 = -4G$.
The magnitude of the potential is $|V_{total}| = 4G$.

(A) The gravitational potential energy $U$ of a satellite of mass $m$ at a distance $r$ from the center of the earth is given by $U = -\frac{GMm}{r}$.
Here,the satellite is at a height $h = R_e$ from the surface of the earth.
Therefore,the distance from the center of the earth is $r = R_e + h = R_e + R_e = 2R_e$.
Substituting $r = 2R_e$ into the formula,we get $U = -\frac{GMm}{2R_e}$.
We know that the acceleration due to gravity at the surface of the earth is $g = \frac{GM}{R_e^2}$,which implies $GM = gR_e^2$.
Substituting $GM = gR_e^2$ into the expression for $U$,we get $U = -\frac{(gR_e^2)m}{2R_e} = -0.5 mgR_e$.
Thus,the correct option is $A$.

(A) The gravitational potential energy of a body of mass $m$ at the surface of the earth (radius $R$) is given by $E = -\frac{GMm}{R}$.
When the body is taken to a height $h = 150\% \text{ of } R$,we have $h = 1.5R$.
The distance of the body from the center of the earth at this height is $R' = R + h = R + 1.5R = 2.5R$.
The new gravitational potential energy $E'$ is given by $E' = -\frac{GMm}{R'} = -\frac{GMm}{2.5R}$.
Substituting $E = -\frac{GMm}{R}$ into the expression for $E'$,we get $E' = \frac{E}{2.5} = 0.4E$.

(A) The gravitational potential at a point outside a spherical body is given by $V = -\frac{GM}{r}$.
Since the point at distance $r = 2.5 a$ is outside both the solid sphere and the spherical shell,both act as point masses at the center.
The total mass of the system is $M_{total} = M + 0.5 M = 1.5 M$.
The gravitational potential energy $U$ of a unit mass $(m = 1)$ at distance $r = 2.5 a$ is given by $U = V = -\frac{G M_{total}}{r}$.
Substituting the values: $U = -\frac{G(1.5 M)}{2.5 a} = -\frac{1.5}{2.5} \frac{GM}{a} = -\frac{3}{5} \frac{GM}{a} = -\frac{3 GM}{5 a}$.

(C) The gravitational potential energy $U$ of a system of point masses is the sum of the potential energies of all pairs of masses.
For three masses $m_1, m_2, m_3$ at distances $r_{12}, r_{23}, r_{31}$,the total potential energy is $U = -G \left( \frac{m_1 m_2}{r_{12}} + \frac{m_2 m_3}{r_{23}} + \frac{m_3 m_1}{r_{31}} \right)$.
Here,$m_1 = m$,$m_2 = 2m$,$m_3 = 3m$,and $r_{12} = r_{23} = r_{31} = a$.
Substituting these values:
$U = -\frac{G}{a} [ (m)(2m) + (2m)(3m) + (3m)(m) ]$
$U = -\frac{G}{a} [ 2m^2 + 6m^2 + 3m^2 ]$
$U = -\frac{G}{a} [ 11m^2 ]$
$U = -11 \frac{Gm^2}{a}$.

(C) The work done by an external agent against a conservative force is equal to the change in potential energy,$W = U_f - U_i$.
The gravitational potential $V$ at a distance $r$ from a sphere of mass $M$ is given by $V = -\frac{GM}{r}$.
The initial potential at the surface $(r = 1 \ m)$ is $V_i = -\frac{G \times 1000}{1} = -6.67 \times 10^{-11} \times 1000 = -6.67 \times 10^{-8} \ J \ kg^{-1}$.
To take the particle to infinity,the final potential is $V_f = -\frac{GM}{\infty} = 0$.
The work done per unit mass is $W = V_f - V_i = 0 - (-6.67 \times 10^{-8} \ J \ kg^{-1}) = 6.67 \times 10^{-8} \ J \ kg^{-1}$.

(A) The gravitational potential energy is the energy possessed by a mass due to its position in a gravitational field.
The expression for gravitational potential energy $U$ of a mass $m$ at a distance $r$ from a mass $M$ is given by $U = -\frac{G M m}{r}$.
Since $U$ is inversely proportional to the distance $r$ with a negative sign,as $r$ increases,the value of $U$ increases.
At $r = \infty$,$U = -\frac{G M m}{\infty} = 0$.
Since the potential energy is negative everywhere else,$0$ is the maximum possible value for gravitational potential energy.

(B) Given,gravitational field $E = (5 \hat{i} + 12 \hat{j}) \text{ N kg}^{-1}$,mass $m = 2 \text{ kg}$,and displacement vector $r = (12 \hat{i} + 15 \hat{j}) \text{ m}$.
The work done by the gravitational field is $W = \int F \cdot dr = \int (mE) \cdot dr$.
The change in gravitational potential energy $\Delta U$ is given by $\Delta U = -W = -m \int E \cdot dr$.
Substituting the values:
$\Delta U = -2 \int_{(0,0)}^{(12,15)} (5 \hat{i} + 12 \hat{j}) \cdot (dx \hat{i} + dy \hat{j})$
$\Delta U = -2 \left[ \int_0^{12} 5 dx + \int_0^{15} 12 dy \right]$
$\Delta U = -2 [5(12 - 0) + 12(15 - 0)]$
$\Delta U = -2 [60 + 180]$
$\Delta U = -2 [240] = -480 \text{ J}$.
Thus,the change in gravitational potential energy is $-480 \text{ J}$.
Therefore,the correct option is $B$.

(D) The work done by an external agent is equal to the change in the gravitational potential energy of the system: $W = U_f - U_i$.
For configuration $1$ (right-angled triangle with sides $a, a, \sqrt{2}a$):
$U_i = -\frac{G(m)(2m)}{a} - \frac{G(m)(3m)}{a} - \frac{G(2m)(3m)}{\sqrt{2}a} = -\frac{G m^2}{a} \left( 2 + 3 + \frac{6}{\sqrt{2}} \right) = -\frac{G m^2}{a} \left( 5 + \frac{6}{\sqrt{2}} \right)$.
For configuration $2$ (equilateral triangle with all sides $a$):
$U_f = -\frac{G(m)(3m)}{a} - \frac{G(m)(2m)}{a} - \frac{G(2m)(3m)}{a} = -\frac{G m^2}{a} (3 + 2 + 6) = -\frac{11 G m^2}{a}$.
Work done $W = U_f - U_i = -\frac{11 G m^2}{a} - \left( -\frac{G m^2}{a} \left( 5 + \frac{6}{\sqrt{2}} \right) \right) = \frac{G m^2}{a} \left( -11 + 5 + \frac{6}{\sqrt{2}} \right) = \frac{G m^2}{a} \left( \frac{6}{\sqrt{2}} - 6 \right) = -\frac{G m^2}{a} \left( 6 - \frac{6}{\sqrt{2}} \right)$.

(B) Let the point where the gravitational field is zero be at a distance $x$ from the mass $m$. The gravitational field due to both masses must be equal in magnitude and opposite in direction.
$E_1 = E_2 \Rightarrow \frac{Gm}{x^2} = \frac{G(9m)}{(r-x)^2}$
Taking the square root on both sides: $\frac{1}{x} = \frac{3}{r-x}$
$r-x = 3x \Rightarrow 4x = r \Rightarrow x = \frac{r}{4}$.
The distance from mass $9m$ is $r-x = r - \frac{r}{4} = \frac{3r}{4}$.
The gravitational potential $V$ at this point is the sum of the potentials due to both masses:
$V = V_1 + V_2 = -\frac{Gm}{x} - \frac{G(9m)}{r-x}$
$V = -\frac{Gm}{r/4} - \frac{9Gm}{3r/4} = -\frac{4Gm}{r} - \frac{12Gm}{r}$
$V = -\frac{16Gm}{r}$.

(B) The change in gravitational potential energy $\Delta U$ is given by the work done against the gravitational field, or $\Delta U = -W = -\int \vec{F} \cdot d\vec{r} = -m \int \vec{E} \cdot d\vec{r}$.
Given $\vec{E} = (5\hat{i} + 12\hat{j}) \text{ N/kg}$ and displacement $\vec{r} = (12\hat{i} + 5\hat{j}) \text{ m}$.
Since the field is uniform, the work done $W = m(\vec{E} \cdot \vec{r})$.
$W = 2 \text{ kg} \times [(5\hat{i} + 12\hat{j}) \cdot (12\hat{i} + 5\hat{j})] \text{ J}$.
$W = 2 \times (5 \times 12 + 12 \times 5) = 2 \times (60 + 60) = 2 \times 120 = 240 \text{ J}$.
The change in gravitational potential energy $\Delta U = -W = -240 \text{ J}$.

(A) Let the point where the gravitational field is zero be at a distance $x$ from the mass $4 \,m$. The distance from the mass $9 \,m$ will be $(r - x)$.
At this point,the magnitudes of the gravitational fields due to both masses are equal:
$\frac{G(4 \,m)}{x^2} = \frac{G(9 \,m)}{(r - x)^2}$
$\frac{4}{9} = \left(\frac{x}{r - x}\right)^2$
Taking the square root on both sides:
$\frac{2}{3} = \frac{x}{r - x}$
$2(r - x) = 3x \Rightarrow 2r - 2x = 3x \Rightarrow 5x = 2r \Rightarrow x = \frac{2r}{5}$
So,the distance from $4 \,m$ is $\frac{2r}{5}$ and from $9 \,m$ is $r - \frac{2r}{5} = \frac{3r}{5}$.
The gravitational potential $V$ at this point is the sum of the potentials due to both masses:
$V = -\frac{G(4 \,m)}{x} - \frac{G(9 \,m)}{r - x}$
$V = -\frac{G(4 \,m)}{\frac{2r}{5}} - \frac{G(9 \,m)}{\frac{3r}{5}}$
$V = -\frac{20Gm}{2r} - \frac{45Gm}{3r} = -\frac{10Gm}{r} - \frac{15Gm}{r} = -\frac{25Gm}{r}$

(B) The gravitational potential energy at the surface of the earth is $PE_1 = -\frac{GMm}{R}$.
At a height $h = R/5$,the potential energy is $PE_2 = -\frac{GMm}{R+h} = -\frac{GMm}{R + R/5} = -\frac{GMm}{6R/5} = -\frac{5GMm}{6R}$.
The increase in potential energy is $\Delta PE = PE_2 - PE_1 = -\frac{5GMm}{6R} - (-\frac{GMm}{R}) = \frac{GMm}{R} (1 - 5/6) = \frac{GMm}{6R}$.
Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$.
Substituting this into the expression,$\Delta PE = \frac{(gR^2)m}{6R} = \frac{mgR}{6}$.
Given $h = R/5$,then $R = 5h$.
Substituting $R = 5h$ into the expression,$\Delta PE = \frac{mg(5h)}{6} = \frac{5}{6} mgh$.

(B) The gravitational potential energy $U$ of a system of particles is given by the sum of the potential energies of all possible pairs: $U = -\sum \frac{Gm_i m_j}{r_{ij}}$.
For a rectangle with sides $a = 3 l_o$ and $b = 4 l_o$,the diagonal $d = \sqrt{(3 l_o)^2 + (4 l_o)^2} = 5 l_o$.
There are $4$ pairs with side length $3 l_o$ (two sides),$4 l_o$ (two sides),and $5 l_o$ (two diagonals).
$U = -\left[ \frac{Gm^2}{3 l_o} \times 2 + \frac{Gm^2}{4 l_o} \times 2 + \frac{Gm^2}{5 l_o} \times 2 \right]$
$U = -\frac{Gm^2}{l_o} \left[ \frac{2}{3} + \frac{2}{4} + \frac{2}{5} \right]$
$U = -\frac{Gm^2}{l_o} \left[ \frac{40 + 30 + 24}{60} \right] = -\frac{94}{60} \frac{Gm^2}{l_o} = -\frac{47}{30} \frac{Gm^2}{l_o}$.
The magnitude is $|U| = \frac{47}{30} \frac{Gm^2}{l_o}$.

(A) Let $m_1 = 100 \,kg$ and $m_2 = 8100 \,kg$ be the masses separated by distance $d = 1 \,m$.
Let $x$ be the distance of the null point (where gravitational field is zero) from $m_1$.
The condition for zero gravitational field is $\frac{G m_1}{x^2} = \frac{G m_2}{(d-x)^2}$.
Taking the square root on both sides: $\frac{\sqrt{m_1}}{x} = \frac{\sqrt{m_2}}{d-x}$.
Substituting values: $\frac{10}{x} = \frac{90}{1-x} \implies 10 - 10x = 90x \implies 100x = 10 \implies x = 0.1 \,m$.
The distance from $m_2$ is $d-x = 0.9 \,m$.
The gravitational potential $V$ at this point is $V = -\frac{G m_1}{x} - \frac{G m_2}{d-x}$.
$V = -G \left( \frac{100}{0.1} + \frac{8100}{0.9} \right) = -G (1000 + 9000) = -10000 G$.
$V = -10000 \times 6.67 \times 10^{-11} = -6.67 \times 10^{-7} \,J/kg$.

(B) The gravitational field is given by $E = (5\hat{i} + 12\hat{j}) \,N/kg$.
The displacement vector $dr$ from the origin $(0, 0)$ to the point $(12 \,m, 5 \,m)$ is $dr = (12\hat{i} + 5\hat{j}) \,m$.
The change in gravitational potential $dV$ is given by $dV = -E \cdot dr$.
$dV = -(5\hat{i} + 12\hat{j}) \cdot (12\hat{i} + 5\hat{j}) = -(5 \times 12 + 12 \times 5) = -(60 + 60) = -120 \,J/kg$.
The change in gravitational potential energy $\Delta U$ is given by $\Delta U = m \cdot dV$.
Given mass $m = 2 \,kg$, we have $\Delta U = 2 \,kg \times (-120 \,J/kg) = -240 \,J$.

(C) The work done by an external agent is equal to the change in the gravitational potential energy of the system: $W_{ext} = \Delta U = U_f - U_i$.
The gravitational potential energy of a system of three masses is given by $U = -G \left( \frac{m_1 m_2}{r} + \frac{m_2 m_3}{r} + \frac{m_1 m_3}{r} \right) = -\frac{G}{r} (m_1 m_2 + m_2 m_3 + m_1 m_3)$.
Given masses: $m_1 = 200 \ kg$,$m_2 = 300 \ kg$,$m_3 = 400 \ kg$.
Sum of products: $(200 \times 300) + (300 \times 400) + (200 \times 400) = 60000 + 120000 + 80000 = 260000 = 2.6 \times 10^5 \ kg^2$.
Initial potential energy $(r_i = 20 \ m)$:
$U_i = -\frac{6.67 \times 10^{-11}}{20} \times 2.6 \times 10^5 = -6.67 \times 10^{-11} \times 0.13 \times 10^5 = -8.671 \times 10^{-7} \ J$.
Final potential energy $(r_f = 25 \ m)$:
$U_f = -\frac{6.67 \times 10^{-11}}{25} \times 2.6 \times 10^5 = -6.67 \times 10^{-11} \times 0.104 \times 10^5 = -6.9368 \times 10^{-7} \ J$.
Work done:
$W = U_f - U_i = (-6.9368 \times 10^{-7}) - (-8.671 \times 10^{-7}) = 1.7342 \times 10^{-7} \ J \approx 1.74 \times 10^{-7} \ J$.

(D) The gravitational potential energy of a body of mass $m$ at a distance $r$ from the center of the Earth is given by $U = -\frac{GMm}{r}$.
At the surface of the Earth,the distance from the center is $r_i = R_e$. Thus,the initial potential energy is $U_i = -\frac{GMm}{R_e}$.
At a height $h = 2R_e$ from the surface,the distance from the center is $r_f = R_e + 2R_e = 3R_e$. Thus,the final potential energy is $U_f = -\frac{GMm}{3R_e}$.
The increase in potential energy is $\Delta U = U_f - U_i = -\frac{GMm}{3R_e} - (-\frac{GMm}{R_e}) = GMm(\frac{1}{R_e} - \frac{1}{3R_e}) = GMm(\frac{2}{3R_e})$.
Since the acceleration due to gravity at the surface is $g = \frac{GM}{R_e^2}$,we have $GM = gR_e^2$.
Substituting this into the expression for $\Delta U$,we get $\Delta U = (gR_e^2)m(\frac{2}{3R_e}) = \frac{2}{3}mgR_e$.