Mix Examples-Gravitation Questions in English

(D) The variation of acceleration due to gravity $g$ with distance $d$ from the center of the earth is given by:
$1$. Inside the earth $(d < R)$:
$g = \frac{GM}{R^3} d$
Since $G, M,$ and $R$ are constants,$g \propto d$. This represents a straight line passing through the origin.
$2$. At the surface of the earth $(d = R)$:
$g = \frac{GM}{R^2} = g_s$ (maximum value).
$3$. Outside the earth $(d > R)$:
$g = \frac{GM}{d^2}$
Here,$g \propto \frac{1}{d^2}$. This represents a rectangular hyperbola.
Combining these,the graph shows a linear increase from the center to the surface,followed by a hyperbolic decrease as distance increases beyond the surface. This matches the graph in option $D$.

(B) Using the Work-Energy Theorem: $\Sigma W = \Delta K$.
Work done by the gravitational force $F = -G \frac{Mm}{r^{2.1}}$ as the satellite moves from $r = a$ to $r = b$ is:
$W = \int_{a}^{b} F \cdot dr = \int_{a}^{b} -\frac{GMm}{r^{2.1}} dr$.
$W = -GMm \int_{a}^{b} r^{-2.1} dr = -GMm \left[ \frac{r^{-1.1}}{-1.1} \right]_{a}^{b}$.
$W = \frac{GMm}{1.1} \left[ \frac{1}{b^{1.1}} - \frac{1}{a^{1.1}} \right]$.
According to the Work-Energy Theorem,$W = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$.
Equating the two expressions:
$\frac{GMm}{1.1} \left[ \frac{1}{b^{1.1}} - \frac{1}{a^{1.1}} \right] = \frac{1}{2}m(v^2 - u^2)$.
Dividing by $m/2$:
$v^2 - u^2 = \frac{2GM}{1.1} \left[ \frac{1}{b^{1.1}} - \frac{1}{a^{1.1}} \right]$.
$v = \sqrt{u^2 + \frac{2}{1.1}GM \left( \frac{1}{b^{1.1}} - \frac{1}{a^{1.1}} \right)}$.

(C) The potential energy $U(r)$ is given by $U(r) = -\int_{\infty}^{r} F dr = -\int_{\infty}^{r} -\frac{GMm}{r^{2.1}} dr = GMm \int_{\infty}^{r} r^{-2.1} dr$.
Evaluating the integral: $U(r) = GMm \left[ \frac{r^{-1.1}}{-1.1} \right]_{\infty}^{r} = -\frac{GMm}{1.1 r^{1.1}}$.
By the law of conservation of energy: $K_i + U_i = K_f + U_f$.
$\frac{1}{2}mu^2 - \frac{GMm}{1.1 a^{1.1}} = \frac{1}{2}mv^2 - \frac{GMm}{1.1 b^{1.1}}$.
Dividing by $m/2$: $u^2 - \frac{2GM}{1.1 a^{1.1}} = v^2 - \frac{2GM}{1.1 b^{1.1}}$.
Solving for $v$: $v^2 = u^2 + \frac{2GM}{1.1} \left( \frac{1}{b^{1.1}} - \frac{1}{a^{1.1}} \right)$.
Thus,$v = \sqrt{u^2 + \frac{2}{1.1} GM \left( \frac{1}{b^{1.1}} - \frac{1}{a^{1.1}} \right)}$.

(D) The center of mass of a hemispherical shell of radius $6R$ is at a distance of $3R$ from its center. Let the mass of the shell be $m_1 = 2M$ and the point mass be $m_2 = M$. The distance between them is $d = 3R$.
For a two-body system,the radii of the circular orbits about the center of mass are $r_1 = \frac{m_2 d}{m_1 + m_2} = \frac{M(3R)}{2M + M} = R$ and $r_2 = \frac{m_1 d}{m_1 + m_2} = \frac{2M(3R)}{2M + M} = 2R$.
The gravitational force between them is $F = \frac{G m_1 m_2}{d^2} = \frac{G(2M)(M)}{(3R)^2} = \frac{2GM^2}{9R^2}$.
For the point mass $M$,the centripetal force is provided by the gravitational force: $M \omega^2 r_2 = F$.
$M \omega^2 (2R) = \frac{2GM^2}{9R^2}$.
$\omega^2 = \frac{GM}{9R^3}$.
Since both bodies rotate with the same angular speed $\omega_1 = \omega_2 = \omega$,we have $\omega = \sqrt{\frac{GM}{9R^3}}$.
Looking at the options,there seems to be a discrepancy in the provided options based on the standard calculation. However,if we re-evaluate the force calculation assuming the distance between the center of mass of the shell and the point mass is $3R$,the result is $\omega = \sqrt{\frac{GM}{9R^3}}$. Given the options,$D$ is the closest in form,but based on the physics,the correct angular speed is $\sqrt{\frac{GM}{9R^3}}$.

(A) Let the linear mass density of the rods be $\lambda = \frac{M}{L}$.
Consider a small element of length $dx$ on the second rod at a distance $x$ from the near end of the first rod.
The mass of this element is $dm = \lambda dx = \frac{M}{L} dx$.
The gravitational force $dF$ between the first rod (of mass $M$) and this element $dm$ is given by $dF = \frac{G M dm}{x^2}$.
Substituting $dm$,we get $dF = \frac{G M (M/L) dx}{x^2} = \frac{GM^2}{L} \frac{dx}{x^2}$.
However,the force is exerted by the entire first rod on the element $dm$. The first rod acts as a point mass $M$ at its center of mass,but since the rods are extended,we must integrate.
Actually,the force between two rods is $F = \int \frac{G M dm}{x(x+L)} = \int_{L}^{2L} \frac{G M (M/L) dx}{x(x+L)} = \frac{GM^2}{L} \int_{L}^{2L} \left( \frac{1}{x} - \frac{1}{x+L} \right) dx$.
$F = \frac{GM^2}{L} [\ln x - \ln(x+L)]_{L}^{2L} = \frac{GM^2}{L} [\ln(\frac{x}{x+L})]_{L}^{2L}$.
$F = \frac{GM^2}{L} [\ln(\frac{2L}{3L}) - \ln(\frac{L}{2L})] = \frac{GM^2}{L} [\ln(\frac{2}{3}) - \ln(\frac{1}{2})] = \frac{GM^2}{L} \ln(\frac{2/3}{1/2}) = \frac{GM^2}{L} \ln(\frac{4}{3})$.
Wait,the options provided suggest a different sign. Let's re-evaluate: $\ln(4/3) = -\ln(3/4)$. Given the options,the magnitude is $\frac{GM^2}{L^2} \ln(4/3)$. If we assume the question implies the force magnitude,and checking the provided solution steps: $\ln(1/2) - \ln(2/3) = \ln(3/4)$. This is mathematically $\ln(0.75)$,which is negative. The force must be positive. The correct magnitude is $\frac{GM^2}{L^2} \ln(4/3)$.

(D) For $r \ge R$,the gravitational acceleration is given by $a_g = \frac{GM}{r^2}$.
Substituting $M = \rho \cdot \frac{4}{3} \pi R^3$,we get $a_g = \frac{G}{r^2} \cdot \rho \cdot \frac{4}{3} \pi R^3 = \frac{4}{3} \pi G \rho \frac{R^3}{r^2}$.
At the surface $(r = R)$,$a_g = \frac{4}{3} \pi G \rho R$.
Since plots $1$ and $2$ coincide for $r \ge R_2$,they represent the same mass $M_1 = M_2$. Since $R_1 < R_2$,the density $\rho = \frac{M}{\frac{4}{3} \pi R^3}$ implies $\rho_1 > \rho_2$.
Similarly,for plots $3$ and $4$,$M_3 = M_4$ and $R_3 < R_4$,so $\rho_3 > \rho_4$.
Comparing the values at the surface,the acceleration is higher for $1$ and $2$ than for $3$ and $4$,implying $M_{1,2} > M_{3,4}$.
Given the curves,the density $\rho \propto \frac{a_g}{R}$. Analyzing the surface values and radii,the descending order of density is $1, 2, 4, 3$.

(D) $1$. Conservation of angular momentum about the black hole:
$mv_0 R = mva \implies v = \frac{v_0 R}{a}$
$2$. Conservation of mechanical energy:
Initial energy = Final energy at closest approach
$\frac{1}{2}mv_0^2 + 0 = \frac{1}{2}mv^2 - \frac{GMm}{a}$
$3$. Substitute $v$ into the energy equation:
$\frac{1}{2}v_0^2 = \frac{1}{2}\left(\frac{v_0 R}{a}\right)^2 - \frac{GM}{a}$
$v_0^2 = \frac{v_0^2 R^2}{a^2} - \frac{2GM}{a}$
$1 = \frac{R^2}{a^2} - \frac{2GM}{a v_0^2}$
$\frac{R^2}{a^2} = 1 + \frac{2GM}{a v_0^2}$
$\frac{R}{a} = \left(1 + \frac{2GM}{a v_0^2}\right)^{1/2}$
$a = R\left(1 + \frac{2GM}{a v_0^2}\right)^{-1/2}$

(B) The gravitational force on a particle of mass $m$ at a distance $r$ from the center inside the core of density $\rho_c = 3\rho$ is given by:
$F = -\frac{G M_{enc} m}{r^2} = -\frac{G (\frac{4}{3} \pi r^3 \cdot 3\rho) m}{r^2}$
$F = -4\pi G \rho m r$
Since $F = ma$,the acceleration is:
$a = -4\pi G \rho r$
This is the equation of simple harmonic motion $a = -\omega^2 r$,where $\omega^2 = 4\pi G \rho$.
Thus,the angular frequency is $\omega = \sqrt{4\pi G \rho}$.
The time period of a full oscillation is $T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{4\pi G \rho}} = \sqrt{\frac{\pi}{G \rho}}$.
The particle travels from $A$ to $B$ in half the time period:
$t = \frac{T}{2} = \frac{1}{2}\sqrt{\frac{\pi}{G \rho}}$.

(A) For an elliptical orbit under a central force,the conservation of energy and angular momentum applies.
At the points of closest approach $(r_1)$ and farthest distance $(r_2)$,the velocity is perpendicular to the radius vector.
Conservation of angular momentum: $L = mv_1r_1 = mv_2r_2 \implies v_1 = \frac{L}{mr_1}$ and $v_2 = \frac{L}{mr_2}$.
Conservation of energy: $\frac{1}{2}mv_1^2 - \frac{Qq}{4\pi\varepsilon_0r_1} = \frac{1}{2}mv_2^2 - \frac{Qq}{4\pi\varepsilon_0r_2}$.
Substituting $v_1$ and $v_2$: $\frac{L^2}{2m}(\frac{1}{r_1^2} - \frac{1}{r_2^2}) = \frac{Qq}{4\pi\varepsilon_0}(\frac{1}{r_1} - \frac{1}{r_2})$.
Using $a^2 - b^2 = (a-b)(a+b)$: $\frac{L^2}{2m} \frac{(r_2-r_1)(r_2+r_1)}{r_1^2r_2^2} = \frac{Qq}{4\pi\varepsilon_0} \frac{(r_2-r_1)}{r_1r_2}$.
Solving for $L$: $L^2 = \frac{2mQq}{4\pi\varepsilon_0} \frac{r_1r_2}{r_1+r_2} = \frac{mr_1r_2Qq}{2\pi\varepsilon_0(r_1+r_2)}$.
Thus,$L = \sqrt{\frac{mr_1r_2Qq}{2\pi\varepsilon_0(r_1+r_2)}}$.

(D) The gravitational force acting on a mass $m$ at a distance $r$ from the center of the Earth is given by $F = -\frac{GMm}{R^3}r$.
This force is of the form $F = -kr$,where $k = \frac{GMm}{R^3}$.
The motion is simple harmonic with an angular frequency $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{GM}{R^3}}$.
The time period is $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{R^3}{GM}}$.
Since the time period $T$ depends only on the mass of the Earth $M$ and its radius $R$,it is independent of the oscillating mass $m$.
Therefore,if the mass is changed to $4m$,the time period remains $T$.

(A) Consider the forces acting on one of the masses $m$. Let this mass be at the bottom position.
The distance to the two adjacent masses is $d = \sqrt{r^2 + r^2} = r\sqrt{2}$.
The gravitational force from each of these two masses is $F_1 = \frac{Gm^2}{(r\sqrt{2})^2} = \frac{Gm^2}{2r^2}$.
The horizontal components of these two forces cancel out,while the vertical components add up towards the center.
The vertical component of each force is $F_1 \cos(45^\circ) = \frac{Gm^2}{2r^2} \times \frac{1}{\sqrt{2}} = \frac{Gm^2}{2\sqrt{2}r^2}$.
Total vertical force from these two masses is $2 \times \frac{Gm^2}{2\sqrt{2}r^2} = \frac{Gm^2}{\sqrt{2}r^2}$.
The distance to the mass at the top is $2r$.
The gravitational force from the mass at the top is $F_2 = \frac{Gm^2}{(2r)^2} = \frac{Gm^2}{4r^2}$.
The total net force towards the center is $F_{net} = \frac{Gm^2}{\sqrt{2}r^2} + \frac{Gm^2}{4r^2} = \frac{Gm^2}{r^2} \left( \frac{1}{\sqrt{2}} + \frac{1}{4} \right) = \frac{Gm^2}{r^2} \left( \frac{2\sqrt{2} + 1}{4} \right)$.
This net force provides the necessary centripetal force: $\frac{mv^2}{r} = F_{net}$.
$\frac{mv^2}{r} = \frac{Gm^2}{r^2} \left( \frac{1 + 2\sqrt{2}}{4} \right)$.
$v^2 = \frac{Gm}{r} \left( \frac{1 + 2\sqrt{2}}{4} \right)$.
$v = \sqrt{\frac{Gm}{r} \left( \frac{1 + 2\sqrt{2}}{4} \right)}$.

(C) The total energy of a body of mass $m$ in a circular orbit of radius $r$ around a planet of mass $M$ is given by $E = -\frac{GMm}{2r}$.
Initial total energy of the body of mass $m$ at radius $R$ is:
$E_i = -\frac{GMm}{2R}$
After splitting,the body divides into two masses,each of mass $m' = \frac{m}{2}$.
The first mass moves in an orbit of radius $r_1 = \frac{R}{2}$,and the second mass moves in an orbit of radius $r_2 = \frac{3R}{2}$.
The final total energy $E_f$ is the sum of the energies of the two masses:
$E_f = -\frac{GM(m/2)}{2(R/2)} - \frac{GM(m/2)}{2(3R/2)}$
$E_f = -\frac{GMm}{2R} - \frac{GMm}{6R}$
$E_f = -\frac{3GMm + GMm}{6R} = -\frac{4GMm}{6R} = -\frac{2GMm}{3R}$
The difference between the final and initial total energies is:
$\Delta E = E_f - E_i = -\frac{2GMm}{3R} - (-\frac{GMm}{2R})$
$\Delta E = -\frac{2GMm}{3R} + \frac{GMm}{2R} = \frac{-4GMm + 3GMm}{6R} = -\frac{GMm}{6R}$

(A) The gravitational force exerted by a body of mass $M$ on a mass $m$ at a distance $r$ is $F = \frac{GMm}{r^2}$.
The difference in force $\Delta F$ between the nearest point $(r-R_e)$ and the farthest point $(r+R_e)$ on Earth is given by $\Delta F = \frac{GMm}{(r-R_e)^2} - \frac{GMm}{(r+R_e)^2} \approx \frac{2GMm}{r^3} (2R_e) = \frac{4GMmR_e}{r^3}$,where $R_e$ is the radius of the Earth.
For the moon: $\Delta F_1 = \frac{4GM_m m R_e}{r_1^3}$.
For the sun: $\Delta F_2 = \frac{4GM_s m R_e}{r_2^3}$.
Taking the ratio:
$\frac{\Delta F_1}{\Delta F_2} = \frac{M_m}{M_s} \times \left( \frac{r_2}{r_1} \right)^3$.
Given values: $M_m = 8 \times 10^{22} \, kg$,$M_s = 2 \times 10^{30} \, kg$,$r_1 = 0.4 \times 10^6 \, km$,$r_2 = 150 \times 10^6 \, km$.
$\frac{\Delta F_1}{\Delta F_2} = \left( \frac{8 \times 10^{22}}{2 \times 10^{30}} \right) \times \left( \frac{150 \times 10^6}{0.4 \times 10^6} \right)^3 = (4 \times 10^{-8}) \times (375)^3$.
$(375)^3 = 52,734,375$.
$\frac{\Delta F_1}{\Delta F_2} = 4 \times 10^{-8} \times 5.27 \times 10^7 \approx 2.1$.
Thus,the closest integer is $2$.

(C) According to Kepler's third law,the time period $T$ of a satellite orbiting a planet of mass $M$ at a radius $r$ is given by:
$T = 2\pi \sqrt{\frac{r^3}{GM}}$
Squaring both sides,we get:
$T^2 = \frac{4\pi^2}{GM} r^3$
Rearranging for the mass $M$:
$M = \frac{4\pi^2 r^3}{GT^2}$
Taking the natural logarithm and differentiating to find the relative uncertainty:
$\ln M = \ln(4\pi^2) + 3\ln r - \ln G - 2\ln T$
$\frac{\Delta M}{M} = 3\frac{\Delta r}{r} - 2\frac{\Delta T}{T}$
Given that the relative uncertainty in the radius $\frac{\Delta r}{r}$ is negligible (i.e.,$\frac{\Delta r}{r} = 0$):
$\left| \frac{\Delta M}{M} \right| = |-2| \frac{\Delta T}{T} = 2 \times 10^{-2}$

(B) The mass $M(r)$ enclosed within a sphere of radius $r$ $(r \leq R)$ is given by:
$M(r) = \int_0^r \rho(r') 4\pi r'^2 dr' = \int_0^r \frac{k}{r'} 4\pi r'^2 dr' = 4\pi k \int_0^r r' dr' = 2\pi k r^2$.
The acceleration $a$ of a test particle at distance $r$ is given by $a = \frac{GM(r)}{r^2}$.
For $r \leq R$:
$a = \frac{G(2\pi k r^2)}{r^2} = 2\pi G k = \text{constant}$.
For $r > R$,the total mass $M = M(R) = 2\pi k R^2$ is constant.
$a = \frac{GM}{r^2} = \frac{G(2\pi k R^2)}{r^2} \propto \frac{1}{r^2}$.
Thus,the acceleration is constant for $r \leq R$ and decreases as $1/r^2$ for $r > R$. The graph corresponding to this behavior is option $(b)$.

(A) For a very long cylindrical mass distribution,the gravitational field $E$ at a distance $r$ $(r > R)$ from the axis is given by Gauss's Law for gravitation: $E = \frac{2GM}{Lr}$.
The gravitational force acting on a star of mass $m$ is $F = mE = \frac{2GMm}{Lr}$.
Since the star is orbiting in a circular path,this gravitational force provides the necessary centripetal force: $\frac{mv^2}{r} = \frac{2GMm}{Lr}$.
Substituting $v = r\omega = r(\frac{2\pi}{T})$,we get: $m r (\frac{2\pi}{T})^2 = \frac{2GMm}{Lr}$.
Simplifying the equation: $r \frac{4\pi^2}{T^2} = \frac{2GM}{Lr}$.
Rearranging for $T^2$: $T^2 = \frac{4\pi^2 L}{2GM} r^2$.
Therefore,$T^2 \propto r^2$,which implies $T \propto r$.

(D) The kinetic energy of a gas molecule is given by $KE = \frac{3}{2} k_B T$.
Given temperature $T = 0^{\circ}C = 273 \ K$.
Substituting the temperature,$KE = \frac{3}{2} k_B (273) = \frac{819 k_B}{2}$.
When the molecule rises to a height $h$,its kinetic energy is converted into potential energy,$PE = Mgh$.
Equating kinetic energy to potential energy: $\frac{819 k_B}{2} = Mgh$.
Solving for height $h$,we get $h = \frac{819 k_B}{2Mg}$.

(D) The orbital velocity of the satellite at height $h$ is given by $v = \sqrt{\frac{GM}{R+h}}$. Since $h \ll R$,we can approximate $R+h \approx R$. Thus,$v = \sqrt{\frac{GM}{R}}$.
Using the relation $g = \frac{GM}{R^2}$,we get $GM = gR^2$. Substituting this,$v = \sqrt{\frac{gR^2}{R}} = \sqrt{gR}$.
The escape velocity $v'$ required to escape the Earth's gravitational field from height $h$ is $v' = \sqrt{\frac{2GM}{R+h}}$.
Again,using $h \ll R$,$v' = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}$.
The minimum increase in speed required is $\Delta v = v' - v$.
$\Delta v = \sqrt{2gR} - \sqrt{gR} = \sqrt{gR}(\sqrt{2} - 1)$.

(B) Let the four particles be at the corners of a square of side $a$. The distance of each particle from the center of the square is $r = \frac{a}{\sqrt{2}}$.
Consider one particle of mass $M$. The gravitational forces acting on it due to the other three particles are:
$1$. Two forces of magnitude $F_1 = \frac{GM^2}{a^2}$ directed along the sides of the square.
$2$. One force of magnitude $F_2 = \frac{GM^2}{(a\sqrt{2})^2} = \frac{GM^2}{2a^2}$ directed along the diagonal.
The resultant force $F_c$ towards the center is the sum of the components of these forces along the diagonal:
$F_c = F_1 \cos(45^\circ) + F_1 \cos(45^\circ) + F_2$
$F_c = 2 \left( \frac{GM^2}{a^2} \right) \frac{1}{\sqrt{2}} + \frac{GM^2}{2a^2} = \frac{GM^2}{a^2} \left( \sqrt{2} + \frac{1}{2} \right)$.
This force provides the necessary centripetal force for circular motion: $F_c = \frac{Mv^2}{r}$.
Substituting $r = \frac{a}{\sqrt{2}}$:
$\frac{Mv^2}{a/\sqrt{2}} = \frac{GM^2}{a^2} \left( \sqrt{2} + 0.5 \right)$
$v^2 = \frac{GM}{a} \left( \frac{\sqrt{2}}{\sqrt{2}} + \frac{0.5}{\sqrt{2}} \right) = \frac{GM}{a} (1 + 0.3535) \approx 1.3535 \frac{GM}{a}$.
$v \approx 1.16 \sqrt{\frac{GM}{a}}$.

(D) For a satellite of mass $m$ in a circular orbit of radius $r$ around the Earth of mass $M_E$:
$1$. Kinetic energy $(K)$: The centripetal force is provided by gravitational force,so $\frac{mv^2}{r} = \frac{GM_Em}{r^2}$,which gives $v^2 = \frac{GM_E}{r}$. Thus,$K = \frac{1}{2}mv^2 = \frac{GM_Em}{2r}$. This matches $(s)$.
$2$. Potential energy $(U)$: The gravitational potential energy is given by $U = -\frac{GM_Em}{r}$. This matches $(r)$.
$3$. Total energy $(E)$: $E = K + U = \frac{GM_Em}{2r} - \frac{GM_Em}{r} = -\frac{GM_Em}{2r}$. This matches $(p)$.
$4$. Orbital velocity $(v)$: $v = \sqrt{\frac{GM_E}{r}}$. This matches $(q)$.
Therefore,the correct matching is $A-s, B-r, C-p, D-q$.

(B) The gravitational force between the two particles provides the necessary centripetal force for circular motion.
The distance between the two particles is $2r$.
The gravitational force is $F = \frac{G \cdot m \cdot m}{(2r)^2} = \frac{Gm^2}{4r^2}$.
The centripetal force required for a particle of mass $m$ moving in a circle of radius $r$ with velocity $v$ is $F_c = \frac{mv^2}{r}$.
Equating the two forces: $\frac{mv^2}{r} = \frac{Gm^2}{4r^2}$.
Solving for $v$: $v^2 = \frac{Gm}{4r} \implies v = \frac{1}{2} \sqrt{\frac{Gm}{r}}$.
The time period $T$ is given by $T = \frac{2\pi r}{v}$.
Substituting the value of $v$: $T = \frac{2\pi r}{\frac{1}{2} \sqrt{\frac{Gm}{r}}} = 4\pi r \sqrt{\frac{r}{Gm}} = \frac{4\pi r^{3/2}}{\sqrt{Gm}}$.

(B) The two particles of mass $m$ are separated by a distance $2R$ and rotate about their common center of mass in a circle of radius $R$.
The gravitational force between them provides the necessary centripetal force for circular motion.
The gravitational force is given by $F = \frac{G m^2}{(2R)^2} = \frac{G m^2}{4R^2}$.
The centripetal force required for a particle of mass $m$ moving in a circle of radius $R$ with speed $v$ is $F_c = \frac{m v^2}{R}$.
Equating the two forces:
$\frac{m v^2}{R} = \frac{G m^2}{4R^2}$
Solving for $v$:
$v^2 = \frac{G m}{4R}$
$v = \sqrt{\frac{Gm}{4R}}$

(C) The gravitational field $F$ at a distance $r$ from the centre of a uniform sphere of mass $M$ and radius $R$ is given by:
$1$. Inside the sphere $(r < R)$: $F = \frac{GM r}{R^3}$,which implies $F \propto r$.
Therefore,$\frac{F_1}{F_2} = \frac{r_1}{r_2}$ for $r_1 < R$ and $r_2 < R$.
$2$. Outside the sphere $(r > R)$: $F = \frac{GM}{r^2}$,which implies $F \propto \frac{1}{r^2}$.
Therefore,$\frac{F_1}{F_2} = \frac{r_2^2}{r_1^2}$ for $r_1 > R$ and $r_2 > R$.
Since both statements $A$ and $B$ are correct,the correct option is $C$.

(D) The gravitational force between the two particles acts as the centripetal force required for their circular motion.
The distance between the two particles is $d = r + r = 2r$.
The gravitational force is $F = \frac{G m m}{(2r)^2} = \frac{G m^2}{4r^2}$.
The centripetal force required for a particle of mass $m$ moving in a circle of radius $r$ with speed $v$ is $F_c = \frac{m v^2}{r}$.
Equating the two forces: $\frac{m v^2}{r} = \frac{G m^2}{4r^2}$.
Simplifying for $v^2$: $v^2 = \frac{G m}{4r}$.
Taking the square root,the speed of each particle is $v = \frac{1}{2} \sqrt{\frac{Gm}{r}}$.

(C) The distance of each particle from the center of the equilateral triangle is $r = \frac{l}{\sqrt{3}}$.
The gravitational force exerted on one particle by the other two is the vector sum of the two forces,each of magnitude $F = \frac{Gm^2}{l^2}$.
The resultant force towards the center is $F_{net} = 2F \cos(30^\circ) = 2 \left( \frac{Gm^2}{l^2} \right) \frac{\sqrt{3}}{2} = \frac{\sqrt{3}Gm^2}{l^2}$.
This force provides the necessary centripetal force for circular motion: $\frac{mv^2}{r} = F_{net}$.
Substituting $r = \frac{l}{\sqrt{3}}$,we get $\frac{mv^2}{l/\sqrt{3}} = \frac{\sqrt{3}Gm^2}{l^2}$,which simplifies to $v^2 = \frac{Gm}{l}$,or $v = \sqrt{\frac{Gm}{l}}$.
The time period $T$ is given by $T = \frac{2\pi r}{v} = \frac{2\pi (l/\sqrt{3})}{\sqrt{Gm/l}} = \frac{2\pi}{\sqrt{3G m}} l^{3/2}$.
Thus,$T \propto l^{3/2}$.

(C) The two particles of mass $m$ are at a distance of $2R$ from each other.
The gravitational force between them provides the necessary centripetal force for circular motion.
The gravitational force is $F = \frac{G m^2}{(2R)^2} = \frac{G m^2}{4R^2}$.
The centripetal force required for a particle of mass $m$ moving with speed $v$ in a circle of radius $R$ is $F_c = \frac{mv^2}{R}$.
Equating the two forces: $\frac{mv^2}{R} = \frac{Gm^2}{4R^2}$.
Simplifying for $v$: $v^2 = \frac{Gm}{4R}$.
Therefore,$v = \sqrt{\frac{Gm}{4R}} = \frac{1}{2} \sqrt{\frac{Gm}{R}}$.

(D) The angular momentum of the Earth-Moon system is conserved because no external torque acts on the system.
From the relation $\tau = \frac{dL}{dt}$,if $\tau = 0$,then $\frac{dL}{dt} = 0$,which implies $L$ is constant.
Therefore,the $Reason$ is incorrect.
Due to tidal friction,the Earth's rotation slows down,meaning its angular velocity $\omega_1$ decreases. Since the total angular momentum $L = I_1\omega_1 + I_2\omega_2$ must remain constant,the angular momentum of the Moon's orbit must increase to compensate for the decrease in the Earth's rotational angular momentum.
As the Moon gains angular momentum,its orbital radius $r_2$ increases,causing the Moon to move away from the Earth. Thus,the $Assertion$ is also incorrect.

(B) Since the bodies are initially at rest,the total momentum of the system is zero. By the law of conservation of momentum,$m_1 v_1 = m_2 v_2$,where $v_1$ and $v_2$ are the speeds of the masses at separation $r$.
By the law of conservation of energy,the decrease in gravitational potential energy equals the increase in kinetic energy: $G m_1 m_2 / r = 1/2 m_1 v_1^2 + 1/2 m_2 v_2^2$.
From the momentum equation,$v_1 = (m_2/m_1) v_2$. Substituting this into the energy equation:
$G m_1 m_2 / r = 1/2 m_1 (m_2/m_1)^2 v_2^2 + 1/2 m_2 v_2^2 = 1/2 (m_2^2/m_1 + m_2) v_2^2 = 1/2 [m_2(m_1 + m_2)/m_1] v_2^2$.
Solving for $v_2$,we get $v_2 = \sqrt{2 G m_1^2 / (r(m_1 + m_2))}$. Similarly,$v_1 = \sqrt{2 G m_2^2 / (r(m_1 + m_2))}$.
The relative velocity of approach is $v_{rel} = v_1 + v_2 = \sqrt{2 G / (r(m_1 + m_2))} (m_1 + m_2) = \sqrt{2 G (m_1 + m_2) / r}$.

(A) The gravitational force on the satellite is always directed towards the centre of the earth. Therefore,the acceleration of the satellite $S$ is always directed towards the centre of the earth.
Since the gravitational force is a central force,the net torque of this force about the centre of the earth is zero. Consequently,the angular momentum (both in magnitude and direction) of $S$ about the centre of the earth remains constant throughout the motion.
Because the gravitational force is conservative in nature,the total mechanical energy of the satellite remains constant.
The speed of $S$ is maximum when it is nearest to the earth (perigee) and minimum when it is farthest (apogee).

(A) The centripetal force required for circular motion is provided by the net gravitational force acting on one particle due to the other three particles.
Let the particles be at the corners of a square of side length $a = \sqrt{2}r$.
The net gravitational force $F_{net}$ on one particle towards the center is:
$F_{net} = 2F \cos(45^{\circ}) + F_1$
Where $F = \frac{Gm^2}{(\sqrt{2}r)^2} = \frac{Gm^2}{2r^2}$ is the force from the two adjacent particles,and $F_1 = \frac{Gm^2}{(2r)^2} = \frac{Gm^2}{4r^2}$ is the force from the diagonally opposite particle.
$F_{net} = 2 \left( \frac{Gm^2}{2r^2} \right) \frac{1}{\sqrt{2}} + \frac{Gm^2}{4r^2} = \frac{Gm^2}{r^2} \left( \frac{1}{\sqrt{2}} + \frac{1}{4} \right) = \frac{Gm^2}{r^2} \left( \frac{2\sqrt{2} + 1}{4} \right)$.
Equating this to the centripetal force $\frac{mv^2}{r}$:
$\frac{mv^2}{r} = \frac{Gm^2}{r^2} \left( \frac{2\sqrt{2} + 1}{4} \right)$
$v^2 = \frac{Gm}{r} \left( \frac{2\sqrt{2} + 1}{4} \right)$
$v = \sqrt{\frac{Gm}{r} \left( \frac{1 + 2\sqrt{2}}{4} \right)}$

(A) The gravitational potential $V$ at a point is the sum of the potential due to the spherical shell and the potential due to the particle at the centre.
$1$. For a spherical shell of mass $M$ and radius $a$,the potential at any point inside the shell (at distance $r < a$) is constant and equal to the potential at its surface: $V_{\text{shell}} = - \frac{GM}{a}$.
$2$. For a particle of mass $M$ at the centre,the potential at a distance $r = \frac{a}{2}$ is given by: $V_{\text{particle}} = - \frac{GM}{r} = - \frac{GM}{a/2} = - \frac{2GM}{a}$.
$3$. The total gravitational potential $V_{\text{total}}$ at the point is:
$V_{\text{total}} = V_{\text{shell}} + V_{\text{particle}}$
$V_{\text{total}} = - \frac{GM}{a} - \frac{2GM}{a} = - \frac{3GM}{a}$.

(C) The length of the day is slowly increasing due to tidal friction caused by the gravitational interaction between the Earth and the Moon,as well as viscous forces between the Earth and its atmosphere.
This process causes a gradual dissipation of the Earth's rotational energy,leading to a slowdown in its rotation.
The gravitational pull of other planets in the solar system is negligible in this context compared to tidal effects.
Therefore,the $Assertion$ is correct,but the $Reason$ is incorrect.

(A) The $Assertion$ is correct: Rockets are launched from west to east to take advantage of the Earth's rotational velocity,which helps in achieving orbital velocity with less fuel.
The $Reason$ is also correct: The effective acceleration due to gravity $g'$ is given by $g' = g - \omega^2 R \cos^2 \lambda$. At the equator,the latitude $\lambda = 0$,so $\cos \lambda = 1$,making $g' = g - \omega^2 R$,which is the minimum value.
However,the $Reason$ does not explain why rockets are launched from west to east. The launch direction is based on the Earth's rotation,not the value of $g$ at the equator. Therefore,both are correct,but the $Reason$ is not the correct explanation for the $Assertion$.

(N/A) The angle between $GC$ and the positive $x$-axis is $30^{\circ}$,and so is the angle between $GB$ and the negative $x$-axis. The individual forces in vector notation are:
$F_{GA} = \frac{G m (2m)}{1^2} \hat{j} = 2Gm^2 \hat{j}$
$F_{GB} = \frac{G m (2m)}{1^2} (-\cos 30^{\circ} \hat{i} - \sin 30^{\circ} \hat{j}) = 2Gm^2 (-\frac{\sqrt{3}}{2} \hat{i} - \frac{1}{2} \hat{j}) = -Gm^2 \sqrt{3} \hat{i} - Gm^2 \hat{j}$
$F_{GC} = \frac{G m (2m)}{1^2} (+\cos 30^{\circ} \hat{i} - \sin 30^{\circ} \hat{j}) = 2Gm^2 (\frac{\sqrt{3}}{2} \hat{i} - \frac{1}{2} \hat{j}) = Gm^2 \sqrt{3} \hat{i} - Gm^2 \hat{j}$
From the principle of superposition,the resultant gravitational force $F_R$ on $(2m)$ is:
$F_R = F_{GA} + F_{GB} + F_{GC} = 2Gm^2 \hat{j} + (-Gm^2 \sqrt{3} \hat{i} - Gm^2 \hat{j}) + (Gm^2 \sqrt{3} \hat{i} - Gm^2 \hat{j}) = 0$
Alternatively,by symmetry,the resultant force is zero.
$(b)$ If the mass at vertex $A$ is doubled,the new force $F'_{GA} = \frac{G (2m) (2m)}{1^2} \hat{j} = 4Gm^2 \hat{j}$.
The forces $F_{GB}$ and $F_{GC}$ remain unchanged.
$F'_R = F'_{GA} + F_{GB} + F_{GC} = 4Gm^2 \hat{j} - Gm^2 \hat{j} - Gm^2 \hat{j} = 2Gm^2 \hat{j}$ (directed towards $A$).

(N/A) Part $(i)$: Using Kepler's third law,$T^{2} = \frac{4 \pi^{2}}{G M_{m}} R^{3}$.
Rearranging for the mass of Mars $(M_{m})$: $M_{m} = \frac{4 \pi^{2} R^{3}}{G T^{2}}$.
Given $T = 7 \text{ hours } 39 \text{ minutes} = (7 \times 60 + 39) \times 60 \text{ s} = 27540 \text{ s}$ and $R = 9.4 \times 10^{6} \text{ m}$.
$M_{m} = \frac{4 \times (3.14)^{2} \times (9.4 \times 10^{6})^{3}}{6.67 \times 10^{-11} \times (27540)^{2}} \approx 6.48 \times 10^{23} \text{ kg}$.
Part $(ii)$: Using Kepler's third law for planetary orbits around the Sun: $\frac{T_{M}^{2}}{T_{E}^{2}} = \frac{R_{M}^{3}}{R_{E}^{3}}$.
Given $\frac{R_{M}}{R_{E}} = 1.52$ and $T_{E} = 365 \text{ days}$.
$T_{M} = T_{E} \times (1.52)^{3/2} = 365 \times 1.873 \approx 684 \text{ days}$.

(N/A) Method $1$: Using the acceleration due to gravity on the surface of the Earth.
$M_E = \frac{g R_E^2}{G}$
$M_E = \frac{9.81 \times (6.37 \times 10^6)^2}{6.67 \times 10^{-11}}$
$M_E \approx 5.97 \times 10^{24} \; kg$
Method $2$: Using the orbital motion of the moon.
From Kepler's third law, $T^2 = \frac{4 \pi^2 R^3}{G M_E}$, so $M_E = \frac{4 \pi^2 R^3}{G T^2}$.
Given $T = 27.3 \; \text{days} = 27.3 \times 24 \times 3600 \; s = 2.3587 \times 10^6 \; s$.
$M_E = \frac{4 \times (3.14)^2 \times (3.84 \times 10^8)^3}{6.67 \times 10^{-11} \times (2.3587 \times 10^6)^2}$
$M_E \approx 6.02 \times 10^{24} \; kg$.
Both methods yield almost the same answer, with a difference of less than $1 \%$.

(N/A) No. Gravitational influence cannot be shielded because gravity is independent of the nature of the medium and the presence of other matter.
$(b)$ Yes. If the space station is large,the astronaut can detect the variation in the gravitational field (tidal forces) across the dimensions of the station.
$(c)$ The tidal effect depends on the gradient of the gravitational field,which is proportional to $1/r^3$,whereas the gravitational force itself is proportional to $1/r^2$. Since the Moon is much closer to the Earth than the Sun,the gradient of the Moon's gravitational field is larger,resulting in a greater tidal effect.

(B, C, D) The correct symptoms are $(b)$,$(c)$,and $(d)$.
In the presence of gravity on Earth,the legs support the entire mass of the body while standing. In space,an astronaut experiences weightlessness due to the absence of gravity,so the feet do not swell as they do on Earth.
$A$ swollen face is a common symptom in space due to the redistribution of body fluids caused by weightlessness. As fluids shift towards the upper part of the body,the face appears puffy.
Headaches are frequently reported by astronauts due to mental strain,changes in intracranial pressure,and the stress of adapting to a microgravity environment.
Orientational problems occur because the vestibular system,which relies on gravity to determine orientation,does not function normally in space,leading to spatial disorientation.

(A) body remains stuck to the surface of a star if the inward gravitational force is greater than the outward centrifugal force caused by the rotation of the star.
Gravitational force,$f_g = \frac{GMm}{R^2}$
Where,
$M = \text{Mass of the star} = 2.5 \times 2 \times 10^{30} = 5 \times 10^{30} \;kg$
$m = \text{Mass of the body}$
$R = \text{Radius of the star} = 12 \;km = 1.2 \times 10^4 \;m$
$G = 6.67 \times 10^{-11} \;N \cdot m^2/kg^2$
$f_g = \frac{6.67 \times 10^{-11} \times 5 \times 10^{30} \times m}{(1.2 \times 10^4)^2} \approx 2.31 \times 10^{11} \;m \;N$
Centrifugal force,$f_c = mR\omega^2$
Where $\omega = 2\pi v$ and $v = 1.2 \;rev/s$
$f_c = m \times (1.2 \times 10^4) \times (2 \times 3.14 \times 1.2)^2$
$f_c = m \times (1.2 \times 10^4) \times (7.54)^2 \approx 6.82 \times 10^5 \;m \;N$
Since $f_g > f_c$,the gravitational force is much stronger than the centrifugal force. Therefore,the object will remain stuck to the surface of the star.

(495 KM) Initial velocity of the rocket,$v = 2 \; km/s = 2 \times 10^3 \; m/s$.
Mass of Mars,$M = 6.4 \times 10^{23} \; kg$.
Radius of Mars,$R = 3395 \; km = 3.395 \times 10^6 \; m$.
Gravitational constant,$G = 6.67 \times 10^{-11} \; N m^2 kg^{-2}$.
Initial kinetic energy $K_i = \frac{1}{2} m v^2$.
Initial potential energy $U_i = -\frac{GMm}{R}$.
Since $20 \%$ of initial kinetic energy is lost,the effective kinetic energy available is $0.8 K_i = 0.8 \times \frac{1}{2} m v^2 = 0.4 m v^2$.
Total initial energy $E_i = 0.4 m v^2 - \frac{GMm}{R}$.
At maximum height $h$,the final velocity is zero,so final energy $E_f = -\frac{GMm}{R+h}$.
By conservation of energy,$E_i = E_f$:
$0.4 m v^2 - \frac{GMm}{R} = -\frac{GMm}{R+h}$.
Dividing by $m$ and rearranging: $0.4 v^2 = GM \left( \frac{1}{R} - \frac{1}{R+h} \right) = \frac{GMh}{R(R+h)}$.
Solving for $h$: $\frac{R+h}{h} = \frac{GM}{0.4 v^2 R} \implies \frac{R}{h} = \frac{GM}{0.4 v^2 R} - 1$.
$h = \frac{0.4 R^2 v^2}{GM - 0.4 v^2 R}$.
Substituting values: $h = \frac{0.4 \times (3.395 \times 10^6)^2 \times (2 \times 10^3)^2}{(6.67 \times 10^{-11} \times 6.4 \times 10^{23}) - (0.4 \times (2 \times 10^3)^2 \times 3.395 \times 10^6)}$.
$h = \frac{18.442 \times 10^{18}}{42.688 \times 10^{12} - 5.432 \times 10^{12}} = \frac{18.442}{37.256} \times 10^6 \approx 495 \; km$.

(A) The earth completes one full rotation of $360^{\circ}$ about its axis in $24\; h$.
Since $1\; h = 60\; min$,the total time in minutes is $24 \times 60 = 1440\; min$.
To find the time taken for a shift of $1^{\circ}$,we divide the total time by $360^{\circ}$:
Time for $1^{\circ} = \frac{1440\; min}{360} = 4\; min$.
Therefore,the sun appears to shift by $1^{\circ}$ in $4\; min$.

(N/A) The Earth's crust is not uniform; it contains discontinuities and dislocations known as fault lines.
These fault lines within the Earth's crust act like compressed springs and store a significant amount of potential energy.
An earthquake occurs when these fault lines suddenly readjust or slip,releasing the stored potential energy.

(N/A) Radius of the Earth $R = 6400 \, km = 6.4 \times 10^6 \, m$.
Time period $T = 1 \, \text{day} = 86400 \, s$.
Centripetal acceleration $a_c = \omega^2 R = R \left( \frac{2\pi}{T} \right)^2 = \frac{4\pi^2 R}{T^2}$.
$a_c = \frac{4 \times (3.14)^2 \times 6.4 \times 10^6}{(86400)^2} \approx 0.034 \, m/s^2$.
At latitude $\theta$, the radius of the circular path is $R \cos \theta$, so $a_c(\theta) = \omega^2 R \cos \theta = 0.034 \cos \theta \, m/s^2$.
Comparing with $g$: $\frac{a_c}{g} = \frac{0.034}{9.8} \approx \frac{1}{288}$, which is much less than $g$.
$(b)$ Orbital radius $R' = 1.5 \times 10^{11} \, m$.
Time period $T' = 1 \, \text{year} = 365 \times 24 \times 3600 \approx 3.15 \times 10^7 \, s$.
Centripetal acceleration $a_c' = \frac{4\pi^2 R'}{T'^2} = \frac{4 \times (3.14)^2 \times 1.5 \times 10^{11}}{(3.15 \times 10^7)^2} \approx 5.97 \times 10^{-3} \, m/s^2$.
Comparing with $g$: $\frac{a_c'}{g} = \frac{5.97 \times 10^{-3}}{9.8} \approx \frac{1}{1642}$, which is much less than $g$.

(C) $(i)$ As the spacecraft moves away from the Earth towards the Moon,the gravitational pull of the Earth decreases,so its weight decreases.
$(ii)$ At a specific point between the Earth and the Moon,known as the neutral point,the gravitational forces exerted by the Earth and the Moon cancel each other out,making the net weight zero.
$(iii)$ As the spacecraft continues its journey towards the Moon,the gravitational pull of the Moon increases,causing the weight to increase from zero until it reaches the surface of the Moon,where its weight becomes $\frac{mg}{6}$ (where $g$ is the acceleration due to gravity on Earth).

(N/A) Every day,the Earth advances in its orbit by approximately $1^{\circ}$.
To have the sun at the zenith point again,the Earth must rotate by $361^{\circ}$ (which we define as $1$ mean solar day).
Since $361^{\circ}$ corresponds to $24\,\text{h}$,we can calculate the time taken for $1^{\circ}$ as follows:
$\because \text{To cover } 361^{\circ}, \text{time taken} = 24\,\text{h}$
$\because \text{To cover } 1^{\circ}, \text{time taken} = t$
$t = \frac{24}{361} \times 1 \approx 0.0664\,\text{h}$
$t = 0.0664 \times 60\,\text{min} \approx 3.99\,\text{min} \approx 4\,\text{min}$.
Hence,the mean solar day is $4\,\text{min}$ longer than the sidereal day,and distant stars rise $4\,\text{min}$ early every successive day.

(N/A) The acceleration due to gravity inside the Earth at a distance $r$ from the center is given by $g' = \frac{GM}{R^3} r$. Thus,$g' \propto r$ (directly proportional).
$(b)$ The weight of an object is $W = mg$. Since $g = \frac{GM}{R^2}$,if the radius $R$ becomes $R' = \frac{R}{2}$,the new acceleration due to gravity $g'$ is:
$g' = \frac{GM}{(R/2)^2} = 4 \frac{GM}{R^2} = 4g$.
Therefore,the weight $W' = mg' = 4mg$,which is $4$ times the original weight.
$(c)$ The orbital velocity $v_0$ is given by $v_0 = \sqrt{\frac{GM_e}{r}}$.
For a geostationary satellite,$r \approx 42260 \times 10^3 \ m$.
Substituting $G = 6.67 \times 10^{-11} \ N \ m^2/kg^2$ and $M_e = 5.98 \times 10^{24} \ kg$:
$v_0 = \sqrt{\frac{6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{42260 \times 10^3}} \approx 3070 \ m/s \approx 3 \ km/s$.

(N/A) At the center of the Earth,the gravitational field intensity is $0$ because the mass enclosed within a sphere of radius $r=0$ is zero.
$(b)$ The binding energy $(BE)$ of a satellite is defined as the negative of its total mechanical energy. Since the total energy $E = -|PE|/2$ is not given,we use the relation $BE = -E$. For a satellite,$PE = 2E$. Given $PE = -8 \times 10^9 \, J$,then $E = -4 \times 10^9 \, J$. Thus,$BE = -(-4 \times 10^9 \, J) = 4 \times 10^9 \, J$. However,if the question implies the total energy is $-8 \times 10^9 \, J$,then $BE = 8 \times 10^9 \, J$. Assuming the standard convention where $PE$ is given,the binding energy is $4 \times 10^9 \, J$. If the value provided is the total energy,it is $8 \times 10^9 \, J$.
$(c)$ Kepler's second law (law of areas) is a direct consequence of the law of conservation of angular momentum.

(A) According to Kepler's third law,$T^2 \propto r^3$,so $r \propto T^{2/3}$. Given $T_M = 8 T_m$,then $r_M = r_m \times (8)^{2/3} = 5.79 \times 10^{10} \times 4 = 23.16 \times 10^{10} \, m$.
$(b)$ Mass is an intrinsic property of matter and remains constant regardless of location. Thus,the mass on the Moon is $m \, kg$.
$(c)$ $A$ geostationary satellite orbits at a height of approximately $35,800 \, km$ above the Earth's surface.
$(d)$ Using Newton's law of gravitation,$F = \frac{G m_1 m_2}{r^2}$. Here $m_1 = 1 \, kg$,$m_2 = 1 \, kg$,$r = 1 \, mm = 10^{-3} \, m$. So,$F = \frac{6.67 \times 10^{-11} \times 1 \times 1}{(10^{-3})^2} = 6.67 \times 10^{-11} \times 10^6 = 6.67 \times 10^{-5} \, N$.

(A) False. The acceleration due to gravity at a depth $d$ is given by $g_d = g(1 - d/R)$,which decreases as depth $d$ increases.
$(b)$ False. Escape velocity is given by $v_e = \sqrt{2GM/R}$. It is independent of the mass of the object $m$. Therefore,for an object of mass $2m$,the escape velocity remains $11.2 \, km/s$.
$(c)$ False. Mass is an intrinsic property of an object and remains constant regardless of gravity. Weight is the force of gravity $(W = mg)$,so if gravity disappears,weight becomes zero,but mass remains unchanged.