Acceleration Due to Gravity and its Variation Questions in English

(B) The acceleration due to gravity $g'$ at a height $h$ above the surface of the earth is given by the formula $g' = g(1 - \frac{2h}{R})$,where $g = 9.8\, m/s^2$ is the acceleration due to gravity at the surface,$h = 12\, km = 1.2 \times 10^4\, m$,and $R = 6400\, km = 6.4 \times 10^6\, m$ is the radius of the earth.
Substituting the values:
$g' = 9.8 \times (1 - \frac{2 \times 12}{6400})$
$g' = 9.8 \times (1 - \frac{24}{6400})$
$g' = 9.8 \times (1 - 0.00375)$
$g' = 9.8 \times 0.99625$
$g' \approx 9.763\, m/s^2$.
Thus,the value is approximately $9.76\, m/s^2$.

(N/A) $1$. At the centre of the Earth,the distance from the centre $r = 0$. The formula for acceleration due to gravity at a depth $d$ is $g_d = g(1 - d/R)$,where $d = R$ at the centre. Thus,$g_{centre} = g(1 - R/R) = 0$. So,the value of $g$ at the centre of the Earth is $0 \ m/s^2$.
$2$. Below the surface: As we go deeper into the Earth (increasing depth $d$),the value of $g$ decreases linearly according to the formula $g_d = g(1 - d/R)$.
$3$. Above the surface: As we go higher above the surface (increasing height $h$),the value of $g$ decreases according to the formula $g_h = g(1 - 2h/R)$ for $h \ll R$,or more generally $g_h = gR^2 / (R + h)^2$. In both cases,$g$ decreases as height increases.

(A) Yes,the acceleration due to gravity $(g)$ is given by the formula $g = \frac{GM}{R^2}$,where $G$ is the universal gravitational constant,$M$ is the mass of the Earth,and $R$ is the radius of the Earth.
Since $g$ depends only on the mass of the Earth and its radius,it is independent of the mass of the falling object.
Therefore,in the absence of air resistance,all objects fall with the same acceleration regardless of their mass.

(B) $1$. At the center of the Earth,the acceleration due to gravity $(g)$ is $0$ because the net gravitational force acting on an object at the center is zero.
$2$. The universal gravitational constant $(G)$ is a fundamental constant of nature and its value remains $6.67 \times 10^{-11} \ Nm^2/kg^2$ everywhere in the universe,including the center of the Earth.
$3$. Acceleration due to gravity $(g)$ is a vector quantity because it has both magnitude and direction (towards the center of the Earth).
$4$. The universal gravitational constant $(G)$ is a scalar quantity as it has only magnitude.

(B) The value of acceleration due to gravity $g$ is higher at the poles of the Earth.
This is because the Earth is not a perfect sphere; it is flattened at the poles and bulges at the equator.
The radius of the Earth at the poles $(R_p)$ is about $21 \, km$ less than the radius at the equator $(R_e)$.
According to the formula $g = \frac{GM}{R^2}$,the acceleration due to gravity is inversely proportional to the square of the radius $(g \propto \frac{1}{R^2})$.
Since the radius at the poles is smaller,the value of $g$ is greater at the poles compared to the equator.

(N/A) The acceleration due to gravity is given by $g = \frac{GM_e}{R_e^2}$,which implies $g \propto \frac{1}{R_e^2}$. The Earth is slightly bulging at the equator,meaning the radius $R_e$ is larger at the equator,resulting in a lower value of $g$. Conversely,the Earth is flattened at the poles,meaning the radius $R_e$ is smaller at the poles,resulting in a higher value of $g$. Since $g$ varies with latitude,we can conclude that the Earth is not a perfect sphere.

(A) The Earth is not a perfect sphere; it is an oblate spheroid.
The equatorial radius $(R_e)$ is approximately $6,378 \ km$.
The polar radius $(R_p)$ is approximately $6,357 \ km$.
The difference between the equatorial radius and the polar radius is $\Delta R = R_e - R_p = 6,378 \ km - 6,357 \ km = 21 \ km$.
Since $1 \ km = 1,000 \ m$,the difference in meters is $21 \times 1,000 \ m = 21,000 \ m$.

(B) The Earth is not a perfect sphere; it is flattened at the poles and bulges at the equator.
As a result,the radius of the Earth at the equator $(R_e)$ is greater than the radius at the poles $(R_p)$.
The formula for acceleration due to gravity is $g = \frac{GM}{R^2}$.
Since $g$ is inversely proportional to the square of the radius $(g \propto \frac{1}{R^2})$,a larger radius at the equator results in a smaller value of $g$.
Therefore,as we move from the poles towards the equator,the value of $g$ decreases.

(B) The acceleration due to gravity at a height $h$ above the Earth's surface is given by the formula $g_h = g \left( 1 + \frac{h}{R_e} \right)^{-2}$,where $g$ is the acceleration due to gravity at the surface,$R_e$ is the radius of the Earth,and $h$ is the height.
As the height $h$ increases,the term $\left( 1 + \frac{h}{R_e} \right)$ increases,which causes the value of $g_h$ to decrease.
Therefore,the value of acceleration due to gravity decreases as we move above the surface of the Earth.

(B) The acceleration due to gravity at the surface of the Earth is given by $g = \frac{GM_e}{R_e^2}$.
At a height $h = R_e$ above the surface,the acceleration due to gravity $g^{\prime}$ is given by the formula $g^{\prime} = \frac{GM_e}{(R_e + h)^2}$.
Substituting $h = R_e$ into the equation,we get $g^{\prime} = \frac{GM_e}{(R_e + R_e)^2} = \frac{GM_e}{(2R_e)^2}$.
This simplifies to $g^{\prime} = \frac{GM_e}{4R_e^2}$.
Since $g = \frac{GM_e}{R_e^2}$,we can write $g^{\prime} = \frac{g}{4}$.

(A) The acceleration due to gravity $g$ on the surface of the Earth is given by the formula $g = \frac{GM}{R^2}$,where $G$ is the universal gravitational constant,$M$ is the mass of the Earth,and $R$ is the radius of the Earth.
Since $G$ and $M$ are constant,we have $g \propto \frac{1}{R^2}$.
If the radius $R$ decreases,the denominator $R^2$ decreases.
Therefore,the value of $g$ will increase.

(B) The acceleration due to gravity on the surface of the Earth is given by $g_e = \frac{G M_e}{R_e^2}$.
Given that the new radius $R'_e = \frac{R_e}{n}$ and the mass $M_e$ remains constant.
The new acceleration due to gravity $g'_e$ is given by $g'_e = \frac{G M_e}{(R'_e)^2}$.
Substituting the value of $R'_e$,we get $g'_e = \frac{G M_e}{(R_e/n)^2} = \frac{G M_e n^2}{R_e^2}$.
Since $g_e = \frac{G M_e}{R_e^2}$,we can write $g'_e = n^2 g_e$.

(B) The effective acceleration due to gravity $g'$ at a latitude $\phi$ is given by $g' = g - \omega^2 R \cos^2 \phi$,where $g$ is the acceleration due to gravity at the poles,$\omega$ is the angular velocity of the Earth,and $R$ is the radius of the Earth.
At the poles,$\phi = 90^\circ$,so $\cos 90^\circ = 0$. Thus,$g' = g$,which is the maximum value.
At the equator,$\phi = 0^\circ$,so $\cos 0^\circ = 1$. Thus,$g' = g - \omega^2 R$,which is the minimum value.

(A) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by the formula: $g' = g - \omega^2 R_e \cos^2 \lambda$.
At the equator,the latitude $\lambda = 0^\circ$,so $\cos 0^\circ = 1$.
Substituting this into the equation,we get $g' = g - \omega^2 R_e$.
For the acceleration due to gravity to be zero at the equator,we set $g' = 0$.
Therefore,$0 = g - \omega^2 R_e$.
Rearranging the terms,we get $\omega^2 R_e = g$.
Solving for $\omega$,we find $\omega = \sqrt{\frac{g}{R_e}}$.

(A) The gravitational force at the surface of the Earth is $F_s = \frac{GM_e m}{R_e^2}$.
The gravitational force at height $h$ is $F_h = \frac{GM_e m}{(R_e + h)^2}$.
According to the problem,$F_h = \frac{1}{3} F_s$.
Substituting the expressions: $\frac{GM_e m}{(R_e + h)^2} = \frac{1}{3} \frac{GM_e m}{R_e^2}$.
Simplifying the equation: $(R_e + h)^2 = 3 R_e^2$.
Taking the square root on both sides: $R_e + h = \sqrt{3} R_e$.
Therefore,$h = \sqrt{3} R_e - R_e = R_e(\sqrt{3} - 1)$.

(C) The weight of an object on the surface of the Earth is given by $W = mg = \frac{GM_e m}{R_e^2}$.
When the radius of the Earth is doubled,the new radius becomes $R_e' = 2R_e$.
The new weight $W'$ is given by $W' = \frac{GM_e m}{(2R_e)^2} = \frac{GM_e m}{4R_e^2}$.
Substituting the original weight $W$ into the equation,we get $W' = \frac{W}{4}$.
Therefore,the weight of the object becomes one-fourth of its original weight.

(D) The weight of an object of mass $m$ on the surface of the Earth is given by $W = mg = \frac{GM_e m}{R_e^2}$.
When the radius of the Earth becomes half $(R' = \frac{R_e}{2})$ while the mass $M_e$ remains constant,the new acceleration due to gravity $g'$ is:
$g' = \frac{GM_e}{(R_e/2)^2} = \frac{GM_e}{R_e^2 / 4} = 4 \left( \frac{GM_e}{R_e^2} \right) = 4g$.
Therefore,the new weight $W'$ is:
$W' = mg' = m(4g) = 4W$.
Thus,the weight of the object becomes $4$ times its original weight.

(B) The acceleration due to gravity $g$ is given by $g = \frac{GM_e}{R_e^2}$.
Since mass $M_e = \text{Volume} \times \text{Density} = \frac{4}{3} \pi R_e^3 \rho$,we have $g = \frac{G}{R_e^2} (\frac{4}{3} \pi R_e^3 \rho) = \frac{4}{3} \pi G R_e \rho$.
This shows that $g \propto R_e$ when density $\rho$ is constant.
If the radius becomes $R_e' = \frac{R_e}{2}$,the new acceleration due to gravity $g'$ will be $g' = \frac{4}{3} \pi G (\frac{R_e}{2}) \rho = \frac{1}{2} g$.
Since weight $W = mg$,the new weight $W'$ is $W' = m g' = m (\frac{g}{2}) = \frac{W}{2}$.

(A) The acceleration due to gravity $g$ on the surface of a planet is given by $g = \frac{GM}{R^2}$.
Since the mass $M$ of a planet can be expressed in terms of its density $\rho$ and radius $R$ as $M = \rho \times V = \rho \times (\frac{4}{3} \pi R^3)$,we substitute this into the formula for $g$:
$g = \frac{G}{R^2} \times (\frac{4}{3} \pi R^3 \rho) = \frac{4}{3} \pi G R \rho$.
Therefore,the ratio of the acceleration due to gravity on the two planets is:
$\frac{g_1}{g_2} = \frac{\frac{4}{3} \pi G R_1 \rho_1}{\frac{4}{3} \pi G R_2 \rho_2} = \frac{R_1 \rho_1}{R_2 \rho_2}$.
Thus,the ratio is $R_1 \rho_1 : R_2 \rho_2$.

(B) The acceleration due to gravity on the surface of a planet is given by $g = \frac{GM}{R^2}$.
For Earth,$g_e = \frac{GM_e}{R_e^2}$.
For the given planet,the mass $M_p = 2M_e$ and the radius $R_p = 2R_e$.
Substituting these values into the formula for the planet's gravity $g_p$:
$g_p = \frac{G(2M_e)}{(2R_e)^2}$
$g_p = \frac{2GM_e}{4R_e^2}$
$g_p = \frac{1}{2} \left( \frac{GM_e}{R_e^2} \right)$
Since $g_e = \frac{GM_e}{R_e^2}$,we get $g_p = \frac{g_e}{2}$.
Thus,the acceleration due to gravity on the planet is half that of Earth.

(N/A) Consider a body of mass $m$ at a point $P$ inside a tunnel dug through the center of the Earth. Let the distance of point $P$ from the center of the Earth be $y$. According to the shell theorem,the gravitational force on the body at distance $y$ from the center is due only to the mass of the Earth contained within a sphere of radius $y$.
The mass of this inner sphere is $M' = \rho \cdot \frac{4}{3} \pi y^3$,where $\rho$ is the density of the Earth. Since $\rho = \frac{M}{\frac{4}{3} \pi R^3}$,we have $M' = M \left( \frac{y^3}{R^3} \right)$.
The gravitational force on the body is $F = -\frac{G M' m}{y^2} = -\frac{G M m y^3}{R^3 y^2} = -\left( \frac{G M m}{R^3} \right) y$.
Since $g = \frac{G M}{R^2}$,we can write $F = -\left( \frac{mg}{R} \right) y$.
This force is of the form $F = -ky$,where $k = \frac{mg}{R}$ is a constant. Since the restoring force is directly proportional to the displacement $y$ and directed towards the center,the body executes simple harmonic motion.

(B) The acceleration due to gravity on the surface of the Earth is given by the formula $g' = g - \omega^2 R \cos^2 \phi$,where $\phi$ is the latitude.
As we move from the equator $(\phi = 0^\circ)$ to the poles $(\phi = 90^\circ)$,the value of $\cos^2 \phi$ decreases.
Consequently,the term $\omega^2 R \cos^2 \phi$ decreases,which causes the effective acceleration due to gravity $(g')$ to increase.
Therefore,$g$ is minimum at the equator and maximum at the poles.

(A) The acceleration due to gravity at the poles $(g_{p})$ is given by $g_{p} = G M / R^{2}$.
At the equator,the effective acceleration due to gravity $(g_{e})$ is given by $g_{e} = g - R_{e} \omega^{2}$,where $R_{e}$ is the radius of the Earth and $\omega$ is the angular velocity of the Earth's rotation.
The difference in the value of $g$ between the poles and the equator is $\Delta g = g_{p} - g_{e}$.
Substituting the expressions: $\Delta g = g - (g - R_{e} \omega^{2})$.
Therefore,$\Delta g = R_{e} \omega^{2}$.

(C) The Earth is slightly flattened at the poles and bulges at the equator.
According to the formula $g = \frac{GM}{R_e^2}$,the acceleration due to gravity is inversely proportional to the square of the radius $(g \propto \frac{1}{R_e^2})$.
Since the radius at the poles $(R_p)$ is less than the radius at the equator $(R_e)$,the value of $g$ is greater at the poles than at the equator $(g_p > g_e)$.

(N/A) The value of $g$ is maximum at the poles.
Reasons:
$(i)$ The radius of the Earth is slightly smaller at the poles compared to the equator. Since $g = \frac{GM}{R^2}$,a smaller $R$ leads to a larger $g$.
$(ii)$ At the poles,the centrifugal force due to the Earth's rotation is zero,which means there is no reduction in the effective gravitational acceleration.

(B) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by $g' = g - \omega^2 R \cos^2 \lambda$,where $g$ is the acceleration due to gravity without rotation,$\omega$ is the angular velocity of the Earth,and $R$ is the radius of the Earth.
Because of the term $\omega^2 R \cos^2 \lambda$,the rotation of the Earth causes a decrease in the effective acceleration due to gravity.
This effect is zero at the poles (where $\lambda = 90^\circ$) and maximum at the equator (where $\lambda = 0^\circ$).

(A) The effective acceleration due to gravity at the equator is given by $g' = g - R\omega^2$,where $g$ is the acceleration due to gravity if the Earth were stationary,$R$ is the radius of the Earth,and $\omega$ is the angular velocity of the Earth's rotation.
If the Earth stops its rotation,then $\omega = 0$.
Therefore,the new value of $g$ becomes $g' = g$.
This means the value of $g$ at the equator increases by an amount equal to $R\omega^2$.

(A) The acceleration due to gravity on the surface of a planet is given by $g = \frac{GM}{R^2}$,where $M$ is the mass and $R$ is the radius of the planet.
From the graph,we observe that at the surfaces,$g_1 = g_3 > g_2$.
For $g_1 = g_3$,we have $\frac{GM_1}{R_1^2} = \frac{GM_3}{R_3^2}$,which implies $\frac{M_1}{R_1^2} = \frac{M_3}{R_3^2}$.
Since $R_1 < R_3$,it follows that $R_1^2 < R_3^2$,and therefore $M_1 < M_3$.
Now,comparing $g_1$ and $g_2$,we have $g_1 > g_2$,so $\frac{GM_1}{R_1^2} > \frac{GM_2}{R_2^2}$.
Given the graph,$R_1 < R_2$ and $g_1 > g_2$,which leads to $M_1 > M_2$.
Combining these results,we get $M_3 > M_1 > M_2$.

(A) Inside a small spaceship orbiting around the earth,the value of acceleration due to gravity $g$ is effectively constant throughout the interior,and the astronaut experiences weightlessness due to free fall.
If the space station orbiting around the earth has a large size,the variation in the gravitational field (tidal forces) across the dimensions of the station becomes significant. In such a case,the astronaut inside the spaceship will experience a non-uniform gravitational force and can detect the presence of gravity.

(A) False. $G$ (Universal Gravitational Constant) is a scalar quantity.
$(b)$ True. The formula $g = \frac{GM}{r^2}$ is derived from Newton's Law of Gravitation and is applicable to all spherical celestial bodies.
$(c)$ False. The effective acceleration due to gravity at the equator is given by $g' = g - R\omega^2$. If the Earth stops rotating,$\omega = 0$,so $g' = g$. Since $g > (g - R\omega^2)$,the value of $g$ at the equator will actually increase.

(C) The acceleration due to gravity $g$ on the surface of the Earth is given by $g = \frac{GM}{R^2}$.
$(1)$ The value of $g$ is maximum at the poles because the radius of the Earth $R$ is minimum at the poles $(R_{pole} < R_{equator})$.
$(2)$ The value of $g$ is minimum at the equator because the radius of the Earth $R$ is maximum at the equator.
$(3)$ The value of $g$ is zero at the center of the Earth because the mass of the Earth enclosed within a sphere of radius $r$ becomes zero as $r \to 0$.
Therefore,the correct matching is $(1-b), (2-c), (3-a)$.

(C) The acceleration due to gravity at the surface of the Earth is given by $g = \frac{GM}{R^2}$,where $G$ is the gravitational constant,$M$ is the mass of the Earth,and $R$ is the radius of the Earth.
At a height $h$ above the surface,the acceleration due to gravity $g_h$ is given by the formula $g_h = \frac{GM}{(R+h)^2}$.
Given that $h = R$,we substitute this into the formula:
$g_h = \frac{GM}{(R+R)^2} = \frac{GM}{(2R)^2} = \frac{GM}{4R^2}$.
Since $g = \frac{GM}{R^2}$,we can write $g_h = \frac{1}{4} \left( \frac{GM}{R^2} \right) = \frac{g}{4}$.
Therefore,the acceleration due to gravity at a height $h = R$ is $g/4$.

(A) The qualitative meaning of 'small' and 'extended' depends on the variation of the acceleration due to gravity $(g)$ over the dimensions of the object.
If the vertical height or dimensions of the object are very small compared to the radius of the earth $(R_e \approx 6400 \ km)$,the gravitational field is considered uniform,and the object is called 'small'. In this case,the centre of gravity coincides with the centre of mass.
If the dimensions of the object are large enough that the variation in $g$ becomes significant,the object is called 'extended'. In this case,the centre of gravity may not coincide with the centre of mass.
$(1)$ $A$ building and a pond are considered 'small' objects because their vertical extent is negligible compared to $R_e$. Thus,for these,the centre of gravity coincides with the centre of mass.
$(2)$ $A$ deep lake and a mountain are examples of 'extended' objects because their vertical extent is significant enough that the variation in $g$ cannot be ignored. Thus,for these,the centre of gravity may not coincide with the centre of mass.

(A) Let $M$ be the mass of the earth and $R$ be its radius.
At a depth $h$ below the surface,the effective mass $M_1$ attracting the body is the mass of the sphere of radius $(R-h)$.
$M_1 = \frac{M}{\frac{4}{3} \pi R^3} \cdot \frac{4}{3} \pi (R-h)^3 = M \frac{(R-h)^3}{R^3}$.
The acceleration due to gravity at depth $h$ is $g_P = \frac{G M_1}{(R-h)^2} = \frac{G M (R-h)^3}{R^3 (R-h)^2} = \frac{G M (R-h)}{R^3}$.
The acceleration due to gravity at height $h$ above the surface is $g_Q = \frac{G M}{(R+h)^2}$.
Given that the weight is the same at both points,$g_P = g_Q$.
$\frac{G M (R-h)}{R^3} = \frac{G M}{(R+h)^2}$.
$(R-h)(R+h)^2 = R^3$.
$(R-h)(R^2 + 2Rh + h^2) = R^3$.
$R^3 + 2R^2h + Rh^2 - hR^2 - 2Rh^2 - h^3 = R^3$.
$R^2h - Rh^2 - h^3 = 0$.
Dividing by $h$ (since $h \neq 0$): $R^2 - Rh - h^2 = 0$.
$h^2 + Rh - R^2 = 0$.
Using the quadratic formula $h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$h = \frac{-R \pm \sqrt{R^2 - 4(1)(-R^2)}}{2} = \frac{-R \pm \sqrt{5R^2}}{2}$.
Since $h > 0$,we take the positive root: $h = \frac{\sqrt{5}R - R}{2}$.

(D) The effective acceleration due to gravity at the equator $(g_e)$ is given by $g_e = g - R\omega^2$,where $g$ is the acceleration due to gravity at the poles.
The acceleration due to gravity at a height $h$ above the poles $(g_h)$ is given by $g_h = g(1 - \frac{2h}{R}) = g - \frac{2gh}{R}$.
Given that the weights are the same at both locations,the effective acceleration due to gravity must be equal: $g_e = g_h$.
Substituting the expressions: $g - R\omega^2 = g - \frac{2gh}{R}$.
Simplifying the equation: $R\omega^2 = \frac{2gh}{R}$.
Solving for $h$: $h = \frac{R^2\omega^2}{2g}$.

(D) The acceleration due to gravity at a height $h$ is given by $g_{h} = \frac{GM}{(R+h)^{2}}$.
Given $h = \frac{R}{2}$,we have $g_{1} = \frac{GM}{(R + R/2)^{2}} = \frac{GM}{(3R/2)^{2}} = \frac{4GM}{9R^{2}} \ldots(1)$
The acceleration due to gravity at a depth $d$ is given by $g_{d} = g(1 - \frac{d}{R}) = \frac{GM}{R^{2}}(1 - \frac{d}{R}) = \frac{GM(R-d)}{R^{3}} \ldots(2)$
Since $g_{1} = g_{d}$,we equate equations $(1)$ and $(2)$:
$\frac{4GM}{9R^{2}} = \frac{GM(R-d)}{R^{3}}$
$\frac{4}{9} = \frac{R-d}{R}$
$\frac{4}{9} = 1 - \frac{d}{R}$
$\frac{d}{R} = 1 - \frac{4}{9} = \frac{5}{9}$

(C) The value of acceleration due to gravity at a depth $d$ below the surface of the Earth is given by the formula:
$g' = g \left(1 - \frac{d}{R}\right)$
where $g$ is the acceleration due to gravity at the surface and $R$ is the radius of the Earth.
According to the problem,the value of acceleration at depth $d$ is $\frac{1}{n}$ times the value at the surface:
$g' = \frac{g}{n}$
Substituting this into the formula:
$\frac{g}{n} = g \left(1 - \frac{d}{R}\right)$
Dividing both sides by $g$:
$\frac{1}{n} = 1 - \frac{d}{R}$
Rearranging the terms to solve for $d$:
$\frac{d}{R} = 1 - \frac{1}{n}$
$\frac{d}{R} = \frac{n-1}{n}$
$d = R \left(\frac{n-1}{n}\right)$

(B) Acceleration due to gravity at altitude $h$ is given by the formula:
$g^{\prime} = \frac{g}{(1 + \frac{h}{R_{e}})^2}$
Given $h = \frac{R_{e}}{2}$,we substitute this into the equation:
$g^{\prime} = \frac{g}{(1 + \frac{R_{e}/2}{R_{e}})^2} = \frac{g}{(1 + \frac{1}{2})^2} = \frac{g}{(\frac{3}{2})^2} = \frac{4g}{9} \dots(i)$
Weight of the body at the earth's surface is $w = mg = 63 \ N \dots(ii)$
Weight of the body at altitude $h = \frac{R_{e}}{2}$ is $w^{\prime} = mg^{\prime} = m(\frac{4g}{9}) = \frac{4}{9}mg \dots(iii)$
Substituting the value from equation $(ii)$ into equation $(iii)$:
$w^{\prime} = \frac{4}{9} \times 63 = 4 \times 7 = 28 \ N$

(A) The acceleration due to gravity $(g)$ near the earth's surface is given by $g = \frac{GM}{R^2} \approx 9.8 \, m/s^2$.
The acceleration of free fall $(g_{\text{eff}})$ considering the rotation of the earth is given by $g_{\text{eff}} = g - \omega^2 R \cos^2 \phi$. Near the equator $(\phi = 0)$,this becomes $g_{\text{eff}} = g - \omega^2 R$.
The difference between the acceleration due to gravity and the acceleration of free fall is $\Delta g = g - g_{\text{eff}} = \omega^2 R$.
Given $R = 6347 \times 10^3 \, m$ and the angular velocity of the earth $\omega = \frac{2\pi}{T}$,where $T = 24 \times 3600 \, s$.
$\Delta g = \left(\frac{2\pi}{86400}\right)^2 \times 6347 \times 10^3$.
$\Delta g = \left(\frac{6.283}{86400}\right)^2 \times 6347000 \approx (7.27 \times 10^{-5})^2 \times 6347000$.
$\Delta g \approx 5.285 \times 10^{-9} \times 6347000 \approx 0.0335 \, m/s^2 \approx 0.0340 \, m/s^2$.

(C) The gravitational field $g'$ inside the Earth at a distance $r$ from the center is given by the formula:
$g' = \frac{G M r}{R^3}$
We know that the acceleration due to gravity at the surface is $g = \frac{G M}{R^2} \approx 9.8\, m/s^2$.
Substituting $g$ into the formula,we get:
$g' = g \times \frac{r}{R}$
Given values:
$g = 9.8\, m/s^2$
$r = 2000\, km$
$R = 6400\, km$
Calculating the value:
$g' = 9.8 \times \frac{2000}{6400}$
$g' = 9.8 \times \frac{20}{64}$
$g' = 9.8 \times 0.3125$
$g' = 3.0625\, m/s^2$
Rounding to two decimal places,we get $3.06\, m/s^2$.

(D) The acceleration due to gravity at the surface of a planet is given by $g = \frac{GM}{R^2}$.
Since $M = V \rho = \frac{4}{3} \pi R^3 \rho$,we can write $g = \frac{G (\frac{4}{3} \pi R^3 \rho)}{R^2} = \frac{4}{3} \pi G \rho R$.
Assuming the radius of the planet is the same as the Earth $(R_p = R_e)$ unless otherwise specified,or evaluating the ratio based on the given constants:
Given $\rho_p = \rho_e$ and $G_p = 2G_e$.
The ratio of acceleration due to gravity is $\frac{g_p}{g_e} = \frac{\frac{4}{3} \pi G_p \rho_p R_p}{\frac{4}{3} \pi G_e \rho_e R_e}$.
Assuming $R_p = R_e$,we get $\frac{g_p}{g_e} = \frac{G_p}{G_e} = \frac{2G_e}{G_e} = 2$.
Thus,the ratio is $2:1$.

(D) For bodies at the equator to start floating,the gravitational force must be equal to the required centripetal force.
$mg = m \omega^2 R$
where $\omega$ is the angular velocity of the Earth and $R$ is the radius of the Earth.
Solving for $\omega$:
$\omega = \sqrt{\frac{g}{R}}$
The duration of the day $T$ is given by $T = \frac{2\pi}{\omega}$.
Substituting $\omega$:
$T = 2\pi \sqrt{\frac{R}{g}}$
Given $g = 10 \, m/s^2$ and $R = 6400 \times 10^3 \, m$:
$T = 2 \times 3.14 \times \sqrt{\frac{6400 \times 10^3}{10}}$
$T = 6.28 \times \sqrt{640000}$
$T = 6.28 \times 800 = 5024 \, s$
To convert the duration into minutes:
$T_{\text{min}} = \frac{5024}{60} \approx 83.73 \, \text{minutes}$
Rounding to the nearest integer,the duration is approximately $84 \, \text{minutes}$.

(B) The weight of the body at the poles is given by $W_p = mg = 49 \, N$.
Since $g = 9.8 \, m/s^2$,the mass of the body is $m = \frac{49}{9.8} = 5 \, kg$.
At the equator,the effective acceleration due to gravity is $g_e = g - R\omega^2$,where $\omega$ is the angular velocity of the Earth.
The angular velocity $\omega = \frac{2\pi}{T}$,where $T = 24 \times 3600 \, s$.
$R\omega^2 = R \left(\frac{2\pi}{T}\right)^2 = 6.4 \times 10^6 \times \left(\frac{2 \times 3.14}{86400}\right)^2 \approx 0.0337 \, m/s^2$.
The weight at the equator is $W_e = m(g - R\omega^2) = 5 \times (9.8 - 0.0337) = 5 \times 9.7663 = 48.8315 \, N$.
Rounding to two decimal places,the weight is $48.83 \, N$.

(C) Let $R$ be the radius of the Earth. Point $A$ is at a distance $r$ from the center $O$ inside the Earth. Point $C$ is at a height $h = 3200 \text{ km} = R/2$ above the surface $B$.
The acceleration due to gravity at point $A$ (inside the Earth) is given by $g_A = \frac{G M r}{R^3}$.
The acceleration due to gravity at point $C$ (outside the Earth) is given by $g_C = \frac{G M}{(R + h)^2} = \frac{G M}{(R + R/2)^2} = \frac{G M}{(3R/2)^2} = \frac{4 G M}{9 R^2}$.
Given that $g_A = g_C$,we equate the expressions:
$\frac{G M r}{R^3} = \frac{4 G M}{9 R^2}$
Solving for $r$:
$r = \frac{4 R}{9}$.
Thus,$OA = r = \frac{4 R}{9}$.
The distance $AB = R - r = R - \frac{4 R}{9} = \frac{5 R}{9}$.
Therefore,the ratio $OA : AB = \frac{4 R}{9} : \frac{5 R}{9} = 4 : 5$.
Since the ratio is $x : y = 4 : 5$,the value of $x$ is $4$.

(D) The acceleration due to gravity at a height $r$ above the Earth's surface is given by $g_{up} = \frac{g}{(1 + \frac{r}{R_{E}})^{2}}$.
The acceleration due to gravity at a depth $r$ below the Earth's surface is given by $g_{down} = g(1 - \frac{r}{R_{E}})$.
The ratio of acceleration due to gravity at depth $r$ to that at height $r$ is:
$\frac{g_{down}}{g_{up}} = \frac{g(1 - \frac{r}{R_{E}})}{\frac{g}{(1 + \frac{r}{R_{E}})^{2}}} = (1 - \frac{r}{R_{E}})(1 + \frac{r}{R_{E}})^{2}$.
Expanding the expression:
$= (1 - \frac{r}{R_{E}})(1 + \frac{2r}{R_{E}} + \frac{r^{2}}{R_{E}^{2}})$
$= 1 + \frac{2r}{R_{E}} + \frac{r^{2}}{R_{E}^{2}} - \frac{r}{R_{E}} - \frac{2r^{2}}{R_{E}^{2}} - \frac{r^{3}}{R_{E}^{3}}$
$= 1 + \frac{r}{R_{E}} - \frac{r^{2}}{R_{E}^{2}} - \frac{r^{3}}{R_{E}^{3}}$.

(D) Given that the density of the planet is equal to the density of Earth,$\rho_p = \rho_e$.
Since density $\rho = \frac{M}{\frac{4}{3}\pi R^3}$,we have $\frac{M_p}{R_p^3} = \frac{M_e}{R_e^3}$.
This implies $\frac{R_p}{R_e} = \left(\frac{M_p}{M_e}\right)^{1/3}$.
Given $M_p = 2 M_e$,we get $\frac{R_p}{R_e} = 2^{1/3}$.
The weight of an object is $W = mg = m \frac{GM}{R^2}$.
Therefore,$\frac{W_p}{W_e} = \frac{M_p}{M_e} \left(\frac{R_e}{R_p}\right)^2$.
Substituting the values,$\frac{W_p}{W_e} = 2 \times \left(\frac{1}{2^{1/3}}\right)^2 = 2 \times 2^{-2/3} = 2^{1 - 2/3} = 2^{1/3}$.
Thus,$W_p = 2^{1/3} W$.

(B) The weight of a body at height $h$ is given by $W' = m g'$,where $g' = g \left( \frac{R}{R+h} \right)^2$.
Given that the weight at height $h$ is $\frac{1}{3}$ of the weight on the surface,we have $m g' = \frac{1}{3} m g$,which implies $g' = \frac{g}{3}$.
Substituting the expression for $g'$,we get $g \left( \frac{R}{R+h} \right)^2 = \frac{g}{3}$.
Taking the square root on both sides,we get $\frac{R}{R+h} = \frac{1}{\sqrt{3}}$.
Rearranging the equation,$R+h = R\sqrt{3}$,so $h = R(\sqrt{3} - 1)$.
Substituting the values $R = 6400 \, km$ and $\sqrt{3} = 1.732$,we get $h = 6400 \times (1.732 - 1) = 6400 \times 0.732$.
$h = 4684.8 \, km \approx 4685 \, km$.

(D) Let $R$ be the radius of the Earth. The diameter of the Earth is $2R$.
Given that the height $h$ of point $P$ above the surface is equal to the diameter of the Earth,so $h = 2R$.
The distance of point $P$ from the center of the Earth is $r = R + h = R + 2R = 3R$.
The acceleration due to gravity at the surface of the Earth is $g = \frac{GM}{R^2}$.
The acceleration due to gravity at point $P$ at a distance $r$ from the center is $g' = \frac{GM}{r^2}$.
Substituting $r = 3R$ into the equation,we get:
$g' = \frac{GM}{(3R)^2} = \frac{GM}{9R^2}$.
Since $g = \frac{GM}{R^2}$,we can write:
$g' = \frac{1}{9} \left( \frac{GM}{R^2} \right) = \frac{g}{9}$.

(D) The effective acceleration due to gravity $(g')$ on the surface of the rotating Earth is given by the vector sum of the gravitational acceleration $(g)$ and the centrifugal acceleration $(rw^2)$.
The magnitude of the effective acceleration due to gravity is $g' = \sqrt{g^2 + (rw^2)^2 - 2g(rw^2)\cos\theta}$,where $\theta$ is the latitude.
As we move from the poles to the equator,the magnitude of the effective acceleration due to gravity changes because the centrifugal component varies with the latitude $\theta$.
Furthermore,the direction of the effective acceleration due to gravity is not always towards the center of the Earth,except at the poles and the equator.
Therefore,Assertion $A$ is false because the magnitude varies and the direction does not always point towards the center of the Earth.
Reason $R$ states that at the equator,the direction of acceleration due to gravity is towards the center of the Earth. This is true because at the equator,the centrifugal acceleration $(rw^2)$ is directed radially outward,and the gravitational acceleration $(g)$ is directed radially inward. Their resultant,the effective acceleration $(g')$,is also directed radially inward towards the center of the Earth.
Thus,$A$ is false but $R$ is true.

(A) The acceleration due to gravity $(g)$ at a distance $(r)$ from the center of the earth is given by:
$1$. Inside the earth $(r \leq R)$: $g = \frac{GMr}{R^3}$. Here,$g$ is directly proportional to $r$ $(g \propto r)$,which represents a straight line passing through the origin.
$2$. Outside the earth $(r \geq R)$: $g = \frac{GM}{r^2}$. Here,$g$ is inversely proportional to the square of the distance $(g \propto 1/r^2)$,which represents a parabolic curve.
Thus,the graph shows a linear increase up to $r = R$ and a non-linear decrease for $r > R$. This corresponds to the graph shown in option $A$.