Gravitational Intensity Questions in English

(A) Consider the circular ring formed by the rod of length $L$ and mass $M$. Let the radius of this circle be $r$. The circumference is $2 \pi r = L$, so $r = \frac{L}{2 \pi}$.
Consider two diametrically opposite small mass segments $dM_1$ and $dM_2$ on the ring.
The gravitational force exerted by segment $dM_1$ on mass $m$ at the centre is $F_1 = \frac{G m dM_1}{r^2}$ directed towards $dM_1$.
The gravitational force exerted by the diametrically opposite segment $dM_2$ on mass $m$ at the centre is $F_2 = \frac{G m dM_2}{r^2}$ directed towards $dM_2$.
Since the segments are diametrically opposite, the forces $F_1$ and $F_2$ are equal in magnitude and opposite in direction $(F_1 = -F_2)$.
Therefore, the net force due to these two segments is $F_1 + F_2 = 0$.
By symmetry, every mass segment on the ring has a corresponding diametrically opposite segment that exerts an equal and opposite gravitational force on the mass $m$ at the centre.
Summing these forces over the entire ring, the net gravitational force acting on the mass $m$ at the centre is zero.

(B) The gravitational field $E$ at a distance $x$ from the centre of a system consisting of a solid sphere (mass $m$,radius $r$) and a concentric shell (mass $m$,radius $2r$) is given by the superposition of the fields of the two objects.
For $r < x < 2r$,the field due to the shell is $0$ (inside the shell),and the field due to the solid sphere is $\frac{Gm}{x^2}$. Thus,$E = \frac{Gm}{x^2}$.
For $x = 1.5r = \frac{3}{2}r$,$E = \frac{Gm}{(1.5r)^2} = \frac{Gm}{2.25r^2} = \frac{4}{9} \frac{Gm}{r^2}$.
For $x > 2r$,both act as point masses at the centre. $E = \frac{G(m+m)}{x^2} = \frac{2Gm}{x^2}$.
For $x = 2.5r = \frac{5}{2}r$,$E = \frac{2Gm}{(2.5r)^2} = \frac{2Gm}{6.25r^2} = \frac{2Gm}{(25/4)r^2} = \frac{8}{25} \frac{Gm}{r^2}$.
Thus,option $B$ is correct.

(C) For a uniform sphere of mass $M$ and radius $R$:
$1$. At a distance $r_1 > R$ (outside the sphere),the gravitational field is given by $E_1 = \frac{GM}{r_1^2}$.
$2$. At a distance $r_2 < R$ (inside the sphere),the gravitational field is given by $E_2 = \frac{GMr_2}{R^3}$.
$3$. The ratio $E_1 : E_2$ is calculated as:
$\frac{E_1}{E_2} = \frac{GM/r_1^2}{GMr_2/R^3} = \frac{GM}{r_1^2} \times \frac{R^3}{GMr_2} = \frac{R^3}{r_1^2 r_2}$.

(C) For a point outside the sphere $(r > R)$,the gravitational field is given by $F = \frac{G M}{r^2}$.
Thus,for $r_1 > R$,$F_1 = \frac{G M}{r_1^2}$.
For a point inside the sphere $(r < R)$,the gravitational field is given by $F = \frac{G M r}{R^3}$.
Thus,for $r_2 < R$,$F_2 = \frac{G M r_2}{R^3}$.
Taking the ratio of $F_1$ to $F_2$:
$\frac{F_1}{F_2} = \frac{G M}{r_1^2} \times \frac{R^3}{G M r_2} = \frac{R^3}{r_1^2 r_2}$.

(A) Let $M_E$ be the mass of the Earth and $M_M$ be the mass of the Moon. Given $M_E = 81 M_M$.
Let $x$ be the distance from the centre of the Earth where the net gravitational force on a test mass $m$ is zero.
At this point,the gravitational pull from the Earth must be equal in magnitude to the gravitational pull from the Moon.
$\frac{G M_E}{x^2} = \frac{G M_M}{(R - x)^2}$
Substituting $M_E = 81 M_M$:
$\frac{G (81 M_M)}{x^2} = \frac{G M_M}{(R - x)^2}$
$\frac{81}{x^2} = \frac{1}{(R - x)^2}$
Taking the square root on both sides:
$\frac{9}{x} = \frac{1}{R - x}$
$9(R - x) = x$
$9R - 9x = x$
$9R = 10x$
$x = \frac{9}{10} R$

(D) The gravitational field intensity $I$ due to a point mass $m$ at a distance $r$ is given by $I = \frac{Gm}{r^2}$.
Since there are infinite masses,the total gravitational field intensity at point $O$ is the sum of the intensities due to each mass:
$I_{total} = \sum \frac{Gm_i}{r_i^2} = G \sum \frac{3}{r_i^2} = 3G \left( \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{8^2} + \dots \right)$.
This is a geometric series with the first term $a = 1$ and common ratio $r = \frac{1}{4}$.
The sum of an infinite geometric series is $S = \frac{a}{1 - r}$.
Substituting the values,$S = \frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}$.
Therefore,$I_{total} = 3G \times \frac{4}{3} = 4G$.

(C) The gravitational field inside a uniform spherical shell is zero at every point.
Let $I_1$ be the gravitational field at point $P$ due to the upper part of the shell and $I_2$ be the gravitational field at point $P$ due to the lower part of the shell.
The total gravitational field at point $P$ due to the entire shell is the vector sum of the fields due to its parts: $\vec{I}_{total} = \vec{I}_1 + \vec{I}_2 = 0$.
Since the point $P$ lies on the plane of the cut,the field vectors $\vec{I}_1$ and $\vec{I}_2$ must be equal in magnitude and opposite in direction to satisfy the condition $\vec{I}_1 + \vec{I}_2 = 0$.
Therefore,the magnitudes of the gravitational fields are equal: $I_1 = I_2$.

(B) Let the two masses be $m_1 = 90 \ kg$ at point $A$ and $m_2 = 160 \ kg$ at point $B$. The point $C$ is at a distance $r_1 = 3 \ m$ from $A$ and $r_2 = 4 \ m$ from $B$. The distance between $A$ and $B$ is $5 \ m$.
Since $3^2 + 4^2 = 5^2$,the triangle $ABC$ is a right-angled triangle with the right angle at $C$.
The gravitational field intensity at $C$ due to mass $m_1$ is $E_A = \frac{G m_1}{r_1^2} = \frac{G \times 90}{3^2} = 10G$.
The gravitational field intensity at $C$ due to mass $m_2$ is $E_B = \frac{G m_2}{r_2^2} = \frac{G \times 160}{4^2} = 10G$.
Since the angle between $E_A$ and $E_B$ is $90^\circ$,the resultant intensity $E$ is given by:
$E = \sqrt{E_A^2 + E_B^2} = \sqrt{(10G)^2 + (10G)^2} = 10G\sqrt{2}$.
Substituting $G = 6.67 \times 10^{-11} \ N \ m^2 \ kg^{-2}$:
$E = 10 \times 6.67 \times 10^{-11} \times 1.414 \approx 9.43 \times 10^{-10} \ N \ kg^{-1}$.

(C) Given, gravitational field intensity, $I = (5 \hat{i} + 12 \hat{j}) \text{ N kg}^{-1}$.
Mass of the object, $m = 3 \text{ kg}$.
The change in gravitational potential energy $\Delta U$ is given by the negative of the work done by the gravitational field, or $\Delta U = -W = -m \int \vec{I} \cdot d\vec{r}$.
However, in many contexts, the magnitude of the change in potential energy is calculated as $\Delta U = m \int_{A}^{B} \vec{I} \cdot d\vec{r}$ considering the work done against the field.
Using the formula $\Delta U = m \int_{(0,0)}^{(8,-2)} \vec{I} \cdot d\vec{r}$:
$\Delta U = 3 \int_{(0,0)}^{(8,-2)} (5 \hat{i} + 12 \hat{j}) \cdot (dx \hat{i} + dy \hat{j})$
$\Delta U = 3 [5x + 12y]_{(0,0)}^{(8,-2)}$
$\Delta U = 3 [5(8) + 12(-2)] - 3 [5(0) + 12(0)]$
$\Delta U = 3 [40 - 24] = 3 [16] = 48 \text{ J}$.

(D) The gravitational field $E$ at a distance $x$ from the centre of a ring of mass $M$ and radius $R$ along its axis is given by:
$E = \frac{GMx}{(R^2 + x^2)^{3/2}}$
To find the point where the gravitational field is strongest,we differentiate $E$ with respect to $x$ and set it to zero:
$\frac{dE}{dx} = 0$
Using the quotient rule:
$\frac{d}{dx} \left[ \frac{GMx}{(R^2 + x^2)^{3/2}} \right] = GM \left[ \frac{(R^2 + x^2)^{3/2}(1) - x \cdot \frac{3}{2}(R^2 + x^2)^{1/2}(2x)}{(R^2 + x^2)^3} \right] = 0$
This simplifies to:
$(R^2 + x^2)^{3/2} - 3x^2(R^2 + x^2)^{1/2} = 0$
$(R^2 + x^2)^{1/2} [R^2 + x^2 - 3x^2] = 0$
$R^2 - 2x^2 = 0$
$2x^2 = R^2$
$x = \frac{R}{\sqrt{2}}$
Thus,the gravitational field is strongest at a distance of $\frac{R}{\sqrt{2}}$ from the centre.

(D) The volume of a spherical shell of thickness $dr$ and radius $r$ is $dV = 4\pi r^2 dr$.
The mass of this shell is $dM = \rho dV = \left( 20 \frac{r^2}{R^2} \right) \cdot 4\pi r^2 dr = \frac{80\pi r^4}{R^2} dr$.
The total mass $M$ of the solid sphere is obtained by integrating from $0$ to $R$:
$M = \int_0^R dM = \int_0^R \frac{80\pi r^4}{R^2} dr = \frac{80\pi}{R^2} \left[ \frac{r^5}{5} \right]_0^R = \frac{80\pi R^5}{5R^2} = 16\pi R^3$.
For a point outside the sphere at a distance $r = 4R$,the sphere acts as a point mass $M$ at its centre. The gravitational field $E$ is given by:
$E = \frac{GM}{r^2} = \frac{G(16\pi R^3)}{(4R)^2} = \frac{16\pi G R^3}{16R^2} = \pi GR$.
Therefore,the ratio $\frac{E}{GR} = \pi$.

(C) The mass $M(r)$ enclosed in a sphere of radius $r$ is given by:
$M(r) = \int_0^r \rho(r') \cdot 4\pi r'^2 dr' = \int_0^r \rho_0 \left(\frac{r'}{R}\right)^\beta \cdot 4\pi r'^2 dr' = \frac{4\pi \rho_0}{R^\beta} \int_0^r r'^{\beta+2} dr' = \frac{4\pi \rho_0}{R^\beta} \cdot \frac{r^{\beta+3}}{\beta+3}$.
For $r = \frac{R}{2}$,the mass enclosed is $M_1 = \frac{4\pi \rho_0}{R^\beta} \cdot \frac{(R/2)^{\beta+3}}{\beta+3} = \frac{4\pi \rho_0 R^3}{\beta+3} \cdot \frac{1}{2^{\beta+3}}$.
The gravitational field at $r = \frac{R}{2}$ is $E_1 = \frac{G M_1}{(R/2)^2} = \frac{G}{(R/2)^2} \cdot \frac{4\pi \rho_0 R^3}{(\beta+3) 2^{\beta+3}} = \frac{16 G \pi \rho_0 R}{(\beta+3) 2^{\beta+3}}$.
For $r = 2R$,the total mass of the sphere is $M_2 = \frac{4\pi \rho_0}{R^\beta} \cdot \frac{R^{\beta+3}}{\beta+3} = \frac{4\pi \rho_0 R^3}{\beta+3}$.
The gravitational field at $r = 2R$ is $E_2 = \frac{G M_2}{(2R)^2} = \frac{G}{4R^2} \cdot \frac{4\pi \rho_0 R^3}{\beta+3} = \frac{G \pi \rho_0 R}{\beta+3}$.
Given $\frac{E_2}{E_1} = 4$:
$\frac{\frac{G \pi \rho_0 R}{\beta+3}}{\frac{16 G \pi \rho_0 R}{(\beta+3) 2^{\beta+3}}} = 4 \Rightarrow \frac{2^{\beta+3}}{16} = 4 \Rightarrow 2^{\beta+3} = 64 = 2^6$.
Therefore,$\beta + 3 = 6$,which gives $\beta = 3$.

(B) The gravitational field $E$ at the origin due to a point mass $m$ at distance $r$ is given by $E = \frac{Gm}{r^2}$.
Since the masses are placed at distances $r = 1, 2, 4, 8, 16, \dots$,the total gravitational field at the origin is the sum of the fields due to each mass:
$E = \frac{Gm}{1^2} + \frac{Gm}{2^2} + \frac{Gm}{4^2} + \frac{Gm}{8^2} + \dots$
$E = Gm \left( 1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \dots \right)$
The term in the bracket is an infinite geometric progression with the first term $a = 1$ and common ratio $r = \frac{1}{4}$.
The sum of an infinite geometric series is $S = \frac{a}{1 - r}$.
$S = \frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}$.
Therefore,the total gravitational field is $E = Gm \left( \frac{4}{3} \right) = \frac{4}{3} Gm$.

(A) Let the distance from the centre to each corner be $r$. Let a test mass $m_{0}$ be placed at the centre.
In the initial configuration,the forces exerted by the masses at the corners on $m_{0}$ are directed towards the corners. Let $k = \frac{Gm_{0}}{r^{2}}$.
The forces are $F_{M} = kM$,$F_{2M} = 2kM$,$F_{3M} = 3kM$,and $F_{4M} = 4kM$.
Opposite pairs are $(M, 3M)$ and $(2M, 4M)$.
The net force along the diagonal with $M$ and $3M$ is $F_{diag1} = (3M - M)k = 2kM$ (towards $3M$).
The net force along the diagonal with $2M$ and $4M$ is $F_{diag2} = (4M - 2M)k = 2kM$ (towards $4M$).
Since the diagonals are perpendicular,$F_{1} = \sqrt{(2kM)^{2} + (2kM)^{2}} = 2\sqrt{2}kM$.
In the new configuration,$3M$ and $4M$ are interchanged. The corners now have $M, 2M, 4M,$ and $3M$ in order.
Opposite pairs are $(M, 4M)$ and $(2M, 3M)$.
The net force along the diagonal with $M$ and $4M$ is $F'_{diag1} = (4M - M)k = 3kM$ (towards $4M$).
The net force along the diagonal with $2M$ and $3M$ is $F'_{diag2} = (3M - 2M)k = kM$ (towards $3M$).
$F_{2} = \sqrt{(3kM)^{2} + (kM)^{2}} = \sqrt{9+1}kM = \sqrt{10}kM$.
Ratio $\frac{F_{1}}{F_{2}} = \frac{2\sqrt{2}kM}{\sqrt{10}kM} = \frac{2\sqrt{2}}{\sqrt{2}\sqrt{5}} = \frac{2}{\sqrt{5}}$.
Comparing with $\frac{\alpha}{\sqrt{5}}$,we get $\alpha = 2$.