Mix Examples-Gravitation Questions in English

(C) Let the mass of the original sphere $A$ be $M$. The force $F$ on particle $B$ of mass $m$ at distance $2R$ is $F = \frac{GMm}{(2R)^2} = \frac{GMm}{4R^2}$.
The mass of the removed spherical portion of radius $r = R/2$ is $M^{\prime} = \rho \cdot \frac{4}{3} \pi (R/2)^3 = \frac{M}{\frac{4}{3} \pi R^3} \cdot \frac{4}{3} \pi \frac{R^3}{8} = \frac{M}{8}$.
The distance of the centre of the removed portion from the particle $B$ is $d = 2R - R/2 = 3R/2$.
The force exerted by the removed portion on $B$ is $F_{removed} = \frac{G M^{\prime} m}{d^2} = \frac{G (M/8) m}{(3R/2)^2} = \frac{GMm}{8 \cdot (9R^2/4)} = \frac{GMm}{18R^2}$.
Since $F = \frac{GMm}{4R^2}$,we have $\frac{GMm}{R^2} = 4F$. Thus,$F_{removed} = \frac{4F}{18} = \frac{2}{9} F$.
The new force $F^{\prime}$ is $F - F_{removed} = F - \frac{2}{9} F = \frac{7}{9} F$.

(C) As the particles move under gravitational influence,the gravitational force provides the necessary centripetal force.
The gravitational force between two bodies of mass $M$ separated by distance $L$ is $F = \frac{G M^2}{L^2}$.
The net force on a particle at vertex $A$ due to particles at $B$ and $C$ is the vector sum of the forces $\vec{F}_{AB}$ and $\vec{F}_{AC}$.
Since the angle between these forces is $60^\circ$,the magnitude of the resultant force is $F_{net} = 2 F \cos(30^\circ) = 2 \left( \frac{G M^2}{L^2} \right) \frac{\sqrt{3}}{2} = \frac{\sqrt{3} G M^2}{L^2}$.
The particles move in a circle of radius $R$. For an equilateral triangle,the distance from the centroid to a vertex is $R = \frac{L}{\sqrt{3}}$.
Equating the net force to the centripetal force,$\frac{M v^2}{R} = F_{net}$.
Substituting the values,$\frac{M v^2}{L/\sqrt{3}} = \frac{\sqrt{3} G M^2}{L^2}$.
Solving for $v$,$v^2 = \frac{\sqrt{3} G M^2}{L^2} \cdot \frac{L}{\sqrt{3} M} = \frac{G M}{L}$.
Therefore,$v = \sqrt{\frac{G M}{L}}$.

(C) The gravitational force is a conservative force. For any conservative force,the work done in a complete closed path (one complete revolution) is always zero.
Thus,statement (iii) is true.
Work done is given by the dot product $W = \int \vec{F} \cdot d\vec{r}$.
Since $\vec{F} = m\vec{a}$,we have $W = \int m\vec{a} \cdot d\vec{r}$.
If the acceleration vector $\vec{a}$ (which is directed towards the sun) is perpendicular to the displacement vector $d\vec{r}$ (which is tangent to the orbit),the work done is zero.
In an elliptical orbit,there are specific points (at perihelion and aphelion) where the velocity is perpendicular to the radius vector,meaning the force is perpendicular to the displacement at those instantaneous points.
Therefore,the work done is zero at these specific points,making statement (ii) true.
Statement $(i)$ is false because work is zero at specific points.
Statement (iv) is false because work is not zero for any arbitrary small part of the orbit.
Hence,(ii) and (iii) are true.

(C) The properties of the four basic forces in nature are given in the following table:
| Force | Approximate Relative Strength | Range | Attraction/Repulsion |
| :--- | :--- | :--- | :--- |
| Gravitational | $10^{-38}$ | Infinite | Attractive only |
| Electromagnetic | $10^{-2}$ | Infinite | Attractive and repulsive |
| Weak nuclear | $10^{-13}$ | $ < 10^{-16} \,m$ | Attractive and repulsive |
| Strong nuclear | $1$ | $ < 10^{-15} \,m$ | Attractive and repulsive |
From the table,we can observe the following:
$1$. Gravitational and electromagnetic forces have an infinite range.
$2$. The range of the weak nuclear force ($ < 10^{-16} \,m$) is smaller than the range of the strong nuclear force ($ < 10^{-15} \,m$),as well as the infinite range of gravitational and electromagnetic forces.
Therefore,option $(c)$ is correct.

(C) Given: Distance between Sun and Earth,$r = 1.6 \times 10^{11} \,m$. Radius of Earth,$R_e = 6.4 \times 10^6 \,m$. Mass of Earth,$M_e = 6.0 \times 10^{24} \,kg$.
Angular momentum of Earth around the Sun $(L_1)$: $L_1 = M_e v r = M_e (\frac{2 \pi r}{T_1}) r = \frac{2 \pi M_e r^2}{T_1}$.
Here,$T_1 = 365 \times 24 \times 3600 \,s \approx 3.15 \times 10^7 \,s$.
$L_1 = \frac{2 \times 3.14 \times 6.0 \times 10^{24} \times (1.6 \times 10^{11})^2}{3.15 \times 10^7} \approx 3.06 \times 10^{40} \,kg \cdot m^2/s$.
Angular momentum of Earth about its own axis $(L_2)$: $L_2 = I \omega = (\frac{2}{5} M_e R_e^2) (\frac{2 \pi}{T_2})$.
Here,$T_2 = 24 \times 3600 \,s = 8.64 \times 10^4 \,s$.
$L_2 = \frac{2}{5} \times 6.0 \times 10^{24} \times (6.4 \times 10^6)^2 \times \frac{2 \times 3.14}{8.64 \times 10^4} \approx 7.15 \times 10^{33} \,kg \cdot m^2/s$.
Ratio $\frac{L_1}{L_2} = \frac{3.06 \times 10^{40}}{7.15 \times 10^{33}} \approx 4.28 \times 10^6 \approx 4.3 \times 10^6$.

(B) The total kinetic energy of the earth is the sum of its rotational kinetic energy about its own axis and its translational kinetic energy due to its orbital motion around the sun.
Rotational kinetic energy $K_{rot} = \frac{1}{2} I \omega_0^2$,where $I = \frac{2}{5} M R^2$ for a solid sphere.
So,$K_{rot} = \frac{1}{2} \times \frac{2}{5} M R^2 \omega_0^2 = \frac{1}{5} M R^2 \omega_0^2$.
Translational kinetic energy $K_{trans} = \frac{1}{2} M v^2$,where $v = D \omega$.
So,$K_{trans} = \frac{1}{2} M (D \omega)^2 = \frac{1}{2} M D^2 \omega^2$.
Total kinetic energy $K_{total} = K_{rot} + K_{trans} = \frac{1}{5} M R^2 \omega_0^2 + \frac{1}{2} M D^2 \omega^2$.
Factoring out $\frac{M R^2 \omega_0^2}{5}$,we get $K_{total} = \frac{M R^2 \omega_0^2}{5} \left[ 1 + \frac{5}{2} \frac{M D^2 \omega^2}{M R^2 \omega_0^2} \right] = \frac{M R^2 \omega_0^2}{5} \left[ 1 + \frac{5}{2} \left( \frac{D \omega}{R \omega_0} \right)^2 \right]$.