Gravitation

To determine the relative rate of change of the separation between the two stars, let's first understand the relationship governing their motion.

Equilibrium of Forces:

$ m_2 \omega^2 r = \frac{G m_1 m_2}{r^2} $

Here, $ \omega $ is the angular velocity. From the balance of gravitational force and the centripetal force, we can solve for $ \omega $:

$ \omega = \sqrt{\frac{G m_1}{r^3}} $

Angular Momentum (L):

$ L = m_2 \omega r^2 = m_2 \sqrt{\frac{G m_1}{r^3}} r^2 $

Simplifying, we find:

$ L = m_2 \sqrt{G m_1 r} = \text{constant} $

Differential Relation:

Taking the natural logarithm of the angular momentum expression:

$ \ln L = \ln m_2 + \ln G + \frac{1}{2} \ln m_1 + \frac{1}{2} \ln r $

Since $ L $ is constant, differentiate both sides:

$ 0 = \frac{dm_2}{m_2} + \frac{1}{2} \frac{dm_1}{m_1} + \frac{1}{2} \frac{dr}{r} $

Rate of Change of Separation (r):

Solving for $\frac{dr}{r}$, we arrive at:

$ \frac{1}{r} \frac{dr}{dt} = -\frac{2}{m_2} \frac{dm_2}{dt} - \frac{1}{m_1} \frac{dm_1}{dt} $

Since mass is transferred at a constant rate $ \gamma $, we have $ \frac{dm_1}{dt} = \gamma $ and $ \frac{dm_2}{dt} = - \gamma $.

Substitute the Variable Rates:

Substituting these values:

$ \frac{1}{r} \frac{dr}{dt} = -\frac{2(-\gamma)}{m_2} - \frac{\gamma}{m_1} \approx -\frac{2\gamma}{m_2} $

Thus, the relative rate of change of the separation between the stars is approximately $-\frac{2\gamma}{m_2}$ s$^{-1}$.

To find the effect of an additional central force on the time period of a particle in a circular orbit, consider the following:

Initial Gravitational Force ($F_1$):

The force due to gravity for a particle in a circular orbit of radius $r_0$ is given by:

$ F_1 = \frac{GMm}{r_0^2} $

where $ G $ is the gravitational constant, $ M $ is the mass of the central body, and $ m $ is the mass of the particle.

Modified Force with Additional Potential ($F_2$):

When an additional force derived from the potential energy $ V_{\mathrm{c}}(r) = \frac{m\alpha}{r^3} $ acts on the particle, the effective central force becomes:

$ F_2 = \frac{GMm}{r_0^2} - \frac{3m\alpha}{r_0^4} $

Relation of Angular Velocities:

The ratio of squares of the angular velocities ($\omega_0$ and $\omega_1$) before and after the additional force is:

$ \frac{\omega_1^2}{\omega_0^2} = \frac{F_2}{F_1} = \frac{\frac{GM}{r_0^2} - \frac{3\alpha}{r_0^4}}{\frac{GM}{r_0^2}} $

Relating Time Periods:

Since the angular velocity is inversely proportional to the time period, $\frac{\omega_1^2}{\omega_0^2} = \frac{T_0^2}{T_1^2}$. Thus, we have:

$ \frac{T_0^2}{T_1^2} = 1 - \frac{3\alpha}{GMr_0^2} $

Find the Change in Time Period:

The expression for the change in the square of the time periods is:

$ \frac{T_1^2 - T_0^2}{T_1^2} = \frac{3\alpha}{GMr_0^2} $

This shows how the additional central force, represented by the potential $ V_{\mathrm{c}}(r) = \frac{m\alpha}{r^3} $, affects the particle’s orbital period when combined with the gravitational field of the mass $ M $.

The formula to calculate the acceleration due to gravity at a height ( h ) from the surface of the Earth is expressed as :

$ g = \frac{G M_e}{(R_e + h)^2} $

To find the ratio of gravitational acceleration at two different heights ( h_P ) and ( h_Q ) above the Earth's surface, use the formula for each and form a ratio :

$ \frac{g_P}{g_Q} = \frac{\frac{G M_e}{(R_e + h_P)^2}}{\frac{G M_e}{(R_e + h_Q)^2}} = \frac{(R_e + h_Q)^2}{(R_e + h_P)^2} $

Given $ h_P = \frac{R_e}{3} $, the ratio simplifies to:

$ \frac{g_P}{g_Q} = \frac{(R_e + h_Q)^2}{(R_e + \frac{R_e}{3})^2} = \frac{36}{25} $

Solving for $ h_Q $ in terms of $ R_e $ yields $ h_Q = \frac{3R_e}{5} $.

Given : ${{{m_1}} \over {{m_2}}} = 2$ and ${{{R_1}} \over {{R_2}}} = {1 \over 4}$

Now,

${{GMm} \over {{R^2}}} = {{m{v^2}} \over R}$

$ \Rightarrow {{GM} \over R} = {v^2} \Rightarrow v = \sqrt {{{GM} \over R}} $

$ \Rightarrow {{{v_1}} \over {{v_2}}} = {{\sqrt {{{GM} \over {{R_1}}}} } \over {\sqrt {{{GM} \over {{R_2}}}} }} \Rightarrow {{{v_1}} \over {{v_2}}} = \sqrt {{{{R_2}} \over {{R_1}}}} $

$ \Rightarrow {{{v_1}} \over {{v_2}}} = \sqrt {{4 \over 1}} \Rightarrow {{{v_1}} \over {{v_2}}} = 2$

Thus, the correct mapping is P $\to$ 3.

Now, L = mvR

$ \Rightarrow {{{L_1}} \over {{L_2}}} = {{{m_1}{v_1}{R_1}} \over {{m_2}{v_2}{R_2}}} = {{{m_1}} \over {{m_2}}} \times {{{v_1}} \over {{v_2}}} \times {{{R_1}} \over {{R_2}}}$

$ \Rightarrow {{{L_1}} \over {{L_2}}} = 2 \times 2 \times {1 \over 4} = 1$

Thus, the correct mapping is Q $\to$ 2.

Now,

$K = {1 \over 2}m{v^2}$

$ \Rightarrow {{{K_1}} \over {{K_2}}} = {{{1 \over 2}{m_1}v_1^2} \over {{1 \over 2}{m_2}v_2^2}} = {{{m_1}} \over {{m_2}}} \times {\left( {{{{v_1}} \over {{v_2}}}} \right)^2}$

$ \Rightarrow {{{K_1}} \over {{K_2}}} = 2 \times {(2)^2} = 8$

Thus, the correct mapping is R $\to$ R.

Now,

$T = {{2\pi R} \over v}$

$ \Rightarrow {{{T_1}} \over {{T_2}}} = {{{{2\pi {R_1}} \over {{v_1}}}} \over {{{2\pi {R_2}} \over {{v_2}}}}} \Rightarrow {{{T_1}} \over {{T_2}}} = {{{R_1}} \over {{R_2}}} \times {{{v_2}} \over {{v_1}}}$

$ \Rightarrow {{{T_1}} \over {{T_2}}} = {1 \over 4} \times {1 \over 2} \Rightarrow {{{T_1}} \over {{T_2}}} = {1 \over 8}$

Thus, the correct mapping is S $\to$ 1.

Therefore, option (B) is correct.

Given : Mass of the sun, M_s = 3 $\times$ 10⁵ $\times$ mass of the earth = 3 $\times$ 10⁵ M_e

Distance between the sun and the earth,

d = 2.5 $\times$ 10⁴ $\times$ radius of the earth = 2.5 $\times$ 10⁴ R_e,

Escape speed, v_e = 11.2 km s^$-$1

$\therefore$ ${v_e} = \sqrt {{{2G{M_e}} \over {{R_e}}}} = 11.2$ km s^$-$1

Minimum velocity required for the rocket to escape the given earth $-$ sun system = v_s

Using energy conservation for the given situation,

${1 \over 2}mv_s^2 = {{G{M_e}m} \over {{R_e}}} + {{G{M_s}m} \over {d + {R_e}}} = {{G{M_e}m} \over {{R_e}}} + {{G \times 3 \times {{10}^5}{M_e}m} \over {(2.5 \times {{10}^4}{R_e} + {R_e})}}$

$\therefore$ $v_s^2 = {{2G{M_e}} \over {{R_e}}} + {{24G{M_e}} \over {{R_e}}}$ ($\because$ 2.5 $\times$ 10⁴ >> 1)

$ \Rightarrow v_s^2 = v_e^2 + 12v_e^2 = 13v_e^2$

or ${v_s} = \sqrt {13} $ v_e = 40.38 km s^$-$1 $\approx$ 42 km s^$-$1

Given R_e = 10R = 6 $\times$ 10⁶ m and g_e = 10 m/s². Consider a wire element of length dr placed at a distance r from the centre O.

Let $\rho$ be the common mass density of the planet and the earth and m be the mass of the planet inside the sphere of radius r i.e., $m = {4 \over 3}\pi {r^3}\rho $. The gravitational force on the wire element by the planet is equal to the force by a point mass of magnitude m placed at the centre O. Thus, force on the wire element is

$dF = {{Gm\lambda dr} \over {{r^2}}} = {4 \over 3}G\pi \rho \lambda rdr$.

Integrate dF from $r = {4 \over 5}R$ to r = R to get

$F = {4 \over 3}G\pi \rho \lambda \int_{{4 \over 5}R}^R {r\,dr = {4 \over 3}G\pi \rho \lambda \left( {{{9{R^2}} \over {50}}} \right)} $

$ = {{{4 \over 3}\pi \rho R_e^3G} \over {R_e^2}}\left( {{{9\lambda {R_e}} \over {5000}}} \right)$

$ = {g_e}{{9\lambda {R_e}} \over {5000}} = {{(10)(9)({{10}^{ - 3}})(6 \times {{10}^6})} \over {5000}} = 108$ N.

A particle escapes from the gravitational pull if its total energy (T) i.e., sum of kinetic energy (K) and potential energy (U), is greater than or equal to zero. The condition for just escape T = K + U = 0 i.e.,

K = $-$ U. .............. (1)

In a circular orbit of radius r, gravitational attraction provides the centripetal acceleration, mV²/r = GMm/r², which gives

$r = {{GM} \over {{V^2}}}$ ........ (2)

From equations (1) and (2), the kinetic energy of the particle at the time of injection is given by

$K = - U = - \left( { - {{GMm} \over r}} \right) = {{GMm} \over {GM/{V^2}}} = m{V^2}$.

We need to find gravitational potential energy of a unit mass placed at the point P.

The surface mass density of the annular disc is

$\sigma = M/(16\pi {R^2} - 9\pi {R^2}) = M/(7\pi {R^2})$.

Consider a small ring of radius r and thickness dr. The mass of the ring is dm = 2$\pi$r$\sigma $dr. As distance of any point of the ring from P is same, the potential at P due to the ring is

${V_P} = - {{G(2\pi r\sigma dr)} \over {\sqrt {16{R^2} + {r^2}} }}$

Integrate from r = 3R to r = 4R to get the potential energy of the unit mass placed at P

${V_P} = \int\limits_{3R}^{4R} { - {{GdM} \over {\sqrt {{{(4R)}^2} + {{(r)}^2}} }} = - {{GM2\pi } \over {7\pi {R^2}}}\int\limits_{3R}^{4R} {{{rdr} \over {\sqrt {16{R^2} + {r^2}} }}} } $

Solving, we get

${V_P} = - {{GM2\pi } \over {7\pi {R^2}}}\left[ {\sqrt {16{R^2} + {r^2}} } \right]_{3R}^{4R} = - {{2GM} \over {7R}}\left( {4\sqrt 2 - 5} \right)$

Work done in moving a unit mass from P to $\infty$ = V_$\infty$ $-$V_P

$ = 0 - \left( {{{ - 2GM} \over {7R}}\left( {4\sqrt 2 - 5} \right)} \right) = {{2GM} \over {7R}}\left( {4\sqrt 2 - 5} \right)$

Case (P) : Since $v$ = Constant, we have

$Mg\sin \theta = \mu Mg\cos \theta = f \Rightarrow \mu = \tan \theta $

Now, $N = Mg\cos \theta $. Therefore,

$R = \sqrt {{N^2} + {f^2}} = Mg\sqrt {{{\cos }^2}\theta + {{\sin }^2}\theta } $

Case (T) : We have $Mg = {F_v} + {F_b}$, where ${F_v}$ is the viscous force and ${F_b}$ is the buoyant force.

Case (S) : We have $Mg - {F_b} = Ma$.

Case (R) : We have $Mg = T$. Therefore,

$T + {M_0}g = {F_y}$ (by the clamp)

$T = {F_X}$ (by the clamp)

Therefore, the force exerted by the clamp X on pulley Y is

$F = \sqrt {F_X^2 + F_Y^2} = g\sqrt {{M^2} + {{(M + {m_0})}^2}} $

Case (Q) : We have

$Mg = {F_m}$ (for Z)

where ${F_m}$ is the magnetic repulsion. Also,

$Mg + {F_m} = N$ (for Y)

Therefore, $N = 2Mg$.

Note :

Option (A) : For option (T), if ${F_b}$ is ignored, then $Mg = {F_v}$; otherwise, no case matches for option (A).

Hence, (A) $\to$ (T), (P).

Option (B) : In option (Q), it is mentioned that the lift is moving up continuously; therefore, the gravitational potential energy of X goes on increasing. In option (B), as Y comes down, X goes up (displaced). The same is applicable for option (T).

Hence, (B) $\to$ (Q), (S), (T).

Option (C) : For option (P), since Y moves down with a constant $v$, the gravitational potential energy of the system X + Y goes on decreasing, similar is the case in Options (R) and (T).

Hence, (C) $\to$ (P), (R), (T).

Option (D) : For option (S), the mass moves down with acceleration. Therefore, the kinetic energy goes on increasing. Since the line of action of Mg of Y phases through point P, as mentioned in option (Q), its torque about P is zero.

Hence, (D) $\to$ (Q).

For $r\ge R$,

Force on test mass m is $F=m\times |E_g|$

where, $E_g$ = Gravitational field intensity at the point of observation

$\therefore$ ${{m{v^2}} \over r} = m \times \left[ {{{GM} \over {{r^2}}}} \right]$,

Where, M = Total mass of spherical system

$\therefore$ $v \propto {1 \over {\sqrt r }}$

For $r < R$,

$F' = m|E{'_g}|$

$\therefore$ ${{m{V^2}} \over r} = m\left[ {{{GM} \over {{R^3}}} \times r} \right]$

Hence, $V \propto r$

1^st Method : The normal force exerted by the astronaut on the orbiting space station is zero. Therefore, the apparent weight of astronaut in the orbiting space station is zero. Astronaut is called in a state of weightlessness. Why because astronaut as well as space ship are free falling bodies. Statement - I is true, Statement - 2 is true and Statement - 2 is the correct explanation of Statement - 1.

2^nd method : For the body to follow circular path, there must be a centripetal force. Here the astronaut is inside the satellites which is revolving around the earth under the influence of the earth's gravitation. Thus, the earth's gravitation acts as a centripetal force and the net force on the astronaut is zero.

Statement - 2 is right explanation of 1.

Reason : Unit of $G{M_e}{M_s} = F{r^2} = N{m^2}$

$ = kg{m \over {{s^2}}} \times {m^2}$

$ = Kg{m^3}{s^{ - 2}}$

Also (volt)(coulomb)(metre) = (Joule)(metre)

= (N $-$ m)(m) = Nm$^2$ = Kgm$^3$s$^{-2}$

A $\to$ P, Q

Reason : ${v_{rms}} = \sqrt {{{3RT} \over M}} $

$ \Rightarrow v_{rms}^2 = {{3RT} \over M}$

Unit of ${{3RT} \over M}$ is m$^2$s$^{-2}$

Also (farad)(volt)$^2$(kg)$^{-1}$ = (joule)(kg$^-1$)

= N-mkg$^{-1}$ = kgms$^{-2}$mkg$^{-1}$ = m$^2$s$^{-2}$

B $\to$ R, S

Reason : $F = qvB \Rightarrow {v^2} = {{{F^2}} \over {{q^2}{B^2}}}$

Unit of v$^2$ is m$^2$s$^{-2}$ which to further equal to FV$^2$kg$^{-1}$

C $\to$ R, S

Reason : Escape velocity ${v_e} = \sqrt {{{2GM} \over R}} $

$v_e^2 = {{2GM} \over R}$

The unit of ${{GM} \over R}$ is ${m^2}{s^{ - 2}}$

D $\to$ R, S

From the centre of mass, we get

$ m_{\mathrm{A}} R_{\mathrm{A}}=m_{\mathrm{B}} R_{\mathrm{B}} $

Force acting on both stars i.e., gravitational force and centripetal force are balanced.

$ \frac{\mathrm{G} m_{\mathrm{A}} m_{\mathrm{B}}}{\mathrm{R}^2}=\frac{m_{\mathrm{A}} v_{\mathrm{A}}^2}{\mathrm{R}_{\mathrm{A}}}=\frac{m_{\mathrm{B}} v_{\mathrm{B}}^2}{\mathrm{R}_{\mathrm{B}}} $

As we know, $v_{\mathrm{A}}=\frac{2 \pi \mathrm{R}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{A}}}, v_2=\frac{2 \pi \mathrm{R}_{\mathrm{B}}}{\mathrm{T}_{\mathrm{B}}}$

Substituting equation (iii) in equation (ii) we get

$ \begin{aligned} \frac{m_{\mathrm{A}} 4 \pi^2 \mathrm{R}_{\mathrm{A}}^2}{\mathrm{R}_{\mathrm{A}}\left(\mathrm{~T}_{\mathrm{A}}\right)^2} & =\frac{m_{\mathrm{B}} 4 \pi^2 \mathrm{R}_{\mathrm{B}}^2}{\mathrm{R}_{\mathrm{B}}\left(\mathrm{~T}_{\mathrm{B}}\right)^2} \\ \frac{m_{\mathrm{A}} \mathrm{R}_{\mathrm{A}}}{\mathrm{~T}_{\mathrm{A}}^2} & =\frac{m_{\mathrm{B}} \mathrm{R}_{\mathrm{B}}}{\mathrm{~T}_{\mathrm{B}}^2} \end{aligned} $

From equation (i), we conclude in equation (iv)

$ \begin{aligned} \left(\mathrm{T}_{\mathrm{A}}\right)^2 & =\left(\mathrm{T}_{\mathrm{B}}\right)^2 \\ \mathrm{~T}_{\mathrm{A}} & =\mathrm{T}_{\mathrm{B}} . \end{aligned} $

Here we will use the law of conservation of energy as there are no non-conservative force involved so the total energy of mass m remains constant.

For minimum initial velocity to escape the gravitational field, the velocity at infinity will be zero.

And we know PE at infinity is zero.

By energy conservation,

$P{E_i} + K{E_i} = P{E_f} + K{E_f}$

$ \Rightarrow {{ - GMm} \over L} - {{GMm} \over L} + {1 \over 2}m{v^2} = 0 + {1 \over 2}m{(0)^2}$

$ \Rightarrow - {{2Gm} \over L}M + {1 \over 2}m{v^2} = 0$

$ \Rightarrow {1 \over 2}{V^2} = {{2Gm} \over L}$

$ \Rightarrow V = 2\sqrt {{{GM} \over L}} $

Hence, options (B) and (D) are correct.

Escape velocity, expressed as

$v_e = \sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} = \sqrt{\frac{2 \mathrm{G} \cdot \frac{4}{3} \pi \mathrm{R}^3 \rho}{\mathrm{R}}}$

can be simplified to

$\sqrt{\frac{8 \pi \mathrm{G} \rho}{3}} \mathrm{R}.$

The surface area, $A_s$, is given by

$A_s = 4 \pi R^2.$

Considering planet Q's surface area is 4 times that of planet P:

$\mathrm{A}_{\mathrm{Q}} = 4 \mathrm{A} = 4 \pi \mathrm{R}_{\mathrm{Q}}^2$

$\mathrm{~A}_{\mathrm{P}} = \mathrm{A} = 4 \pi \mathrm{R}_{\mathrm{P}}^2$

therefore,

$4 \mathrm{R}_{\mathrm{P}}^2 = \mathrm{R}_{\mathrm{Q}}^2.$

Solving for $R_Q$ :

$\Rightarrow \quad R_{\mathrm{Q}} = 2 \mathrm{R}_{\mathrm{P}} \quad \text{........(i)}.$

For planet R, with mass $\mathrm{M}_{\mathrm{R}} = \mathrm{M}_{\mathrm{P}} + \mathrm{M}_{\mathrm{Q}}$:

$\rho \cdot \frac{4}{3} \pi \mathrm{R}_{\mathrm{R}}^3 = \rho \cdot \frac{4}{3} \pi (\mathrm{R}_{\mathrm{P}}^3 + \mathrm{R}_{\mathrm{Q}}^3)$

Since

$\mathrm{R}_{\mathrm{Q}} = 2\mathrm{R}_{\mathrm{P}},$

we have :

$\mathrm{R}_{\mathrm{R}}^3 = \mathrm{R}_{\mathrm{P}}^3 + 8 \mathrm{R}_{\mathrm{P}}^3 = 9 \mathrm{R}_{\mathrm{P}}^3.$

Thus,

$\mathrm{R}_{\mathrm{R}} = (9)^{1/3} \mathrm{R}_{\mathrm{P}} \quad \text{........(ii)}.$

From (i) and (ii), it follows that :

$ R_R > R_Q > R_P.$

Since

$V_e \propto \mathrm{R},$

it implies that :

$\mathrm{V}_{\mathrm{R}} > \mathrm{V}_{\mathrm{Q}} > \mathrm{V}_{\mathrm{P}}.$

Also,

$\frac{\mathrm{V}_{\mathrm{R}}}{\mathrm{V}_{\mathrm{P}}} = \frac{\mathrm{R}_{\mathrm{R}}}{\mathrm{R}_{\mathrm{P}}} = 9^{1/3},$

and

$\frac{\mathrm{V}_{\mathrm{P}}}{\mathrm{V}_{\mathrm{Q}}} = \frac{\mathrm{R}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{Q}}} = \frac{1}{2}.$