Gravitation

Dimensional Analysis:

$ \mathrm{T} \propto \mathrm{m}^{\mathrm{x}} \mathrm{G}^{\mathrm{y}} \mathrm{a}^{\mathrm{z}} $

Using dimensional analysis for gravitational interactions, we have:

$ \mathrm{T} \propto \mathrm{M}^{\mathrm{x}} G^{\mathrm{y}} a^{\mathrm{z}} $

Where $ G $ has dimensions $\left[\mathrm{M}^{-1} \mathrm{L}^3 \mathrm{T}^{-2}\right]$.

Solving for Exponents:

$ \mathrm{T} \propto \mathrm{M}^{\mathrm{x-y}} \mathrm{L}^{3y+z} \mathrm{T}^{-2y} $

Equating dimensions, we solve:

$ \mathrm{x}-\mathrm{y}=0 \Rightarrow \mathrm{x}=\mathrm{y} $

$ -2\mathrm{y}=1 \Rightarrow \mathrm{y}=-\frac{1}{2}, \mathrm{x}=-\frac{1}{2} $

$ 3\mathrm{y}+\mathrm{z}=0 \implies \mathrm{z}=-3\mathrm{y}=\frac{3}{2} $

Time Proportionality:

$ \mathrm{T} \propto \mathrm{~m}^{-1/2} \mathrm{G}^{-1/2} \mathrm{a}^{3/2} $

Which simplifies to:

$ \mathrm{T} \propto \left(\frac{a^3}{m}\right)^{1/2} $

Applying New Conditions:

When the side length becomes $2a$ and mass becomes $2m$:

$ \mathrm{T} = 4 \times \left(\frac{(2a)^3}{2m}\right)^{1/2} $

Simplifying:

$ \mathrm{T} = 4 \times \left(\frac{8a^3}{2m}\right)^{1/2} = 4 \times (4)^{1/2} = 8 \text{ seconds} $

Thus, the spheres will collide after 8 seconds under the new conditions.

The kinetic energy (KE) of a satellite revolving around the Earth can be expressed with the following formula:

$ \mathrm{KE} = \frac{1}{2} m v^2 = \frac{1}{2} m \frac{GM_e}{r} = \frac{GM_e m}{2r} $

For this satellite, the radius $ r $ is the sum of the Earth's radius $ R_E $ and the height $ h $ above the Earth's surface:

$ r = R_E + h $

Given:

Mass of the satellite, $ m = 1000 \, \text{kg} $

Mass of the Earth, $ M_e = 6 \times 10^{24} \, \text{kg} $

Radius of the Earth, $ R_E = 6.4 \times 10^6 \, \text{m} $

Height of orbit above the Earth's surface, $ h = 270 \, \text{km} = 2.7 \times 10^5 \, \text{m} $

Gravitational constant, $ G = 6.67 \times 10^{-11} \, \text{Nm}^2 \, \text{kg}^{-2} $

Substitute these values into the equation:

$ \mathrm{KE} = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 1000}{2(6.4 \times 10^6 + 2.7 \times 10^5)} $

$ = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 1000}{2 \times 6.67 \times 10^6} $

$ = 3 \times 10^{10} \, \text{J} $

Hence, the kinetic energy of the satellite in its orbit is $ 3 \times 10^{10} \, \text{J} $.

Let a planet of mass m orbits a star of mass M in circular orbit of radius r with speed v.

For this circular motion, centripetal force is provided by the gravitational force between both masses.

Hence, ${F_C} = {F_4}$

$ \Rightarrow {{m{v^2}} \over r} = {{GMm} \over {{r^2}}}$

$ \Rightarrow v = \sqrt {{{Gm} \over r}} $

Angular momentum of the planet, $L = mvr$

$ \Rightarrow L = m\sqrt {{{Gm} \over r}} r$

$ \Rightarrow L = \sqrt {GM} m\sqrt r $

Given, ${m_B} = 4\sqrt 2 {m_A}$ and ${R_B} = 2{R_A}$

M is same for both.

So, ${{{L_B}} \over {{L_A}}} = \left( {{{{m_B}} \over {{m_A}}}} \right)\sqrt {{{{R_B}} \over {{R_A}}}} $

$ = \left( {4\sqrt 2 } \right)\sqrt 2 = 4 \times 2$

$ \Rightarrow {{{L_B}} \over {{L_A}}} = 8$.

$\because$ acceleration due to gravity on surface is given by

$\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}_{\mathrm{e}}^2}$

Now since diameter is reduced to $1 / 3^{\text {rd }}$, radius also reduces to $1 / 3^{\text {rd }}$, keeping mass constant

New value of acceleration due to gravity on Earth's surface is

$g^{\prime}=\frac{\mathrm{GM}}{\left(\frac{\mathrm{R}_{\mathrm{e}}}{3}\right)^2}=9 \frac{\mathrm{GMe}}{\mathrm{R}_{\mathrm{e}}^2}=9 \mathrm{~g}$

$ r = R + \frac{R}{3} = \frac{4R}{3} $

For a circular orbit, the gravitational force provides the required centripetal force:

$ \frac{GMm}{r^2} = \frac{mv^2}{r} $

which simplifies to

$ v = \sqrt{\frac{GM}{r}}. $

The angular momentum $L$ of the satellite is given by

$ L = mvr. $

Substitute the values:

Satellite mass: $ m = \frac{M}{2} $

Orbital radius: $ r = \frac{4R}{3} $

Speed: $ v = \sqrt{\frac{GM}{\frac{4R}{3}}} = \sqrt{\frac{3GM}{4R}} $

Thus,

$ L = \frac{M}{2} \cdot \frac{4R}{3} \cdot \sqrt{\frac{3GM}{4R}}. $

Simplify the expression:

$ L = \frac{M \cdot 4R}{6} \sqrt{\frac{3GM}{4R}} = \frac{2MR}{3} \sqrt{\frac{3GM}{4R}}. $

Notice that the square root can be combined as:

$ \sqrt{\frac{3GM}{4R}} = \sqrt{\frac{3}{4}} \sqrt{\frac{GM}{R}}. $

Therefore,

$ L = \frac{2MR}{3} \cdot \sqrt{\frac{3}{4}} \sqrt{\frac{GM}{R}} = \frac{2MR \sqrt{3}}{3 \cdot 2} \sqrt{\frac{GM}{R}} = \frac{M \sqrt{3}}{3} \sqrt{GMR}, $

where we used the fact that $ R \sqrt{\frac{1}{R}} = \sqrt{R} $ to form $ \sqrt{GMR} $.

So the final expression is:

$ L = \frac{M}{\sqrt{3}} \sqrt{GMR}. $

It is given that the angular momentum can also be expressed as:

$ L = M \sqrt{\frac{GMR}{x}}. $

Equate the two expressions:

$ \frac{M}{\sqrt{3}} \sqrt{GMR} = M \sqrt{\frac{GMR}{x}}. $

Cancel $M$ and $\sqrt{GMR}$ on both sides (assuming they are nonzero):

$ \frac{1}{\sqrt{3}} = \sqrt{\frac{1}{x}}. $

Taking squares on both sides:

$ \frac{1}{3} = \frac{1}{x}. $

Thus,

$ x = 3. $

Given that the radius of the Earth is decreased to three-fourths of its current value while its mass remains unchanged, we need to determine the new duration of the Earth's day.

First, using the principle of conservation of angular momentum, we have:

$ \tau_{\text{ext}} = 0 \implies \text{Angular momentum is conserved} $

Therefore,

$ \frac{2}{5} M R^2 \cdot \omega_i = \frac{2}{5} M \left(\frac{3R}{4}\right)^2 \cdot \omega_f $

Simplifying this equation, we find:

$ \omega_f = \frac{16}{9} \omega $

Since the period $ T $ of rotation is given by:

$ T = \frac{2\pi}{\omega} $

For the new period $ T_1 $, we have:

$ T_1 = \frac{2\pi}{\omega_f} = \frac{2\pi}{\frac{16}{9}\omega} = \frac{9}{16} \times T $

Given that the initial period $ T $ is 24 hours:

$ T_1 = \frac{9}{16} \times 24 \text{ hours} $

$ T_1 = 13 \text{ hours} ~30 \text{ minutes} $

Thus, the new duration of the Earth's day would be 13 hours and 30 minutes.

Acceleration due to gravity g' $=\frac{g}{4}$

$\begin{aligned} & \mathrm{T}=2 \pi \sqrt{\frac{4 \ell}{\mathrm{g}}} \\ & \mathrm{T}=2 \pi \sqrt{\frac{4 \times 4}{\mathrm{~g}}} \\ & \mathrm{~T}=2 \pi \frac{4}{\pi}=8 \mathrm{~s} \end{aligned}$

From the conservation of angular momentum, we have:

$ \frac{2}{5}MR^2\omega_1 = \frac{2}{5}M\left(\frac{R}{4}\right)^2\omega_2 $

This simplifies to:

$ MR^2\omega_1 = \frac{MR^2}{16}\omega_2 $

From this, we can derive the ratio of the initial and final angular velocities:

$ \frac{\omega_1}{\omega_2} = \frac{1}{16} $

Since the angular velocity (\omega) is inversely proportional to the period of rotation (T) ((\omega = \frac{2\pi}{T})), we can write:

$ \frac{T_2}{T_1} = \frac{1}{16} $

We can express this ratio in terms of the variable (x):

$ \frac{T_1}{T_2} = \frac{16}{1} = \frac{24}{x} $

Solving this equation for (x) gives:

$ x = 16 $

So, if the Earth suddenly shrinks to ( $\frac{1}{64}$ )th of its original volume with its mass remaining the same, the period of rotation of Earth becomes ( $\frac{24}{16}$ )h, or 1.5 hours. Therefore, the value of (x) is 16.

$g\left( {1 - {{2h} \over R}} \right) = g\left( {1 - {d \over R}} \right)$

$ \Rightarrow 2h = d$

$ \Rightarrow \alpha = 2$

$v = \sqrt {{{GM} \over R}} $

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

${{{v_2}} \over {{v_1}}} = \sqrt {{{3200} \over {800}}} = 2$

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

$x = 2$

We know that binding energy of earth,

$BE = - {3 \over 5}{{G{M^2}} \over R}$

$\therefore$ Energy required to break the earth into pieces

$ = - BE = {3 \over 5}{{G{M^2}} \over R}$ ...... (i)

According to question, the amount of energy that needs to be supplied is ${x \over 5}{{G{M^2}} \over R}$.

Comparing it with value in Eq. (i), we get,

$x = 3$

For satellites, the time period $ T $ is proportional to the distance to the power of $ \frac{3}{2} $:

$ T \propto r^{\frac{3}{2}} $

This gives us the ratio of the time periods for the second and first satellites:

$ \frac{T_2}{T_1} = \left(\frac{r_2}{r_1}\right)^{\frac{3}{2}} $

Converting this to angular velocity, $ \omega $, which is inversely proportional to the time period, we have:

$ \frac{\omega_2}{\omega_1} = \left(\frac{r_1}{r_2}\right)^{\frac{3}{2}} = (1.21)^{\frac{3}{2}} $

Calculating $ (1.21)^{\frac{3}{2}} $, we find:

$ \omega_2 = \omega_1 \times 1.331 $

The combined angular displacement of both satellites is $ 2\pi $ radians in the same time period $ t_0 $:

$ (\omega_2 + \omega_1) t_0 = 2\pi $

Solving for $ t_0 $:

$ t_0 = \frac{2\pi}{\omega_2 + \omega_1} = \frac{2\pi}{\left(\frac{4}{3} + 1\right) \omega_1} = \frac{6\pi}{7\omega_1} $

Since $ T_{GSS} $, the period of the geostationary satellite, is 24 hours, we substitute:

$ t_0 = \frac{6\pi}{2\pi} \frac{T_{GSS}}{7} = \frac{3 \times 24 \text{ hours}}{7} = \frac{24}{p} \text{ hours} $

Therefore, solving for $ p $:

$ p = \frac{7}{3} = 2.33 $

Due to an interaction process, star A losses some of it's mass and radius becomes ${R \over 2}$. Let new mass of star A is M'_A. Here in both cases density of star A remains same ${\rho _A}$.

$\therefore$ Initially $({\rho _A}) = {{{M_A}} \over {{4 \over 3}\pi {R^3}}}$

Finally $({\rho _A}) = {{M{'_A}} \over {{4 \over 3}\pi {{\left( {{R \over 2}} \right)}^3}}}$

Density remains same,

$\therefore$ ${{{M_A}} \over {{4 \over 3}\pi {R^3}}} = {{M{'_A}} \over {{4 \over 3}\pi {{\left( {{R \over 2}} \right)}^3}}}$

$ \Rightarrow 8M{'_A} = {M_A}$

$ \Rightarrow M{'_A} = {{{M_A}} \over 8}$

$\therefore$ Lost mass by $A = {M_A} - M{'_A} = {M_A} - {{{M_A}} \over 8} = {{7{M_A}} \over 8}$

This lost mass ${{7{M_A}} \over 8}$ is attached on the star B and density of the attached mass stay ${\rho _A}$. So new radius of star B is ${R_2}$.

Density of the removed part from star A is,

${\rho _A} = {{{{7{M_A}} \over 8}} \over {{4 \over 3}\pi \left( {{R^3} - {{\left( {{R \over 2}} \right)}^3}} \right)}}$

Density of the added part in star B stay's same as ${\rho _A}$,

$\therefore$ ${\rho _A} = {{{{7{M_A}} \over 8}} \over {{4 \over 3}\pi \left( {R_2^3 - {R^3}} \right)}}$

$\therefore$ ${{{{7{M_A}} \over 8}} \over {{4 \over 3}\pi \left( {{R^3} - {{\left( {{R \over 2}} \right)}^3}} \right)}} = {{{{7{M_A}} \over 8}} \over {{4 \over 3}\pi \left( {R_2^3 - {R^3}} \right)}}$

$ \Rightarrow {R^3} - {{{R^3}} \over 8} = R_2^3 - {R^3}$

$ \Rightarrow R_2^3 = 2{R^3} - {{{R^3}} \over 8}$

$ \Rightarrow R_2^3 = {{15{R^3}} \over 8}$

$ \Rightarrow {R_2} = {(15)^{{1 \over 3}}} \times {R \over 2}$

Escape velocity from star A after interaction process,

${V_A} = \sqrt {{{2G\left( {{{{M_A}} \over 8}} \right)} \over {{R \over 2}}}} $

And escape velocity from star B after interaction process,

${V_B} = \sqrt {{{2G\left( {{{23{M_A}} \over 8}} \right)} \over {{{(15)}^{{1 \over 3}}} \times {R \over 2}}}} $

$\therefore$ ${{{V_B}} \over {{V_A}}} = \sqrt {{{{{23{M_A}} \over 8}} \over {{{(15)}^{{1 \over 3}}} \times {R \over 2}}} \times {{{R \over 2}} \over {{{{M_A}} \over 8}}}} $

$ = \sqrt {{{23} \over {{{(15)}^{{1 \over 3}}}}}} $ ..... (1)

Given,

${{{V_B}} \over {{V_A}}} = \sqrt {{{10n} \over {{{(15)}^{{1 \over 3}}}}}} $ ..... (2)

Comparing equation (1) and (2), we get,

$10n = 23$

$ \Rightarrow n = 2.3$

Given situation is shown in the figure. Let acceleration due to gravity at the surface of the planet be g. At height h above planet's surface v = 0.

According to question, acceleration due to gravity of the planet at height h above its surface becomes g/4.

${g_h} = {g \over 4} = {g \over {{{\left( {1 + {h \over R}} \right)}^2}}}$

$4 = {\left( {1 + {h \over R}} \right)^2} \Rightarrow 1 + {h \over R} = 2$

${h \over R} = 1 \Rightarrow h = R$.

So, velocity of the bullet becomes zero at h = R.

Also, ${v_{esc}} = v\sqrt N \Rightarrow \sqrt {{{2GM} \over R}} = v\sqrt N $ ...... (i)

Applying energy conservation principle,

Energy of bullet at surface of earth = Energy of bullet at highest point

${{ - GMm} \over R} + {1 \over 2}m{v^2} = {{ - GMm} \over {2R}}$

${1 \over 2}m{v^2} = {{GMm} \over {2R}}$ $\therefore$ $v = \sqrt {{{GM} \over R}} $

Putting this value in eqn. (i), we get

$\sqrt {{{2GM} \over R}} = \sqrt {{{NGM} \over R}} $

$\therefore$ N = 2

Let stars A and B are rotating about their centre of mass with angular velocity $\omega$.

Let distance of stars A and B from the centre of mass be r_A and r_B respectively as shown in the figure.

Total angular momentum of the binary stars about the centre of mass is

$L = {M_A}r_A^2\omega + {M_B}r_B^2\omega $

Angular momentum of the star B about centre of mass is

${L_B} = {M_B}r_B^2\omega $

$\therefore$ ${L \over {{L_B}}} = {{({M_A}r_A^2 + {M_B}r_B^2)\omega } \over {{M_B}r_B^2\omega }} = \left( {{{{M_A}} \over {{M_B}}}} \right){\left( {{{{r_A}} \over {{r_B}}}} \right)^2} + 1$

Since ${M_A}{r_A} = {M_B}{r_B}$

or, ${{{r_A}} \over {{r_B}}} = {{{M_B}} \over {{M_A}}}$

$\therefore$ ${L \over {{L_B}}} = {{{M_B}} \over {{M_A}}} + 1 = {{11{M_S}} \over {2.2{M_S}}} + 1 = {{11 + 2.2} \over {2.2}} = 6$

On the planet,

${g_p} = {{G{M_p}} \over {R_p^2}} = {G \over {R_p^2}}\left( {{4 \over 3}\pi R_p^3{\rho _p}} \right) = {4 \over 3}G\pi R_p^{}{\rho _p}$

On the earth,

${g_e} = {{G{M_e}} \over {R_e^2}} = {G \over {R_e^2}}\left( {{4 \over 3}\pi R_e^3{\rho _e}} \right) = {4 \over 3}G\pi R_e^{}{\rho _e}$

$\therefore$ ${{{g_p}} \over {{g_e}}} = {{R_p^{}{\rho _p}} \over {R_e^{}{\rho _e}}}$ or ${{{R_p}} \over {{R_e}}} = {{g_p^{}{\rho _p}} \over {g_e^{}{\rho _e}}}$ ..... (i)

On the planet, ${v_p} = \sqrt {2{g_p}{R_p}} $

On the earth, ${v_e} = \sqrt {2{g_e}{R_e}} $

$\therefore$ ${{{v_p}} \over {{v_e}}} = \sqrt {{{{g_p}{R_p}} \over {{g_e}{R_e}}}} = {{{g_p}} \over {{g_e}}}\sqrt {{{{\rho _e}} \over {{\rho _p}}}} $ (Using (i))

Here, ${\rho _p} = {2 \over 3}{\rho _e}$, ${g_p} = {{\sqrt 6 } \over {11}}{g_e}$

$\therefore$ ${{{v_p}} \over {{v_e}}} = {{\sqrt 6 } \over {11}}\sqrt {{3 \over 2}} $

or, ${v_p} = 11 \times {{\sqrt 6 } \over {11}} \times \sqrt {{3 \over 2}} $ ($\because$ v_e = 11 km s^$-$1 (Given))

= 3 km s^$-$1