Gravitation

As we know that,

$g^{\prime}=g-\omega^2 r \cos \theta$

where, $g^{\prime}$ is effective acceleration, $g$ is acceleration due to gravity, $\omega$ is angular frequency, $r$ is radius of earth and $\theta$ is angle between equator and body.

If $\omega=0 \mathrm{~rads}^{-1}$ (i.e. earth stops rotation)

$g^{\prime}=g$

i.e., weight remains same everywhere.

Given, height, H = 1600 km

Let depth be $d$.

Acceleration due to gravity at height, depth and surface be $g_h, g_d$ and $g$.

Also, $g_d=\frac{g_h}{2}$ ...... (i)

As we know that,

$\begin{aligned} g_h & =g\left(1-\frac{2 h}{R}\right) \\ \Rightarrow \quad g_h & =g\left(1-\frac{2 \times 1600}{6400}\right)=\frac{g}{2} \end{aligned}$

and $g_d=g\left(1-\frac{d}{R}\right)$

From Eq. (i),

$\begin{aligned} \frac{1}{2} \times \frac{g}{2} & =g\left(1-\frac{d}{R}\right) \\ \Rightarrow \quad \frac{d}{R} & =\frac{3}{4} \\ \Rightarrow \quad d & =\frac{3}{4} R=\frac{3}{4} \times 6400=4.8 \times 10^6 \mathrm{~m} \end{aligned}$

As we know that,

Gravitational potential energy, $(U)=-\frac{GM_e m}{r}$

If $\gamma=\infty,U=0$ which is maximum.

Let, Initial radius $=R$

Final radius $=2 R$

Initial time period, $T=24 \mathrm{~h}$

Final time period, $T^{\prime}=$ ?

By using third Kepler's law of planetary motion,

$T^2 \propto R^3$

$\Rightarrow \left(\frac{T_1}{T_2}\right)^2=\left(\frac{R_1}{R_2}\right)^3 \Rightarrow\left(\frac{24}{T^{\prime}}\right)^2=\left(\frac{R}{2 R}\right)^3$

$\Rightarrow \quad \frac{24 \times 24}{T^{\prime 2}}=\frac{1}{8} \Rightarrow T^{\prime 2}=24 \times 24 \times 8$

$\Rightarrow \quad T^{\prime}=24 \times 2 \sqrt{2}=48 \sqrt{2} \mathrm{~h}$

Given, radius of earth, $R=6400 \mathrm{~km}$

Let gravity on surface of earth be $g$.

$\therefore$ Gravity at depth $d$ be $g_d=g-40 \%$ of $g=\frac{60}{100} g$

As we know that, $g_d=g\left(1-\frac{d}{R}\right)$

$\begin{aligned} \therefore \quad \frac{60}{100} g & =g\left(1-\frac{d}{R}\right) \\ \Rightarrow \quad 0.6 & =1-\frac{d}{R} \Rightarrow \frac{d}{R}=0.4 \\ \Rightarrow \quad d & =0.4 \times R \Rightarrow d=\frac{4}{10} \times 6400 \\ & =2560 \mathrm{~km} \end{aligned}$

Given, mass of body, m = 3 kg

Position of masses are shown below.

Let, gravitational field intensity be E.

As we know that, $E=\frac{Gm}{R^2}$

$\begin{aligned} E & =\frac{3 G}{1^2}+\frac{3 G}{2^2}+\frac{3 G}{4^2}+\frac{3 G}{8^2}+\ldots \ldots \\\\ & =3 G\left(1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\ldots\right)\quad ....(\text{i}) \\\\ & =3 G\left[\frac{1}{1-\frac{1}{4}}\right]\left[\because S=\frac{a}{1-r}\right] \\\\ E & =3 G \frac{4}{3}=4 G\end{aligned}$

Given, mass of sphere = 1000 kg

Radius of sphere = 3m

Since, work (W) = $\frac{GM_1M_2}{r}$

where, G is gravitational constant i.e., $6.67\times10^{-11}~\mathrm{N-m^2kg^{-1}}$

M$_1$, M$_2$ are mass of particle and sphere respectively.

$\therefore$ ${W \over {{m_1}}} = {{G{M_2}} \over r}$

$ = {{6.67 \times {{10}^{ - 11}} \times 1000} \over 1}$

$ = 6.67 \times {10^{ - 8}}$ J/kg

Given, let the radius of earth = R

Height above earth, H = R/20

Acceleration due to gravity at H, g$_H$ = 9 ms$^{-2}$

Acceleration due to gravity on the surface of earth, g = 10 ms$^{-2}$

Acceleration due to gravity at depth d from the surface of earth,

$\begin{aligned} g_d & =g\left(1-\frac{d}{R}\right) \\ \Rightarrow \quad g_d & =10\left(1-\frac{R}{20 \times R}\right)=10\left(\frac{19}{20}\right)=9.5 \mathrm{~ms}^{-2}\end{aligned}$

Potential at point P due to the solid sphere,

${V_1} = - {{GM} \over {2{R^3}}}\left[ {3{R^2} - {{\left( {{R \over 2}} \right)}^2}} \right]$

$ = - {{11GM} \over {8R}}$

Potential at point P due to the cavity part,

${V_2} = - {3 \over 2}{{G\left( {{M \over 8}} \right)} \over {{R \over 2}}} = - {{3GM} \over {8R}}$

Potential due to the remaining part at point P,

$V = {V_1} - {V_2} = {{ - 11GM} \over {8R}} + {{3GM} \over {8R}}$

$ \Rightarrow V = - {{GM} \over R}$

The value of g at equator is expressed as

${g_e} = g - R{\omega ^2}\cos \phi = g - R{\omega ^2}$ [$\because$ $\cos \phi = 1$]

$\Rightarrow$ Change in value of g is $g - {g_e}$

$ = R{\omega ^2} = R{\left( {{{2\pi } \over T}} \right)^2}$

$ = 6.4 \times {10^6} \times {\left[ {{{2\pi } \over {24 \times 60 \times 60}}} \right]^2}$

$ = 3.37 \times {10^{ - 2}}$ m/s²

$ \simeq 3.4$ cm/s²

The long range force experienced by a neutral particle with a finite mass is gravitational force.

In this universe each body attracts other body with a force that is directly proportional to the product of their masses ( $m_1$ and $m_2$ ) and inversely proportional to the square of distance $(r)$ between them.

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

This force is called gravitational force.

Acceleration due to gravity,

$ g_1=\frac{G M}{R_1^2} $

As, $M$ remain constant and $R$ becomes $R_2=R_1-1 \%$ of $R_1=R_1-\frac{R_1}{100}=\frac{99}{100} R_1$

Therefore, the new gravitational acceleration becomes,

$ \begin{aligned} g_2 & =\frac{G M}{R_2^2}=\frac{G M \times(100)^2}{(99)^2 R_1^2} \\ & =\frac{(100)^2}{(99)^2} g_1=1.02 g_1 \end{aligned} $

Change in the gravitational acceleration,

$ g_2-g_1=1.02 g_1-g_1=0.02 g_1 $

Hence, the percentage increase is

$ \frac{g_2-g_1}{g_1} \times 100=0.02 \times 100=2 \% $

So, if the radius of earth shrinks by $1 \%$, the value of $g$ would increase by $2 \%$.

$ \text { According to given situation, } $

Again, $m_0$ and $M-m_0$ are kept at a distance $D$

gravitational force,

$ F=\frac{G m_0\left(M-m_0\right)}{D^2} $

$F$ will be maximum, when $\frac{d F}{d m_0}=0$

$ \begin{array}{lc} \Rightarrow & \frac{d}{d m_0} \frac{G m_0\left(M-m_0\right)}{D^2}=0 \\ \Rightarrow & \frac{d}{d m_0}\left(\frac{G m_0 M-G m_0^2}{D^2}\right)=0 \\ \Rightarrow & \frac{G M-2 G m_0}{D^2}=0 \\ \Rightarrow & G M-2 G m_0=0 \\ \Rightarrow & M-2 m_0=0 \Rightarrow 2 m_0=M \\ \Rightarrow & \frac{m_0}{M}=\frac{1}{2}=0.5 \end{array} $

$ \begin{aligned} &\text { At depth } d \text {, }\\ &g^{\prime}=g\left(1-\frac{d}{R_e}\right)=g\left(\frac{R_e-d}{R_e}\right)=\frac{g r}{R_e} \end{aligned} $

$\Rightarrow g^{\prime} \propto r$ (where, $r=$ distance from centre of earth)

At height $h$,

$ g^{\prime}=g\left(\frac{R_e^2}{\left(R_e+h\right)^2}\right)=\frac{g R_e^2}{r^2} \Rightarrow g^{\prime} \propto \frac{1}{r^2} $

So, correct graph is

Acceleration due to gravity at altitude $h$ is

$ g_h=\frac{G M}{(R+h)^2} $

$\therefore g_h$ decreases with altitude.

$ g=\frac{G M}{R^2}, g \propto M $

Geostationary satellite has a time period of $24 \mathrm{~h} . g$ at depth $d$ is

$ g_d=g\left(\frac{R-d}{R}\right) $

$\therefore g$ decreases with increasing depth.

The two spheres are touching each other as shown below.

Let m be the mass of each sphere.

Gravitational force between the spheres,

$F = G{{m.m} \over {{{(2r)}^2}}} = G.{{{m^2}} \over {4{r^2}}}$

$ = G.{{{{\left[ {{4 \over 3}\pi {r^3}.\rho } \right]}^2}} \over {4{r^2}}}$ [Mass = volume $\times$ density]

$ = {{4G{\pi ^2}{r^6}{\rho ^2}} \over {3 \times 4{r^2}}}$

$ \Rightarrow F = \left( {{{4G{\pi ^2}{\rho ^2}} \over 3}} \right){r^4} \Rightarrow F \propto {r^4}$

As, orbital velocity of a satellite, $v = \sqrt {{{GM} \over r}} $

Hence, $v \propto {1 \over {\sqrt r }}$

% increase in speed = ${1 \over 2}$ (% decrease in radius)

$={1 \over 2}(2\% ) = 1\% $

Thus, the speed of satellite will increase by 1%.

Net force on mass M at position B towards centre of circle is

${F_{BO\,net}} = {F_{BD}} + {F_{BA}}\sin 45^\circ + {F_{BC}}\sin 45^\circ $

$ = {{G{M^2}} \over {{{(\sqrt 2 a)}^2}}} + {{G{M^2}} \over {{a^2}}}\left( {{1 \over {\sqrt 2 }}} \right) + {{G{M^2}} \over {{a^2}}}\left( {{1 \over {\sqrt 2 }}} \right)$

[where, diagonal length BD is $\sqrt2$a]

$ = {{G{M^2}} \over {2{a^2}}} + {{G{M^2}} \over {{a^2}}}\left( {{2 \over {\sqrt 2 }}} \right) = {{G{M^2}} \over {{a^2}}}\left( {{1 \over 2} + \sqrt 2 } \right)$

This force will act as centripetal force.

Distance of particle from centre of circle is ${a \over {\sqrt 2 }}$.

Here, ${F_{centripetal}} = {{M{v^2}} \over r} = {{M{v^2}} \over {{a \over {\sqrt 2 }}}} = {{\sqrt 2 M{v^2}} \over a}$ ($\because$ $r = {a \over {\sqrt 2 }}$)

So, for rotation about the centre,

${F_{centripetal}} = {F_{BO(net)}}$

$ \Rightarrow \sqrt 2 {{M{v^2}} \over a} = {{G{M^2}} \over {{a^2}}}\left( {{1 \over 2} + \sqrt 2 } \right)$

$ \Rightarrow {v^2} = {{GM} \over a}\left( {1 + {1 \over {2\sqrt 2 }}} \right) = {{GM} \over a}(1.35)$

$ \Rightarrow v = 1.16\sqrt {{{GM} \over a}} $