Gravitation

Escape velocity, $V_e=\sqrt{\frac{2 G M}{R}}$

From conservation of energy

$ \begin{aligned} \mathrm{KE}_i+\mathrm{PE}_i & =\mathrm{KE}_f+\mathrm{PE}_f \\ \frac{-G M m}{R}+\frac{1 m}{2} \frac{V_e^2}{9} & =\frac{-G M m}{R+h} \\ \frac{G M}{R+h} & =\frac{G M}{R}-\frac{V_e^2}{18} \quad\left(\therefore V=\frac{V_e}{3}\right) \\ \frac{G M}{R+h} & =\frac{G M}{R}-\frac{G M}{9 R} \quad\left(\therefore V_e=\sqrt{\frac{2 G M}{R}}\right) \\ \frac{G M}{R+h} & =\frac{8 G M}{9 R} \\ \frac{1}{R+h} & =\frac{8}{9 R} \end{aligned} $

                                    $ \begin{aligned} & R=8 h \Rightarrow h=\frac{R}{8} \\ & h=\frac{6400}{8}=800 \mathrm{~km} \end{aligned} $

Let $m_1=m$, then $m_2=M-m$

$ =\frac{G}{r^2}\left(M m-m^2\right) $

Differentiating w.r.t. m , we get

$ \frac{d F}{d m}=\frac{G}{r^2}(M-2 m) $

For $F$ to be maximum,

$ \begin{aligned} \frac{d F}{d m} & =0 \\ \frac{G}{r^2}(M-2 m) & =0 \\ M-2 m & =0 \\ 2 m & =M \\ m & =\frac{M}{2} \end{aligned} $

So, $m_1=\frac{M}{2}$

$ m_2=M-\frac{M}{2}=\frac{M}{2} $

To determine the velocity of a body falling from rest at a height $ R $ above the Earth's surface, we must equate the increase in kinetic energy to the decrease in potential energy.

Given:

Initial height $ h = R $

Radius of Earth $ R $

Acceleration due to gravity $ g $

The change in potential energy when the body falls to the Earth's surface can be given by:

$ \Delta U = \frac{mgh}{1 + \frac{h}{R}} $

Substituting $ h = R $, we get:

$ \Delta U = \frac{mgR}{1 + \frac{R}{R}} = \frac{mgR}{2} $

By the conservation of energy principle, the increase in kinetic energy equals the decrease in potential energy:

$ \frac{1}{2} mv^2 = \frac{mgR}{2} $

Solving for $ v $:

$ mv^2 = mgR $

$ v^2 = gR $

$ v = \sqrt{gR} $

Thus, the velocity of the body when it reaches the Earth's surface is $ v = \sqrt{gR} $.

When the man is outside the planet and close to its surface, the gravitational force is determined by the entire mass of the planet, $M$. The force of gravity at the surface can be given by Newton's law of gravitation:

$ F = \frac{G M m}{R^2} $

However, as the man passes through the hole in the planet, the gravitational force he experiences changes due to the unique mass distribution of the planet. The planet consists of a spherical shell of mass $\frac{3M}{4}$ and negligible thickness, and a point mass $\frac{M}{4}$ at the center.

Let's analyze the gravitational force when the man is inside the planet:

1. Spherical Shell Contribution:

A spherical shell of uniform density exerts no net gravitational force on a particle located inside it. Thus, the man inside the shell experiences no force from the spherical shell of mass $\frac{3M}{4}$.

2. Point Mass Contribution:

The point mass at the center, which is $\frac{M}{4}$, exerts a gravitational force on the man. According to Newton's law of gravitation, this force is given by:

$ F' = \frac{G \left(\frac{M}{4}\right) m}{R^2} $

To determine the change in the force of gravity experienced by the man, we subtract the force due to the point mass from the original force at the planet's surface:

Original force at the surface:

$ F = \frac{G M m}{R^2} $

Force inside due to point mass:

$ F' = \frac{1}{4} \frac{G M m}{R^2} $

Change in the force:

$ \Delta F = F - F' $

$ \Delta F = \frac{G M m}{R^2} - \frac{1}{4} \frac{G M m}{R^2} $

Simplifying the expression:

$ \Delta F = \frac{G M m}{R^2} \left(1 - \frac{1}{4}\right) $

$ \Delta F = \frac{G M m}{R^2} \cdot \frac{3}{4} $

$ \Delta F = \frac{3}{4} \frac{G M m}{R^2} $

Hence, the change in the force of gravity experienced by the man is

$ \frac{3}{4} \frac{G M m}{R^2} $

The correct answer is Option A: $\frac{3}{4} \frac{G M m}{R^2}$.

Acceleration due to gravity at the surface of earth,

$g = {{GM} \over {{R^2}}} = {4 \over 3}\pi \rho GR$ ...... (i)

[$\because$ $M = {4 \over 3}\pi {R^3}\,.\rho $]

Where, $\rho$ = density of earth.

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

$g' = {4 \over 3}\pi \rho G(R - d)$ ...... (ii)

From Eqs. (i) and (ii), we get,

$g' = g\left[ {1 - {d \over R}} \right]$

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 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%.