Gravitation

The force of mutual attraction between any two objects by virtue of their masses is gravitation force.

Gravitational force is conservative, attractive, non-contact and a central force.

Escape speed from Earth's surface,

$ v_e=\sqrt{\frac{2 G M}{R}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

where, $M$ is mass of Earth and $R$ its radius.

Now, apply conservation of mechanical energy for Earth-meteor system.

$ \begin{aligned} & \mathrm{KE}_i+\mathrm{PE}_i=\mathrm{KE}_f+\mathrm{PE}_f \\ & \Rightarrow \quad \frac{1}{2} m v^2+0=\frac{1}{2} m v^{\prime 2}-\frac{G M m}{R} \\ & \Rightarrow \quad m\left[v^2+\frac{2 G M}{R}\right]=m v^{\prime 2} \\ & \Rightarrow \quad v^{\prime}=\sqrt{v^2+v_e^2} \end{aligned} $

Orbital speed at radius $r$ from Earth's centre

$ v=\sqrt{\frac{G M}{r}} $

So, $\frac{v_2}{v_1}=\sqrt{\frac{r_1}{r_2}}$

$ \begin{aligned} r_1 & =R=6400 \mathrm{~km} \\ r_2 & =R+h \\ & =6400+19200=25600 \mathrm{~km} \\ \frac{v_2}{8} & =\sqrt{\frac{6400}{25600}}=\sqrt{\frac{1}{4}}=\frac{1}{2} \\ v_2 & =8 \times \frac{1}{2}=4 \mathrm{~km} / \mathrm{s} \end{aligned} $

Step 1: Use the conservation of energy

When an object moves in Earth's gravity, its total energy (kinetic plus potential) stays the same if no energy is lost.

Step 2: Write the energy equation

The equation is: $ \frac{1}{2} m V_i^2 - \frac{G M m}{R} = \frac{1}{2} m V_f^2 + 0 $

Here, $V_i$ is the speed at the start, and $V_f$ is the speed far away from Earth (where gravity is almost zero).

Step 3: Substitute the initial speed

The body is thrown with speed $V_i = \sqrt{5} V_e$, where $V_e$ is escape velocity.

Step 4: Plug in values and simplify

$ \frac{1}{2} m (\sqrt{5} V_e)^2 - \frac{g R^2 m}{R} = \frac{1}{2} m V_f^2 $

$ \Rightarrow \frac{5}{2} m V_e^2 - g R m = \frac{1}{2} m V_f^2 $

$ \Rightarrow \frac{5}{2} V_e^2 - g R = \frac{1}{2} V_f^2 $

Step 5: Express $g R$ in terms of $V_e$

Remember, $V_e = \sqrt{2 g R}.$ Square both sides to get $V_e^2 = 2 g R.$

So, $g R = \frac{V_e^2}{2}.$

Step 6: Substitute and solve for $V_f$

Plug $g R = \frac{V_e^2}{2}$ into the previous equation:

$ \frac{5}{2} V_e^2 - \frac{V_e^2}{2} = \frac{1}{2} V_f^2 $

$ (5 - 1)\frac{V_e^2}{2} = \frac{1}{2} V_f^2 $

$ 4\frac{V_e^2}{2} = \frac{1}{2} V_f^2 $

Multiply both sides by $2$:

$ 4 V_e^2 = V_f^2 $

Take square root:

$ V_f = 2 V_e $

Time Period of a Simple Pendulum:

The time period (how long it takes for one swing) of a simple pendulum is given by: $ T = 2\pi \sqrt{\frac{l}{g}} $

This means $T$ depends on $1/\sqrt{g}$, where $g$ is gravity at the pendulum's location.

Gravity at a Height:

Gravity changes as you go away from the Earth's surface. At height $h$ above Earth, gravity is: $ g' = \frac{g}{\left(1+\frac{h}{R_E}\right)^2} $ where $R_E$ is Earth's radius.

Time Period Ratio at Two Heights:

Let $T_1$ be the time period at height $2R_E$ and $T_2$ at $3R_E$.

So, $h_1 = 2R_E$ and $h_2 = 3R_E$.

The ratio is: $ \frac{T_1}{T_2} = \sqrt{\frac{g_2}{g_1}} $

Calculate Gravity at Each Height:

At $h_1 = 2R_E$, gravity is $ g_1 = \frac{g}{(1+2)^2} = \frac{g}{9} $

At $h_2 = 3R_E$, gravity is $ g_2 = \frac{g}{(1+3)^2} = \frac{g}{16} $

Find the Ratio:

So, $ \frac{T_1}{T_2} = \sqrt{\frac{g_2}{g_1}} = \sqrt{\frac{g/16}{g/9}} = \sqrt{\frac{9}{16}} = \frac{3}{4} $

Therefore, the ratio of the time periods is $3:4$.

Applying conservation of energy

$ \frac{1}{2} m v^2-\frac{G M m}{R}=-\frac{G M m}{R+h} $

But $G M=g R^2$

$ \therefore \quad \frac{1}{2} m v^2-\frac{g R^2 m}{R}=\frac{-g R^2 m}{R+h} $

$ \begin{aligned} & \Rightarrow \quad \frac{v^2}{2}-g R=-\frac{g R^2}{R+h} \\ & \Rightarrow \quad h=R\left(\frac{v^2}{2 g R-v^2}\right) \\ & =6.4 \times 10^6\left[\frac{(8000)^2}{2 \times 10 \times 6.4 \times 10^6-(8000)^2}\right] \\ & =6.4 \times 10^6 \mathrm{~m}=6400 \times 10^3 \mathrm{~m} \\ & =6400 \mathrm{~km} \end{aligned} $

Gravitational force is one of the basic force in nature which is weakest force but has infinite range.

Initial energy at distance $ 20 R $,

Kinetic energy, $ \mathrm{KE}_{i}=\frac{1}{2} m u^{2} $

Gravitational potential energy,

$ U_{i}=-\frac{G M m}{20 R} $

Total initial energy, $ E_{i}=K_{i}+U_{i} $

$ =\frac{1}{2} m u^{2}-\frac{G M m}{20 R} $

Final energy at the surface of the planet,

$ \begin{aligned} E_{f} & =\mathrm{KE}_{f}+U_{f} \\\\ & =\frac{1}{2} m u_{f}^{2}-\frac{G M m}{R} \end{aligned} $

According to conservation of energy,

$ \begin{array}{l} E_{i}=E_{f} \\\\ \frac{1}{2} m u^{2}-\frac{G M m}{20 R}=\frac{1}{2} m u_{f}^{2}-\frac{G M m}{R} \\\\ m u_{f}^{2}=m u^{2}-\frac{G M m}{10 R}+\frac{2 G M m}{R} \\\\ m u_{f}^{2}=m u^{2}+\frac{19}{10} \frac{G M m}{R} \end{array} $

$ u_{f}^{2}=u^{2}+\frac{19}{10} \frac{G M}{R} $

$ u_{f}=\left[u^{2}+\frac{19}{10} \frac{G M}{R}\right]^{\frac{1}{2}} $

To determine the acceleration due to gravity on the surface of a planet given certain ratios, let's start with the provided details:

The ratio of the radii of the planet and Earth is $ \frac{r_1}{r_2} = \frac{1}{2} $.

The ratio of their mean densities is $ \frac{\rho_1}{\rho_2} = \frac{4}{1} $.

The acceleration due to gravity on Earth is $ g_2 = 9.8 \, \mathrm{m/s}^2 $.

The formula for gravitational acceleration $ g $ on the surface of a celestial body is:

$ g = \frac{4}{3} \pi G \rho R $

where $ G $ is the universal gravitational constant.

Applying this formula:

For the planet, the gravitational acceleration $ g_1 $ is given by:

$ g_1 = \frac{4}{3} \pi G \rho_1 r_1 $

For Earth, its gravitational acceleration $ g_2 $ is:

$ g_2 = \frac{4}{3} \pi G \rho_2 r_2 $

To find the ratio of $ g_1 $ to $ g_2 $, we divide the equations:

$ \frac{g_1}{g_2} = \frac{\rho_1 r_1}{\rho_2 r_2} $

Substitute the given ratios:

$ \frac{g_1}{g_2} = \frac{4}{1} \times \frac{1}{2} = 2 $

Since $ g_2 = 9.8 \, \mathrm{m/s}^2 $, we calculate $ g_1 $:

$ g_1 = 9.8 \times 2 = 19.6 \, \mathrm{m/s}^2 $

Thus, the acceleration due to gravity on the planet's surface is $ 19.6 \, \mathrm{m/s}^2 $.

Mass of 1st star $ m_{1}=M $

Mass of 2nd star $ m_{2}=2 M $

Distance between two star $ =d $

Let $ r_{1} $ and $ r_{2} $ be the orbital radius of 1 st and 2 nd star.

We know, $ r_{1}+r_{2}=d $

$ \Rightarrow \quad r_{1}=d-r_{2} $

Gravitational force between two masses

$ F_{G}=\frac{G m_{1} m_{2}}{r^{2}} $

They revolve are each other due to centripetal force

$ F_{C}=m r \omega^{2} $

From the centre of mass fomula,

$ \begin{aligned} M r_{1} & =2 M r_{2} \\\\ r_{1} & =2\left(d-r_{1}\right) \\\\ r_{1} & =\frac{2}{3} d \end{aligned} $

Equate gravitation force acting on lst mass and centripetal force on lst mass

$ \begin{aligned} F_{C} & =F_{G} \\\\ m_{1} r_{1} \omega^{2} & =\frac{G m_{1} m_{2}}{d^{2}} \\\\ M \frac{2 d}{3} \omega^{2} & =\frac{G M 2 M}{d^{2}} \\\\ \omega^{2} & =\frac{3 G M}{d^{3}} \\\\ \omega & =\sqrt{\frac{3 G M}{d^{3}}} \end{aligned} $

Given, $h_1=1280 \mathrm{~km}$

$ h_2=3200 \mathrm{~km} $

Acceleration due to gravity at height $h$,

$ \begin{aligned} & g^{\prime}=g\left(\frac{R_r}{R_t+h}\right)^2 \\\\ & \therefore g_1=g\left(\frac{R_r}{R_e+h}\right)^2=g\left(\frac{6400}{6400+1280}\right)^2 \\\\ & \text { Similarly, } g_2=g\left(\frac{6400}{6400+3200}\right)^2 \\\\ & \therefore \frac{g_1}{g_2}=\frac{\left(\frac{6400}{6400+1280}\right)^2}{\left(\frac{6400}{6400+3200}\right)^2}=\frac{\left(\frac{1}{7680}\right)^2}{\left(\frac{1}{9600}\right)^2} \end{aligned} $

$ \begin{aligned} & =\frac{9600 \times 9600}{7680 \times 7680} \\\\ & =\frac{5 \times 5}{4 \times 4}=\frac{25}{16} \end{aligned} $

The energy required to take a body from the surface of the earth to a height equal to the radius of the earth ( $W$ ).

The gravitational potential energy,

$ \begin{aligned} & W=\frac{G M m}{R}-\frac{G M m}{2 R} \\\\ & W=\frac{G M m}{2 R} \quad \ldots \text { (i) } \end{aligned} $

To take body from the surface of the earth to a height to twice of the radius, the energy required ( $W^{\prime}$ ) is

$ \begin{aligned} W^{\prime} & =\frac{G M m}{R}-\frac{G M m}{3 R} \\\\ & =\frac{2 G M m}{3 R} \quad \ldots \text { (ii) } \end{aligned} $

Comparing Eq. (i) and Eq. (ii)

$ W^{\prime}=\frac{4}{3} W $

So, the energy required to take the body from the surface of the earth to twice of the radius of the earth is $4 / 3 \mathrm{~W}$ times the energy required.

Given,

Mass of body $=m$

Radius $=R$

Initial height $=R$ from surface

Final height $=3 R$ from surface

Gravitational potential energy is given by at height, $h$

$ U=\frac{m g h}{\left[1+\frac{h}{R}\right]}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) $

Potential energy for $h=R$

$ U_1=\frac{m g R}{1+\frac{R}{R}}=\frac{m g R}{2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Potential energy for $h=3 R$

$ U_2=\frac{m g 3 R}{1+\frac{3 R}{R}}=\frac{3}{4} m g R \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

From Eq. (iii) and Eq. (ii) increase in potential energy is

$ \begin{aligned} U_2-U_1 & =\frac{3}{4} m g R-\frac{m g R}{2} \\ \Delta U & =\frac{m g R}{4} \end{aligned} $

We know that, orbital speed of any planet

$ v_0=\sqrt{g R} $

escape speed of any planet, $v_e=\sqrt{2 g R}$

Gravitational forces operate among all objects in the universe. All object having mass. We affected by gravitational forces exerted by another objects having mass. It has to be noted that photon are massless particles (quanta of light) yet recent discoveries of gravitational waves and lensing showed that light (photons) is also affecteded by gravitational forces.

Binding energy, $f=$ satellite is given by

$ \text { PE }=\frac{1}{2} \frac{G M m}{r}(\text { for orbit } r) $

and $\mathrm{PE}^{\prime}=\frac{1}{3} \frac{G M m}{r}$ (For orbit $\frac{3 r}{2}$ )

Increase in energy of satellite will be

$ \begin{aligned} \Delta E & =\frac{G M m}{r}\left(\frac{1}{2}-\frac{1}{3}\right) \cdot \frac{G M m}{6 r} \\ \therefore \% \text { increase } & =\frac{\frac{G M m}{6 r}}{\frac{G M m}{2 r}} \times 100 \%=33.33 \% \end{aligned} $

To determine the ratio of the escape velocities from two planets, consider the following:

Given Ratios:

Ratio of the radii of the two planets: $ r $

Ratio of the accelerations due to gravity: $ x $

Escape Velocity Formula:

The escape velocity, $ v $, is given by the formula:

$ v = \sqrt{2Rg} $

Where $ R $ is the radius of the planet and $ g $ is the acceleration due to gravity.

Applying to Two Planets:

For Planet 1:

$ v_1 = \sqrt{2R_1g_1} $

For Planet 2:

$ v_2 = \sqrt{2R_2g_2} $

Finding the Ratio of Escape Velocities:

The ratio of the escape velocities is:

$ \frac{v_1}{v_2} = \sqrt{\frac{2R_1g_1}{2R_2g_2}} $

Substitute the Given Ratios:

Use the given ratios:

$ r = \frac{R_1}{R_2} $

$ x = \frac{g_1}{g_2} $

Final Calculation:

The ratio becomes:

$ \frac{v_1}{v_2} = \sqrt{r \cdot x} $

Thus, the ratio of the escape velocities is $\sqrt{rx}$.

Given,

Mass of ring $=m$

Radius $=R$

We have to calculate the distance from its geometrical axisGravitational field is given by at point $P$.$ E_P=\frac{G M x}{\left(R^2+x^2\right)^{3 / 2}} $

To get the maximum value of $E_P$

$ \begin{aligned} & \frac{d E_p}{d x}=0 \\ & \Rightarrow \quad \frac{G M}{\left(R^2+x^2\right)^{3 / 2}}-G M x \\ & \times \frac{3}{2\left(R^2+x^2\right)^{1 / 2}} \times 2 x=0 \end{aligned} $

Solving above expression,

$ \begin{aligned} 3 x^2 & =R^2+x^2 \\ 2 x^2 & =R^2 \\ \Rightarrow \quad x & =\frac{R}{\sqrt{2}} \end{aligned} $

Inside a hollow spherical shell, net gravitational field is zero at all the points. Hence, force of attraction on a point mass inside shell is always zero. Spherical objects like shells acts like a point mass at centre. This is shell theorem. So statement II is correct.

Escape speed is $v_0$,

$ \Rightarrow \quad v_0=\sqrt{\frac{2 G M}{R}} $

Now, if velocity of object at infinity is $v$, then by conservation of energy we have,

Energy at surface $=$ Energy at infinity

$ \frac{-G M m}{R}+\frac{1}{2} m\left(5 v_0\right)^2=\frac{1}{2} m v^2+0 $

Now, $\quad \frac{G M}{R}=\frac{1}{2} v_0^2$

[from Eq. (i)]

So, $-\frac{1}{2} m v_0^2+\frac{1}{2} m\left(5 v_0\right)^2=\frac{1}{2} m v^2$

$ \begin{array}{rlrl} \Rightarrow &\ -v_0^2+25 v_0^2 & =v^2 \\ \text { or } & & v & =\sqrt{24\left(v_0^2\right)} \\ \text { or } & & v & =2 \sqrt{6} v_0 \end{array} $

The given situation is shown below.

$ A C=B D=\sqrt{\left(4 l_0\right)^2+\left(3 l_0\right)^2}=5 l_0 $

∴ Potential energy of the system,

$ \begin{aligned} U=U_{A B}+U_{B C}+U_{C D} & +U_{D A} +U_{A C}+U_{B D} \end{aligned} $

Applying the formula,

$U=-G \frac{m_1 m_2}{r}$, we get

$ \begin{aligned} U & =-\frac{G \cdot m \cdot m}{4 l_0}-\frac{G \cdot m \cdot m}{3 l_0}-\frac{G \cdot m \cdot m}{4 l_0} \\ & \quad-\frac{G \cdot m \cdot m}{3 l_0}-\frac{G \cdot m \cdot m}{5 l_0}-\frac{G \cdot m \cdot m}{5 l_0} \\ & =-\frac{G m^2}{l_0}\left[\frac{1}{4}+\frac{1}{3}+\frac{1}{4}+\frac{1}{3}+\frac{1}{5}+\frac{1}{5}\right] \\ & =-\frac{G m^2}{l_0}\left[\frac{1}{2}+\frac{2}{3}+\frac{2}{5}\right] \\ & =-\frac{G m^2}{l_0}\left[\frac{15+20+12}{30}\right] \\ & =-\frac{G m^2}{l_0}\left(\frac{47}{30}\right) \\ |U| & =\frac{47}{30}\left(\frac{G m^2}{l_0}\right) \end{aligned} $

The force of attraction between the complete sphere of mass $M$ and particle of mass $m$ is

$ F=\frac{G m M}{4 R^2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,..(i)$

Mass of the complete sphere,

$ M=\frac{4}{3} \pi R^3 \rho $

where, $\rho$ is density of sphere. Mass of the cut out portion,

$ m^{\prime}=\frac{4}{3} \pi\left(\frac{R}{2}\right)^3 \rho=\frac{M}{8} $

Now, the distance between the centre of the cut out portion and mass $m$,

$ =2 R-\frac{R}{2}=\frac{3 R}{2} $

Hence, the force of attraction between the cut out portion and mass $m$ is

$ \begin{aligned} F^{\prime} & =\frac{G m^{\prime} M}{\left(\frac{3 R}{2}\right)^2}=\frac{G\left(\frac{M}{8}\right) m}{\frac{9 R^2}{4}} \\ & =\frac{G m M}{4 R^2} \times \frac{2}{9}=\frac{2}{9} F[\text { from Eq }(\text { i) }] \end{aligned} $

Thus, the force of attraction between the remaining part of the sphere and

$ \text { mass } m=F-F^{\prime}=F-\frac{2}{9} F=\frac{7}{9} F $

The given situation is shown below.

When the rocket will reach at maximum height, then the total gravitational potential energy at this height will be equal to the sum of kinetic energy at the surface of earth and the potential energy when the rocket was at the earth's surface.

$ \begin{array}{cc} \therefore & -\frac{G m M_c}{(R+h)}=\frac{1}{2} m v^2+\left[\frac{-G m M_c}{R}\right] \\ \Rightarrow & \frac{-G m M_c}{(R+h)}=\frac{1}{2} m v^2-\frac{G m M_c}{R} \\ \Rightarrow & \frac{1}{2} m v^2=\frac{G m M_c}{R}-\frac{G m M_c}{R+h} \\ \Rightarrow & \frac{1}{2} v^2=G M_c\left[\frac{1}{R}-\frac{1}{R+h}\right] \\ \Rightarrow & \frac{1}{2} \times(4000)^2=\frac{G M_c}{R^2} \times R^2\left[\frac{R+h-R}{R(R+h)}\right] \\ \Rightarrow & 8 \times 10^6=g R^2\left[\frac{h}{R(R+h)}\right] \\ & =\frac{g h R}{(R+h)}=\frac{10 h R}{(R+h)} \end{array} $

$ \begin{aligned} & \Rightarrow \quad 8 \times 10^5=\frac{h R}{R+h} \\ & \Rightarrow 8 \times 10^5(R+h)=h R \\ & \Rightarrow 8 \times 10^5\left[6.4 \times 10^6+h\right]=h \times 6.4 \times 10^6 \\ & \Rightarrow 51.2 \times 10^{11}=\left(6.4 \times 10^6-8 \times 10^5\right) h \\ & \Rightarrow 51.2 \times 10^{11}=\left(64 \times 10^5-8 \times 10^5\right) h \\ & \Rightarrow \quad h=\frac{51.2 \times 10^{11}}{56 \times 10^5}=0.914 \times 10^6 \mathrm{~m} \\ & \Rightarrow \quad h=\frac{0.914 \times 10^6}{10^3} \mathrm{~km}=914 \mathrm{~km} \end{aligned} $

The situation given in the question is depicted below

Let the speed of particle is $v$ for circular motion.

Clearly the gravitational force acting all the 3 masses will be same due to symmetry and let it be $F$.

Now, considering the mass $M$ placed at point $B$. Since, it is performing circular motion in radius say $R$, hence the net gravitational pull on it due to the other two masses are providing the required centripetal force.

Now, net force on $M$ at $B$,

$ \begin{aligned} & F^{\prime} =\sqrt{F^2+F^2+2 F^2 \cos 60^{\circ}} \\ \Rightarrow F^{\prime} & =\sqrt{3} F \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \text { and } & F =\frac{G M^2}{l^2} \end{aligned} $

(gravitational force of Newton)

Also, by symmetry of forces the $F^{\prime}$ will be acting along the bisector of angle at point $B$ i.e. at $30^{\circ}$ from each force $F$.

Now, in $\triangle O B P$,

$ \begin{array}{rlrl} \cos 30^{\circ} & =\frac{\frac{l}{2}}{R} \left(\because B P=\frac{B C}{2}\right) \\ \frac{\sqrt{3}}{2} & =\frac{l}{2 R} \\ \Rightarrow \quad R & =\frac{l}{\sqrt{3}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

Now, since $F^{\prime}=F_{\text {centripetal }}$

i.e. $\quad F^{\prime}=F_c$

$ \begin{array}{ll} \Rightarrow \quad \sqrt{3} F=\frac{M v^2}{R} & \text { [from Eq. (i)] } \\ & {\left[\because F_c=\frac{M v^2}{R}\right]} \end{array} $

$ \Rightarrow \quad \sqrt{3} \frac{G M^2}{l^2}=\frac{M v^2}{R} $

$ \begin{aligned} & \sqrt{3} \frac{G M^2}{l^2} =\frac{M v^2}{l} \sqrt{3} \\ \Rightarrow & v^2 =\frac{G M}{l} \\ \Rightarrow & v =\sqrt{\frac{G M}{l}} \end{aligned} $

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.