Gravitation

$ \begin{aligned} &\text {}\\ &\begin{aligned} g^{\prime} & =\frac{g}{\left(1+\frac{h}{R}\right)^2}=\frac{g}{\left[1+\frac{(\sqrt{2}-1) R}{R}\right]^2} \\ & =\frac{g}{(\sqrt{2})^2}=\frac{g}{2}=\frac{10}{2}=5 \mathrm{~m} / \mathrm{s}^2 \end{aligned} \end{aligned} $

$ \begin{aligned} V_{R_1} & =\sqrt{\frac{2 G M_1}{R_1}}=\sqrt{\frac{2 G M}{R}}=14 \mathrm{~km} \mathrm{~s}^{-1} \\ V_{P_2} & =\sqrt{2 G \cdot \frac{M_2}{R_2}}=\sqrt{\frac{2 G \cdot M}{4 R}} \\ & \quad\left[\because M_2=M \text { and } R_2=4 R\right] \\ & =\frac{1}{2} \sqrt{\frac{2 G M}{R}}=\frac{1}{2} \times 14 \\ & =7 \mathrm{~km} \mathrm{~s}^{-1} \end{aligned} $

Step 1: Formula for Potential Energy

The potential energy ($U$) of a satellite of mass $m$ at a distance $r$ from the center of Earth is given by: $ U = -\frac{G M_e m}{r} $ where $G$ is the gravitational constant, and $M_e$ is the mass of the Earth.

Step 2: Substitute the Satellite's Height

The satellite is at a height $h=R_e$ above the surface of the Earth. The total distance from the Earth's center is $r = R_e + h = R_e + R_e = 2R_e$.

Step 3: Put this Distance Into the Formula

So, the potential energy is: $ U = -\frac{G M_e m}{2R_e} $

Step 4: Use the Relationship $G M_e = g R_e^2$

We know that $G M_e = g R_e^2$, where $g$ is the acceleration due to gravity. Substituting, we get: $ U = -\frac{g R_e^2 m}{2 R_e} $

Step 5: Simplify

Divide $R_e^2$ by $R_e$: $ U = -\frac{g R_e m}{2} $

Final Answer

The potential energy of the satellite is: $ U = -0.5 m g R_e $

Time period of pendulum

$ T=2 \pi \sqrt{\frac{l}{g}} $

At a height $h=\frac{R}{2}$ above Earth's surface,

$ g_h=g \cdot \frac{R^2}{(R+h)^2}=\frac{g \cdot R^2}{\frac{9}{4} R^2}=\frac{4}{9} g $

$ \begin{aligned} &\text { Time period at this height is, }\\ &\begin{aligned} & T^{\prime}=2 \pi \sqrt{\frac{l}{g h}}=2 \pi \sqrt{\frac{l}{\frac{4}{9} g}} \\ & =\frac{3}{2}\left(2 \pi \sqrt{\frac{l}{g}}\right)=\frac{3}{2} T \end{aligned} \end{aligned} $

Step 1: Escape velocity from Earth

The escape velocity from the Earth's surface is given by:

$ v_{\text{escape}} = \sqrt{\frac{2GM}{R}} $

Here, $ G $ is the gravitational constant, $ M $ is Earth's mass, and $ R $ is Earth's radius.

Step 2: Orbital velocity at height $h = R$

The orbital velocity for a satellite at a distance $ r $ from the center of the Earth is:

$ v_{\text{orbit}} = \sqrt{\frac{GM}{r}} $

If the satellite is at a height equal to the Earth's radius ($ h = R $), the total distance from Earth's center is $ r = R + R = 2R $.

So, the orbital velocity becomes:

$ v_{\text{orbit}} = \sqrt{\frac{GM}{2R}} $

Step 3: Find the relationship between escape velocity and orbital velocity

Let's compare the orbital and escape velocities:

$ \frac{v_{\text{orbit}}}{v_{\text{escape}}} = \frac{\sqrt{\frac{GM}{2R}}}{\sqrt{\frac{2GM}{R}}} $

Simplifying this gives:

$ \frac{v_{\text{orbit}}}{v_{\text{escape}}} = \frac{1}{2} $

Step 4: Calculate the final answer

This means the orbital velocity is half the escape velocity. Given that the escape velocity is $11.2~\mathrm{km}~\mathrm{s}^{-1}$:

$ v_{\text{orbit}} = \frac{11.2}{2} = 5.6~\mathrm{km}~\mathrm{s}^{-1} $

$ \begin{aligned} & \text { } T \propto a^{\frac{3}{2}} g^x R^y \Rightarrow T=k a^{\frac{3}{2}} g^x R^y \\ & {\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^1\right]=\left[\mathrm{L}^{3 / 2}\right]\left[\mathrm{LT}^2\right]^x\left[\mathrm{~L}^y\right]} \\ & {\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^1\right]=\left[\mathrm{L}^{\frac{3}{2}+x+y} \mathrm{~T}^{-2 x}\right]} \\ & \therefore-2 x=1 \\ & \Rightarrow \quad x=\frac{-1}{2} \\ & \frac{3}{2}+x+y=0 \\ & \frac{3}{2}-\frac{1}{2}+y=0 \\ & 1+y=0 \Rightarrow y=-1 \end{aligned} $

Escape velocity is given as

$ v=\sqrt{2 g R}=\sqrt{\frac{2 G M}{R}} $

$M \rightarrow$ Mass of sphere, $R \rightarrow$ Radius of sphere

$ \begin{aligned} & 3 \times 10^4=\sqrt{\frac{2 \times 6.66 \times 10^{-11} \times 6 \times 10^{24}}{R}} \\ & \quad R=\frac{2 \times 6.66 \times 10^{-11} \times 6 \times 10^{24}}{9 \times 10^8} \\ & R=8.88 \times 10^5 \text { metre } \\ & R=888 \times 10^3 \text { metre }=888 \mathrm{~km} \end{aligned} $

Given data

• Height of satellite $A$ above Earth's surface: $h_A = 1.25 R_E$

• Height of satellite $B$ above Earth's surface: $h_B = 19.25 R_E$

Hence,

the orbital radii (distance from Earth's center):

$ r_A = R_E + h_A = R_E + 1.25R_E = 2.25R_E $

$ r_B = R_E + h_B = R_E + 19.25R_E = 20.25R_E $

Formula for orbital speed

For a satellite in a circular orbit around Earth:

$ v = \sqrt{\frac{GM_E}{r}} $

where $G$ is the gravitational constant and $M_E$ is Earth’s mass.

Thus, for two satellites $A$ and $B$:

$ \frac{v_A}{v_B} = \sqrt{\frac{r_B}{r_A}} $

Substitute $r_A$ and $r_B$

$ \frac{v_A}{v_B} = \sqrt{\frac{20.25R_E}{2.25R_E}} = \sqrt{\frac{20.25}{2.25}} = \sqrt{9} = 3 $

✅ Therefore,

$ \boxed{v_A : v_B = 3 : 1} $

Correct Option: D) $3 : 1$

$ \begin{aligned} &\text { Mass of each, sphere }\\ &M=\frac{4}{2} \pi R^3 \times \rho \end{aligned} $

$ \begin{array}{ll} \Rightarrow & M \propto R^3 \\ \therefore & F \propto \frac{M_1 M_2}{(2 R)^2} \\ & \propto \frac{R^3 \times R^3}{R^2} \\ \Rightarrow & \propto R^4 \end{array} $

$ \begin{aligned} & \text { } T^2 \propto R^3 \\ & \Rightarrow \quad\left(\frac{2 \pi}{\omega}\right)^2 \propto R^3 \\ & \Rightarrow \quad \frac{\omega^2}{4 \pi^2} \propto \frac{1}{R^3} \end{aligned} $

$ \begin{array}{ll} \Rightarrow & \omega^2 \propto \frac{1}{R^3} \\ \Rightarrow & \left(\frac{\omega_2}{\omega_1}\right)^2=\left(\frac{R_1}{R_2}\right)^3 \\ \Rightarrow & \left(\frac{\omega_{1 / 2}}{\omega_1}\right)^2=\left(\frac{R_1}{2^n R_1}\right)^3 \\ \Rightarrow & \left(\frac{1}{2}\right)^2=\left(\frac{1}{2^n}\right)^3 \\ \Rightarrow & \frac{1}{2^2}=\frac{1}{2^{3 n}} \\ \Rightarrow & 2^2=2^{3 n} \Rightarrow 2=3 n \\ & n=\frac{2}{3} \end{array} $

Gravitational potential

$ V=-\frac{G m}{r} $

Since, the mass of each object is 1 kg and the distance are $1,2,4,8, \ldots$ on both sides of $x=0$. Thus, the total potential at $x=0$

$ \begin{aligned} V_{\text {total }}= & -G\left(\frac{m_1}{r_1}+\frac{m_2}{r_2}+\frac{m_3}{r_3}+\ldots\right) \\ & \quad-G\left(\frac{m_1}{r_1}+\frac{m_2}{r_2}+\frac{m_3}{r_3}+\ldots\right) \\ = & -2 G\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots\right) \\ = & -2 G\left(\frac{1}{1-\frac{1}{2}}\right)=-4 G \end{aligned} $

According to Kepler's law, $T^2 \propto r^3$, where $T$ is the orbital period and $r$ is the orbital radius.

For satellites orbiting close to the surface of a planet, we use:

$T_1$: the orbital time period of the first satellite.

$R$: the radius of the planet.

Since the first satellite is close to the planet's surface, its orbital radius is approximately equal to the planet's radius, $R$.

For the second satellite, which orbits at a height of 3 times the planet's radius, its orbital radius becomes $R + 3R = 4R$.

Let $T_2$ denote the orbital period of this second satellite. From Kepler's law, we have:

$ \frac{T_2^2}{T_1^2} = \frac{(4R)^3}{R^3} \quad \Rightarrow \quad \frac{T_2^2}{T_1^2} = 4^3 = 64 $

Thus,

$ \frac{T_2}{T_1} = 8 $

Given $T_1$ is 80 minutes, we can calculate $T_2$:

$ T_2 = 8 \times T_1 = 8 \times 80 \text{ minutes} = 640 \text{ minutes} $

Therefore, the orbital period of the second satellite, which is at a height of 3 times the planet's radius from its surface, is 640 minutes.

The gravitational potential energy $ U $ at a distance $ r $ from the center of the Earth is given by the formula:

$ U = -\frac{G M m}{r} $

where:

$ M $ = mass of the Earth

$ m $ = mass of the body

$ r $ = distance from the center of the Earth

On the surface of the Earth ($ r = R $), the gravitational potential energy is expressed as:

$ U = E = -\frac{G M m}{R} \quad \text{...(i)} $

When the body is taken to a height equal to $ 150\% $ of the Earth's radius, the new distance from the center of the Earth, $ r' $, becomes:

$ r' = R + \frac{150}{100} \times R = R + 1.5R = 2.5R $

The new gravitational potential energy $ U' $ at this height is calculated as:

$ U' = -\frac{G M m}{r'} = -\frac{G M m}{2.5R} $

Rewriting this in terms of $ E $, we have:

$ U' = \frac{1}{2.5} \left(-\frac{G M m}{R}\right) $

Substituting from Eq. (i):

$ U' = \frac{E}{2.5} $

Simplifying this gives:

$ U' = 0.4E $

To understand the energy required for a satellite to escape Earth's gravitational pull, let's break down the process:

Orbital Velocity of the Satellite:

The orbital velocity ($v_0$) of a satellite moving in a circular orbit around Earth is given by:

$ v_0 = \sqrt{\frac{G M}{r}} $

where $G$ is the gravitational constant, $M$ is the mass of the Earth, and $r$ is the radius of the orbit.

Kinetic Energy of the Satellite:

The kinetic energy ($E$) of the satellite in orbit is:

$ E = \frac{1}{2} m v_0^2 = \frac{1}{2} \cdot \frac{G M m}{r} $

where $m$ is the mass of the satellite.

Kinetic Energy for Escape Velocity:

To escape Earth's gravitational field, the satellite needs to achieve escape velocity ($v_e$), which is calculated as:

$ v_e = \sqrt{\frac{2 G M}{r}} $

Therefore, the kinetic energy required for escape ($E'$) is:

$ E' = \frac{1}{2} m v_e^2 = \frac{1}{2} m \cdot \frac{2 G M}{r} = \frac{G M m}{r} $

Simplifying, we find:

$ E' = 2E $

Additional Energy Required:

The additional kinetic energy needed to achieve escape velocity is:

$ E' - E = 2E - E $

Therefore, the minimum additional energy required is:

$ E $

Given:

Particle velocity, $ v_p = 2v_{\text{escape}} $

The total energy of the projectile on Earth is expressed as:

$ \frac{1}{2} m v_p^2 - \frac{1}{2} m v_{\text{escape}}^2 $

When the particle is very far from the Earth, its gravitational potential energy becomes zero. Therefore, the total energy of the particle far from Earth is:

$ \frac{1}{2} m v_f^2 $

where $ v_f $ is the velocity of the particle when it is very far away.

By the conservation of energy, we have:

$ \frac{1}{2} m v_p^2 - \frac{1}{2} m v_{\text{escape}}^2 = \frac{1}{2} m v_f^2 $

Solving for $ v_f $:

$ v_f = \sqrt{v_p^2 - v_{\text{escape}}^2} $

Substituting $ v_p = 2v_{\text{escape}} $:

$ v_f = \sqrt{(2v_{\text{escape}})^2 - v_{\text{escape}}^2} $

$ v_f = \sqrt{4v_{\text{escape}}^2 - v_{\text{escape}}^2} $

$ v_f = \sqrt{3v_{\text{escape}}^2} $

$ v_f = \sqrt{3} v_{\text{escape}} $

The time period of a satellite's revolution, denoted as $ T $, depends on the radius of the circular orbit $ R $, the mass of the Earth $ M $, and the universal gravitational constant $ G $. Using dimensional analysis, we derive the expression for $ T $ as follows:

Given:

Time period of revolution is $ T $.

Radius of circular orbit is $ R $.

We use the relation:

$ T \propto G^a M^b R^c $

This can be expressed as:

$ T = K G^a M^b R^c $

where $ K $ is a constant of proportionality.

The dimensional expressions for the relevant quantities are:

$ [T] = [T] $

$ [G] = [M^{-1}L^3T^{-2}] $

$ [M] = [M] $

$ [R] = [L] $

Thus, the dimensional equation becomes:

$ [T] = [M^{-a}L^{3a+c}T^{2a}] $

Comparing dimensions on both sides, we obtain the following equations:

$ -a + b = 0 $

$ 3a + c = 0 $

$ 2a = -1 $

Solving these equations:

From $ 2a = -1 $, we find $ a = -\frac{1}{2} $.

Using $ -a + b = 0 $, we substitute for $ a $ to get $ b = -\frac{1}{2} $.

From $ 3a + c = 0 $, substituting for $ a $ gives $ c = \frac{3}{2} $.

Substituting these values back into the relation for $ T $:

$ T = K G^{-\frac{1}{2}} M^{-\frac{1}{2}} R^{\frac{3}{2}} $

This simplifies to:

$ T = K \frac{R^{\frac{3}{2}}}{G^{\frac{1}{2}} M^{\frac{1}{2}}} $

Therefore:

$ T = K \sqrt{\frac{R^3}{G M}} $

To solve this, let's start with the time period of the satellite's revolution, denoted as $ T $. According to Kepler's third law, the relationship between the time period $ T $ and the radius $ r $ of the satellite's orbit can be expressed as:

$ T^2 \propto r^3 $

This implies:

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

Thus, rearranging gives:

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

Next, let's consider the kinetic energy (KE) of the satellite. It's known that the kinetic energy is inversely proportional to the radius $ r $:

$ \text{KE} \propto \frac{1}{r} $

Substituting the expression for $ r $ from above, we find:

$ \text{KE} \propto \frac{1}{T^{2/3}} $

Therefore, the kinetic energy is proportional to:

$ \text{KE} \propto T^{-2/3} $

To find the height above the Earth's surface where the acceleration due to gravity is one-fourth of its value at the surface, we can follow these steps:

Given Condition:

$ g_h = \frac{g_0}{4} \quad \text{(equation 1)} $

Formula for Gravity at Height $ h $:

$ g_h = g \left(\frac{R_E}{R_E + h}\right)^2 $

Here, $ R_E = 6400 \, \text{km} $.

Set the Equation:

$ \frac{g}{4} = g \left(\frac{R_E}{R_E + h}\right)^2 $

Simplify the Equation:

$ \frac{1}{2} = \frac{R_E}{R_E + h} $

Solve for $ h $:

$ R_E + h = 2 R_E $

$ h = R_E $

Therefore, $ h = 6400 \, \text{km} $.

Thus, the height where the acceleration due to gravity is one-fourth of its surface value is 6400 km.

We know the gravitational acceleration follows an inverse-square law, meaning :

$ g \propto \frac{1}{r^2} $

So for two positions with distances $r_1$ and $r_2$ from the Earth's center, we have :

$ \frac{g_2}{g_1} = \left(\frac{r_1}{r_2}\right)^2 $

The Earth's radius is given as 6400 km.

At a height of 6400 km above the Earth's surface, the distance from the center is :

$r_1 = 6400\, \text{km} + 6400\, \text{km} = 12800\, \text{km}$

At a height of 12800 km above the Earth's surface, the distance from the center is :

$r_2 = 6400\, \text{km} + 12800\, \text{km} = 19200\, \text{km}$

Given that the acceleration at $r_1$ is $g_1 = 2.5\, \text{m/s}^2$, we can find $g_2$ by :

$ g_2 = g_1 \times \left(\frac{r_1}{r_2}\right)^2 $

Substitute the values :

$ g_2 = 2.5 \times \left(\frac{12800}{19200}\right)^2 $

Notice that :

$ \frac{12800}{19200} = \frac{2}{3} $

So,

$ g_2 = 2.5 \times \left(\frac{2}{3}\right)^2 = 2.5 \times \frac{4}{9} = \frac{10}{9} \approx 1.11\, \text{m/s}^2. $

Thus, the acceleration due to gravity at a height of 12800 km from the Earth's surface is approximately $1.11\, \text{m/s}^2$.

Answer: Option A.

The maximum height reached by the rocket occurs when its kinetic energy is completely converted into gravitational potential energy, meaning the rocket momentarily stops.

Using the conservation of energy principle:

$ \frac{-GMm}{R} + \frac{1}{2}mv^2 = \frac{-GMm}{R+H} $

Here, $v$ is the velocity of the rocket when launched, given as $ \frac{v_e}{2} $, where $ v_e $ is the escape velocity. The escape velocity is given by:

$ v_e = \sqrt{\frac{2GM}{R}} $

Thus, the initial velocity of the rocket is:

$ v = \frac{v_e}{2} = \frac{1}{2} \sqrt{\frac{2GM}{R}} = \sqrt{\frac{GM}{2R}} $

Substituting into the energy equation:

$ \frac{-GMm}{R} + \frac{1}{2}m \left(\sqrt{\frac{GM}{2R}}\right)^2 = \frac{-GMm}{R+H} $

Solving this equation:

$ \frac{-GMm}{R} + \frac{1}{2}m \cdot \frac{GM}{2R} = \frac{-GMm}{R+H} $

This simplifies to:

$ \frac{-GMm}{R} + \frac{GMm}{4R} = \frac{-GMm}{R+H} $

$ \frac{-3GMm}{4R} = \frac{-GMm}{R+H} $

Setting the fractions equal:

$ \frac{3}{4} \cdot \frac{1}{R} = \frac{1}{R+H} $

Cross-multiplying gives:

$ 3(R + H) = 4R $

Solving for $H$:

$ 3R + 3H = 4R $

$ 3H = R $

Thus:

$ H = \frac{R}{3} $

Therefore, the maximum height $H$ attained by the rocket is $\frac{R}{3}$.

Given:

Mass of first satellite, $ m_1 = m $

Mass of second satellite, $ m_2 = 1.5m $

Radius of Earth, $ R_E $

Calculating Distances:

For the first satellite:

Distance from Earth's center, $ r_1 = R_E + R_E = 2R_E $

For the second satellite:

Distance from Earth's center, $ r_2 = R_E + 2R_E = 3R_E $

Gravitational Forces:

Gravitational force by the first satellite:

$ F_1 = \frac{G M \cdot m}{(2R_E)^2} = \frac{G M m}{4R_E^2} $

Gravitational force by the second satellite:

$ F_2 = \frac{1.5 \times G M m}{(3R_E)^2} = \frac{1.5 G M m}{9R_E^2} $

Since $ r_2 > r_1 $, it follows that $ F_1 > F_2 $.

Ratio of Forces:

$ \frac{F_2}{F_1} = \frac{\frac{1.5 \cdot G M m}{9 R_E^2}}{\frac{G M m}{4 R_E^2}} = \frac{1.5}{9} \times \frac{4}{1} = \frac{6}{9} = \frac{2}{3} $

Thus, the ratio $ F_2 : F_1 = 2 : 3 $.

Time period of satellite moving around a planet with angular velocity $\omega$ is given as,

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

Since, the two artificial satellites revolving in the same circular orbit, then their angular velocities will be same.

Hence, both satellites revolve with same period of revolution.

Orbital velocity of revolving satellite, $v=\sqrt{\frac{G m}{r}}$ i.e. $v \propto \frac{1}{\sqrt{r}}$

Where, $m=$ mass of the planet

$G=$ Universal gravitational constant

$r=$ radius of circular orbit.

Escape velocity depends on the gravitational potential at the point from where the body is launched. Since, this potential depends on the height of the point of projection, hence escape velocity also depends on the height (altitude) of the point of projection.

Since, uniform solid sphere of radius $R$ produces a gravitational acceleration $a_0$ on its surface, hence $a_0=\frac{G m}{R^2}$

where, $m$ is the mass of solid sphere.

Let $x$ be the distance of the point from the surface of sphere where the gravitational acceleration becomes $\frac{a_0}{4}$.

i. e. $\frac{a_0}{4}=\frac{a_0}{\left(1+\frac{x}{R}\right)^2} \Rightarrow\left(\frac{1}{2}\right)^2=\frac{1}{\left(1+\frac{x}{R}\right)^2}$

$\begin{aligned} \Rightarrow & & \frac{1}{2} & =\frac{1}{1+\frac{x}{R}} \Rightarrow 1+\frac{x}{R}=2 \\ \Rightarrow & & x & =R \end{aligned}$

Hence, distance of the point from the centre of sphere $=x+R=R+R=2 R$

Initial speed of projectile from the surface of earth, $v=\alpha v_{\text {e }}$

Where, $\alpha$ is constant and $v_e$ is escape velocity of projectile.

Radius of earth, $R_E=6400 \mathrm{~km}$

$\qquad=6.4 \times 10^6 \mathrm{~m}$

Maximum height attained by the projectile,

$h=800 \mathrm{~km}=8 \times 10^5 \mathrm{~m}$

We know that, escape velocity on the surface of earth, $v_e=11.2 \mathrm{~km} / \mathrm{s}$

$\begin{aligned} & =11.2 \times 10^3 \mathrm{~m} / \mathrm{s} \\ v & =\alpha v_e \\ v & =\alpha \times 11.2 \times 10^3 \quad \text{... (i)} \end{aligned}$

According to given situation, applying the law of conservation of energy

$\begin{aligned} & \frac{1}{2} m v^2=m g h \\ \Rightarrow & v^2=2 g h \Rightarrow\left(\alpha v_e\right)^2=2 g h \\ \Rightarrow \quad & \alpha=\frac{\sqrt{2 g h}}{v_e}=\frac{\sqrt{2 \times 10 \times 8 \times 10^5}}{11.2 \times 10^3}=\frac{4}{11.2} \approx \frac{1}{3} \end{aligned}$

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,

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