Gravitation

To solve this question, we first need to understand how gravitational force (and hence weight) changes with depth under the surface of the Earth. The gravitational force at a depth $d$ is given by the formula:

$ F = F_0 \left(1 - \frac{d}{R}\right) $

where $F_0$ is the gravitational force (or the weight) at the surface, $R$ is the radius of the Earth, and $d$ is the depth below the Earth's surface.

In your question, the body weighs 300 N on the surface, so $F_0 = 300$ N. It is taken to a depth of $\frac{R}{4}$ under the surface. Therefore, $d = \frac{R}{4}$.

Substituting these values into our formula, we get:

$ F = 300 \left(1 - \frac{\frac{R}{4}}{R}\right) $

$ F = 300 \left(1 - \frac{1}{4}\right) $

$ F = 300 \left(\frac{3}{4}\right) $

$ F = 225 \, \text{N} $

Therefore, at a depth of $\frac{R}{4}$ under the surface of the Earth, the body would weigh 225 N. The correct option is Option D.

The kinetic energy required to project a body of mass $m$ from the Earth's surface to infinity, also known as the escape kinetic energy, can be calculated using the concept of gravitational potential energy. The escape velocity $v_e$ is the velocity a body must have to escape the gravitational field of the Earth without any further propulsion. The formula for escape velocity is:

$v_e = \sqrt{2gR_E}$

Where $g$ is the acceleration due to gravity on the surface of Earth and $R_E$ is the radius of the Earth. The kinetic energy $K$ required for this is given by:

$K = \frac{1}{2}mv_e^2$

Substituting the escape velocity formula into the kinetic energy formula gives:

$K = \frac{1}{2}m\left(2gR_E\right)$

$K = \frac{1}{2} \times 2 \times mgR_E$

$K = mgR_E$

Therefore, the required kinetic energy to project a body of mass $m$ from Earth's surface to infinity is $mgR_E$. So, the correct answer is:

Option C: $mgR_E$

$\begin{aligned} & \frac{T_1}{T_2}=\left(\frac{r_1}{r_2}\right)^{3 / 2} \sqrt{\frac{m_2}{m_1}} \\ & \Rightarrow \quad \frac{24}{6}=\left(\frac{4.2 \times 10^4}{r_2}\right)^{3 / 2} \sqrt{\frac{m / 4}{m}} \\ & \Rightarrow r_2=1.05 \times 10^4 \mathrm{~km} \end{aligned}$

To determine the dimension of the quantity $\sqrt{uG}$, we first need to understand the dimensions of both the gravitational constant (G) and the energy density (u).

The gravitational constant $G$ has dimensions given by:

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

where $M$ stands for mass, $L$ for length, and $T$ for time.

Energy density $u$ is defined as the energy per unit volume. Since energy has dimensions of $ML^{2}T^{-2}$ (from the dimension of work or energy, which is force times distance, and force itself has dimension $MLT^{-2}$), and volume has dimensions of $L^{3}$, the dimensions of energy density would be:

$[u] = \frac{ML^{2}T^{-2}}{L^{3}} = M L^{-1} T^{-2}$

Now, we find the dimensions of $\sqrt{uG}$ by multiplying the dimensions of $u$ and $G$, and then taking the square root:

$[\sqrt{uG}] = \sqrt{[u][G]} = \sqrt{(M L^{-1} T^{-2})(M^{-1}L^{3}T^{-2})} = \sqrt{L^{2}T^{-4}} = LT^{-2}$

So, the dimension of $\sqrt{uG}$ is $LT^{-2}$, which corresponds to acceleration (length per square time).

Now, let's match this with the provided options:

• Option A (Gravitational potential) has dimensions of $[L^{2}T^{-2}]$, not matching our target of $LT^{-2}$.
• Option B (Pressure gradient per unit mass) would have dimensions of $[M^{-1}L^{-2}T^{-2}][L^{-1}]$ ($Pressure\ Gradient = \frac{Pressure}{Length} = \frac{ML^{-1}T^{-2}}{L}$, and then divided by mass, $M$), which simplifies to $L^{-3}T^{-2}M^{-1}$, not matching.
• Option C (Energy per unit mass) has dimensions $ML^{2}T^{-2}M^{-1}$ which simplifies to $L^{2}T^{-2}$, also not a match for the target dimension.
• Option D (Force per unit mass) has dimensions $MLT^{-2}M^{-1}$ which simplifies directly to $LT^{-2}$, an exact match for our target dimension.

Thus, the correct answer is Option D (Force per unit mass), which has the same dimensions as that of $\sqrt{\mathrm{uG}}$.

$\text { K.E }=\frac{G M m}{2 a} \quad \text{(II)}$

$U_G=\frac{-G M m}{a} \quad \text{(I)}$

$M . E=\frac{-G M m}{2 a} \quad \text{(IV)}$

$\text { and Escape Energy }=\frac{G m}{r} \quad \text{(III)}$

The gravitational force $F$ that an object of mass $m$ placed at a distance $r$ (from the center of Earth) experiences can be calculated using the universal law of gravitation, given by:

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

Where:

• $G$ is the gravitational constant ($6.674 \times 10^{-11} \text{Nm}^2/\text{kg}^2$)
• $M$ is the mass of the Earth (approximately $5.972 \times 10^{24} \text{kg}$)
• $m$ is the mass of the object
• $r$ is the distance from the center of the Earth to the object

However, when the object is at a distance $2R$ from the surface of the Earth, the total distance from the center of the Earth $r'$ becomes $R + 2R = 3R$. This is because the radius of the Earth $R$ is the distance from the Earth's center to its surface, so if the object is $2R$ above the surface, the total distance from the center is $3R$.

At the surface of the Earth, the gravitational force ($F_{earth}$) that acts on an object is given by its weight, which can be calculated using the formula $F_{earth} = m \cdot g$, where $g$ is the acceleration due to gravity on the surface of the Earth. Given that $g = 10 \text{m/s}^2$ and the mass of the body $m = 90 \text{kg}$, we get:

$F_{earth} = 90 \text{kg} \cdot 10 \text{m/s}^2 = 900 \text{N}$

To find the gravitational pull at a distance $2R$ from the Earth's surface, we use the fact that gravitational force varies inversely as the square of the distance from the center of the Earth. Since the distance triples ($3R$ from $R$), the gravitational force becomes $\frac{1}{3^2} = \frac{1}{9}$ of the force at the surface.

Therefore, the gravitational pull $F_{2R}$ on the body when placed at $2R$ from the Earth's surface is:

$F_{2R} = \frac{F_{earth}}{9} = \frac{900 \text{N}}{9} = 100 \text{N}$

Thus, the gravitational pull on the body when it is placed at a distance of $2R$ from the Earth's surface is 100 N. So, the correct option is:

Option C: 100 N

$\begin{aligned} & T=2 \pi \sqrt{\frac{r^3}{G M}} \\ & T^2=\frac{4 \pi^2}{G M}(R+h)^3 \\ & h=\left(\frac{G M T^2}{4 \pi^2}\right)^{\frac{1}{3}}-R \\ & =\left(\frac{G M \cdot R}{R^2} \cdot \frac{T^2}{4 \pi^2}\right)^{\frac{1}{3}}-R \\ & =\left(\frac{T^2 R^2 g}{4 \pi^2}\right)^{\frac{1}{3}}-R \end{aligned}$

Field at centre due to arc, $I=\frac{2 G M \pi}{L^2}$

$\therefore \quad$ Net force on mass, $F=\frac{2 G M m \pi}{L^2}$

To find the correct option for the relationship between the period of revolution T and the radius of the orbit R, we will consider the force of attraction and its proportionality to $\mathrm{R}^{-3 / 2}$.

According to Newton's law of universal gravitation, the force of attraction $F$ between two masses $m_1$ and $m_2$ separated by a distance $r$ is given by $ F = \frac{G m_1 m_2}{r^2}, $ where $G$ is the gravitational constant.

However, in this particular case, the force of attraction is given to be proportional to $\mathrm{R}^{-3 / 2}$, so we can write $ F \propto \frac{1}{R^{3/2}}. $

Since the planet is in a circular orbit around the star, the centripetal force required to keep the planet in orbit must be provided by this gravitational force. Hence, we can write that $ \frac{m v^2}{R} \propto \frac{1}{R^{3/2}}, $ where $m$ is the mass of the planet and $v$ is its orbital speed.

Simplifying this, we get $ v^2 \propto \frac{1}{R^{1/2}}. $

Now, the speed $v$ can be related to the period T through the circumference of the orbit, which is given by $2\pi R$. The orbital speed is the circumference divided by the period: $ v = \frac{2\pi R}{T}. $

Substituting this into our proportionality, we get

$ \left( \frac{2\pi R}{T} \right)^2 \propto \frac{1}{R^{1/2}}, $

which simplifies to

$ \frac{4\pi^2 R^2}{T^2} \propto \frac{1}{R^{1/2}}. $

Solving for $T^2$, we get

$ T^2 \propto \frac{R^{2 + 1/2}}{4\pi^2}, $

so $ T^2 \propto R^{5/2}. $

Therefore, the correct option is Option C

$ T^2 \propto R^{5/2}. $

To find the length of the second's pendulum at a height $h = 2R$ from the surface of the Earth, we must first understand that the length of a second's pendulum, $L$, is related to the gravitational acceleration, $g$, and the period, $T$, by the formula: $ T = 2\pi\sqrt{\frac{L}{g}} $ Since we are talking about a second's pendulum, the period, $T$, is 2 seconds (since it takes one second for the pendulum to swing in one direction and another second to swing back), thus $T = 2 \text{ seconds}$.

Now let's find the gravitational acceleration at height $h = 2R$ where $R$ is the radius of the earth. The general formula for gravitational acceleration at a height $h$ above the surface is: $ g_h = \frac{g}{{\left(1 + \frac{h}{R}\right)}^2} $ Plugging $h = 2R$ into the formula, we get:

$ g_h = \frac{g}{{(1 + \frac{2R}{R})}^2} $

$ g_h = \frac{g}{{(1 + 2)}^2} $

$ g_h = \frac{g}{3^2} = \frac{g}{9} $

So the gravitational acceleration at height $h$ is one-ninth of the gravitational acceleration at the surface of the Earth. Given that $g = \pi^2 \text{ m/s}^2$, we get: $ g_h = \frac{\pi^2}{9} \text{ m/s}^2 $

Now knowing the gravitational acceleration at height $h$ and with the period $T$ of 2 seconds, we can rearrange the formula for the second's pendulum to solve for the length $L_h$:

$ 2 = 2\pi\sqrt{\frac{L_h}{g_h}} $

$ 1 = \pi\sqrt{\frac{L_h}{g_h}} $

$ \frac{1}{\pi} = \sqrt{\frac{L_h}{g_h}} $

Squaring both sides, we get:

$ \frac{1}{\pi^2} = \frac{L_h}{g_h} $

Multiplying both sides by $g_h$ gives us the length $L_h$:

$ L_h = \frac{g_h}{\pi^2} $

Substituting $g_h$ into the equation yields:

$ L_h = \frac{\pi^2}{9\pi^2} $

$ L_h = \frac{1}{9} \text{ m} $

Therefore, the length of the second's pendulum at a height $h = 2R$ from the surface of the Earth is $\frac{1}{9}$ meters. The correct answer is Option A.

$\begin{aligned} & \mathrm{V}_{\text {escape }}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} \\ & \mathrm{V}_{\text {planet }}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}=\mathrm{V} \\ & \mathrm{V}_{\text {Moon }}=\sqrt{\frac{2 \mathrm{GM} \times 16}{144 \mathrm{R}}}=\frac{1}{3} \sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} \\ & \mathrm{V}_{\text {Moon }}=\frac{\mathrm{V}_{\text {Planet }}}{3}=\frac{\mathrm{V}}{3} \end{aligned}$

$\begin{aligned} & F_{\text {net }}=\sqrt{2} F+F^{\prime} \\ & F=\frac{G m^2}{a^2} \text { and } F^{\prime}=\frac{G^2}{(\sqrt{2} \mathrm{a})^2} \\ & F_{\text {net }}=\sqrt{2} \frac{\mathrm{Gm}^2}{\mathrm{a}^2}+\frac{\mathrm{Gm}^2}{2 \mathrm{a}^2} \\ & \left(\frac{2 \sqrt{2}+1}{32}\right) \frac{\mathrm{Gm}^2}{\mathrm{~L}^2}=\frac{\mathrm{Gm}^2}{\mathrm{a}^2}\left(\frac{2 \sqrt{2}+1}{2}\right) \\ & \mathrm{a}=4 \mathrm{~L} \end{aligned}$

$\begin{aligned} & \mathrm{R}_{\mathrm{P}}=\frac{\mathrm{R}_{\mathrm{E}}}{3}, \mathrm{M}_{\mathrm{P}}=\frac{\mathrm{M}_{\mathrm{E}}}{6} \\ & \mathrm{~V}_{\mathrm{c}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}} \quad \text{.... (i)}\\ & \mathrm{V}_{\mathrm{P}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{P}}}} \quad \text{.... (ii)}\\ & \frac{\mathrm{V}_{\mathrm{e}}}{\mathrm{V}_{\mathrm{p}}}=\sqrt{2} \\ & \mathrm{~V}_{\mathrm{P}}=\frac{\mathrm{V}_{\mathrm{e}}}{\sqrt{2}}=\frac{11.2}{\sqrt{2}}=7.9 \mathrm{~km} / \mathrm{sec} \end{aligned}$

$-\frac{G M_E}{R_E+h}=-5.12 \times 10^{-7}$ .... (i)

$\frac{G M_E}{\left(R_E+h\right)^2}=6.4$ ..... (ii)

By (i) and (ii)

$\Rightarrow h=16 \times 10^5 \mathrm{~m}=1600 \mathrm{~km}$

$\begin{aligned} & \mathrm{T}^2 \propto \mathrm{r}^3 \\ & \frac{\mathrm{T}_1^2}{\mathrm{r}_1^3}=\frac{\mathrm{T}_2^2}{\mathrm{r}_2^3} \\ & \frac{(200)^2}{\mathrm{r}^3}=\frac{\mathrm{T}_2^2}{\left(\frac{\mathrm{r}}{4}\right)^3} \\ & \frac{200 \times 200}{4 \times 4 \times 4}=\mathrm{T}_2^2 \\ & \mathrm{~T}_2=\frac{200}{4 \times 2} \\ & \mathrm{~T}_2=25 \text { days } \end{aligned}$

$\begin{aligned} & g_p=\frac{g R^2}{(R+h)^2} \\ & g_q=g\left(1-\frac{h}{R}\right) \\ & g_p=g_q \\ & \frac{g}{\left(1+\frac{h}{R}\right)^2}=g\left(1-\frac{h}{R}\right) \\ & \left(1-\frac{h^2}{R^2}\right)\left(1+\frac{h}{R}\right)=1 \end{aligned}$

Take $\frac{\mathrm{h}}{\mathrm{R}}=\mathrm{x}$

So

$\begin{aligned} & \mathrm{x}^3-\mathrm{x}+\mathrm{x}^2=0 \\ & \mathrm{x}=\frac{\sqrt{5}-1}{2} \\ & \mathrm{~h}=\frac{\mathrm{R}}{2}(\sqrt{5}-1) \end{aligned}$

The angular speed $ \omega $ of an object in circular motion is given by the equation

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

where $ T $ is the period of the motion - the time it takes to make one complete revolution.

Assertion (A) states that the angular speed of the Moon in its orbit around the Earth is more than the angular speed of the Earth in its orbit around the Sun. We can compare these angular speeds by comparing their periods.

The Moon takes approximately 27.3 days to orbit the Earth (this is its sidereal period, not its synodic period, which accounts for the Earth's motion around the Sun as well). The Earth takes approximately 365.25 days to orbit the Sun. Since the period of the Moon’s orbit is much less than the period of the Earth’s orbit, the angular speed of the Moon is much larger than that of the Earth. This makes Assertion (A) correct.

Reason (R) states that the Moon takes less time to move around the Earth than the time taken by the Earth to move around the Sun. This is also correct, as the previously mentioned periods demonstrate: 27.3 days for the Moon to orbit Earth vs. 365.25 days for Earth to orbit the Sun.

Furthermore, the reason (R) directly explains why assertion (A) is true. Since angular speed is inversely proportional to the period of orbit, the fact that the Moon takes less time to complete an orbit implies it has a higher angular speed relative to the Earth’s angular speed in its orbit around the Sun.

Therefore, the correct answer is:

Option C Both (A) and (R) are correct and (R) is the correct explanation of (A)

$\begin{aligned} & g=\frac{G M}{R^2} \Rightarrow g \propto \frac{1}{R^2} \\ & \frac{g_2}{g_1}=\frac{R_1^2}{R_2^2} \\ & g_2=4 g_1\left(R_2=\frac{R_1}{2}\right) \end{aligned}$

To find the effect of an additional central force on the time period of a particle in a circular orbit, consider the following:

Initial Gravitational Force ($F_1$):

The force due to gravity for a particle in a circular orbit of radius $r_0$ is given by:

$ F_1 = \frac{GMm}{r_0^2} $

where $ G $ is the gravitational constant, $ M $ is the mass of the central body, and $ m $ is the mass of the particle.

Modified Force with Additional Potential ($F_2$):

When an additional force derived from the potential energy $ V_{\mathrm{c}}(r) = \frac{m\alpha}{r^3} $ acts on the particle, the effective central force becomes:

$ F_2 = \frac{GMm}{r_0^2} - \frac{3m\alpha}{r_0^4} $

Relation of Angular Velocities:

The ratio of squares of the angular velocities ($\omega_0$ and $\omega_1$) before and after the additional force is:

$ \frac{\omega_1^2}{\omega_0^2} = \frac{F_2}{F_1} = \frac{\frac{GM}{r_0^2} - \frac{3\alpha}{r_0^4}}{\frac{GM}{r_0^2}} $

Relating Time Periods:

Since the angular velocity is inversely proportional to the time period, $\frac{\omega_1^2}{\omega_0^2} = \frac{T_0^2}{T_1^2}$. Thus, we have:

$ \frac{T_0^2}{T_1^2} = 1 - \frac{3\alpha}{GMr_0^2} $

Find the Change in Time Period:

The expression for the change in the square of the time periods is:

$ \frac{T_1^2 - T_0^2}{T_1^2} = \frac{3\alpha}{GMr_0^2} $

This shows how the additional central force, represented by the potential $ V_{\mathrm{c}}(r) = \frac{m\alpha}{r^3} $, affects the particle’s orbital period when combined with the gravitational field of the mass $ M $.

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.

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 $.

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} $.

To find the additional velocity required to overcome the gravitational pull and launch the spaceship into orbit, we first need to find the orbital velocity. The formula for orbital velocity (v) is given by:

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

where G is the gravitational constant, M is the mass of Earth, and r is the distance from the center of the Earth to the spaceship (which is the sum of the Earth's radius and the altitude of the spaceship's orbit).

In this problem, the spaceship is orbiting close to Earth's surface, so we can approximate r as the Earth's radius. Given that g = 10 m/s² and Earth's radius R = 6400 km, we can relate the gravitational constant G and the mass of Earth M through the formula:

$g = \frac{GM}{R^2}$

Now, we can find the orbital velocity:

$v = \sqrt{\frac{gR^2}{R}} = \sqrt{gR}$

Converting the Earth's radius to meters:

$R = 6400 \times 10^3 \mathrm{~m}$

Plugging in the values:

$v = \sqrt{10 \times 6400 \times 10^3} = 8 \times 10^3 \mathrm{~m/s}$

Now, we need to find the additional velocity required to overcome the gravitational pull. To do this, we can use the formula for escape velocity:

$v\text{escape} = \sqrt{2} \times v\text{orbital}$

Finding the additional velocity required:

$\Delta v = v\text{escape} - v\text{orbital} = (\sqrt{2} - 1) \times v_\text{orbital}$

Plugging in the values:

$\Delta v = (\sqrt{2} - 1) \times 8 \times 10^3 \mathrm{~m/s}$

Converting the velocity to km/s:

$\Delta v = (\sqrt{2} - 1) \times 8 \mathrm{~km/s}$

Thus, the additional velocity required to overcome the gravitational pull and launch the spaceship into orbit is:

$8(\sqrt{2}-1) \mathrm{~km/s}$

For a satellite orbiting the Earth at a height equal to Earth's radius, its distance from the center of the Earth will be 2R, where R is the radius of the Earth.

Using the formula for the gravitational force:

$F = G\frac{Mm}{r^2}$

where F is the gravitational force, G is the gravitational constant, M is the mass of the Earth, m is the mass of the satellite, and r is the distance from the center of the Earth.

The centripetal force acting on the satellite is given by:

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

Equating the gravitational force and the centripetal force, we get:

$G\frac{Mm}{(2R)^2} = \frac{mv^2}{2R}$

Solving for the orbital speed v, we get:

$v^2 = \frac{GM}{2R}$

The circumference of the satellite's orbit is given by:

$C = 2\pi(2R) = 4\pi R$

The time period T of the satellite's orbit can be calculated as the ratio of the circumference to the orbital speed:

$T = \frac{C}{v} = \frac{4\pi R}{\sqrt{\frac{GM}{2R}}}$

Given that the acceleration due to gravity at the Earth's surface is $g = \pi^2 \,\mathrm{m/s^2}$, we can express the gravitational constant G in terms of the Earth's radius R and mass M:

$g = \frac{GM}{R^2} \Rightarrow GM = gR^2 = \pi^2 R^2$

Substituting the expression for GM into the equation for the time period T, we get:

$T = \frac{4\pi R}{\sqrt{\frac{\pi^2 R^2}{2R}}} = \frac{4\pi R}{\sqrt{\frac{\pi^2 R}{2}}}$

$T = \frac{4\pi R}{\sqrt{\pi^2}\sqrt{\frac{R}{2}}} = \frac{4\pi R}{\pi\sqrt{\frac{R}{2}}} = \frac{4R}{\sqrt{\frac{R}{2}}}$

Multiplying the numerator and denominator by $\sqrt{2}$, we get:

$T = \frac{4R\sqrt{2}}{\sqrt{R}} = 4\sqrt{2R} = \sqrt{32R}$