Gravitation

The work done ( W ) by an external agent to rearrange a system of masses against gravitational force is equal to the change in the Gravitational Potential Energy $(U)$ of the system.

$ \mathrm{W}=\Delta \mathrm{U}=\mathrm{U}_{\text {final }}-\mathrm{U}_{\text {initial }} $

For a system of three masses $m_1, m_2, m_3$ separated by distances $r_{12}, r_{23}, r_{31}$, the total potential energy is the sum of the potential energies of all pairs :

$ \mathrm{U}=\mathrm{U}_{12}+\mathrm{U}_{23}+\mathrm{U}_{31}=-\mathrm{G}\left[\frac{\mathrm{~m}_1 \mathrm{~m}_2}{\mathrm{r}_{12}}+\frac{\mathrm{m}_2 \mathrm{~m}_3}{\mathrm{r}_{23}}+\frac{\mathrm{m}_3 \mathrm{~m}_1}{\mathrm{r}_{31}}\right] $

Since the masses are at the vertices of an equilateral triangle, all distances are equal ( $r_{12}=r_{23}=r_{31}=L$ ).

$ \mathrm{U}=-\frac{\mathrm{G}}{\mathrm{~L}}\left(\mathrm{~m}_1 \mathrm{~m}_2+\mathrm{m}_2 \mathrm{~m}_3+\mathrm{m}_3 \mathrm{~m}_1\right) $

Given masses are,

• 1. $\mathrm{m}_1=200 \mathrm{~kg}$

• 2. $\mathrm{m}_2=300 \mathrm{~kg}$

• 3. $\mathrm{m}_3=400 \mathrm{~kg}$

$ U=-\frac{G}{L}(200 \times 300+300 \times 400+400 \times 200)=-\frac{G\left(2.6 \times 10^5\right)}{L} $

So, the work done by external agent is,

$ \mathrm{W}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}} $

$\Rightarrow $ $\mathrm{W}=\left[-\frac{\mathrm{G}\left(2.6 \times 10^5\right)}{\mathrm{L}_{\mathrm{f}}}\right]-\left[-\frac{\mathrm{G}\left(2.6 \times 10^5\right)}{\mathrm{L}_{\mathrm{i}}}\right]$

$ \mathrm{W}=\mathrm{G}\left(2 . \times 10^5\right)\left(\frac{1}{\mathrm{~L}_{\mathrm{i}}}-\frac{1}{\mathrm{~L}_{\mathrm{f}}}\right) $

The initial side is $=\mathrm{L}_{\mathrm{i}}=20 \mathrm{~m}$

The final side is $\mathrm{L}_{\mathrm{f}}=25 \mathrm{~m}$

Putting the values,

$ W=\left(6.7 \times 10^{-11}\right) \times\left(2.6 \times 10^5\right) \times\left(\frac{1}{20}-\frac{1}{25}\right) $

$\Rightarrow $ $W=\left(6.7 \times 10^{-11}\right) \times\left(2.6 \times 10^5\right) \times\left(10^{-2}\right)$

$\Rightarrow $ $ \mathrm{W}=1.742 \times 10^{-7} \mathrm{~J} $

Therefore, the work done in this process is $1.74 \times 10^{-7} \mathrm{~J}$.

Hence, the correct option is (B).

For a satellite revolving in a circular orbit of radius $r$ around the Earth (mass $M$ ), the time period ( $T$ ) is given by Kepler's Third Law

For the equilibrium along the radial direction,

$ F_g=\frac{m v^2}{r} \Rightarrow \frac{G M_e m}{r^2}=\frac{m v^2}{r} \Rightarrow v=\sqrt{\frac{G M_e}{r}} $

So, the time period of the revolution is,

$ \mathrm{T}=\frac{2 \pi \mathrm{r}}{\mathrm{v}}=\frac{2 \pi \mathrm{r}}{\sqrt{\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{r}}}} $

$\Rightarrow $ $ \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{r}^3}{\mathrm{GM}_{\mathrm{e}}}} $

where G is the universal gravitational constant and $\mathrm{M}_{\mathrm{e}}$ is the mass of Earth.

If the satellite is orbiting very close to the Earth's surface, its orbital radius $r$ is approximately equal to the Earth's radius, $\mathrm{R}_{\mathrm{e}}$.

$ \mathrm{T} \approx 2 \pi \sqrt{\frac{\mathrm{R}_{\mathrm{e}}^3}{\mathrm{GM}_{\mathrm{e}}}} $

Assuming the Earth is a uniform sphere, its mass M is the product of its volume and density:

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

Substitute this mass into our time period equation :

$ T=2 \pi \sqrt{\frac{R_e^3}{G\left(\frac{4}{3} \pi R_e^3 \rho\right)}}=2 \pi \sqrt{\frac{3}{4 \pi G \rho}} $

$\Rightarrow $ $ T=\sqrt{\frac{3 \pi}{G \rho}} $

Since $3, \pi$, and G are all constants, this equation proves that the time period of a satellite orbiting very close to the Earth's surface depends entirely on the density ( $\rho$ ) of the Earth.

Therefore, Statement I is true.

We also know the formula for the acceleration due to gravity (g) on the surface of the Earth :

$ \mathrm{g}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^2} $

$ \Rightarrow \mathrm{GM}=\mathrm{gR}_{\mathrm{e}}^2 $

Substitute this expression for $\mathrm{GM}_{\mathrm{e}}$ into the time period equation $\mathrm{T} \approx 2 \pi \sqrt{\frac{\mathrm{R}_{\mathrm{e}}^3}{\mathrm{GM}_{\mathrm{e}}}}$

$ \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}_{\mathrm{e}}^3}{\mathrm{gR}_{\mathrm{e}}^2}}=2 \pi \sqrt{\frac{\mathrm{R}_{\mathrm{e}}}{\mathrm{~g}}} $

Therefore, Statement II is also true.

As both statement I and statement II are true, so the correct option is (C).

Escape velocity ( $\mathrm{v}_{\mathrm{e}}$ ) is the minimum initial velocity an object needs to completely escape a planet's gravitational field, meaning it reaches an infinite distance with zero remaining kinetic energy.

Let an object of mass $m$ be on the surface of a planet with mass $M$ and radius $R$.

Initial Energy at the surface:

The object has kinetic energy $\left(\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^2\right)$ and gravitational potential energy $\left(-\frac{\mathrm{GMm}}{\mathrm{R}}\right)$.

Total Initial Energy $=\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^2-\frac{\mathrm{GMm}}{\mathrm{R}}$

Final Energy at infinity:

The object just escapes, meaning it arrives with zero velocity (zero kinetic energy) and at an infinite distance, potential energy is also zero.

Total Final Energy $=0$

By the law of conservation of energy, Initial Energy = Final Energy:

$ \frac{1}{2} m v_e^2-\frac{G M m}{R}=0 $

$\Rightarrow $ $\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^2=\frac{\mathrm{GMm}}{\mathrm{R}}$

$ \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} $

For a spherical planet, Volume (V) is $\frac{4}{3} \pi R^3$.

$ \operatorname{Mass}(M)=\operatorname{Volume}(V) \times \text { Density }(\rho) $

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

Substituting this into escape velocity formula :

$ \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{G}}{\mathrm{R}} \cdot\left(\frac{4}{3} \pi \mathrm{R}^3 \rho\right)} $

$\Rightarrow $ $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{8}{3} \pi \mathrm{GR}^2 \rho}$

$\Rightarrow $ $\mathrm{v}_{\mathrm{e}}=\mathrm{R} \sqrt{\frac{8}{3} \pi \mathrm{G} \rho}$

$ \Rightarrow \mathrm{v}_{\mathrm{e}} \propto \mathrm{R} \sqrt{\rho} $

Escape velocity of planet $\mathrm{A}, \mathrm{v}_{\mathrm{eA}}=10 \mathrm{~km} / \mathrm{s}=10000 \mathrm{~m} / \mathrm{s}$

Radius of planet $B$ is $10 \%$ of radius of planet $A, R_B=10 \%$ of $R_A=0.1 R_A$

Density of planet B is $10 \%$ of that of planet A , $\rho_{\mathrm{B}}=10 \%$ of $\rho_{\mathrm{A}}=0.1 \rho_{\mathrm{A}}$ Using proportionality relationship,

$ \frac{\mathrm{v}_{\mathrm{eB}}}{\mathrm{v}_{\mathrm{eA}}}=\frac{\mathrm{R}_{\mathrm{B}} \sqrt{\rho_{\mathrm{B}}}}{\mathrm{R}_{\mathrm{A}} \sqrt{\rho_{\mathrm{A}}}} $

$\Rightarrow $ $\frac{\mathrm{v}_{\mathrm{eB}}}{\mathrm{v}_{\mathrm{eA}}}=\frac{0.1 \mathrm{R}_{\mathrm{A}} \sqrt{0.1 \rho_{\mathrm{A}}}}{\mathrm{R}_{\mathrm{A}} \sqrt{\rho_{\mathrm{A}}}}$

$\Rightarrow $ $\frac{\mathrm{v}_{\mathrm{eB}}}{\mathrm{v}_{\mathrm{eA}}}=0.1 \sqrt{0.1}$

$\Rightarrow $ $\frac{\mathrm{v}_{\mathrm{eB}}}{\mathrm{v}_{\mathrm{eA}}}=\frac{1}{10} \times \frac{1}{\sqrt{10}}=\frac{1}{10 \sqrt{10}}$

$\Rightarrow $ $\mathrm{v}_{\mathrm{eB}}=10000 \times\left(\frac{1}{10 \sqrt{10}}\right)$

$ \mathrm{v}_{\mathrm{eB}}=\frac{1000}{\sqrt{10}}=100 \sqrt{10} \mathrm{~m} / \mathrm{s} $

Therefore, the escape velocity of another planet B is $100 \sqrt{10} \mathrm{~m} / \mathrm{s}$.

Hence, the correct option is (D).

Newton's Law of Universal Gravitation states that every point mass attracts every other point mass by a force acting along the line intersecting both points. The magnitude of this force is,

$ \mathrm{F}=\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{r}^2} $

where G is the gravitational constant, $\mathrm{m}_1$ and $\mathrm{m}_2$ are the masses, and r is the distance between their centers.

Let's consider square of side length a and the distance from any corner to the centre be r .

Let's place a test mass $\mathrm{m}_0$ at the center. The gravitational force exerted by a mass m at a corner on this test mass will be directed towards that corner with a magnitude:

$ \mathrm{F}=\frac{\mathrm{Gm} \cdot \mathrm{~m}_0}{\mathrm{r}^2} $

Let the force on $\mathrm{m}_0$ by another mass M is $\mathrm{F}_0=\frac{\mathrm{GM} \cdot \mathrm{m}_0}{\mathrm{r}^2}$

So, a mass of 2 M will exert a force of $2 \mathrm{~F}_0, 3 \mathrm{M}$ will exert $3 \mathrm{~F}_0$, and 4 M will exert $4 \mathrm{~F}_0$.

Case 1 : Initial arrangement

The magnitudes of forces on the mass kept at centre of square are;

$ \mathrm{F}_1=4 \mathrm{~F}_0, \mathrm{~F}_2=3 \mathrm{~F}_0, \mathrm{~F}_3=2 \mathrm{~F}_0 \text { and } \mathrm{F}_4=\mathrm{F}_0 $

Since $F_2$ and $F_4$ pull in opposite directions, the net force along this diagonal is $3 F_0-F_0=2 F_0$ directed along $\mathrm{F}_2$.

The net force of $F_1$ and $F_3$ is along $F_1$ is $4 F_0-2 F_0=2 F_0$.

Now two net forces, each of magnitude $2 \mathrm{~F}_0$, acting at a $90^{\circ}$ angle to each other.

Then the resultant magnitude $(F)_1$ :

$ (F)_1=\sqrt{\left(2 F_0\right)^2+\left(2 F_0\right)^2}=\sqrt{8 F_0^2} $

$\Rightarrow $ $ (\mathrm{F})_1=2 \sqrt{2} \mathrm{~F}_0 $

Case 2 : Positions of 3 M and 4 M interchanged

The magnitudes of forces on the mass kept at centre of square are;

$\mathrm{F}_1=4 \mathrm{~F}_0, \mathrm{~F}_2=3 \mathrm{~F}_0, \mathrm{~F}_3=2 \mathrm{~F}_0$ and $\mathrm{F}_4=\mathrm{F}_0$

Since $\mathrm{F}_2$ and $\mathrm{F}_3$ pull in opposite directions, the net force along this diagonal is $3 \mathrm{~F}_0-2 \mathrm{~F}_0=\mathrm{F}_0$ directed along $F_2$.

The net force of $\mathrm{F}_1$ and $\mathrm{F}_3$ is along $\mathrm{F}_1$ is $4 \mathrm{~F}_0-\mathrm{F}_0=3 \mathrm{~F}_0$.

Now two net forces are acting at a $90^{\circ}$ angle to each other.

Then the resultant magnitude $(F)_2$ :

$ (F)_2=\sqrt{\left(F_0\right)^2+\left(3 F_0\right)^2}=\sqrt{10 F_0^2} $

$\Rightarrow $ $ (\mathrm{F})_2=\sqrt{10} \mathrm{~F}_0 $

The ratio $\frac{(\mathrm{F})_1}{(\mathrm{~F})_2}$ is :

$ \frac{(\mathrm{F})_1}{(\mathrm{~F})_2}=\frac{2 \sqrt{2} \mathrm{~F}_0}{\sqrt{10} \mathrm{~F}_0}=\frac{2}{\sqrt{5}} $

$\Rightarrow $ $ \Rightarrow \frac{2}{\sqrt{5}}=\frac{\alpha}{\sqrt{5}} \Rightarrow \alpha=2 $

Therefore, the value of $\alpha$ is 2 .

Hence, the correct option is (A).

The total energy ( $E$ ) of a satellite of mass $m$ orbiting a planet of mass $M$ at a radius $r$ is the sum of its Kinetic Energy (K) and Potential Energy (U).

Potential Energy is, $U=-\frac{G M m}{r}$

For a circular orbit, the centripetal force is provided by gravity:

$ \frac{\mathrm{mv}^2}{\mathrm{r}}=\frac{\mathrm{GMm}}{\mathrm{r}^2} \Rightarrow \frac{1}{2} \mathrm{mv}^2=\frac{\mathrm{GMm}}{2 \mathrm{r}} $

So, the total energy is,

$ \mathrm{E}=\mathrm{K}+\mathrm{U}=\frac{\mathrm{GMm}}{2 \mathrm{r}}-\frac{\mathrm{GMm}}{\mathrm{r}}=-\frac{\mathrm{GMm}}{2 \mathrm{r}} $

We also know that at the Earth's surface, $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}_{\mathrm{E}}^2}$, so $\mathrm{GM}=\mathrm{gR}_{\mathrm{E}}^2$.

$ \Rightarrow \mathrm{E}=-\frac{\mathrm{mgR}_{\mathrm{E}}^2}{2 \mathrm{r}} $

The mass of satellite is, $\mathrm{m}=100 \mathrm{~kg}$

The initial radius is $\mathrm{r}_1=1.5 \mathrm{R}_{\mathrm{E}}$

The final radius is $r_2=3 R_E$

The energy supplied ( $\Delta \mathrm{E}$ ) is the difference between the final total energy and the initial total energy :

$ \Delta \mathrm{E}=\mathrm{E}_2-\mathrm{E}_1 $

$\Rightarrow $ $\Delta \mathrm{E}=\left(-\frac{\mathrm{mgR}_{\mathrm{E}}^2}{2 \mathrm{r}_2}\right)-\left(-\frac{\mathrm{mgR}_{\mathrm{E}}^2}{2 \mathrm{r}_1}\right)$

$\Rightarrow $ $ \Delta \mathrm{E}=\frac{\mathrm{mgR}_{\mathrm{E}}^2}{2}\left[\frac{1}{\mathrm{r}_1}-\frac{1}{\mathrm{r}_2}\right] $

Substituting the radii in terms of $\mathrm{R}_{\mathrm{E}}$ :

$ \Delta \mathrm{E}=\frac{\mathrm{mgR}_{\mathrm{E}}^2}{2}\left[\frac{1}{1.5 \mathrm{R}_{\mathrm{E}}}-\frac{1}{3 \mathrm{R}_{\mathrm{E}}}\right] $

$\Rightarrow $ $\Delta \mathrm{E}=\frac{\mathrm{mgR}_{\mathrm{E}}}{2}\left[\frac{2}{3}-\frac{1}{3}\right]$

$\Rightarrow $ $ \Delta \mathrm{E}=\frac{\mathrm{mgR}_{\mathrm{E}}}{2} \times \frac{1}{3}=\frac{\mathrm{mgR}_{\mathrm{E}}}{6} $

Putting the values, $\mathrm{m}=100, \mathrm{~g}=10, \mathrm{R}_{\mathrm{E}}=6 \times 10^6$ :

$ \Delta \mathrm{E}=\frac{100 \cdot 10 \cdot 6 \times 10^6}{6} $

$\Rightarrow $ $\Delta \mathrm{E}=1000 \times 10^6 \mathrm{~J}=\alpha \times 10^6 \mathrm{~J}$

$\Rightarrow $ $ \alpha \times 10^6=1000 \times 10^6 \Rightarrow \alpha=1000 $

The value of $\alpha$ required to move the satellite to a higher orbit is 1000 .

Hence, the correct option is (B).

The height to which the body is taken is

$ h = 2R_e $

So, its final distance from the centre of the earth is

$ r_2 = R_e + 2R_e = 3R_e $

Initial distance from the centre of the earth is

$ r_1 = R_e $

Now, gravitational potential energy at a distance $r$ from the centre of earth is

$ U = -\frac{GMm}{r} $

Therefore, increase in potential energy is

$ \Delta U = U_2 - U_1 $

$ \Delta U = \left(-\frac{GMm}{3R_e}\right) - \left(-\frac{GMm}{R_e}\right) $

$ \Delta U = \frac{GMm}{R_e} - \frac{GMm}{3R_e} $

$ \Delta U = \frac{2GMm}{3R_e} $

Now use the relation

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

So,

$ GM = gR_e^2 $

Substituting in the expression:

$ \Delta U = \frac{2}{3} \cdot \frac{gR_e^2 m}{R_e} $

$ \Delta U = \frac{2}{3} mgR_e $

Hence, the increase in potential energy is

$ \boxed{\frac{2}{3}mgR_e} $

So, the correct option is:

$ \boxed{\text{Option D}} $

For small heights and depths compared to the radius of Earth, NCERT uses these approximations:

• At a depth $d$ below the surface:

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

• At a height $h$ above the surface:

$ g_h = g\left(1-\frac{2h}{R}\right) $

Here,

$ d = 16\ \text{km}, \quad h = 16\ \text{km}, \quad R = 6400\ \text{km} $

Now,

$ g_{\text{below}} = g\left(1-\frac{16}{6400}\right) $

and

$ g_{\text{above}} = g\left(1-\frac{2\times 16}{6400}\right) $

So the change in $g$ while moving from 16 km below the surface to 16 km above the surface is

$ \Delta g = g_{\text{below}} - g_{\text{above}} $

$ = g\left(1-\frac{16}{6400}\right) - g\left(1-\frac{32}{6400}\right) $

$ = g\left(\frac{16}{6400}\right) $

Hence percentage change is

$ \frac{\Delta g}{g}\times 100 = \frac{16}{6400}\times 100 $

$ = 0.25\% $

So,

$ \alpha = 0.25 $

Therefore, the correct option is:

$ \boxed{0.25} $

For a point at height $h$ above the surface of the earth, acceleration due to gravity is

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

Given that

$ g_h = \frac{g}{9} $

So,

$ g\left(\frac{R}{R+h}\right)^2 = \frac{g}{9} $

Cancel $g$ from both sides:

$ \left(\frac{R}{R+h}\right)^2 = \frac{1}{9} $

Taking square root on both sides,

$ \frac{R}{R+h} = \frac{1}{3} $

So,

$ 3R = R + h $

$ h = 2R $

Hence, the correct answer is:

$ \boxed{2R} $

So, Option C is correct.

For a body falling from infinity to the surface of the earth, the gain in kinetic energy is equal to the loss in gravitational potential energy.

For a mass $m$ falling from infinity:

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

Also, we know that

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

Substitute this into the energy equation:

$ \frac{1}{2}mv^2 = \frac{gR^2 m}{R} = mgR $

So,

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

For $m = 1\,\text{kg}$:

$ \frac{1}{2}v^2 = gR $

Given:

$ R = 6400\,\text{km} = 6.4 \times 10^6\,\text{m} $

and

$ g = 9.8\,\text{m/s}^2 $

Now,

$ v = \sqrt{2gR} $

$ v = \sqrt{2 \times 9.8 \times 6.4 \times 10^6} $

$ v = \sqrt{125.44 \times 10^6} $

$ v = 1.12 \times 10^4\,\text{m/s} $

$ v = 11.2\,\text{km/s} $

Now kinetic energy:

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

$ k = 1 \times 9.8 \times 6.4 \times 10^6 $

$ k = 62.72 \times 10^6\,\text{J} = 6.27 \times 10^7\,\text{J} $

So, the correct values are:

$ v = 11.2\,\text{km/s}, \qquad k = 6.27 \times 10^7\,\text{J} $

Therefore, the correct option is:

$ \boxed{\text{Option A}} $

The assertion (A) states that the “radius vector from the Sun to a planet sweeps out equal areas in equal intervals of time” which is just another way of stating Kepler's second law. This implies that the areal velocity is constant.

The reason (R) states that "For a central force field the angular momentum is a constant." In a central force system (like that of a planet orbiting the Sun under gravity), the force always acts along the line joining the two bodies. This means there is no torque acting on the planet, and thus, its angular momentum is conserved.

The connection between angular momentum and areal velocity is given by:

$\text{Areal velocity} = \frac{1}{2}r^2\dot{\theta} = \frac{L}{2m}$

where $L$ is the angular momentum and $m$ is the mass of the planet. Since the angular momentum $L$ is constant, the areal velocity must also remain constant.

Therefore, both the assertion and the reason are correct, and the reason correctly explains why the areal velocity (as stated in the assertion) is constant.

So, the most appropriate answer is:

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

$\begin{aligned} &\begin{aligned} & \mathrm{P}_{\mathrm{P}}+\mathrm{k}_{\mathrm{P}}=\mathrm{P}_{\mathrm{o}}+\mathrm{k}_0 \\ & -\frac{\mathrm{GMm}}{4 \mathrm{R}}+\frac{1}{2} \mathrm{mV}_{\mathrm{P}}^2=0 \\ & \mathrm{~V}_{\mathrm{P}}=\sqrt{\frac{\mathrm{GM}}{2 \mathrm{R}}} \end{aligned}\\ &\text { Choice } 1 \end{aligned}$

Let's analyze both the Assertion and the Reason step by step.

Calculation of the Required Kinetic Energy:

The gravitational potential energy of a body of mass $m$ at a distance $r$ from the center of the Earth is given by:

$U(r) = -\frac{GMm}{r},$

where $G$ is the gravitational constant and $M$ is the mass of the Earth.

At the Earth's surface (where $r = R$), the potential energy is:

$U(R) = -\frac{GMm}{R}.$

At infinity, the potential energy is defined as:

$U(\infty) = 0.$

The energy needed to move the body from the Earth’s surface to infinity is the change in potential energy:

$\Delta U = U(\infty) - U(R) = 0 - \Bigg(-\frac{GMm}{R}\Bigg) = \frac{GMm}{R}.$

Using the relation $g = \frac{GM}{R^2},$ we find:

$\frac{GMm}{R} = mgR.$

So, the required kinetic energy to just reach infinity (with zero speed at infinity) is:

$\text{Kinetic Energy} = mgR.$

The Assertion A states that the required kinetic energy is $\frac{1}{2}mgR$, which is only half of the actual value.

Verification of the Reason:

The Reason states: "The maximum potential energy of a body is zero when it is projected to infinity from earth surface."

By convention in gravitational problems, the potential energy at infinity is taken to be zero. This is a standard and correct statement.

Conclusion:

Assertion A is false because it underestimates the needed energy; the correct kinetic energy should be $mgR$.

Reason R is true since the gravitational potential energy is indeed defined to be zero at infinity.

Therefore, the correct answer is:

Option A: $\mathbf{A}$ is false but $\mathbf{R}$ is true.

Here’s how each quantity lines up with its dimensional formula:

Gravitational constant, $G$

• From Newton’s law $F = G\frac{m_1m_2}{r^2}$

• $[G] = \frac{[F]\,[r]^2}{[m]^2} = \frac{(MLT^{-2})\;L^2}{M^2} = M^{-1}L^3T^{-2}$

⇒ IV

Gravitational potential energy, $U$

• As work or $U = mgh$

• $[U] = [F]\,[h] = (MLT^{-2})\;L = ML^2T^{-2}$

⇒ III

Gravitational potential, $\phi$

• Energy per unit mass: $\phi = \frac{U}{m}$

• $[\phi] = \frac{ML^2T^{-2}}{M} = L^2T^{-2}$

⇒ II

Acceleration due to gravity, $g$

• Just an acceleration

• $[g] = LT^{-2}$

⇒ I

Matching them gives

A–IV, B–III, C–II, D–I

That corresponds to Option A.

The escape velocity ($v_{\text{esc}}$) for a planet with mass $M$ and radius $R$ is given by the formula:

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

where $G$ is the gravitational constant.

For Earth, let's denote its mass as $M_E$ and its radius as $R_E$. The escape velocity is given as 11.2 km/s. Therefore:

$ v_{\text{esc, Earth}} = \sqrt{\frac{2G M_E}{R_E}} = 11.2 \text{ km/s} $

For the planet in question, its mass $M_P$ is $M_E/8$ and its radius $R_P$ is $R_E/2$. The escape velocity from the planet is:

$ v_{\text{esc, Planet}} = \sqrt{\frac{2G M_P}{R_P}} = \sqrt{\frac{2G (M_E/8)}{R_E/2}} $

Simplifying the expression:

$ v_{\text{esc, Planet}} = \sqrt{\frac{2G (M_E/8)}{R_E/2}} = \sqrt{\frac{2G M_E \cdot 2}{8R_E}} = \sqrt{\frac{G M_E}{2R_E}} $

Now, relate it to the escape velocity from Earth:

$ v_{\text{esc, Planet}} = \frac{1}{\sqrt{4}} \cdot v_{\text{esc, Earth}} $

Calculating further:

$ v_{\text{esc, Planet}} = \frac{1}{2} \cdot 11.2 \text{ km/s} = 5.6 \text{ km/s} $

Thus, the escape velocity from the planet is 5.6 km/s.

Option C: 5.6 km/s is the correct answer.

$ T \propto R^{\frac{3}{2}} $

For a satellite in a circular orbit with radius $ R $, the time period is given by

$ T = 2\pi \sqrt{\frac{R^3}{GM}}. $

If the radius is increased to $ 1.03R $, the new time period $ T' $ becomes

$ T' = 2\pi \sqrt{\frac{(1.03R)^3}{GM}} = T \times (1.03)^{\frac{3}{2}}. $

For a small change, we can approximate

$ (1.03)^{\frac{3}{2}} \approx 1 + \frac{3}{2} \times 0.03 = 1 + 0.045 = 1.045. $

This means the new time period is approximately 4.5% larger than the original time period.

Thus, the time period of the second satellite is larger by approximately 4.5%.

To find the time period of the satellite, we can use Kepler’s third law, which states that the square of the orbital period is proportional to the cube of the orbital radius. Mathematically, this is expressed as:

$\frac{T^2}{R^3} = \text{constant}$

Here’s how we solve the problem step by step:

Let:

The Moon's distance from Earth be $R_{\text{moon}}$.

The Moon's orbital period be $T_{\text{moon}} = 27$ days.

The satellite's orbital radius be $R_{\text{sat}} = \frac{R_{\text{moon}}}{9}$ (since it is 9 times closer).

According to Kepler's law for both orbits, we have:

$\frac{T_{\text{sat}}^2}{R_{\text{sat}}^3} = \frac{T_{\text{moon}}^2}{R_{\text{moon}}^3}$

Substitute $R_{\text{sat}}$:

$\frac{T_{\text{sat}}^2}{\left(\frac{R_{\text{moon}}}{9}\right)^3} = \frac{27^2}{R_{\text{moon}}^3}$

Simplify the left side by noting that:

$\left(\frac{R_{\text{moon}}}{9}\right)^3 = \frac{R_{\text{moon}}^3}{9^3} = \frac{R_{\text{moon}}^3}{729}$

So the equation becomes:

$\frac{T_{\text{sat}}^2}{\frac{R_{\text{moon}}^3}{729}} = \frac{27^2}{R_{\text{moon}}^3}$

Multiply both sides by $\frac{R_{\text{moon}}^3}{729}$:

$T_{\text{sat}}^2 = 27^2 \times \frac{1}{729}$

Recognize that $27^2 = 729$:

$T_{\text{sat}}^2 = \frac{729}{729} = 1$

Take the square root:

$T_{\text{sat}} = 1 \text{ day}$

Thus, the satellite’s orbital period is 1 day, which corresponds to Option D: 1 day.

Given mass of whole sphere = $M$

$M_2$ = mass of removed sphere

$ = {M \over {{4 \over 3}\pi {R^3}}}{4 \over 3}\pi {\left( {{R \over 3}} \right)^3} = {M \over {27}}$

Due to whole sphere,

${F_1} = {{GMm} \over {{{(2R)}^2}}}$ (using Newton's gravitational law)

${F_2}$ =Force due to whole sphere $-$ Force due to removed sphere

$ = {{GMm} \over {{{(2R)}^2}}} - {{G{M_2}m} \over {{{\left( {{{4R} \over 3}} \right)}^2}}}$

$ = {{GMm} \over {4{R^2}}} - {{G\left( {{M \over {27}}} \right)m} \over {{{16} \over 9}{R^2}}}$

$ = {{GMm} \over {4{R^2}}}\left[ {1 - {1 \over {12}}} \right] = {{4Mm} \over {4{R^2}}} \times {{11} \over {12}}$

So, $ = {{{F_1}} \over {{F_2}}} = {{GMm} \over {4{R^2}}}/{{GMm} \over {4{R^2}}} \times {{11} \over {12}} = {{12} \over {11}}$

$ \Rightarrow {F_1}:{F_2} = 12:11$

To determine the relative rate of change of the separation between the two stars, let's first understand the relationship governing their motion.

Equilibrium of Forces:

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

Here, $ \omega $ is the angular velocity. From the balance of gravitational force and the centripetal force, we can solve for $ \omega $:

$ \omega = \sqrt{\frac{G m_1}{r^3}} $

Angular Momentum (L):

$ L = m_2 \omega r^2 = m_2 \sqrt{\frac{G m_1}{r^3}} r^2 $

Simplifying, we find:

$ L = m_2 \sqrt{G m_1 r} = \text{constant} $

Differential Relation:

Taking the natural logarithm of the angular momentum expression:

$ \ln L = \ln m_2 + \ln G + \frac{1}{2} \ln m_1 + \frac{1}{2} \ln r $

Since $ L $ is constant, differentiate both sides:

$ 0 = \frac{dm_2}{m_2} + \frac{1}{2} \frac{dm_1}{m_1} + \frac{1}{2} \frac{dr}{r} $

Rate of Change of Separation (r):

Solving for $\frac{dr}{r}$, we arrive at:

$ \frac{1}{r} \frac{dr}{dt} = -\frac{2}{m_2} \frac{dm_2}{dt} - \frac{1}{m_1} \frac{dm_1}{dt} $

Since mass is transferred at a constant rate $ \gamma $, we have $ \frac{dm_1}{dt} = \gamma $ and $ \frac{dm_2}{dt} = - \gamma $.

Substitute the Variable Rates:

Substituting these values:

$ \frac{1}{r} \frac{dr}{dt} = -\frac{2(-\gamma)}{m_2} - \frac{\gamma}{m_1} \approx -\frac{2\gamma}{m_2} $

Thus, the relative rate of change of the separation between the stars is approximately $-\frac{2\gamma}{m_2}$ s$^{-1}$.

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} $

$ \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} $

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 new radius of the orbit after the energy is supplied to the satellite, we need to compare the initial and final energies of the satellite in its orbit around the Earth. We will use the formula for total energy (the sum of kinetic and potential energy) of a satellite in a circular orbit:

The total energy, $E$, of a satellite of mass $m$ orbiting at a distance $r$ from the center of the Earth is given by:

$ E = -\frac{GMm}{2r} $

where:

• $G$ is the gravitational constant
• $M$ is the mass of the Earth
• $r$ is the radius of the orbit

Given, the initial radius of orbit is $2R$. Therefore, the initial total energy $E_i$ is:

$ E_i = -\frac{GMm}{2 \times 2R} = -\frac{GMm}{4R} $

Now, the energy supplied to the satellite is given as $\frac{10^4 R}{6} \mathrm{~J}$. The new total energy $E_f$ will be the sum of the initial energy and the supplied energy:

$ E_f = E_i + \text{Energy Supplied} $

Substituting the values:

$ E_f = -\frac{GMm}{4R} + \frac{10^4 R}{6} $

The final energy formula for a new orbit radius $r_f$ is similar to the initial energy formula:

$ E_f = -\frac{GMm}{2r_f} $

By equating the two expressions for $E_f$, we get:

$ -\frac{GMm}{2r_f} = -\frac{GMm}{4R} + \frac{10^4 R}{6} $

Rearrange the equation to solve for $r_f$:

$ \frac{GMm}{2r_f} = \frac{GMm}{4R} - \frac{10^4 R}{6} $

Since $GMm = gR^2 m$ (using gravitational acceleration $g$ and radius of Earth $R$), we can substitute this to simplify the expression:

$ \frac{gR^2 m}{2r_f} = \frac{gR^2 m}{4R} - \frac{10^4 R}{6} $

By cancelling the common terms and rearranging:

$ \frac{1}{2r_f} = \frac{1}{4R} - \frac{10^4 R}{6gR^2 m} $

Recall that $g = 10 \mathrm{~m/s^2}$ and the mass of the satellite $m = 10^3 \mathrm{~kg}$:

$ \frac{1}{2r_f} = \frac{1}{4R} - \frac{10^4 \times R}{6 \times 10 \times R^2 \times 10^3} $

Simplifying the second term further:

$ \frac{10^4 \times R}{60 \times R^2 \times 10^3} = \frac{10 \times R}{60 \times R^2} = \frac{1}{6R} $

So, the final equation becomes:

$ \frac{1}{2r_f} = \frac{1}{4R} - \frac{1}{6R} $

Finding a common denominator for the right-hand side:

$ \frac{1}{2r_f} = \frac{3 - 2}{12R} = \frac{1}{12R} $

Therefore:

$ 2r_f = 12R $

Thus:

$ r_f = 6R $

Hence, the new radius of the circular orbit is 6R.

The correct answer is:

Option B: 6R

At earth surface, $E_1=-\frac{G M_m}{R_e}$

in the orbit, $E_2=-\frac{G M_m}{2 r}$

$\begin{aligned} \Delta E & =G M_m\left[\frac{1}{R_e}-\frac{1}{2 r}\right] \\\\ & =G M_m\left[\frac{1}{R_e}-\frac{1}{2.1 R_e}\right] \\\\ & =\frac{11}{21} \frac{G M_m}{R_e} \\\\ \Rightarrow & x=11 \end{aligned}$

To solve this question, we will use the fact that for an object in a circular orbit, the centripetal force required to keep the object in orbit is provided by the gravitational force between the object and the planet it is orbiting. This principle gives us the relationship between the speed of the satellite, its orbital radius, and the mass of the planet.

The formula for the gravitational force is given by:

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

where:

• $F$ is the gravitational force between the two masses,
• $G$ is the gravitational constant,
• $M$ is the mass of the planet,
• $m$ is the mass of the satellite, and
• $r$ is the distance between the center of the planet and the satellite (radius of the orbit).

The centripetal force required to keep the satellite in orbit is given by:

$ F_c = \frac{m v^2}{r} $

where:

• $F_c$ is the centripetal force,
• $v$ is the orbital speed of the satellite, and
• $r$ is the radius of the orbit.

Since the gravitational force is providing the centripetal force, we can set the two forces equal to each other to find the relationship between speed and radius:

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

By simplifying this, we find the equation for the orbital speed of a satellite:

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

Now, we can compare the speeds of satellites $A$ and $B$ based on their radii. Satellite $A$ orbits at a radius of $4R$ with a speed of $3v$, and we need to find the speed of satellite $B$ which orbits at a radius of $R$.

Substituting the respective radii into the speed equation:

• For satellite $A$, $v_A = 3v = \sqrt{\frac{G M}{4R}}$
• For satellite $B$, we want to find $v_B$ given that its radius is $R$: $v_B = \sqrt{\frac{G M}{R}}$

To find the ratio of $v_B$ to $3v$ (or the speed of $A$), we can write:

$ v_B = \sqrt{\frac{G M}{R}} = \sqrt{4} \sqrt{\frac{G M}{4R}} = 2 \cdot \sqrt{\frac{G M}{4R}} $

Given that $\sqrt{\frac{G M}{4R}}$ is the speed of satellite $A$ divided by 3 ($3v$), we find that:

$ v_B = 2 \cdot 3v = 6v $

Therefore, the correct answer is Option A: $6 v$.

$\begin{aligned} & \frac{v_1}{v_2}=\sqrt{\frac{r_2}{r_1}} \quad \text{.... (i)}\\ & m_1 v_1 r_1=h \\ & m_2 v_2 r_2=32 \\ & \Rightarrow \frac{v_1}{v_2}=\frac{1}{3} \frac{m_2}{m_1} \frac{r_2}{r_1} \quad \text{.... (ii)} \end{aligned}$

From (i) & (ii)

$\begin{aligned} & \sqrt{\frac{r_2}{r_1}}=\frac{1}{3} \frac{m_2}{m_1} \frac{r_2}{r_1} \\ & \frac{3 m_1}{m_2}=\sqrt{\frac{r_2}{r_1}} \\ & \frac{T_1}{T_2}=\left(\frac{r_1}{r_2}\right)^{3 / 2}=\left(\frac{m_2}{3 m_1}\right)=\frac{1}{27}\left(\frac{m_2}{m_1}\right)^3 \end{aligned}$

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