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 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 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

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

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 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:

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

Let's analyze both statements:

Statement I: Rotation of the earth shows an effect on the value of acceleration due to gravity (g).

This is true. The value of $ g $ is affected by the rotation of the Earth. The centripetal force due to the Earth's rotation causes a reduction in the perceived gravitational acceleration for objects on the surface. This effect is zero at the poles and increases towards the equator. The formula for the effective acceleration due to gravity at a latitude $ \theta $ is:

where:

• $ g $ is the theoretical acceleration due to gravity without Earth's rotation,
• $ R $ is the radius of the Earth,
• $ \omega $ is the angular velocity of the Earth's rotation, and
• $ \theta $ is the latitude.

Statement II: The effect of rotation of the earth on the value of 'g' at the equator is minimum and that at the pole is maximum.

This is false. For clarification, the effect of the Earth's rotation on the value of $ g $ is maximum at the equator and minimum at the poles. At the equator, the centrifugal force is highest because of the maximum velocity due to Earth's rotation, which decreases the effect of gravity more than at any other latitude. At the poles, the centrifugal force is zero since there is no rotational velocity contributing to a centripetal effect; therefore, the effect of rotation on $ g $ is minimum (zero).

With these considerations, the correct option is:

Option B: Statement I is true but Statement II is false.

The orbital speed of an object moving in a circular orbit around Earth (or any other celestial body) is given by the formula:

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

where (v) is the orbital speed, (G) is the gravitational constant, (M) is the mass of the central body (Earth, in this case), and (r) is the radius of the orbit.

For the two satellites of masses (m) and (3m) in orbits of radii (r) and (3r) respectively, the ratio of their orbital speeds ($v_1/v_2$) is:

$ \frac{v_1}{v_2} = \sqrt{\frac{\frac{GM}{r}}{\frac{GM}{3r}}} = \sqrt{\frac{3r}{r}} = \sqrt{3} $

So, the ratio of the orbital speeds of the satellites is ($\sqrt{3} : 1$), which corresponds to Option B.

The inclusion of two different masses for the satellites, $m$ and $3m$, in the problem might initially seem to suggest that the masses would influence their orbital speeds. However, when it comes to circular orbital motion, especially around a much larger body like the Earth, the mass of the orbiting satellite does not directly affect its orbital speed. This is because the orbital speed equation:

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

only takes into account the mass of the central body (in this case, Earth's mass $M$), and the radius of the orbit $(r)$, where $G$ is the gravitational constant.

The gravitational field strength, or equivalently, the weight of an object, decreases linearly from its surface value to zero at the center of a sphere of uniform mass density. This is because only the mass inside the radius at which the object is located contributes to the gravitational force at that location.

If ( $d = \frac{R}{2}$ ) is the depth below the surface of the Earth, then the radius of the sphere contributing to the gravitational force at that depth is ( $R - d = R - \frac{R}{2} = \frac{R}{2}$ ).

Since the gravitational force (or weight) decreases linearly with the radius in a sphere of uniform density, the weight of the object at depth ( $d = \frac{R}{2} $) is half its weight at the surface of the Earth.

So, if the weight of the body on the surface of the Earth is 200 N, its weight at depth ( $d = \frac{R}{2} $) is half of that, or 100 N.

Therefore, the correct answer is 100 N.

The acceleration due to gravity ($g$) at the surface of Earth is approximately $9.81 \, \text{m/s}^2$. This acceleration decreases as we move away from the Earth's surface because gravity is a force that attracts objects towards the center of the Earth. This force decreases with the square of the distance from the center of the Earth, due to the inverse square law.

However, if we are at a height ($h$) much less than the radius of the Earth ($R$), we can use a linear approximation to calculate the new acceleration due to gravity ($g'$) at this height. This is because for small heights compared to the radius of Earth, the decrease in $g$ can be approximated to be linear. This is represented by the following formula:

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

Here's how this formula is derived:

The force of gravity ($F$) is given by the universal law of gravitation:

$F = G \frac{m_1 m_2}{d^2}$

where:

• $G$ is the gravitational constant,
• $m_1$ and $m_2$ are the masses of the two objects (in this case, the Earth and the object we're considering),
• $d$ is the distance between the centers of the two objects.

If we consider an object of mass $m$ at the surface of Earth, the force it experiences due to gravity is:

$F = G \frac{mM}{R^2} = mg$

where:

• $M$ is the mass of Earth,
• $R$ is the radius of Earth,
• $g = \frac{GM}{R^2}$ is the acceleration due to gravity at the Earth's surface.

Now, if the object is at a height $h$ above the Earth's surface, the force it experiences is:

$F' = G \frac{mM}{(R + h)^2}$

We can write $(R + h)^2$ as $R^2 (1 + \frac{h}{R})^2$, and because $h << R$, we can use the binomial approximation $(1 + x)^2 \approx 1 + 2x$ for small $x$ to get:

$F' \approx G \frac{mM}{R^2} \left(1 - \frac{2h}{R}\right) = mg \left(1 - \frac{2h}{R}\right)$

Equating the forces $F = mg$ and $F' = mg'$ gives us:

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