Gravitation

If we take the velocity (v) of the satellite to be $v = \sqrt{GM/r}$, where $G$ is the gravitational constant, $M$ is the mass of the Earth, and $r$ is the distance from the center of the Earth to the satellite, then the orbital angular momentum (L) is:

$ L = mvr = m \sqrt{GMr} $

If the distance (r) from the Earth's center is increased by a factor of 8, the new angular momentum $L'$ is:

$ L' = m \sqrt{GM(8r)} = m \sqrt{8GMr} = 2 \sqrt{2} L $

However, the distance from the earth's surface is given, so we have to take into account the radius of the Earth ($R$) in our calculations. When the height from the earth's surface is increased eight times, the distance from the earth's center is $r = R + 8h$, which is approximately $9R$ (because the height of the satellite above the Earth is generally much less than the radius of the Earth).

So, the new angular momentum (L'') is:

$ L'' = m \sqrt{GM(9R)} = 3 \sqrt{GM(R)} = 3L $

A satellite in orbit around a planet is subject to two main forces: gravitational force, which is trying to pull it towards the planet, and its own kinetic energy or inertia, which is trying to keep it moving in a straight line. The balance of these two forces results in the satellite moving in a circular or elliptical orbit.

The gravitational potential energy ($U$) of the satellite is given by the formula:

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

where $G$ is the gravitational constant, $M$ is the mass of the Earth, $m$ is the mass of the satellite, and $R$ is the radius of the orbit. The negative sign indicates that work would have to be done to remove the satellite from the Earth's gravitational influence.

The kinetic energy ($K$) of the satellite is given by the formula:

$K = \frac{GMm}{2R}$

This is obtained from the fact that for a satellite in stable orbit, the gravitational force must be equal to the centripetal force required to keep the satellite moving in a circle. From this, we can derive an expression for the velocity of the satellite, and hence its kinetic energy.

The total mechanical energy ($E$) of the satellite, which is the sum of its kinetic and potential energy, is therefore:

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

So the potential energy $U$ is $-2E$, and the kinetic energy $K$ is $-E$. Thus, the statement "If $E$ be the total energy of a satellite moving around the earth, then its potential energy will be $2E$" is correct, and the statement "The kinetic energy of a satellite revolving in an orbit is equal to the half the magnitude of total energy $E$" is incorrect.

The gravitational field inside a uniform spherical body varies linearly with distance from the center. If we consider the Earth to be such a body, then the gravitational field strength (and hence weight) of an object would decrease linearly as we go deeper inside the Earth.

The weight of an object at a depth $d$ from the Earth's surface is given by:

$W_d = W_e (1 - \frac{d}{R})$,

where:

• $W_d$ is the weight at depth $d$,
• $W_e$ is the weight at the Earth's surface (i.e., the weight of the object), and
• $R$ is the radius of the Earth.

In this case, we're given that $W_e = 400 \, \text{N}$, and we're asked to find the weight at a depth of $d = \frac{R}{2}$. Substituting these values into the formula, we get:

$W_d = 400 \, \text{N} (1 - \frac{1}{2}) = 200 \, \text{N}$.

Therefore, the weight of the body when taken to a depth half of the radius of the Earth is 200 N.

To find the gravitational force on the body when taken at a height equal to one-fourth the radius of the Earth, we can use the formula for gravitational force:

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

where $F$ is the gravitational force, $G$ is the gravitational constant, $m_1$ and $m_2$ are the masses of the two objects, and $r$ is the distance between their centers.

The weight of the body on the surface of the Earth is given as $100\,\text{N}$, which is also the gravitational force acting on it:

$F_\text{surface} = G \frac{m_\text{body} m_\text{earth}}{R_\text{earth}^2}$

When the body is taken to a height equal to one-fourth the radius of the Earth, the distance between the centers of the body and the Earth becomes $r_\text{new} = R_\text{earth} + \frac{1}{4} R_\text{earth} = \frac{5}{4} R_\text{earth}$.

Now the gravitational force acting on the body at this new height is:

$F_\text{new} = G \frac{m_\text{body} m_\text{earth}}{r_\text{new}^2}$

$F_\text{new} = G \frac{m_\text{body} m_\text{earth}}{\left(\frac{5}{4} R_\text{earth}\right)^2}$

To find the ratio between the new gravitational force and the original force on the surface, we can write:

$\frac{F_\text{new}}{F_\text{surface}} = \frac{G \frac{m_\text{body} m_\text{earth}}{\left(\frac{5}{4} R_\text{earth}\right)^2}}{G \frac{m_\text{body} m_\text{earth}}{R_\text{earth}^2}}$

Canceling out the common terms, we get:

$\frac{F_\text{new}}{F_\text{surface}} = \frac{R_\text{earth}^2}{\left(\frac{5}{4} R_\text{earth}\right)^2} = \frac{1}{\left(\frac{5}{4}\right)^2} = \frac{1}{\left(\frac{25}{16}\right)} = \frac{16}{25}$

Now, since the weight of the body on the surface is $100\,\text{N}$, we can find the new gravitational force as:

$F_\text{new} = \frac{16}{25} \times 100\,\text{N} = 64\,\text{N}$

So, the gravitational force on the body when taken at a height equal to one-fourth the radius of the Earth is $64\,\text{N}$.

The assertion (A) is true: Earth does have an atmosphere, while the Moon does not have a significant atmosphere.

The reason (R) is also true: The escape velocity on the Moon is indeed smaller than that on Earth. The escape velocity is the minimum velocity an object must have to escape the gravitational pull of a planet or moon. A smaller escape velocity means it's easier for particles (such as the particles that make up an atmosphere) to escape into space.

Additionally, the reason (R) is a correct explanation for the assertion (A). The fact that the Moon's escape velocity is smaller than the Earth's is a major reason why the Moon doesn't have a significant atmosphere. Over time, particles that could have made up an atmosphere have escaped the Moon's gravitational pull and dispersed into space.

The weight of an object on a planet is given by the equation $W = mg$, where $m$ is the mass of the object and $g$ is the acceleration due to gravity.

The acceleration due to gravity on a planet is given by the equation $g = \frac{GM}{R^2}$, where $G$ is the gravitational constant, $M$ is the mass of the planet, and $R$ is the radius of the planet.

In this case, the mass of the planet is double that of Earth ($M = 2M_E$), but the density is the same. Density is defined as mass divided by volume, so if the mass is doubled and the density stays the same, the volume must also double.

Since the volume of a sphere (like a planet) is given by the equation $V = \frac{4}{3}\pi R^3$, a doubling of the volume implies that the radius of the planet is $R = 2^{1/3}R_E$.

Substituting these values back into the equation for $g$, we get:

$g_{\text{planet}} = G * \frac{2M_E}{(2^{1/3}R_E)^2} = 2^{1/3} * g_E$

So the weight of the object on the planet is $W_{\text{planet}} = m*g_{\text{planet}} = m2^{1/3} * g_E = 2^{1/3}W_E$

Statement I is true. The acceleration due to gravity varies slightly over the surface of the Earth, being affected by factors such as latitude (because of the Earth's oblate shape or its equatorial bulge), altitude (it decreases with height above the Earth's surface), and local geological variations in the Earth's density.

Statement II is false. The acceleration due to gravity actually decreases as we go down below the Earth's surface. This is because the mass that is "above" or outside the location begins to pull the object away from the Earth's center, and the net gravitational acceleration decreases.

Therefore, Option D: Statement I is true but Statement II is false, is the correct answer.

The weight of an object varies with altitude due to the change in gravitational force. The force of gravity decreases with the square of the distance from the center of the Earth.

The gravitational force (weight) at a height h from the surface of the Earth is given by:

$W' = W \left(\frac{R}{{R + h}}\right)^2$

where:

• W' is the weight at height h,
• W is the weight at the Earth's surface,
• R is the radius of the Earth,
• h is the height above the Earth's surface.

In this problem, the height h is given as nine times the radius of the Earth (h = 9R). Substituting h = 9R into the equation gives:

$W' = W \left(\frac{R}{{R + 9R}}\right)^2 = W \left(\frac{1}{10}\right)^2$

Simplifying this gives:

$W' = \frac{W}{100}$

Therefore, the weight of the body at that height will be 1/100th of its weight on the surface of the Earth.

The correct answer is Option C: $\frac{W}{100}$.

${U_P} = - {{GMm} \over {2R}}$

${U_S} = - {{GMm} \over R}$

$\Rightarrow$ Energy conservation

${1 \over 2}m{v^2} - {{GMm} \over R} = - {{GMm} \over {2R}}$

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

$v = \sqrt {{{GM} \over R}} = \sqrt {gR} $

$E = - {K \over {{r^2}}}$

$\Delta V = - \int\limits_{r = 2\,cm}^{3\,cm} {E.\,dr} $

$ = \int\limits_2^3 {{k \over {{r^2}}}dr} $

$ = \left[ { - {K \over r}} \right]_2^3 = \left( {{K \over 6}} \right) = {6 \over 6} = 1$ J/kg

${V_f} - {V_i} = 1$

$ \Rightarrow {V_f} - 10 = 1$

${V_f} = 11$ J/kg

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

$\therefore$ ${{T_1^2} \over {T_2^2}} = {{R_1^3} \over {R_2^3}}$

${{{{24}^2}} \over {T_2^2}} = {{R_1^3} \over {{{\left( {{{{R_1}} \over 4}} \right)}^3}}}$

${{{{24}^2}} \over {T_2^2}} = {4^3}$

${T_2} = {{24} \over {{2^3}}} = 3$ hours

A. The force acting on a planet is inversely proportional to the square of the distance from the sun. This is known as the inverse square law and is described by Newton's law of gravitation.

B. Force acting on a planet is inversely proportional to the product of the masses of the planet and the sun. This is also described by Newton's law of gravitation, but it is not directly related to the planet's elliptical orbit.

C. The centripetal force acting on the planet is directed towards the sun, not away from it.

D. The square of the time period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of the elliptical orbit. This is known as Kepler's third law.

The most appropriate answer is Statement I is correct but statement II is incorrect.

Statement I is correct as acceleration due to Earth's gravity decreases as you move away from its surface either upward or downward. This is because the gravity of the Earth follows an inverse square law, which means that the gravitational force decreases as the square of the distance between two objects increases.

However, statement II is incorrect because acceleration due to Earth's gravity is not the same at a height and depth from Earth's surface, even if they are equal in magnitude. This is because the Earth is not a perfect sphere and has a non-uniform distribution of mass, which causes variations in the strength of gravity at different locations. Therefore, the acceleration due to Earth's gravity will be different at a certain height and depth, even if they are equal.

The weight of an object at a height h from the Earth's surface is given by:

$W' = W \left(\frac{R}{{R + h}}\right)^2$

where:

• W' is the weight at height h,
• W is the weight at the Earth's surface,
• R is the radius of the Earth,
• h is the height above the Earth's surface.

In this problem, the weight at the Earth's surface W is given as 18 N, the radius of the Earth R is given as 6400 km, and the height h is given as 3200 km. Substituting these values into the equation gives:

$W' = 18 N \left(\frac{6400~km}{{6400~km + 3200~km}}\right)^2 = 18 N \left(\frac{2}{3}\right)^2 = 18 N \cdot \frac{4}{9} = 8 N$

Therefore, the weight of the body at an altitude of 3200 km above the Earth's surface is 8 N.

The acceleration due to gravity (g) on the surface of a planet is given by the formula:

$g = \frac{G M}{R^2}$

where:

• G is the gravitational constant,
• M is the mass of the planet,
• R is the radius of the planet.

If the radius of the Earth shrinks by 2% but its mass remains the same, the new acceleration due to gravity (g') will be:

$g' = \frac{G M}{(0.98R)^2} = \frac{G M}{0.9604 R^2} = \frac{g}{0.9604}$

This implies that g' is approximately 1.0412 times g, or an increase of approximately 4.12%.

Therefore, the acceleration due to gravity on the Earth's surface will approximately increase by 4%.

Using energy conservation

$ - {{G{M_e}m} \over {{R_e}}} + {1 \over 2}m{\left( {\lambda \sqrt {{{2G{M_e}} \over {{R_e}}}} } \right)^2} = - {{G{M_e}m} \over r}$

${{G{M_e}m} \over r} = {{G{M_e}m} \over {{R_e}}} - {{G{M_e}m} \over {{R_e}}}{\lambda ^2}$

$r = {{{R_e}} \over {1 - {\lambda ^2}}}$

$U = - {{G{M_e}m} \over {2r}}$

So, ${{{U_A}} \over {{U_B}}} = {{{m_A}} \over {{m_B}}} \times {{{r_B}} \over {{r_A}}}$

$ = {4 \over 3} \times {4 \over 3} = {{16} \over 9}$

Applying conservation of energy

$ - {{G{M_e}m} \over {{R_e}}} + {1 \over 2}m{\left( {{1 \over 3}\sqrt {{{2G{m_e}} \over {{R_e}}}} } \right)^2} = - {{G{M_e}m} \over {{R_e} + h}}$

$ - {{G{M_e}m} \over {{R_e}}} + {{G{M_e}m} \over {9{R_e}}} = - {{G{M_e}m} \over {{R_e} + h}}$

${8 \over {9{R_e}}} = {1 \over {{R_e} + h}}$

$ \Rightarrow h = {{{R_e}} \over 8} = {{6400} \over 8} = 800$ km

$\because$ $g = {{GM} \over {{r^2}}}$

$ \Rightarrow {{\Delta g} \over g} = 2{{\Delta r} \over r}$

$ \Rightarrow {{\Delta g} \over g} \times 100 = 2 \times {{32} \over {6400}} \times 100\% = 1\% $

$\Rightarrow$ % decrease in weight = 1%

$g = {{GM} \over {{{(R + h)}^2}}} = {{GM} \over {9{R^2}}} = {{{g_0}} \over 9}$

$ \Rightarrow T = 2\pi \sqrt {{l \over g}} = 2\pi \sqrt {{l \over {{{{g_0}} \over 9}}}} $

$ \Rightarrow 2 = 2\pi \sqrt {{{9l} \over {{g_0}}}} $

$ \Rightarrow l = {{{g_0}} \over {9{\pi ^2}}} = {1 \over 9}\,m$

The weight of an object at a distance d from the center of the Earth is given by:

$W' = W \left(\frac{R}{d}\right)^2$

where:

• W' is the weight at distance d,
• W is the weight at the Earth's surface,
• R is the radius of the Earth,
• d is the distance from the center of the Earth.

In this problem, we are asked to find the percentage decrease in weight, which can be calculated by:

$\Delta W = \frac{W - W'}{W} \times 100\%$

where ΔW is the percentage change in weight. Substituting the weight formula into the percentage change formula gives:

$\Delta W = \left(1 - \left(\frac{R}{d}\right)^2\right) \times 100\%$

The radius of the Earth R is 6400 km and the distance d from the center of the Earth is given as $\frac{5}{4}R$. Substituting these values into the equation gives:

$\Delta W = \left(1 - \left(\frac{6400~km}{\frac{5}{4} \cdot 6400~km}\right)^2\right) \times 100\% = \left(1 - \left(\frac{4}{5}\right)^2\right) \times 100\% = 36\% $

Therefore, the percentage decrease in the weight of the object when taken to a height of $\frac{1}{4}R$ above the surface of the Earth is 36%.

m = 100 kg

${F_{AP}} = {{G{m^2}} \over {{{\left( {13\sqrt 2 } \right)}^2}}}$

${F_{BP}} = {{G{m^2}} \over {{{13}^2}}}$

${F_{CP}} = {{G{m^2}} \over {{{\left( {13\sqrt 2 } \right)}^2}}}$

${F_{net}} = {F_{BP}} + {F_{AP}}\cos 45^\circ + {F_{CP}}\cos 45^\circ $

$ = {{G{m^2}} \over {{{13}^2}}}\left( {1 + {1 \over {\sqrt 2 }}} \right)$

$ = {{G{{100}^2}} \over {169}}(1 + 0.707)$

$ \simeq 100\,G$

Given,

${{{r_A}} \over {{r_B}}} = {2 \over 3}$

${{{\rho _A}} \over {{\rho _B}}} = {3 \over 5}$

We know,

Acceleration due to gravity

$g = {{GM} \over {{r^2}}} = {G \over {{r^2}}} \times {4 \over 3}\pi {r^3} \times \rho $

$ = {{4\pi Gr\rho } \over 3}$

$\therefore$ $g \propto \rho r$

$\therefore$ ${{{g_A}} \over {{g_B}}} = {{{\rho _A}} \over {{\rho _B}}} \times {{{r_A}} \over {{r_B}}}$

$ = {3 \over 5} \times {2 \over 3}$

$ = {2 \over 5}$

$T_2^2 = {\left( {{{{R_2}} \over {{R_1}}}} \right)^3}T_1^2$

$ \Rightarrow {T_2} = {(3)^{3/2}} \times 7 \approx 5.2 \times 7$

${T_2} \cong 36$ hrs

${v_{esc}} - \sqrt {{{2GM} \over R}} = \sqrt {{{2G} \over R} \times \rho \times {4 \over 3}\pi {R^3}} $

$ \Rightarrow {v_{esc}} \propto R\sqrt \rho $

$ \Rightarrow {{{{({v_{esc}})}_B}} \over {{{({v_{esc}})}_A}}} = 1$

$ \Rightarrow {({v_{esc}})_B} = 12$ km/s

Total gravitational PE of water per second $ = {{mgh} \over T}$

$ = {{9 \times {{10}^4} \times 10 \times 40} \over {3600}} = {10^4}$ J/sec

50% of this energy can be converted into electrical energy so total electrical energy $ = {{{{10}^4}} \over 2} = 5000$ W

So total bulbs lit can be $ = {{5000\,W} \over {100\,W}} = 50$ bulbs

Let the masses are m and distance between them is l, then $F = {{G{m^2}} \over {{l^2}}}$.

When 1/3^rd mass is transferred to the other then masses will be ${{4m} \over 3}$ and ${{2m} \over 3}$. so new force will be

$F' = {{G{{4m} \over 3} \times {{2m} \over 3}} \over {{l^2}}} = {8 \over 9}{{G{m^2}} \over {{l^2}}} = {8 \over 9}F$

${T_A} = 2{T_B}$

Now $T_A^2 \propto r_A^3$

$ \Rightarrow {\left( {{{{r_A}} \over {{r_B}}}} \right)^3} = {\left( {{{{T_A}} \over {{T_B}}}} \right)^2}$

$ \Rightarrow r_A^3 = 4r_B^3$

Diameter = r $\times$ $\delta$

$ = 1.5 \times {10^{11}} \times (2000) \times \left( {{1 \over {3600}}} \right) \times \left( {{\pi \over {180}}} \right)$

$ = 1.45 \times {10^9}$ m

Total gravitational potential energy

$ = - \left\{ {{{4GMm} \over {d/\sqrt 2 }} + {{4G{m^2}} \over d} + {{2G{m^2}} \over {\sqrt 2 d}}} \right\}$

$ = - {{Gm} \over d}\left\{ {M4\sqrt 2 + (4 + \sqrt 2 )m} \right\}$

$ = - {{Gm} \over d}\left\{ {4\sqrt 2 M + (4 + \sqrt 2 )m} \right\}$

Statement I is true. Newton's law of universal gravitation states that every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between their centers. This law applies to any pair of bodies in the universe.

Statement II is also true. The weight of an object is the force of gravity acting on it. If a person were at the center of the Earth, the gravitational pull from all the surrounding mass of the Earth would cancel out, resulting in zero net gravitational force and therefore zero weight.

Therefore, Option A: Both Statement I and Statement II are true, is the correct answer.

Assertion A is false. While the direction of acceleration due to gravity does always point towards the center of the Earth, its magnitude actually varies from the poles to the equator. This is due to the Earth's rotation and the fact that Earth is not a perfect sphere but an oblate spheroid, meaning it's slightly flattened at the poles and bulging at the equator.

Reason R is true. At the equator, the direction of acceleration due to gravity is towards the center of the Earth.

The acceleration due to gravity (g) at a distance (r) from the center of a planet is given by:

$g = \frac{G M}{r^2}$

where:

• G is the gravitational constant,
• M is the mass of the planet,
• r is the distance from the center of the planet.

If the height h of a point P above the surface of the Earth is equal to the diameter of the Earth, then the distance r from the center of the Earth to the point P is 3 times the radius of the Earth. Substituting this into the equation for g gives:

$g' = g \left(\frac{R}{3R}\right)^2 = \frac{g}{9}$

where:

• g' is the acceleration due to gravity at the point P,
• R is the radius of the Earth.

Therefore, the value of acceleration due to gravity at point P is g/9.

We know that

T² $\propto$ R³

$ \Rightarrow {\left( {{{T'} \over T}} \right)^2} = {\left( {{{3R} \over R}} \right)^3}$

$ \Rightarrow {{T'} \over T} = 3\sqrt 3 $

$ \Rightarrow T' = 3\sqrt 3 $ years

According to the given information

${{GM} \over {{{(R + h)}^2}}} = {1 \over 3} \times {{GM} \over {{R^2}}}$

$ \Rightarrow R + h = \sqrt 3 R$

$ \Rightarrow h = (\sqrt 3 - 1)R \simeq 4685$ km

Let us consider the gravitational force acting on each mass M by adjacent particles be F.

and the gravitational force acting on each mass M diagonally be F₁.

Along the centre of circle,