Gravitation

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:

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

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 formula to calculate the acceleration due to gravity at a height ( h ) from the surface of the Earth is expressed as :

$ g = \frac{G M_e}{(R_e + h)^2} $

To find the ratio of gravitational acceleration at two different heights ( h_P ) and ( h_Q ) above the Earth's surface, use the formula for each and form a ratio :

$ \frac{g_P}{g_Q} = \frac{\frac{G M_e}{(R_e + h_P)^2}}{\frac{G M_e}{(R_e + h_Q)^2}} = \frac{(R_e + h_Q)^2}{(R_e + h_P)^2} $

Given $ h_P = \frac{R_e}{3} $, the ratio simplifies to:

$ \frac{g_P}{g_Q} = \frac{(R_e + h_Q)^2}{(R_e + \frac{R_e}{3})^2} = \frac{36}{25} $

Solving for $ h_Q $ in terms of $ R_e $ yields $ h_Q = \frac{3R_e}{5} $.

Given,

Mass of body $=m$

Radius $=R$

Initial height $=R$ from surface

Final height $=3 R$ from surface

Gravitational potential energy is given by at height, $h$

$ U=\frac{m g h}{\left[1+\frac{h}{R}\right]}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) $

Potential energy for $h=R$

$ U_1=\frac{m g R}{1+\frac{R}{R}}=\frac{m g R}{2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Potential energy for $h=3 R$

$ U_2=\frac{m g 3 R}{1+\frac{3 R}{R}}=\frac{3}{4} m g R \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

From Eq. (iii) and Eq. (ii) increase in potential energy is

$ \begin{aligned} U_2-U_1 & =\frac{3}{4} m g R-\frac{m g R}{2} \\ \Delta U & =\frac{m g R}{4} \end{aligned} $

We know that, orbital speed of any planet

$ v_0=\sqrt{g R} $

escape speed of any planet, $v_e=\sqrt{2 g R}$

Gravitational forces operate among all objects in the universe. All object having mass. We affected by gravitational forces exerted by another objects having mass. It has to be noted that photon are massless particles (quanta of light) yet recent discoveries of gravitational waves and lensing showed that light (photons) is also affecteded by gravitational forces.

Binding energy, $f=$ satellite is given by

$ \text { PE }=\frac{1}{2} \frac{G M m}{r}(\text { for orbit } r) $

and $\mathrm{PE}^{\prime}=\frac{1}{3} \frac{G M m}{r}$ (For orbit $\frac{3 r}{2}$ )

Increase in energy of satellite will be

$ \begin{aligned} \Delta E & =\frac{G M m}{r}\left(\frac{1}{2}-\frac{1}{3}\right) \cdot \frac{G M m}{6 r} \\ \therefore \% \text { increase } & =\frac{\frac{G M m}{6 r}}{\frac{G M m}{2 r}} \times 100 \%=33.33 \% \end{aligned} $

To determine the ratio of the escape velocities from two planets, consider the following:

Given Ratios:

Ratio of the radii of the two planets: $ r $

Ratio of the accelerations due to gravity: $ x $

Escape Velocity Formula:

The escape velocity, $ v $, is given by the formula:

$ v = \sqrt{2Rg} $

Where $ R $ is the radius of the planet and $ g $ is the acceleration due to gravity.

Applying to Two Planets:

For Planet 1:

$ v_1 = \sqrt{2R_1g_1} $

For Planet 2:

$ v_2 = \sqrt{2R_2g_2} $

Finding the Ratio of Escape Velocities:

The ratio of the escape velocities is:

$ \frac{v_1}{v_2} = \sqrt{\frac{2R_1g_1}{2R_2g_2}} $

Substitute the Given Ratios:

Use the given ratios:

$ r = \frac{R_1}{R_2} $

$ x = \frac{g_1}{g_2} $

Final Calculation:

The ratio becomes:

$ \frac{v_1}{v_2} = \sqrt{r \cdot x} $

Thus, the ratio of the escape velocities is $\sqrt{rx}$.

Given,

Mass of ring $=m$

Radius $=R$

We have to calculate the distance from its geometrical axisGravitational field is given by at point $P$.$ E_P=\frac{G M x}{\left(R^2+x^2\right)^{3 / 2}} $

To get the maximum value of $E_P$

$ \begin{aligned} & \frac{d E_p}{d x}=0 \\ & \Rightarrow \quad \frac{G M}{\left(R^2+x^2\right)^{3 / 2}}-G M x \\ & \times \frac{3}{2\left(R^2+x^2\right)^{1 / 2}} \times 2 x=0 \end{aligned} $

Solving above expression,

$ \begin{aligned} 3 x^2 & =R^2+x^2 \\ 2 x^2 & =R^2 \\ \Rightarrow \quad x & =\frac{R}{\sqrt{2}} \end{aligned} $

When the man is outside the planet and close to its surface, the gravitational force is determined by the entire mass of the planet, $M$. The force of gravity at the surface can be given by Newton's law of gravitation:

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

However, as the man passes through the hole in the planet, the gravitational force he experiences changes due to the unique mass distribution of the planet. The planet consists of a spherical shell of mass $\frac{3M}{4}$ and negligible thickness, and a point mass $\frac{M}{4}$ at the center.

Let's analyze the gravitational force when the man is inside the planet:

1. Spherical Shell Contribution:

A spherical shell of uniform density exerts no net gravitational force on a particle located inside it. Thus, the man inside the shell experiences no force from the spherical shell of mass $\frac{3M}{4}$.

2. Point Mass Contribution:

The point mass at the center, which is $\frac{M}{4}$, exerts a gravitational force on the man. According to Newton's law of gravitation, this force is given by:

$ F' = \frac{G \left(\frac{M}{4}\right) m}{R^2} $

To determine the change in the force of gravity experienced by the man, we subtract the force due to the point mass from the original force at the planet's surface:

Original force at the surface:

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

Force inside due to point mass:

$ F' = \frac{1}{4} \frac{G M m}{R^2} $

Change in the force:

$ \Delta F = F - F' $

$ \Delta F = \frac{G M m}{R^2} - \frac{1}{4} \frac{G M m}{R^2} $

Simplifying the expression:

$ \Delta F = \frac{G M m}{R^2} \left(1 - \frac{1}{4}\right) $

$ \Delta F = \frac{G M m}{R^2} \cdot \frac{3}{4} $

$ \Delta F = \frac{3}{4} \frac{G M m}{R^2} $

Hence, the change in the force of gravity experienced by the man is

$ \frac{3}{4} \frac{G M m}{R^2} $

The correct answer is Option A: $\frac{3}{4} \frac{G M m}{R^2}$.

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