Answer (d) If we assume the earth as a sphere of uniform density, then it can be treated as point mass placed at its centre. In this case acceleration due to gravity $g=0$, at the centre.

It is not so, if the earth is considered as a sphere of non-uniform density, in that case value of $g$ will be different at different points and cannot be zero at any point.

Answer (c) As observed from the earth, the sun appears to move in an approximate circular orbit. The gravitational force of attraction between the earth and the sun always follows inverse square law.

Due to relative motion between the earth and mercury, the orbit of mercury, as observed from the earth will not be approximately circular, since the major gravitational force on mercury is due to the sun.

Answer (c) As the total energy of the earth satellite bounded system is negative $(\frac{-G M}{2 a})$. where, $a$ is radius of the satellite and $M$ is mass of the earth.

Due to the viscous force acting on satellite, energy decreases continuously and radius of the orbit or height decreases gradually.

Thinking Process

The particle $B$ will move towards the greater force, between forces by $A$ and $B$.

Answer (c) Force on $B$ due to $A=F_{B A}=\frac{G(2 M m)}{(A B)^{2}}$ towards $B A$

Force on $B$ due to $C=F_{B C}=\frac{G M m}{(B C)^{2}}$ towards $B C$

As,

$(B C)=2 A B$

$\Rightarrow \quad F_{B C}=\frac{G M m}{(2 A B)^{2}}=\frac{G M m}{4(A B)^{2}}<F_{B A}$

Hence, $m$ will move towards $B A$ (i.e., $2 M$ )

Thinking Process

Acceleration due to gravity is maximum on the surface of the earth, it decreases in both cases while going upward or at a depth.

Answer $(a, c, d)$

Acceleration due to gravity at altitude $h, g_{h}=\frac{g}{(1+h / R)^{2}} \approx g(1-\frac{2 h}{R})$

At depth $d$,

$$ g_{d}=g(1-\frac{d}{R}) $$

In both cases with increase in $h$ and $d, g$ decreases.

At latitude $\phi, \quad g_{\phi}=g-\omega^{2} R \cos ^{2} \phi$

As $\phi$ increases $g_{\phi}$ increases.

Also, we can conclude from the formulae, that it is independent of mass.

Answer (a, c)

If the law of gravitation becomes an inverse cube law, then we can write, for a planet of mass $m$ revolving around the sun of mass $M$,

$$ \begin{aligned} & \qquad F=\frac{G M m}{a^{3}}=\frac{m v^{2}}{a} \quad \text { (where } a \text { is radius of orbiting planet) } \\ & \Rightarrow \quad v=\text { orbital speed }=\frac{\sqrt{G M}}{a} \Rightarrow v \propto \frac{1}{a} \\ & \text { Time period of revolution of a planet } T=\frac{2 \pi a}{v}=\frac{2 \pi a}{\frac{\sqrt{G M}}{a}}=\frac{2 \pi a^{2}}{\sqrt{G M}} \\ & \Rightarrow \quad T^{2} \propto a^{4} \end{aligned} $$

Hence, orbit will not be elliptical.

[for elliptical orbit $T^{2} \propto a^{3}$ ]

As force

where,

$$ \begin{aligned} F & =(\frac{G M}{a^{3}}) m=g^{\prime} m \\ g^{\prime} & =\frac{G M}{a^{3}} \end{aligned} $$

As $g^{\prime}$, acceleration due to gravity is constant, hence path followed by a projectile will be approximately parabolic. (as $T \propto a^{2}$ )

Answer $(a, c, d)$

Given, $\quad G^{\prime}=10 G$

Consider the adjacent diagram.

Force on the object due to the earth $=\frac{G^{\prime} M_{e} m}{R^{2}}=\frac{10 G M_{e} m}{R^{2}} \quad[\because G^{\prime}=10 G.$ given $]$

$$ \begin{aligned} & =10(\frac{G M_{e} m}{R^{2}}) \\ & =(10 g) m=10 mg \quad[\because g=\frac{G M_{e}}{R^{2}}] \end{aligned} $$

Force on the object due to the sun $F=\frac{G M_s^{\prime} m}{r^{2}}$

$$ =\frac{G(M_{s}) m}{10 r^{2}} \quad[\because M_s^{\prime}=\frac{M_{s}}{10} \text { (given) }] $$

As $r \gg R$ (radius of the earth) $\Rightarrow F$ will be very small.

So, the effect of the sun will be neglected.

Now, as $g^{\prime}=10 g$

Hence, weight of person $=m g^{\prime}=10 mg$

[from Eq. (i)]

i.e., gravity pull on the person will increase. Due to it, walking on ground would become more difficult.

Critical velocity, $v_{c}$ is proportional to $g$ i.e.

As,

$$ v_{c} \propto g $$

$\Rightarrow$

$$ g^{\prime}>g $$

Hence, rain drops will fall much faster.

To overcome the increased gravitational force of the earth, the aeroplanes will have to travel much faster.

Thinking Process

Electrostatic force of attraction acts between two opposite charges.

Answer $(a, c, d)$

Due to huge amounts of opposite charges on the sun and the earth electrostatic force of attraction will be large. Gravitational force is also attractive in nature have both forces will be added.

Both the forces obey inverse square law and are central forces. As both the forces are of same nature, hence all the three Kepler’s laws will be valid.

Answer $(c, d)$

We know that gravitational force between the earth and the sun.

$F_{G}=\frac{G M m}{r^{2}}$, where $M$ is mass of the sun and $m$ is mass of the earth.

When $G$ decreases with time, the gravitational force $F_{G}$ will become weaker with time. As $F_{G}$ is changing with time. Due to it, the earth will be going around the sun not strictly in closed orbit and radius also increases, since the attraction force is getting weaker.

Hence, after long time the earth will leave the solar system.

Answer $(a, c, d)$

Given

$F_1=-F_2=\frac{-r_{12}}{r_12^{3}} G M_0^{2}(\frac{m_1 m_2}{M_0^{2}})^{n} $

$r_{12}=r_1-r_2$

Acceleration due to gravity, $g=\frac{|F|}{\text { mass }}$

$$ =\frac{G M_0^{2}(m_1 m_2)^{n}}{r_12^{2}(M_0)^{2 n}} \times \frac{1}{(\text { mass })} $$

Since, $g$ depends upon position vector, hence it will be different for different objects. As $g$ is not constant, hence constant of proportionality will not be constant in Kepler’s third law. Hence, Kepler’s third law will not be valid.

As the force is of central nature.

$$ [\because \text { force } \propto \frac{1}{r^{2}}] $$

Hence, first two Kepler’s laws will be valid.

For negative $n$,

$$ \begin{aligned} g & =\frac{G M_0^{2}(m_1 m_2)^{-n}}{r_12^{2}(M_0)^{-2 n}} \times \frac{1}{(\text { mass })} \\ & =\frac{G M_0^{2(1+n)}}{r_12^{2}} \frac{(m_1 m_2)^{-n}}{(\text { mass })} \\ g & =\frac{G M_0^{2}}{r_12^{2}}(\frac{M_0^{2}}{m_1 m_2})^{n} \times \frac{1}{\text { mass }} \\ M_0 & >m_1 \text { or } m_2 \end{aligned} $$

As

$g>0$, hence in this case situation will reverse i.e., object lighter than water will sink in water.

Answer $(a, c)$

A geostationary satellite is having same sense of rotation as that of earth i.e., west-east direction.

A polar satellite goes around the earth’s pole in north-south direction.

Polar satellite

satellite

Answer (d)

For small objects, say of sizes less than $100 m$ centre of mass is very close with the centre of gravity of the body. But when the size of object increases, its weight changes and its CM and $C G$ become far from each other.

Answer Air molecules in the atmosphere are attracted vertically downward by gravitational force of the earth just like an apple falling from a tree. Air molecules move randomly due to their thermal velocity and hence the resultant motion of air molecules is not exactly in the vertical downward direction.

But in case of apple, only vertical motion dominates because of being heavier than air molecules. But due to gravity, the density of atmosphere increases near to the earth’s surface.

Answer Example of central force Gravitational force, electrostatic force etc.

Example of non-central force Nuclear force, magnetic force acting between two current carrying loops etc.

Answer Areal velocity of a planet revolving around the sun is constant with time. Therefore, graph between areal velocity and time is a straight line ( $A B$ ) parallel to time axis. (Kepler’s second law).

Answer Areal velocity of the earth around the sun is given by

$$ \frac{d A}{d t}=\frac{L}{2 m} $$

where, $\mathbf{L}$ is the angular momentum and $m$ is the mass of the earth.

But angular momentum

$$ \begin{aligned} \mathbf{L} & =\mathbf{r} \times \mathbf{p}=\mathbf{r} \times m \mathbf{v} \\ (\frac{d A}{d t}) & =\frac{1}{2 m}(\mathbf{r} \times m \mathbf{v})=\frac{1}{2}(\mathbf{r} \times \mathbf{v}) \end{aligned} $$

$\therefore$ Areal velocity

Therefore, the direction of areal velocity $(\frac{d A}{d t})$ is in the direction of $(r \times v)$, i.e., perpendicular to the plane containing $r$ and $v$ and directed as given by right hand rule.

Answer Yes, a body can have inertia (i.e., mass) but no weight. Everybody always have inertia (i.e., mass) but its weight $(m g)$ can be zero, when it is taken at the centre of the earth or during free fall under gravity.

e.g., In the tunnel through the centre of the earth, the object moves only due to inertia at the centre while its weight becomes zero.

Answer A body cannot be shielded from the gravitational influence of nearby matter, because gravitational force between two point mass bodies is independent of the intervening medium between them.

It is due to the above reason, we cannot shield a body from the gravitational influence of nearby matter by putting it either inside a hollow sphere or by some other me

Answer Inside a small spaceship orbitting around the earth, the value of acceleration due to gravity $g$, can be considered as constant and hence astronaut feels weightlessness.

If the space station orbitting around the earth has a large size, such that variation in $g$ matters in that case astronaut inside the spaceship will experience gravitational force and hence can detect gravity. e.g., On the moon, due to larger size gravity can be detected.

Answer Consider the diagram, density of the shell is constant. Let it is $\rho$.

Mass of the shell $=($ density $) \times($ volume $)$

$$ =(\rho) \times \frac{4}{3} \pi R^{3}=M $$

As the density of the shell is uniform, it can be treated as a point mass placed at its centre. Therefore, $F=$ gravitational force between $M$ and $m=\frac{G M m}{r^{2}}$

$$ \begin{aligned} F & =0 \text { for } r<R \quad \text { (i.e., force inside the shell is zero) } \\ & =\frac{G M}{r^{2}} \text { for } r \geq R \end{aligned} $$

The variation of $F$ versus $r$ is shown in the diagram.

Note When $r$ tends to infinity, force tends to zero, also force is maximum on the surface of the hollow spherical shell.

Answer Aphelion is the location of the earth where it is at the greatest distance from the sun and perihelion is the location of the earth where it is at the nearest distance from the sun.

The areal velocity $(\frac{1}{2} \mathbf{r} \times \mathbf{v})$ of the earth around the sun is constant (Kepler’s Ilnd law).

Therefore, the speed of the earth is more at the perihelion than at the aphelion.

Answer Consider the diagram where plane of geostationary and polar satellite are shown.

Clearly

(a) Angle between the equatorial plane and orbital plane of a polar satellite is $90^{\circ}$.

(b) Angle between equatorial plane and orbital plane of a geostationary satellite is $0^{\circ}$.

Answer Consider the diagram below, the earth moves from the point $P$ to $Q$ in one solar day.

Every day the earth advances in the orbit by approximately $1^{\circ}$. Then, it will have to rotate by $361^{\circ}$ (which we define as 1 day) to have the sun at zenith point again.

$\because 361^{\circ}$ corresponds to $24 h$.

$\therefore \quad 1^{\circ}$ corresponds to $\frac{24}{361} \times 1=0.066 h=3.99 min \approx 4 min$

Hence, distant stars would rise 4 min early every successive day.

Thinking Process

To determine the nature of equilibrium, we have to displace the object through a small distance, from the middle point and then force will be calculated in displaced position.

Answer Let the mass and radius of each identical heavy sphere be $M$ and $R$ respectively. An object of mass $m$ be placed at the mid-point $P$ of the line joining their centres.

Force acting on the object placed at the mid-point,

$$ F_1=F_2=\frac{G M m}{(5 R)^{2}} $$

The direction of forces are opposite, therefore net force acting on the object is zero.

To check the stability of the equilibrium, we displace the object through a small distance $x$ towards sphere $A$.

Now, force acting towards sphere $A, F_1^{\prime}=\frac{G M m}{(5 R-x)^{2}}$

Force acting towards sphere $B, \quad F_2^{\prime}=\frac{G M m}{(5 R+x)^{2}}$

As $F_1{ }^{\prime}>F_2{ }^{\prime}$, therefore a resultant force $(F_1{ }^{\prime}-F_2{ }^{\prime})$ acts on the object towards sphere $A$, therefore object start to move towards sphere $A$ and hence equilibrium is unstable.

Answer Consider the diagram, where a satellite of mass $m$, moving around the earth in a circular orbit of radius $R$.

Orbital speed of the satellite orbitting the earth is given by $v_0=\sqrt{\frac{G M}{R}}$ where, $M$ and $R$ are the mass and radius of the earth.

(a) $\therefore$ KE of a satellite of mass $m, E_{K}=\frac{1}{2} m v_0^{2}=\frac{1}{2} m \times \frac{G M}{R}$

$\therefore \quad E_{K} \propto \frac{1}{R}$

It means the KE decreases exponentially with radius.

The graph for KE versus orbital radius $R$ is shown in figure.

(b) Potential energy of a satellite $E_{P}=-\frac{G M m}{R}$

$$ E_{P} \propto-\frac{1}{R} $$

The graph for PE versus orbital radius $R$ is shown in figure.

(c) Total energy of the satellite $E=E_{K}+E_{P}=\frac{G M m}{2 R}-\frac{G M m}{R}$

$$ =-\frac{G M m}{2 R} $$

The graph for total energy versus orbital radius $R$ is shown in the figure.

Note We should keep in mind that Potential Energy (PE) and Kinetic Energy (KE) of the satellite-earth system is always negative.

Answer The trajectory of a particle under gravitational force of the earth will be a conic section (for motion outside the earth) with the centre of the earth as a focus. Only (c) meets this requirement.

Note The trajectory of the particle depends upon the velocity of projection. Depending upon the magnitude and direction of velocity it may be parabolic or elliptical.

Answer Consider the diagram where an object of mass $m$ is raised from the surface of the earth to a distance (height) equal to the radius of the earth $(R)$.

Potential energy of the object at the surface of the earth $=-\frac{G M m}{R}$

$P E$ of the object at a height equal to the radius of the earth $=-\frac{G M m}{2 R}$

$\therefore \quad$ Gain in PE of the object $=\frac{-G M m}{2 R}-(-\frac{G M m}{R})$

$$ \begin{aligned} & =\frac{-G M m+2 G M m}{2 R}=+\frac{G M m}{2 R} \\ & =\frac{g R^{2} \times m}{2 R}=\frac{1}{2} m g R \end{aligned} $$

Answer Consider the diagram, in which a system consisting of a ring and a point mass is shown.

Gravitational force acting on an object of mass $m$, placed at point $P$ at a distance $h$ along the normal through the centre of a circular ring of mass $M$ and radius $r$ is given by

$$ F=\frac{G M m h}{(r^{2}+h^{2})^{3 / 2}} $$

When mass is displaced upto distance $2 h$, then

$$ \begin{matrix} F^{\prime} & =\frac{G M m \times 2 h}{[r^{2}+(2 h)^{2}]^{3 / 2}} & {[\because h=2 r]} \\ & =\frac{2 G M m h}{(r^{2}+4 h^{2})^{3 / 2}} & \ldots \text { (ii) } \end{matrix} $$

When $h=r$, then from Eq.(i)

and

$$ F=\frac{G M m \times r}{(r^{2}+r^{2})^{3 / 2}} \Rightarrow F=\frac{G M m}{2 \sqrt{2 r^{2}}} $$

$$ \begin{matrix} \therefore & \frac{F^{\prime}}{F}=\frac{4 \sqrt{2}}{5 \sqrt{5}} \\ \Rightarrow & F^{\prime}=\frac{4 \sqrt{2}}{5 \sqrt{5}} F \end{matrix} $$

$$ F^{\prime}=\frac{2 G M m r}{(r^{2}+4 r^{2})^{3 / 2}}=\frac{2 G M m}{5 \sqrt{5 r^{2}}} \quad \text { [From Eq. (ii) substituting } h=r \text { ] } $$

Answer The situation is shown in the diagram, where a body of mass $m$ is revolving around a star of mass $M$.

Linear velocity of the body

$$ \Rightarrow \quad v \propto \frac{1}{\sqrt{r}} $$

Therefore, when $r$ increases, $v$ decreases.

Angular velocity of the body

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

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

According to Kepler’s law of period,

$$ T^{2} \propto r^{3} \Rightarrow T=k r^{3 / 2} $$

where $k$ is a constant

$$ \therefore \quad \omega=\frac{2 \pi}{k r^{3 / 2}} \Rightarrow \omega \propto \frac{1}{r^{3 / 2}} \quad(\because \omega=\frac{2 \pi}{T}) $$

Therefore, when $r$ increases, $\omega$ decreases.

Kinetic energy of the body $K=\frac{1}{2} m v^{2}=\frac{1}{2} m \times \frac{G M}{r}=\frac{G M m}{2 r}$

$$ \therefore \quad K \propto \frac{1}{r} $$

Therefore, when $r$ increases, KE decreases.

Gravitational potential energy of the body,

$$ U=-\frac{G M m}{r} \Rightarrow U \propto-\frac{1}{r} $$

Therefore, when $r$ increases, PE becomes less negative i.e., increases.

Total energy of the body

$$ E=KE+PE=\frac{GMm}{2 r}+(-\frac{GMm}{r})=-\frac{GMm}{2 r} $$

Therefore, when $r$ increases, total energy becomes less negative, i.e., increases.

Angular momentum of the body $L=m v r=m r \sqrt{\frac{G M}{r}}=m \sqrt{G M r}$

$$ L \propto \sqrt{r} $$

Therefore, when $r$ increases, angular momentum $L$ increases.

Note In this case, we have not considered the sun-object system as isolated and the force on the system is not zero. So, angular momentum is not conserved.

Thinking Process

To calculate resultant force, we will apply principle of superposition i.e., net force will be equal to sum of individual forces by each point mass (m).

Answer Consider the diagram below, in which six point masses are placed at six vertices $A, B, C, D, E$ and $F$.

Force on mass $m$ at $A$ due to mass $m$ at $B$ is, $f_1=\frac{G m m}{l^{2}}$ along $A B$.

Force on mass $m$ at $A$ due to mass $m$ at $C$ is, $f_2=\frac{G m \times m}{(\sqrt{3} l)^{2}}=\frac{G m^{2}}{3 l^{2}}$ along $A C$.

$[\because A C=\sqrt{3 l}]$

Force on mass $m$ at $A$ due to mass $m$ at $D$ is, $f_3=\frac{G m \times m}{(2 l)^{2}}=\frac{G m^{2}}{4 l^{2}}$ along $A D \cdot[\because A D=2 l]$

Force on mass $m$ at $A$ due to mass $m$ at $E$ is, $f_4=\frac{G m \times m}{(\sqrt{3} l)^{2}}=\frac{G m^{2}}{3 l^{2}}$ along $A E$.

Force on mass $m$ at $A$ due to mass $m$ at $F$ is, $f_5=\frac{G m \times m}{l^{2}}=\frac{G m^{2}}{l^{2}}$ along $A F$.

Resultant force due to $f_1$ and $f_5$ is, $F_1=\sqrt{f_1^{2}+f_5^{2}+2 f_1 f_5 \cos 120^{\circ}}=\frac{G m^{2}}{l^{2}}$ along $A D$.

$[\because.$ Angle between $f_1$ and $.f_5=120^{\circ}]$

Resultant force due to $f_2$ and $f_4$ is, $F_2=\sqrt{f_2^{2}+f_4^{2}+2 f_2 f_4 \cos 60^{\circ}}$

$$ =\frac{\sqrt{3} G m^{2}}{3 l^{2}}=\frac{G m^{2}}{\sqrt{3} l^{2}} \text { along } A D $$

So, net force along $A D=F_1+F_2+F_3=\frac{G m^{2}}{l^{2}}+\frac{G m^{2}}{\sqrt{3} l^{2}}+\frac{G m^{2}}{4 l^{2}}=\frac{G m^{2}}{l^{2}}(1+\frac{1}{\sqrt{3}}+\frac{1}{4})$.

Answer Consider the adjacent diagram

Given,

$$ \text { mass of the earth } M=6 \times 10^{24} kg $$

Radius of the earth, $R=6400 km=6.4 \times 10^{6} m$

Time period $T=24 h$

$$ \begin{aligned} & =24 \times 60 \times 60=86400 s \\ G & =6.67 \times 10^{-11} N-m^{2} / kg^{2} \end{aligned} $$

(a) Time period

$$ T=2 \pi \sqrt{\frac{(R+h)^{3}}{G M}} \quad[\because v_0=\sqrt{\frac{G M}{R+h}} \text { and } T=\frac{2 \pi(R+h)}{v_0}] $$

$$ \begin{aligned} \Rightarrow \quad T^{2} & =4 \pi^{2} \frac{(R+h)^{3}}{G M} \Rightarrow(R+h)^{3}=\frac{T^{2} G M}{4 \pi^{2}} \\ \Rightarrow \quad R+h & =(\frac{T^{2} G M}{4 \pi^{2}})^{1 / 3} \Rightarrow h=(\frac{T^{2} G M}{4 \pi^{2}})^{1 / 3}-R \\ \Rightarrow \quad h & =[\frac{(24 \times 60 \times 60)^{2} \times 6.67 \times 10^{-11} \times 6 \times 10^{24}}{4 \times(3.14)^{2}}]^{1 / 3}-6.4 \times 10^{6} \\ & =4.23 \times 10^{7}-6.4 \times 10^{6} \\ & =(42.3-6.4) \times 10^{6} \\ & =35.9 \times 10^{6} m \\ & =3.59 \times 10^{7} m \end{aligned} $$

(b) If satellite is at height $h$ from the earth’s surface, then according to the diagram.

$ \begin{aligned} \cos \theta & =\frac{R}{R+h}=\frac{1}{(1+\frac{h}{R})}=\frac{1}{(1+\frac{3.59 \times 10^{7}}{6.4 \times 10^{6}})} \\ & =\frac{1}{1+5.61}=\frac{1}{6.61}=0.1513=\cos 81918^{\prime} \\ \therefore \quad \theta & =81918^{\prime} \\ \therefore \quad 2 \theta & =2 \times(81^{\prime} 18^{\prime})=162^{\circ} 36^{\prime} \end{aligned} $

If $n$ is the number of satellites needed to cover entire the earth, then

$$ n=\frac{360^{\circ}}{2 \theta}=\frac{360^{\circ}}{162^{\circ} 36^{\prime}}=2.31 $$

$\therefore \quad$ Minimum 3 satellites are required to cover entire the earth.

Thinking Process

As the earth orbits the sun, the angular momentum is conserved and areal velocity is constant.

Answer Consider the diagram. Let $m$ be the mass of the earth, $v_{p}, v_{a}$ be the velocity of the earth at perigee and apogee respectively. Similarly, $\omega_{p}$ and $\omega_{a}$ are corresponding angular velocities.

Angular momentum and areal velocity are constant as the earth orbits the sun.

At perigee, $r_p^{2} \omega_{p}=r_a^{2} \omega_{a}$ at apogee

If $a$ is the semi-major axis of the earth’s orbit, then $r_{p}=a(1-e)$ and $r_{a}=a(1+e)$

$$ \begin{matrix} \therefore & \frac{\omega_{p}}{\omega_{a}}=(\frac{1+e}{1-e})^{2}, e=0.0167 & \text { [from Eqs. (i) and (ii)] } \\ \therefore & \frac{\omega_{p}}{\omega_{a}}=1.0691 \end{matrix} $$

Let $\omega$ be angular speed which is geometric mean of $\omega_{p}$ and $\omega_{a}$ and corresponds to mean solar day,

$$ \begin{aligned} & \therefore & (\frac{\omega_{p}}{\omega})(\frac{\omega}{\omega_{a}}) & =1.0691 \\ & \therefore & \frac{\omega_{p}}{\omega} & =\frac{\omega}{\omega_{a}}=1.034 \end{aligned} $$

If $\omega$ corresponds to $1^{\circ}$ per day (mean angular speed), then $\omega_{p}=1.034^{\circ}$ per day and $\omega_{a}=0.967^{\circ}$ per day. Since, $361^{\circ}=24$, mean solar day, we get $361.034^{\circ}$ which corresponds to $24 h, 8.14^{\prime \prime}$ (8.1" longer) and $360.967^{\circ}$ corresponds to 23 h 59 min 52 " ( $7.9^{\prime \prime}$ smaller).

This does not explain the actual variation of the length of the day during the year.

Answer Given,

$$ r_{p}=\text { radius of perihelion }=2 R $$

$$ r_{a}=\text { radius of apnelion }=6 R $$

Hence, we can write

$$ \begin{aligned} & r_{a}=a(1+e)=6 R \\ & r_{p}=a(1-e)=2 R \end{aligned} $$

Solving Eqs. (i) and (ii), we get

$$ \text { eccentricity, } e=\frac{1}{2} $$

By conservation of angular momentum, angular momentum at perigee $=$ angular momentum at apogee

$$ \begin{aligned} & \therefore \quad m v_p r_p=m v_a r_a \ & \therefore \quad \frac{v_a}{v_p}=\frac{1}{3} \ & \end{aligned} $$

where $m$ is mass of the satellite.

Applying conservation of energy, energy at perigee $=$ energy at apogee

$$ \frac{1}{2} m v_p^{2}-\frac{G M m}{r_{p}}=\frac{1}{2} m v_a^{2}-\frac{G M m}{r_{a}} $$

where $M$ is the mass of the earth.

For circular orbit of radius $r$,

$$ \begin{aligned} v_p^{2}(1-\frac{1}{9}) & .=-2 G M(\frac{1}{r_{a}}-\frac{1}{r_{p}})=2 G M(\frac{1}{r_{p}}-\frac{1}{r_{a}}) \quad \text { (By putting } v_{a}=\frac{v_{p}}{3}) \\ v_{p} & =\frac{[2 G M(\frac{1}{r_{p}}-\frac{1}{r_{a}})]^{1 / 2}}{[1-(v_{a} / V_{p})^{2}]^{1 / 2}}=[\frac{\frac{2 G M}{R}(\frac{1}{2}-\frac{1}{6})}{(1-\frac{1}{9})}]^{1 / 2} \\ & =(\frac{2 / 3}{8 / 9} \frac{G M}{R})^{1 / 2}=\sqrt{\frac{3}{4}} \frac{G M}{R}=6.85 km / s \\ v_{p} & =6.85 km / s, v_{a}=2.28 km / s \end{aligned} $$

For

$$ \begin{aligned} v_{c} & =\text { orbital velocity }=\sqrt{\frac{G M}{r}} \\ r & =6 R, v_{c}=\sqrt{\frac{G M}{6 R}}=3.23 km / s . \end{aligned} $$

Hence, to transfer to a circular orbit at apogee, we have to boost the velocity by $\Delta=(3.23-2.28)=0.95 km / s$. This can be done by suitably firing rockets from the satellite.