NCERT Exemplar Class 11 Physics Solutions Chapter 8 Gravitation

Question:8.2

As observed from earth, the sun appears to move in an approximately circular orbit. For the motion of another planet like mercury as observed from earth, this would
a) be similarly true
b) not be true because the force between Earth and Mercury is not the inverse square law
c) not be true because the major gravitational force on Mercury is due to Sun
d) not be true because mercury is influenced by forces other than gravitational forces

Answer:

The answer is option (c). The universal law of force of attraction is inversely dependent on the distance between them. For mercury, this force due to the sun is exceptionally larger than that due to Earth. Due to the relative motion of Sun and Earth with Mercury, this would not be true.

Question:8.3

Different points in the earth are at slightly different distances from the sun and hence experience different forces due to gravitation. For a rigid body, we know that if various forces act at various points in it, the resultant motion is as if a net force acts on the COM, causing translation and a net torque at the COM causing rotation around an axis through the COM. For the Earth-Sun system
a) the torque is zero
b) the torque causes the earth to spin
c) the rigid body result is not applicable since the earth is not even approximately a rigid body
d) the torque causes the earth to move around the sun

Answer:

The answer is option (a). The direction of the gravitational force F and the line joining the centre of mass of the Earth (point of application of force) both lie in the same direction. So, the angle between them is 0.
In the formula :$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}=rF\; \sin\; \theta$
$\sin\; 0=0$
Hence, torque is zero.

Question:8.4

Satellites orbiting the earth have a finite life and sometimes debris of satellites fall to the earth. This is because
a) the solar cells and batteries in satellites run out
b) the laws of gravitation predict a trajectory spiralling inwards
c) of viscous forces causing the speed of the satellite and hence height to gradually decrease
d) of collisions with other satellites

Answer:

The answer is option (c)
P.E is given by $\frac{-GM}{2r}$
The negative sign shows the force of attraction between the satellite and Earth.
P.E gets gradually reduced due to atmospheric friction, and the radius/ height of the satellite keeps decreasing until it comes back to earth with increasing speed and burns in the atmosphere.

Question:8.5

Both earth and moon are subject to the gravitational force of the sun. as observed from the sun, the orbit of the moon
a) will be elliptical
b) will not be strictly elliptical because the total gravitational force on it is not central
c) is not elliptical but will necessarily be a closed curve
d) deviates considerably from being elliptical due to the influence of planets other than earth

Answer:

The answer is option (b) Moon is not always on the line joining the Sun and Earth. Therefore, the gravitational force of attraction due to Earth and Sun have different lines of action, or let us say the forces are not central. So, the orbit of the moon will not be strictly elliptical.

Question:8.6

In our solar system, the inter-planetary region has chunks of matter called asteroids. They
a) will not move around the sun since they have very small masses compared to the sun
b) will move in an irregular way because of their small masses and will drift away outer space
c) will move around the sun in closed orbits but not obey Kepler’s laws
d) will move in orbits like planets and obey Kepler’s laws

Answer:

The answer is option (d). Central gravitational forces act upon asteroids. Hence, they will move in orbits and obey Kepler's law.

Question:8.7

Choose the wrong option.
a) inertial mass is a measure of the difficulty of accelerating a body by an external force whereas the gravitational mass is relevant in determining the gravitational force on it by an external mass
b) that the gravitational mass and inertial mass are equal is an experimental result
c) that the acceleration due to gravity on earth is the same for all bodies is due to the equality of gravitational mass and inertial mass
d) gravitational mass of a particle-like proton can depend on the presence of neighbouring heavy objects, but the inertial mass cannot

Answer:

The answer is option (d). The gravitational mass of a proton is equivalent to its inertial mass & is independent of the presence of neighbouring heavy objects.

Question:8.8

Particles of masses 2M, m and M are respectively at points A, B, and C with $AB=\frac{1}{2}(BC).$. M is much-much smaller than 2M and at time $t=0$, they are all at rest. At subsequent times before any collision takes place.
a) m will remain at rest
b) m will move towards M
c) m will move towards 2M
d) m will have oscillatory motion

Answer:

The answer is option (c) m will move towards 2M.
Let $F_{BC}$ be the force experienced by m at point B due to M at point C, and $F_{BA}$ be the force experienced by m at pointA due to 2 M at point A
$F_{Bc}=\frac{GmM}{Bc^{2}}$
$F_{AB}=\frac{Gm2M}{AB^{2}}$
If $AB=x$ then $BC=2x$
$BC^{2}>AB^{2}$
$F_{BA}>F_{BC}$
Thus, m will move towards point A.

Question:8.10

If the law of gravitation, instead of being an inverse-square law, becomes an inverse-cube law
a) planets will not have elliptic orbits
b) circular orbits of planets is not possible
c) projectile motion of a stone thrown by hand on the surface of the earth will be approximately parabolic
d) there will be no gravitational force inside a spherical shell of uniform density

Answer:

The correct answers are options (a), (b), and (c).
We know that the Force of gravitation = Centrifugal force for a body in a circular motion
If the law of gravitation becomes an inverse cube law, then
$\frac{GmM}{r^{3}}=\frac{v^{2}m}{r}$
where m = mass of the planet, M = mass of the sun and r = radius of the orbit
Thus, v is inversely proportional to r
The time period of a planet around the sun $=\frac{2\pi r}{v}$
$T^2\propto r^4$
So the path is not elliptical. (By Kepler's law $T^2\propto r^3$)
So, a is correct
When $\frac{GmM}{r^{3}}=mg'$
$g'=\frac{GM}{r^{2}}$
Hence, b is incorrect
Since $T\alpha r^{2}$, the path of the projectile will be parabolic. Thus, (c) is correct
Gravitational force is a universal force as it acts everywhere. Hence, (d) is incorrect

Question:8.11

If the mass of sun were ten times smaller and gravitational constant G were ten times larger in magnitudes
a) walking on ground would become more difficult
b) the acceleration due to gravity on earth will not change
c) raindrops will fall much faster
d) aeroplanes will have to travel much faster

Answer:

The correct answers are options (a), (c), and (d).
If $G'=10G$
And $M'_{s}=\frac{M_{s}}{10}$
Then $g'=\frac{G'M_{E}}{r^{2}}=\frac{10GM}{r^{2}}=10g$
Weight of a person =$mg'=10 mg$
Hence, a is correct, and b is incorrect
F on man due to sun $=\frac{GM'_{s}m}{r^{2}}=\frac{GM_{s}m}{10r^{2}}$
The terminal velocity $v_{T}\alpha g'\; \; \; or\; \; \; \; v_{T}\; \alpha 10g$
Hence, raindrops fall 10 times faster. Option c is correct
In order to maintain the speed after the increase in gravitational force, the aeroplane will have to travel faster.

Question:8.12

If the sun and the planets carried huge amounts of opposite charges,
a) all three of Kepler’s laws would still be valid
b) only the third law will be valid
c) the second law will not change
d) the first law will still be valid

Answer:

The correct answers are options (a), (c), and (d). Due to opposite charges, large electrostatic forces of attraction will be produced in addition to the gravitational forces. Both forces, when added, will be radial in nature. Since they are central forces and obey the inverse square law, all 3 Kepler's laws will be valid.

Question:8.13

There have been suggestions that the value of the gravitational constant G becomes smaller when considered over an exceptionally large time period in the future. If that happens for our earth,
a) nothing will change
b) we will become hotter after billions of years
c) we will be going around but not strictly in closed orbits
d) after a sufficiently long time we will leave the solar system

Answer:

The correct answers are options (c) and (d)
Force of gravitation = Centrifugal force for a body in a circular motion
$F=\frac{GmM}{r^{2}}$
If G becomes smaller, the centripetal force also decreases
We know that $v_{o}\alpha \sqrt{G}$. Thus, the centrifugal force becomes larger than the centripetal force $(F\; \alpha \; G)$
Distance from the sun increases, Earth will move in a larger orbit over time.
C, D are correct.

Question:8.14

Supposing Newton’s law of gravitation for gravitation forces F₁ and F₂ between two masses m₁ and m₂ at positions r₁ and r₂ read
$\mathbf{F_{1}}=\mathbf{-F_{2}}=\frac{\mathbf{r_{12}}}{r_{12}^{3}}GM_{0}^{2}\left ( \frac{m_{1}m_{2}}{M_{1}^{2}} \right )^{n}$ where Mo is a constant of the dimension of mass, $\mathbf{r_{12}} = \mathbf{r_{1}} - \mathbf{r_{2}}$ and n is a number. In such a case,
a) the acceleration due to gravity on earth will be different for different object
b) none of the three laws of Kepler will be valid
c) only the third law will become invalid
d) for n negative, an object lighter than water will sink in water

Answer:

The correct answers are options (a) and (c)
$\overrightarrow{F_{1}}=\overrightarrow{F_{2}}=\frac{\overrightarrow{r_{12}}}{r_{12}^{3}}GM_{0}^{2}\left [ \frac{m_{1}m_{2}}{M_{0}^{2}} \right ]^{n}$
given, $\overrightarrow{r_{12}}=\overrightarrow{r_{1}}-\overrightarrow{r_{2}}$
Acceleration due to gravity (g) =
$\frac{\left | F \right |}{mass (m)}$
$g=\left | \frac{GM_{0}^{2}(m_{1}m_{2})^{n}}{r_{12}^{2}M_{0}^{2n}}\times\frac{\hat{r}_{12}}{m} \right |$
$\hat{r}_{12}$ tells the direction of g from 1 to 2
$g= \frac{GM_{0}^{2}(m_{1}m_{2})^{n}}{r_{12}^{2}M_{0}^{2n}}\; \frac{1}{n}$
Since g is dependent on the position vector

$\overrightarrow{r_{12}}$, mass of body m. So, g on eath will be different for different bodies of different mass and their
Position. Hence, a and c are correct
As the force is central in nature, Kepler's first and second laws are valid, rejecting option b.
$g= \frac{GM_{0}^{2}(m_{1}m_{2})^{-n}}{r_{12}^{2}M_{0}^{-2n}}\; \frac{1}{n}=\frac{GM_{0}^{2}M_{0}^{2n}}{(m_{1}m_{2})^{n}r_{12}^{2}}\; \frac{1}{m}=\frac{GM_{0}^{2}}{r_{12}^{2}}\left [ \frac{M_{0}^{2}}{m_{1}m_{2}} \right ]\; \frac{1}{m}$
$g>0$
So $M_{0}>m_{1}\; or \; m_{2}$
The lighter object can sink in water. Hence, d is correct

Question:8.15

Which of the following are true?
a) a polar satellite goes around the earth’s pole in a north-south direction
b) a geostationary satellite goes around the earth in an east-west direction
c) a geostationary satellite goes around the earth in a west-east direction
d) a polar satellite goes around the earth in an east-west direction

Answer:

The correct answers are options (a) and (c). A geostationary satellite appears stationary relative to the earth and revolves around the earth from west to east with the same angular velocity as the earth's rotation about its axis from west to east. A polar satellite revolves around the Earth from north to south, independent of the Earth's rotation.

Question:8.16

The centre of mass of an extended body on the surface of the earth and its centre of gravity
a) are always at the same point for any size of the body
b) are always at the same point only for spherical bodies
c) can never be at the same point
d) is close to each other for objects, say of sizes less than 100 m
e) both can change if the object is taken deep inside the earth

Answer:

The correct answer is option (d). For smaller bodies, the COM (Centre of mass) and COG (Centre of gravity) usually coincide. For extended objects like mountains and lakes, they are far away. COG is the point in a body where the net torque due to gravity is 0. The COM is where the whole mass of a body is supposed to be concentrated.

Question:8.18

Give one example each of central force and non-central force.

Answer:

Gravitational force, electrostatic force and spring force are examples of central force. Magnetic forces between 2 current-carrying loops are an example of a non-central force

Question:8.19

Draw areal velocity versus time graph for mars.

Answer:

According to Kepler's second law, a planet sweeps an equal area in equal times. So the areal velocity is constant with respect to time.

Question:8.20

What is the direction of areal velocity of the earth around the sun?

Answer:

Real velocity of the earth around the sun is given by $\frac{dA}{dt}=\frac{L}{2m}$
Where L is the angular momentum, and m isthe mass of Earth
Since $\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}=\overrightarrow{r}\times m\overrightarrow{v}$
$\frac{dA}{dt}=\frac{\vec{r}\times m\vec{v}}{2m}=\frac{(\vec{r}\times\vec{v})}{2}$
Thus, the direction of areal velocity is perpendicular to the plane of $\vec{r}$ and $\vec{v}$ (Maxwell's right-hand rule)

Question:8.21

How is the gravitational force between two point masses affected when they are dipped in the water keeping the separation between them the same?

Answer:

The force of attraction varies inversely with the distance between two objects. Since the separation between two point masses is the same as the force of attraction also remains the same.

Question:8.22

Is it possible for a body to have inertia but no weight?

Answer:

Everybody has some mass or inertia. The weight of a body is equal to mg. It is possible for a body to have no weight when g=0 and hence mg=0, which is possible in space where there is no gravity.

Question:8.23

We can shield a charge from electric fields by putting it inside a hollow conductor. Can we shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?

Answer:

The electric force depends on the nature of the medium, whereas the gravitational force is independent of the medium. So, we cannot shield a body from the gravitational force.

Question:8.24

An astronaut inside a small spaceship orbiting around the Earth cannot detect gravity. If the space station orbiting around the Earth has a large size, can it hope to detect gravity?

Answer:

If the space station orbiting around the Earth has a large size, then the astronaut inside can experience acceleration due to gravity at a very close proximity to the moon.

Question:8.25

The gravitational force between a hollow spherical shell and a point mass is F. Show the nature of F vs r graph where r is the distance of the point from the centre of the hollow spherical shell of uniform density.

Answer:

The gravitational force inside the hollow spherical shell is 0, and on the surface is $F_{0}$
$F_{0}=\frac{GmM}{r^{2}}$ for $r\ge R$ and $F_{0}=0$ for $0<r<R$

Question:8.27

What is the angle between the equatorial plane and the orbital plane of
a) polar satellite?
b) geostationary satellite?

Answer:

(a) The orbital plane of a polar satellite makes $90^{o}$ with the equatorial plane.
(b) A geostationary satellite makes $0^{o}$ with the equatorial plane as it revolves from west to east.

Question:8.26

Out of aphelion and perihelion, where is the speed of the earth more and why?

Answer:

According to Kepler's second law
$\frac{dA}{dt}=\frac{L}{2m}=\frac{r\; *\; p}{2m}=\frac{r\; *\: mv}{2m}=\frac{r\; *\; v}{2}$
Since “r” at aphelion is larger, the velocity decreases, and “r” at perihelion is smaller, the velocity increases.

Question:8.29

Two identical heavy spheres are separated by a distance 10 times their radius. Will an object placed at the midpoint of the line joining their centres be in stable equilibrium or unstable equilibrium? Give a reason for your answer.

Answer:

$m_{1}=m_{2}=M$
$r=10\; R$

Let m be placed at the midpoint P of the line joining the centres of spheres A and B
$\left | F_{2} \right |=\left | F_{1} \right |=\frac{GMm}{(5R)^{2}}=\frac{GMm}{25R^{2}}$
Net force $F_{1}+F_{2}=0$
$F_{1}=-F_{2}$
If m is displaced slightly by x from P to A
$PA=(5R-x)\; and\; PB=(5R+x)$
$F_{1}=\frac{GMm}{(5R-x)^{2}}$
$F_{2}=\frac{GMm}{(5R+x)^{2}}$
$F_{2}<F_{1}$
This results in an unstable equilibrium.

Question:8.30

Show the nature of the following graph for a satellite orbiting the earth.
a) KE vs orbital radius R
b) PE vs orbital radius R
c) TE vs orbital radius R

Answer:

Mass of Earth = $M_{E}$
Radius of orbit of satellite = R
mass of satellite=m
$v_{0}=\sqrt{\frac{GM}{R}}$
$E_{k}=\frac{1}{2}mv_{0}^{2}=\frac{1}{2}m\frac{GM}{R}=\frac{GMm}{2R}$
$E_{k}\; \alpha \; \frac{1}{R}$
$E_{p}=-\frac{GMm}{R}$
$E_{p}\; \alpha \; -\frac{1}{R}$
$TE=\frac{1}{2}\left ( \frac{GMm}{R} \right )-\frac{GMm}{R}=-\frac{1}{2}\left ( \frac{GMm}{R} \right )$

Question:8.31

Shown are several curves. Explain with reason, which ones amongst them can be possible trajectories traced by a projectile

Answer:

The trajectory of a projectile under the gravitational force of Earth should be a conic section or parabolic, or elliptical with focus at the centre of the Earth. Hence, c is correct.

Question:8.32

An object of mass m is raised from the surface of the earth to a height equal to the radius of the earth, that is, taken from a distance R to 2R from the centre of the earth. What is the gain in its potential energy?

Answer:

Let $PE_{i}$ be the initial P.E when mass is at the surface of Earth
Let $PE_{f}$ be the final P.E when mass is lifted to a height equal to R
$PE_{i}=-\frac{GMm}{R}$
$PE_{f}=-\frac{GMm}{2R}$
Gain in $PE=PE_{f}-PE_{i}=-\frac{GMm}{2R}-\left (-\frac{GMm}{R} \right )=\frac{GMm}{2R}$
$GM=gR^{2}$
Gain in $PE=\frac{1}{2}mg\; R$

Question:8.33

A mass m is placed at P a distance h along the normal through the centre O of a thin circular ring of mass M and radius r. If the mass is removed further away such that OP becomes 2h, by what factor will the force of gravitation decrease, if h = r?

Answer:

Let us consider a small element of a ring of mass dM at point A, and the distance between dM and m is x.
$x^{2}=r^{2}+h^{2}$
The gravitational force between dM and m, $dF=\frac{G(dM)m}{x^{2}}$
dF has 2 components: one along PO and another perpendicular to PO
$\int \; \sin \theta =0$ (due to symmetry of ring)
$F=\int dF\; \cos \theta =\int \frac{G(dM)m}{x^{2}}\frac{h}{x}$
$=\frac{Gmh}{x^{3}}\int dM=\frac{GMmh}{x^{3}}=\frac{GMmh}{(r^{2}+h^{2})^{\frac{3}{2}}}$
When the mass is displaced to a distance of 2 h $F'=\frac{GMm(2h)}{(r^{2}+(2h)^{2})^{\frac{3}{2}}}=\frac{2GMmh}{(r^{2}+4h^{2})^{\frac{3}{2}}}$
Where h =r $F=\frac{GMmr}{(r^{2}+r^{2})^{\frac{3}{2}}}=\frac{GMm}{2\sqrt{2}r^{2}}$
$F'=\frac{GMmr}{(5r^{2})^{\frac{3}{2}}}=\frac{2GMm}{5\sqrt{5}r^{2}}$
$\frac{F^{'}}{F}=\frac{4\sqrt{2}}{5\sqrt{5}}$
$F'=\frac{4\sqrt{2}}{5\sqrt{5}}F$

Question:8.35

Six-point masses of mass m each are at the vertices of a regular hexagon of side l. Calculate the force on any of the masses.

Answer:

Force on A due to B$=f_{1}=\frac{Gmm}{l^{2}}=\frac{Gm^{2}}{l^{2}}$ along B to A

Force on A due to C$=f_{2}=\frac{Gmm}{(\sqrt{3}l)^{2}}=\frac{Gm^{2}}{3l^{2}}$ along C to A

Force on A due to D $=f_{3}=\frac{Gmm}{(2l)^{2}}=\frac{Gm^{2}}{4l^{2}}$ along D to A

Force on A due to E $=f_{4}=\frac{Gmm}{(\sqrt{3}l)^{2}}=\frac{Gm^{2}}{3l^{2}}$ along E to A

$F=F_{1}+F_{2}+F_{3}=\frac{Gm^{2}}{l^{2}}+\frac{Gm^{2}}{\sqrt{3}l^{2}}+\frac{Gm^{2}}{4l^{2}}$ along DA

Question:8.36

A satellite is to be placed in equatorial geostationary orbit around the earth for communication
a) calculate height of such a satellite
b) find out the minimum number of satellites that are needed to cover entire earth so that at least one satellites is visible from any point on the equator

Answer:

$M=6\times10^{24}\; kg$
$R=6400\; km=6.4\times10^{6}m$
$Time \; Period \; (T)=24h=24\times3600s=24\times36\times10^{2}s$
$G=6.67\times10^{-11}Nm^{2}kg^{-2}$
Orbital Radius = R+h
Orbital velocity is $v_{0}=\sqrt{\frac{GM}{R+h}}\Rightarrow v_{0}^{2}=\frac{GM}{R+h}$
$T= \frac{2\pi (R+h)}{v_{0}}$
$T^{2}= \frac{4\pi^{2} (R+h)^{2}}{v_{0}^{2}}= \frac{4\pi^{2} (R+h)^{3}}{GM}$
$R+h=\left [ \frac{GT^{2}M}{4\pi^{2}} \right ]^{\frac{1}{3}}$
$h=\left [ \frac{GT^{2}M}{4\pi^{2}} \right ]^{\frac{1}{3}}-R$
$h=\left [ \frac{6.67\times10^{-11}\times(24\times36)^{2}\times(10^{2})^{2}\times6\times10^{24}}{4\times3.14\times3.14} \right ]^{\frac{1}{3}}-6.4\times10^{6}$
$h=35940\; km$

Let a satellite S be at h m above the Earth's surface. Let the angle subtended by it at the centre of the earth be 2 $\theta$

$\cos\; \theta =\frac{R}{R+h}=\frac{1}{[1+\frac{h}{R}]}$

$h=3.59\times10^{7}m$ (height of geostationary satellite)

$\cos\; \theta =\frac{1}{[1+\frac{3.59\times10^{7}}{6.4\times10^{6}}]}=0.1515$

$\theta =81.28$

$2\; \theta =2\times 81.28$

Number of satellites required to cover $360^{0}=\frac{360}{2\times 81.28}=2.21$

Number of satellites required=3

Question:8.37

Earth’s orbit is an ellipse with eccentricity 0.0167. Thus, the earth’s distance from the sun and speed as it moves around the sun varies from day to day. This means that the length of the solar day is not constant throughout the year. Assume that earth’s spin axis is normal to its orbital plane and find out the length of the shortest and the longest day. A day should be taken from noon to noon. Does this explain the variation of the length of the day during the year?

Answer:

Perihelion distance for an ellipse of eccentricity e and semi-major axis a,$r_{p}=a(1-e)$
Aphelion distance $r_{a}=a(1+e)$
$\frac{dA}{dt}=\frac{L}{2m}=const$ (Kepler's law)

$L_{a}=L_{p}$
$r_{a}\times p_{a}=r_{p}\times p_{p}$
$r_{a}\times mv_{a}=r_{p}\times mv_{p}$
$r_{a}\times m\omega _{a}r_{a}=r_{p}\times m\omega _{p}r_{p}$
$\omega _{a}r_{a}^{2}=\omega _{p}r_{p}^{2}$
$\frac{\omega _{p}}{\omega _{a}}=\left ( \frac{r_{a}}{r_{p}} \right )^{2}$
$\frac{\omega _{p}}{\omega _{a}}=\left ( \frac{a(1+e)}{a(1-e)} \right )^{2}$
$e=0.0167(given)$
$\frac{\omega _{p}}{\omega _{a}}=\frac{(1+0.0167)^{2}}{(1-0.0167)^{2}}=\left ( \frac{1.0167}{0.9833} \right )^{2}=1.0691$
$\frac{\omega _{p}}{\omega }\frac{\omega }{\omega _{a}}=1.0691$ where $\omega$ is the meanangular velocity of earth
$\frac{\omega _{p}}{\omega }=\frac{\omega }{\omega _{a}}=\sqrt{1.0691}=1.034$
If $\omega$ is $1^{o}$ it corresponds to 1 day i.e., avg angular speed$=1.034^{0}$ per day
and $\omega _{a}0.9833^{0},\omega _{p}=1.0167^{0}$
Since $361^{0}$ = 24 hrs mean solar day. We get $361.034^{0}$ which corresponds to 24 hrs 8.14'
This does not explain the actual variation in the length of the day during the year.

Question:8.38

A satellite is in an elliptic orbit around the earth with an aphelion of 6R and perihelion of 2R where R = 6400 km is the radius of the earth. Find eccentrically of the orbit. Find the velocity of the satellite at apogee and perigee. What should be done if this satellite has to be transferred to a circular orbit of radius 6R?

Answer:

$r_{p}= 2R$
$r_{a}= 6R$
$r_{p}= a(1-e)=2R$
$r_{a}= a(1+e)=6R$
$\frac{1-e}{1+e}=\frac{2}{6}$
$3-3e=1+e$
$4e=2\Rightarrow e=\frac{1}{2}$
$L_{1}=L_{2}$
$m_{a}v_{a}r_{a}=m_{p}v_{p}r_{p}$
$m_{a}=m_{p}=m=mass\; of\; satellite$
$\frac{v_{a}}{v_{p}}=\frac{r_{p}}{r_{a}}=\frac{2R}{6R}=\frac{1}{3}$
$v_{p}=3v_{a}$
Applying conservation of energy at apogee and perigee
$\frac{1}{2}mv_{p}^{2}-\frac{GMm}{r_{p}}=\frac{1}{2}mv_{a}^{2}-\frac{GMm}{r_{a}}$
Multiplying $\frac{2}{m}$ to both side and putting $r_{p}=2R$ and $r_{a}=6R$
$v_{p}^{2}-\frac{2GM}{2R}=v_{a}^{2}-\frac{2GM}{6R}$
$v_{a}=\frac{v_{p}}{3}$
$v_{p}^{2}-v_{a}^{2}=\frac{GM}{R}-\frac{1}{3}\frac{GM}{R}$
$v_{p}^{2}-\left ( (\frac{v_{p}}{3})^{2} \right )=\frac{2GM}{3R}$
$v_{p}^{2}\left [ 1-\frac{1}{9} \right ]=\frac{2GM}{3R}$
$\frac{8}{9}v_{p}^{2}=\frac{2GM}{3R}$
$v_{p}^{2}=\frac{3GM}{4R}$
$v_{p}=\sqrt{\frac{3GM}{4R}}=\sqrt{\frac{3\times6.67\times10^{-11}\times6\times10^{24}}{4\times6.6\times10^{6}}}=\sqrt{46.89\times10^{6}}$
$v_{p}=6.85\; km/s$
$v_{a}=\frac{v_{p}}{3}=2.28\; km/s$
$v_{c}=\sqrt{\frac{GM}{r}}=\sqrt{\frac{GM}{6R}}=\sqrt{\frac{6.67\times10^{-11}\times6\times10^{24}}{6\times6.6\times10^{6}}}$
$v_{c}=3.23\; km/s$
Thus, to transfer to a circular orbit at apogee, we have to boost the velocity by
$v_{c}-v_{a}=(3.23-2.28)=0.95\; km/s$