Physics Ncert Exemplar 2019 Solutions for Class 11 Science Physics Chapter 8 Gravitation are provided here with simple step-by-step explanations. These solutions for Gravitation are extremely popular among class 11 Science students for Physics Gravitation Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Physics Ncert Exemplar 2019 Book of class 11 Science Physics Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Physics Ncert Exemplar 2019 Solutions. All Physics Ncert Exemplar 2019 Solutions for class 11 Science Physics are prepared by experts and are 100% accurate.

Page No 57:

Question 8.1:

The earth is an approximate sphere. If the interior contained matter which is not of the same density everywhere, then on the surface of the earth, the acceleration due to gravity
(a) will be directed towards the centre but not the same everywhere.
(b) will have the same value everywhere but not directed towards the centre.
(c) will be same everywhere in magnitude directed towards the centre.
(d) cannot be zero at any point.

Answer:

Acceleration due to gravity varies with the density as g=43πρGR
If the density of the earth is non-uniform and non-zero then g at different points on the surface of the earth will be different. If we substitute any non-zero value of density then g can not be zero at any point on the surface of the earth.

Hence, the correct answer is option (d).

Page No 57:

Question 8.2:

As observed from earth, the sun appears to move in an approximate 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 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:

As observed from the earth, sun appears to move around the earth in a circular orbit though in reality earth moves around the sun due to gravitational force between them. All planets move around the sun due to huge gravitational force of the sun acting on them. The gravitational force on mercury due to earth is much smaller as compared to that acting on it due to sun and hence it revolves around the sun and not around the earth.
Hence, the correct answer is option (c).



Page No 58:

Question 8.3:

Different points in 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 c.m. (centre of mass) causing translation and a net torque at the c.m. causing rotation around an axis through the c.m. For the earth-sun system (approximating the earth as a uniform density sphere)
(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:

We know τ=r×F
Gravitational force by Sun and the position vector of the center of mass of the earth with respect to the sun are anti-parallel. Hence the Torque on the earth will be zero.

Hence, the correct answer is option (a).

Page No 58:

Question 8.4:

Satellites orbiting the earth have 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 satellite and hence height to gradually decrease.
(d) of collisions with other satellites.

Answer:

The energy of the satellite is continuously reduced because of the friction of the atmosphere. This reduction in energy leads to a change in speed of satellite and reduction in the radius of its circular orbit and it finally falls to the earth.

Hence, the correct answer is option (c).

Page No 58:

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 influence of planets other than earth.

Answer:

The major force of gravitational attraction on the moon is because of the earth. Hence the moon will have an elliptical orbit around the earth. The total gravitational force on the moon is also not directed along the center of the Sun.

Hence, the correct answer is option (b).

Page No 58:

Question 8.6:

In our solar system, the inter-planetary region has chunks of matter (much smaller in size compared to planets) called asteroids. They
(a) will not move around the sun since they have very small masses compared to sun.
(b) will move in an irregular way because of their small masses and will drift away into 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:

Asteroids will also move in circular orbits as the planets move around the Sun. Asteroids and planets, both experience centripetal forces because of the sun and obey Kepler's laws.

Hence, the correct answer is option (d).

 



Page No 59:

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 neighboring heavy objects but the inertial mass cannot.

Answer:

The gravitational mass of a particle-like proton is equivalent to its inertial mass and is independent of the presence of neighboring heavy objects.

Hence, the correct answer is option (d).

Page No 59:

Question 8.8:

Particles of masses 2M, m and M are respectively at points A, B, and C with AB = 12(BC). m is much-much smaller than M and at time t = 0, they are all at rest (Fig. 8.1). 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:

FAB = Force by the particle of mass 2M placed at A on the particle of mass m placed at B.
FC= Force by the particle of mass M placed at C on the particle of mass m placed at B.
Let AB = x, then BC = 2x

FAB = G(2M)m(AB)2 =2GMmx2FCB =G(M)m(CB)2=GMm4x2

As FAB > FCB, m will move towards point A (2M).
Hence, the correct answer is option (c).

Page No 59:

Question 8.9:

Which of the following options are correct?
(a) Acceleration due to gravity decreases with increasing altitude.
(b) Acceleration due to gravity increases with increasing depth (assume the earth to be a sphere of uniform density).
(c) Acceleration due to gravity increases with increasing latitude.
(d) Acceleration due to gravity is independent of the mass of the earth.

Answer:

(a) The acceleration due to gravity at an altitude (height) is given by
gh =g(1+hR)2
As we can see in the expression above if the value of altitude (height) increases, the value of gh decreases. Hence option (a) is correct.

(b) Assuming the earth to be a sphere of uniform density, the acceleration due to gravity at a particular depth (d) is given by
gd =g(1-dR)
As we can see in the expression above if the value of depth increases, the value of gd decreases. Hence option (b) is incorrect.

(c) If λ is the latitude at a point on earth then, gλ = g-ω2 Rcos2(λ)
The value of cos(λ) decreases from 1 to 0 when λ increases from 0° to 90°. The acceleration due to gravity increases from the
equator(λ=0°) to the poleλ=90°. Hence option (c) is correct

(d) The acceleration due to gravity on the surface of the earth is given by
gS = GMeRe2
So acceleration due to gravity on earth depends on the mass of earth. Hence option (d) is incorrect.

Hence, the correct options are (a) and (c).

Page No 59:

Question 8.10:

If the law of gravitation, instead of being 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:

If the law of gravitation becomes an inverse cube law instead of the inverse square law, then for a planet of mass MP revolving around the sun of mass MS, we will have:
Force of gravitation = Centripetal force
GMSMPr3= MPvP2r
The velocity of the planet = GMsr
Time period of the satellite = 2πr×rGMs= 2πr2GMs
Hence T2r4
For a planet to have an elliptical orbit the condition is T2r3 as given by Kepler’s law.
Hence, the orbit of the planetary motion will not be elliptical as T2r4. Option (a) is correct.
The new acceleration due to gravity can be given as:
g' = GMR3
Since g' is constant, hence path followed by the projectile will be approximately parabolic when thrown. So, option (c) is correct.
Gravitational force is a universal force that acts everywhere. So, option (d) is incorrect.
Hence, the correct options are (a) and (c).



Page No 60:

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) airplanes will have to travel much faster.

Answer:

If the gravitational constant G becomes 10 times larger in magnitude.
G'=10G

Then acceleration due to gravity becomes 10g as shown below
g'=10GMer2=10g
​Weight of a person on earth will become 10mg

Force on the person by the Sun will be reduced by 10 times if its mass is reduced 10 times.
Fs=G(Ms)m10rs2 
(a) Walking on the ground would become more difficult
Weight of the person becomes 10 times larger so it will be more difficult to walk. Option (a) is correct.

(b) The acceleration due to gravity on earth will not change.
As g’ = 10g, the acceleration due to gravity changes. Option (b) is incorrect.

(c) Raindrops will fall much faster.
The terminal velocity vT g' (g' = 10g), the terminal velocity increases 10 times. Hence the raindrops fall 10 times faster. Option (c) is correct.

(d) Airplanes will have to travel much faster.
As the g’ = 10g, to overcome the increased gravitational force and in order to maintain the speed, the airplane will have to travel much faster. So, option (d) is correct.
Hence, the correct options are (a), (c) and (d).

Page No 60:

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:

Due to opposite charges the attractive electrostatic forces of attraction produced will be large, along with the gravitational forces. Both the forces will be added and would be radial in nature. Since both the forces are central forces and obey the inverse square law, all the three Kepler's laws will be valid.
Hence, the correct options are (a), (c) and (d).

Page No 60:

Question 8.13:

There have been suggestions that the value of the gravitational constant G becomes smaller when considered over very large time period (in billions of years) 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 sufficiently long time we will leave the solar system.

Answer:

We know that the centripetal force is provided by the gravitational.

GMsMeR2=Mev2Rv=GMsR
Now if G becomes smaller, it will reduce the centripetal force and hence the radius of the orbit will increase i.e., distance of the earth from the sun is increased, so the temperature drops further and the orbit will also not be closed one. And after sufficiently long time the earth will leave the solar system.
Hence, the correct options are (c) and (d).

Page No 60:

Question 8.14:

Supposing Newton’s law of gravitation for gravitation forces F1and F2 between two masses m1 and m2 at positions r1 and r2 read F1=-F2=-r12r123GM02m1m2M02nwhere M0 is a constant of dimension of mass, r12 = r1 r2 and n is a number. In such a case,
(a) the acceleration due to gravity on earth will be different for different objects.
(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:

Let, m1=M= mass of the earth and, m2=m= mass of objectF=GMo2(1-n)MnR2mn=Kmn ( Where, K is constant)g=Fm=Kmnm=Kmn-1g depends on mass of the object
Since, acceleration due to gravity depends upon the mass of the object, hence it will be different for different objects. 
Hence, option (a) is correct.
As g is not constant, so the constant of proportionality will not be constant in Kepler’s third law expression. Hence, Kepler’s third law will not be valid.
But, the force is central in nature, so, the first two Kepler’s laws will be valid.
Hence, option (b) is incorrect but (c) is correct
For negative value of n,
g=Kmn-1g varies inversally with mass of the object
It means Force F is inversely proportional to the mass of the object. Hence, lighter bodies will experience a greater force than the heavier bodies and vice versa. 
So, objects lighter than water will sink in water. So, option (d) is also correct.

Hence, the correct options are (a), (c) and (d).

 

Page No 60:

Question 8.15:

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

Answer:

The satellite which appears stationary relative to the earth is called the geostationary satellite. It revolves around the earth in the west-east direction with the same angular velocity as done by the earth about its own axis in the west-east direction. Whereas a polar satellite revolves around the earth's pole in the north-south direction. It is independent of the earth's rotation.
Hence, the correct options are (a) and (c).



Page No 61:

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:

Center of mass (COM) depends on the orientation and distribution of mass particles of an object and its distance from a point. The center of gravity is based on weight. So, when the gravitational field across an object is uniform, COM and COG are identical. But, when the object enters in a varying gravitational field region, the center of gravity (COG)  will move closer to regions of the object in a stronger field, whereas the COM will remain unchanged .
Hence, the correct answer is option (d).

Page No 61:

Question 8.17:

Molecules in air in the atmosphere are attracted by gravitational force of the earth. Explain why all of them do not fall into the earth just like an apple falling from a tree.

Answer:

Air molecules in the atmosphere are attracted by the gravitational force of the earth but they do not fall into the earth as the molecules in the air have random motion due to temperature (thermal motion of molecules), so their resultant velocity is not exactly in the vertically downward direction.

Page No 61:

Question 8.18:

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

Answer:

An example of a central force is Gravitational force. Whereas an example of non-central force is the frictional force.
 

Page No 61:

Question 8.19:

Draw areal velocity versus time graph for mars.

Answer:

As per Kepler's second law, the areal velocity of any planet around the sun is constant with respect to time.

Page No 61:

Question 8.20:

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

Answer:

Areal velocity = dAdt = L2m where, L is the angular momentum of earth wrt to the sun and m is the mass of earth.
So, dAdt = L2m=r×mv2m=12r×v

Hence, the direction of the areal velocity is perpendicular to the plane of r and vin accordance with the right-hand grip rule of Maxwell.

Page No 61:

Question 8.21:

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

Answer:

Gravitational force is independent of the medium between the point masses. So dipping the point masses in water will not change the gravitational force between the point masses.

Page No 61:

Question 8.22:

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

Answer:

Inertia is a measure of the mass of a body hence inertia can not be zero whereas weight i.e mg can be zero when acceleration due to gravity is zero. Hence it is possible for a body to have inertia but no weight.

Page No 61:

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:

Since the gravitational force is independent of the medium in which the masses are placed hence we can not shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means which is not the case with electric charges and electric fields.

Page No 61:

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 he hope to detect gravity?

Answer:

Inside a small spaceship orbiting around the earth the value of acceleration due to gravity g, can be considered as constant and hence astronaut feels weightless. If the space station orbiting 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.

Page No 61:

Question 8.25:

The gravitational force between a hollow spherical shell (of radius R and uniform density) 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 on any point mass placed inside the spherical shell and when placed at the center of the hollow spherical shell is zero.

Let MS be the mass of the spherical shell and MP be the mass of point mass, then the gravitational force between them when the point mass is on the surface is F = GMSMPR2
If the point mass is placed outside the spherical shell at any distance then the force between them will be

F = GMSMPr2

Page No 61:

Question 8.26:

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

Answer:

The earth revolves around the sun in an elliptical orbit as per Kepler’s first law and the sun remains at one of its foci. The position of the earth at the shortest and longest distance is called perihelion and aphelion respectively. According to Kepler's second law, the areal velocity of the planet around the sun is constant.
Since dAdt = L2m = r×v2
r is maximum, so speed (v) is minimum at aphelion and vice-versa for perihelion.

Page No 61:

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 polar satellite revolves along the north to the south pole. So its plane makes 90° with the equatorial plane.
(b) Geostationary satellite revolves in the west to the east direction along the equatorial plane. So the angle between geostationary and equatorial plane is zero.



Page No 62:

Question 8.28:

Mean solar day is the time interval between two successive noon when sun passes through zenith point (meridian). Sidereal day is the time interval between two successive transit of a distant star through the zenith point (meridian).

By drawing appropriate diagram showing earth’s spin and orbital motion, show that mean solar day is four minutes longer than the sidereal day. In other words, distant stars would rise 4 minutes early every successive day.

Answer:


Consider that on a day at noon sun passes through the zenith (meridian). After one revolution (360°) of the earth about its own axis the sun again passes through the zenith. During this time when the earth revolves at its own axis by 360° it changes its angle by 1°.
So 361° rotation by the earth is considered as one solar day.

Now if 361° corresponds to 24 hrs
therefore 1°=24hrs361°=24×3600361=3 min 59 sec4 mins
Hence, a distant star rises 4 min. early every day.

Page No 62:

Question 8.29:

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

Answer:

Let the masses of the spheres be M and their radiuses be R. Let a point mass be placed at the midpoint having mass m.
Let us displace the point mass by distance x towards right.

So force on object of mass m,
F1= GMm5R-x2      towards rightF2= GMm5R+x2       towards left

As F1>F2 point mass m will be attracted to right.
Hence, it is an unstable equilibrium.

Page No 62:

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 = Me
The radius of the orbit of satellite = R
Mass of satellite = m
Orbital velocity = vo=GMeR

(a) KE vs orbital radius R

KE=12mvo2=GMem2RKE 1R



(b) PE vs orbital radius R

PE=-GMemR



(c) TE vs orbital radius R

TE=-GMem2R

We can compare all the above graphs as given below:


 

Page No 62:

Question 8.31:

Shown are several curves (Fig. 8.2). Explain with reason, which ones amongst them can be possible trajectories traced by a projectile (neglect air friction).

Answer:

The trajectory or the path followed by a projectile under the gravitational force of the earth, will be a conic section (for motion outside the earth) with the center of the earth as a focus.
Hence, the correct answer is option (c).



Page No 63:

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:


Gravitational Potential Energy of the object when it is at the surface of the earth (initial) = -GMmR

(where R is the radius of the earth and M is the mass of the earth)

Gravitational Potential Energy of the object when it is raised to height R from the surface of the earth (final) =-GMm2R

∴ Gain in potential energy =Final PE - Initial PE

                                          =-GMm2R --GMmR =GMmR - GMm2R  =GMm2R
               
                                           =12mGMR2R   =12mgR   g=GMR2

                                            



 

Page No 63:

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 (Fig. 8.3).

If the mass is removed further away such that OP becomes 2h, by what factor the force of gravitation will decrease, if h = r ?

Answer:

Let a small element on the ring of mass dM be such that the distance between elementary mass dM and mass m is x.
x=r2+h2Net force will be along the axis,F=Gmr2+h22×hr2+h2dM=GmMhr2+h23/2      ...1F'=GmM(2h)r2+4h23/2             ...2The ratio,FF'=r2+4r23/22r2+r23/2=523/2=5542      (if h=r)F'=4255F

Page No 63:

Question 8.34:

A star like the sun has several bodies moving around it at different distances. Consider that all of them are moving in circular orbits. Let r be the distance of the body from the centre of the star and let its linear velocity be v, angular velocity ω

, kinetic energy K, gravitational potential energy U, total energy E and angular momentum l. As the radius r of the orbit increases, determine which of the above quantities increase and which ones decrease.

Answer:

Gravitation pull force will provide the necessary centripetal force to revolve in a circular path. 
(i) Let a body of mass m revolving around a star of mass M.
So, F=mv2r=GMmr2v=GMrv1rSo, v decreases with increase in r.
(ii) angular velocity,
ω=vr=GMr3ω will decrease if r will increase
(iii) Kinetic energy of the body,
KE=12mv2=GMm2rKE will decrease with increase in r.
(iv) Potential energy,
PE=-GmMrPotential energy will increase with increase in r.
(v) Total energy,
TE=PE+KE=-GMmr+GMm2r=-GMm2rTotal energy will increase with increase in r.
(vi) angular momentum,
L=mvr=mrGMr=GMm2rHence, angular momentum will increase with increase in r. 

 

Page No 63:

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:

Let, six point masses are placed at six vertices A, B, C, D, E and F of a regular hexagon ABCDEF.
From figure, 

AB=AF=lAC=AE=l3AD=2l
So, force on mass m at A due to mass m at B, C, D, E and F,
FAB=Gm2l2 along ABFAF=Gm2l2 along AFFAC=Gm23l2 along ACFAE=Gm23l2 along AEFAD=Gm24l2 along AD​ along AB
Now, finding the vector sum of forces acting on the mass at vertex A by the remaining five masses placed at other 5 vertices, we get:
Fnet=Gm2l21+13+14

Page No 63:

Question 8.36:

A satellite is to be placed in equatorial geostationary orbit around 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.

[M = 6 × 1024 kg, R = 6400 km, T = 24h, G = 6.67 × 10–11 SI units]

Answer:

(a) Time period of geostationary orbit is T = 24 h. 
So, radius of orbit of the geostationary satellite is,
r=GMT24π21/3If h be the height of the setellite from the surface of the earth,then,r=R+h=GMT24π21/3h=GMT24π21/3-R 
(b) Let angle subtended by the satellite (height h) at the centre of earth is 2θ ,  

cosθ=11+hR=16.6θ=81.26oNumber of satellites to cover 360o = 360o2θ= 360o2×81.26o=2.21So, the minimum number of satellites=3
 

Page No 63:

Question 8.37:

Earth’s orbit is an ellipse with eccentricity 0.0167. Thus, 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 through 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 variation of length of the day during the year?

Answer:

Let a be the length of semi major axis and e be the eccentricity of the elliptical path.
So, rp= a(1-e) and  ra= a(1+e)

According to Kepler's law, areal velocity of the earth remains constant.
Also angular momentum of it will also conserved.
Hence, at perigee, ωprp2=ωara2  at apogee.
ωpωa=1+e1-e2ωpωa=1.069                  Given eccentricity of the eliptical path, e=0.0167
If we analyse these values to find the mean solar day, we find the data does not clearly explain the actual value of the day during the year.
 

Page No 63:

Question 8.38:

A satellite is in an elliptic orbit around the earth with aphelion of 6R and perihelion of 2R where R = 6400 km is the radius of the earth. Find eccentricity 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?
[G = 6.67 × 10–11 SI units and M = 6 × 1024 kg]

Answer:

Let a be the length of semimajor axis and e be the eccentricity of the elliptical path.
So,
ra=a1+e=6R and rp=a1-e=2RSolving we get, Eccentricity, e=12As, no external torque is acting on the system ,so angular momentum must be conserved at apogee and perigee,mvprp=mvaravavp=rpra=(1-e)(1+e)=(1-12)(1+12)=13
Total energy at apogee and perigee will also be conserved,
so,
Total energy=KE+PE12mvp2-GMmrp=12mva2-GMmravp=3GM4R=6.85 km/sAnd, va=vp3=6.853 km/s=2.28 km/s
If radius of the circular orbit, r = 6R,
Orbital velocity,
 vc=GM6R=3.23 km/sHence the required change in velocity,=vc-va=3.23-2.28 km/s=0.95 km/s
By firing of rocket from the satellite, such transfer can be done.
 



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