Chapter 17 Gravitation Part 3 – Physics free study material by TEACHING CARE online tuition and coaching classes
Chapter 17 Gravitation Part 3 – Physics free study material by TEACHING CARE online tuition and coaching classes
Time Period of Satellite.
It is the time taken by satellite to go once around the earth.
\ T = Circumference of the orbit orbital velocity
Þ T = 2pr = 2pr v
Þ T = 2p = 2p
[As v = ] [As GM = gR 2 ]
R æ h ö3 / 2
Þ T = 2p
= 2p
g ç1 + R ÷
[As r = R + h ]
è ø
(i) From T = 2p it is clear that time period is independent of the mass of orbiting body and depends on
the mass of central body and radius of the orbit
(ii)
Þ
T = 2p
T 2 = 4p 2 r 3
GM
i.e., T 2 µ r 3
This is in accordance with Kepler’s third law of planetary motion r becomes a (semi major axis) if the orbit is elliptic.
 Time period of nearby satellite,
From
T = 2p
= 2p
= 2p
[As h = 0
and GM = gR 2 ]
For earth
R = 6400km and
g = 9.8m / s 2
T = 84.6 minute » 1.4 hr
 Time period of nearby satellite in terms of density of planet can be given as
2p (R 3 )1 / 2
T = 2p
= 2p
=
éG.
êë
1 / 2 =



pR 3 r
3 û
1
n+1

 If the gravitational force of attraction of the sun on the planet varies as F µ r n then the time period varies as T µ r
 If there is a satellite in the equatorial plane rotating in the direction of earth’s rotation from west to east, then for an observer, on the earth, angular velocity of satellite will be (w S – w E ). The time interval between the two consecutive appearances overhead will be
T = 2p =
TSTE
éAs T = 2p ù
w s – w E
TE – TS
ëê w úû
If w _{S} = w _{E} , T = ¥ i.e. satellite will appear stationary relative to earth. Such satellites are called geostationary satellites.
Problem 64. A satellite is launched into a circular orbit of radius ‘R’ around earth while a second satellite is launched into an orbit of radius 1.02 R. The percentage difference in the time periods of the two satellites is [EAMCET 2003]
(a) 0.7 (b) 1.0 (c) 1.5 (d) 3
Solution : (d) Orbital radius of second satellite is 2% more than first satellite
So from T µ (r)^{3} ^{/} ^{2} , Percentage increase in time period = 3
2
= 3
2
(Percentage increase in orbital radius)
(2%) = 3%.
Problem 65. Periodic time of a satellite revolving above Earth’s surface at a height equal to R, where R the radius of Earth, is [g is acceleration due to gravity at Earth’s surface] [MP PMT 2002]
 2p
4 2p
 2p
 8p
Solution : (b)
T = 2p
= 2p
= 2p
= 4 2p
[As h = R (given)].
Problem 66. An earth satellite S has an orbit radius which is 4 times that of a communication satellite C. The period of revolution of S is [MP PMT 1994; DCE 1999]
(a) 4 days (b) 8 days (c) 16 days (d) 32 days
Solution : (b) Orbital radius of satellite rs
= 4rc
(given)
T æ r
ö 3 / 2
From Kepler’s law T µ r ^{3} ^{/} ^{2} \
s = ç s ÷
= (4)^{3} ^{/} ^{2} Þ Ts
= 8Tc
= 8 ´ 1 day = 8 days.
Tc è sc ø
Problem 67. One project after deviation from its path, starts moving round the earth in a circular path at radius equal to nine times the radius at earth R, its time period will be [RPET 1987, 88]
(a) 2p
(b)
27 ´ 2p
 p
(d)
8 ´ 2p
Solution : (b)
T = 2p
= 2p
= 2p (9)^{3} ^{/} ^{2}
= 27 ´ 2p
[As r = 9R (given)].
Problem 68. A satellite A of mass m is revolving round the earth at a height ‘r’ from the centre. Another satellite B of mass 2m is revolving at a height 2r. The ratio of their time periods will be [CBSE PMT 1993]
(a) 1 : 2 (b) 1 : 16 (c) 1 : 32 (d) 1: 2
T æ r
ö 3 / 2
æ r ö ^{3} ^{/} ^{2} 1
Solution : (d) Time period depends only upon the orbital radius ^{1} = ç 1 ÷ = ç ÷ = .
T2 è r2 ø
è 2r ø
Height of Satellite.
As we know, time period of satellite T = 2p
= 2p
By squaring and rearranging both sides

æ T 2 g R 2 ö1 / 3
g R 2 T 2
4p 2
= (R + h)3
Þ h = ç
è
4p 2 ÷ – R
By knowing the value of time period we can calculate the height of satellite the surface of the earth.
Problem 69. Given radius of earth ‘R’ and length of a day ‘T’ the height of a geostationary satellite is


[G – Gravitational constant, M – Mass of earth] [MP PMT 2002]
æ 4p ^{2}GM ö1 / 3
æ 4pGM ö^{1/} ^{3}
æ GMT 2 ö1 / 3
æ GMT 2 ö1 / 3


 ç ÷
è T ø
 ç ÷
è R ^{2} ø
 R (c)
ç ÷


ç 4p 2 ÷
(d)
ç ÷


ç 4p 2 ÷
æ T 2 gR 2 ö1 / 3
æ GMT 2 ö1 / 3
Solution : (c) From the expression h = ç
÷ – R
\ h = ç ÷ – R
[As gR ^{2} = GM ]




ç 4p 2 ÷ ç 4p 2 ÷
Problem 70. A satellite is revolving round the earth in circular orbit at some height above surface of earth. It takes
5.26 ´ 10^{3} seconds to complete a revolution while its centripetal acceleration is 9.92m / s ^{2} . Height of satellite above surface of earth is (Radius of earth 6.37 ´ 10^{6} m ) [MP PET 1993]
(a) 70 km (b) 120 km (c) 170 km (d) 220 km
Solution : (c) Centripetal acceleration (a ) = v2 and T = 2pr
c r
ac T ^{2}
v
9.32 ´ (5.26 ´ 10^{3} )^{2}
From equation (i) and (ii) r = Þ R + h =
4p ^{2}
4 ´ p ^{2}
h = 6.53 ´ 10^{6} – R
= 6.53 ´ 10^{6} – 6.37 ´ 10 ^{6} = 160 ´ 10^{3} m
= 160km
» 170km .
Geostationary Satellite.
The satellite which appears stationary relative to earth is called geostationary or geosynchronous satellite, communication satellite.
A geostationary satellite always stays over the same place above the earth such a satellite is never at rest. Such a satellite appears stationary due to its zero relative velocity w.r.t. that place on earth.
The orbit of a geostationary satellite is known as the parking orbit.
 It should revolve in an orbit concentric and coplanar with the equatorial
 It sense of rotation should be same as that of earth about its own axis e., in anticlockwise direction (from west to east).
 Its period of revolution around the earth should be same as that of earth about its own axis.
\ T = 24 hr = 86400 sec
 Height of geostationary satellite
As T = 2p
Þ 2p
= 24hr
Substituting the value of G and M we get R + h = r = 42000 km = 7R
\ height of geostationary satellite from the surface of earh h = 6R = 36000 km
 Orbital velocity of geo stationary satellite can be calculated by v =
Substituting the value of G and M we get v = 3.08 km / sec
Angular Momentum of Satellite.
Angular momentum of satellite L = mvr
Þ L = m r [As v = ]
\ L =
i.e., Angular momentum of satellite depend on both the mass of orbiting and central body as well as the radius of orbit.
 In case of satellite motion, force is central so torque = 0 and hence angular momentum of satellite is
conserved i.e., L = constant
 In case of satellite motion as areal velocity
dA = 1 (r)(vdt) = 1 rv
dt 2 dt 2
Þ dA = L
[As
L = mvr ]
dt 2 m
But as L = constant, \ areal velocity (dA/dt) = constant which is Kepler’s II law
i.e., Kepler’s II law or constancy of areal velocity is a consequence of conservation of angular momentum.
Problem 71. The orbital angular momentum of a satellite revolving at a distance r from the centre is L. If the distance is increased to 16r, then new angular momentum will be [MP PET 2003]
(a) 16 L (b) 64 L (c) L
4
Solution : (d) Angular momentum L = \ L µ
 4 L
L2 = = = 4
L1
L2 = 4 L1 = 4 L
Problem 72. Angular momentum of a planet of mass m orbiting around sun is J, areal velocity of its radius vector will be
Solution : (b)
(a)
1 mJ
2
 J
2m
 m
2J
1 2mJ
Energy of Satellite.
When a satellite revolves around a planet in its orbit, it possesses both potential energy (due to its position against gravitational pull of earth) and kinetic energy (due to orbital motion).
 Potential energy : U = mV = –GMm = – L2
éAs V = – GM , L2 = m2 GMrù
r mr 2
 Kinetic energy : K = 1mv2 = GMm = L2
ëê r úû
é = ù
2 2 r 2 mr 2
êAs v ú

êë û
 Total energy :
E = U + K = – GMm + GMm = – GMm = – L^{2}
r 2r 2r 2mr ^{2}
 Kinetic energy, potential energy or total energy of a satellite depends on the mass of the satellite and the central body and also on the radius of the
 From the above expressions we can say that Kinetic energy (K) = – (Total energy) Potential energy (U) = 2 (Total energy) Potential energy (K) = – 2 (Kinetic energy)
 Energy graph for a satellite (iv) Energy distribution in elliptical orbit
+
Satellite
Perigee

K.E. = max
P.E. = min
Apogee
K.E. = min
P.E. = max
–
(A)
(B)
 If the orbit of a satellite is elliptic then
 Total energy (E) = –GMm = constant ; where a is semimajor axis .
2a
 Kinetic energy
(K)
will be maximum when the satellite is closest to the central body (at perigee) and
maximum when it is farthest from the central body (at apogee)
 Potential energy (U) will be minimum when kinetic energy = maximum e., the satellite is closest to the
central body (at perigee) and maximum when kinetic energy = minimum i.e., the satellite is farthest from the central body (at apogee).
 Binding Energy : Total energy of a satellite in its orbit is Negative energy means that the satellite is bound to the central body by an attractive force and energy must be supplied to remove it from the orbit to infinity. The energy required to remove the satellite from its orbit to infinity is called Binding Energy of the system, i.e.,
Binding Energy (B.E.) = –E = GMm
2r
Chanae in the Orbit of Satellite.
When the satellite is transferred to a higher orbit (r2 > r1 )
by the following table
then variation in different quantities can be shown
Quantities  Variation Relation with  r  
Orbital velocity  Decreases  v µ 1
r 

Time period  Increases  T µ r ^{3} ^{/} ^{2}
P µ 1 r L µ r K µ 1 r U µ – 1 r E µ – 1 r BE µ 1 r 

Linear momentum  Decreases  
Angular momentum  Increases  
Kinetic energy  Decreases  
Potential energy  Increases  
Total energy  Increases  
Binding energy  Decreases  
Note:: @ Work done in changing the orbit W = E2 – E1
æ GMm ö æ GMm ö
W = ç
2r
÷ – ç ÷
2r
è 2 ø è 1 ø
W = GMm é 1 – 1 ù
2 êr r ú
ë 1 2 û
Problem 73. Potential energy of a satellite having mass ‘m’ and rotating at a height of 6.4 ´ 10^{6} m from the earth centre is
[AIIMS 2000; CBSE PMT 2001; BHU 2001]
(a)
0.5 mgRe
(b)
– mgRe
(c)
2mgRe
4 mgRe
Solution : (a) Potential energy = – GMm
= – GMm
= – GMm
[As h = R

(given)]
r
gR ^{2}m
Re + h
2Re
2
\ Potential energy = – ^{e}
2Re
= 0.5mgRe
[As GM = gR ]
Problem 74. In a satellite if the time of revolution is T, then kinetic energy is proportional to [BHU 1995]
 1
T
 1
T 2
 1
T 3
T 2 / 3
Solution : (d) Time period T µ r ^{3} ^{/} ^{2}
Þ r µ T ^{2} ^{/} ^{3}
and Kinetic energy µ 1 µ
r
1
T 2 / 3
µ T 2 / 3 .
Problem 75. Two satellites are moving around the earth in circular orbits at height R and 3R respectively, R being the radius of the earth, the ratio of their kinetic energies is
(a) 2 (b) 4 (c) 8 (d) 16
Solution : (a)
r1 = R + h1 = R + R = 2R and r2 = R + h2 = R + 3R = 4 R
Kinetic energy µ 1
\ (KE)1
= r2
= 4 R = 2 .
Weightlessness.
r (KE)_{2} r_{1}
2R 1
The weight of a body is the force with which it is attracted towards the centre of earth. When a body is stationary with respect to the earth, its weight equals the gravity. This weight of the body is known as its static or true weight.
We become conscious of our weight, only when our weight (which is gravity) is opposed by some other object. Actually, the secret of measuring the weight of a body with a weighing machine lies in the fact that as we place the body on the machine, the weighing machine opposes the weight of the body. The reaction of the weighing machine to the body gives the measure of the weight of the body.
The state of weightlessness can be observed in the following situations.
 When objects fall freely under gravity : For example, a lift falling freely, or an airship showing a feat in which it falls freely for a few seconds during its flight, are in state of
 When a satellite revolves in its orbit around the earth : Weightlessness poses many serious problems to the astronauts. It becomes quite difficult for them to control their movements. Everything in the satellite has to be kept tied Creation of artificial gravity is the answer to this problem.
 When bodies are at null points in outer space : On a body projected up, the pull of the earth goes on decreasing, but at the same time the gravitational pull of the moon on the body goes on increasing. At one particular position, the two gravitational pulls may be equal and opposite and the net pull on the body becomes This is zero gravity region or the null point and the body in question is said to appear weightless.
Weightlessness in a Satellite.
A satellite, which does not produce its own gravity moves around the earth in a circular orbit under the action
of gravity. The acceleration of satellite is
GM towards the centre of earth.
r 2
If a body of mass m placed on a surface inside a satellite moving around the earth. Then force on the body are
 The gravitational pull of earth = GMm
r 2
 The reaction by the surface = R
GmM
By Newton’s law r 2 – R = m a


GmM – R = m æ GM ö
r 2
\ R = 0
ç r 2 ÷
Thus the surface does not exert any force on the body and hence its apparent weight is zero.
A body needs no support to stay at rest in the satellite and hence all position are equally comfortable. Such a state is called weightlessness.
 One will find it difficult to control his movement, without weight he will tend to float To get from one spot to the other he will have to push himself away from the walls or some other fixed objects.
 As everything is in free fall, so objects are at rest relative to each other, e., if a table is withdrawn from below an object, the object will remain where it was without any support.
 If a glass of water is tilted and glass is pulled out, the liquid in the shape of container will float and will not flow because of surface
 If one tries to strike a match, the head will light but the stick will not This is because in this situation convection currents will not be set up which supply oxygen for combustion
 If one tries to perform simple pendulum experiment, the pendulum will not It is because there will
not be any restoring torque and so T = 2p
= ¥ . [As
g¢ = 0 ]
 Condition of weightlessness can be experienced only when the mass of satellite is negligible so that it does not produce it own
e.g. Moon is a satellite of earth but due to its own weight it applies gravitational force of attraction on the body placed on its surface and hence weight of the body will not be equal to zero at the surface of the moon.
Problem 76. The time period of a simple pendulum on a freely moving artificial satellite is [CPMT 1984; AFMC 2002]
 Zero (b) 2 sec (c) 3 sec (d) Infinite
Solution : (d)
T = 2p
= 2p = ¥
[As g ¢ = 0
in the satellite]
Problem 77. The weight of an astronaut, in an artificial satellite revolving around the earth, is [BHU 1999]
(a) Zero (b) Equal to that on the earth
(c) More than that on the earth (d) Less than that on the earth
Solution : (a)
Problem 78. A ball is dropped from a spacecraft revolving around the earth at a height of 120 km. What will happen to the ball
 It will continue to move with velocity v along the original orbit of spacecraft
 If will move with the same speed tangentially to the spacecraft
 It will fall down to the earth gradually
 It will go very far in the space
Solution : (a) Because ball possess same initial tangential speed as that of space craft.. So it also feels the condition of weightlessness.
Problem 79. Two particles of equal mass go round a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is [CBSE PMT 1995]
 v =
 v =
 v =
 v =
Solution : (c) Both the particles moves diametrically opposite position along the circular path of radius R and the gravitational force provides required centripetal force
mv ^{2}
R
= Gmm Þ
(2R)^{2}
v = 1
2
Problem 80. Two types of balances, the beam balance and the spring balance are commonly used for measuring weight in shops. If we are on the moon, we can continue to use [AIIMS 1992]
 Only the beam type balance without any change
 Only the spring balance without any change
 Both the balances without any change
 Neither of the two balances without making any change
Solution : (a) Because in beam type balance effect of less gravitation force works on both the Pans. So it is neutralizes but in spring balance weight of the body decreases so apparent weight varies with actual weight.
Problem 81. During a journey from earth to the moon and back, the greatest energy required from the spaceship rockets is to overcome [CBSE PMT 1991]
 The earth’s gravity at take off
 The moon’s gravity at lunar landing
 The moon’s gravity at lunar take off
 The point where the pull of the earth and moon are equal but opposite
Solution : (a)
Problem 82. If the radius of earth contracts 1
n
of its present value, the length of the day will be approximately
(a)
24 h n
(b)
24 h
n2
(c)
24 n h
(d)
24 n^{2} h
Solution : (b) Conservation of angular momentum L = Iw = = 2 MR ^{2} ´ 2p
= constant \ T µ R ^{2}
[If M remains same]
5 T
T æ R ö 2 æ R / n ö^{2} 1 24
Þ 2 = ç 2 ÷ = ç
÷ = Þ T_{2} = hr
[As T_{1} = 24 hr ].
T1 ç R1 ÷
è R ø n ^{2} n^{2}
è ø
Problem 83. A body released from a height h takes time t to reach earth’s surface. The time taken by the same body released from the same height to reach the moon’s surface is
 t (b) 6t (c) 6t
(d) t
6
Solution : (c) If body falls from height h then time of descent t =
Þ tmoon =
tearth
= Þ t
moon = t .
Problem 84. A satellite is revolving round the earth with orbital speed v_{0} . If it stops suddenly, the speed with which it will
strike the surface of earth would be ( ve
v 2
= escape velocity of a particle on earth’s surface)
(a)
e (b) v0
v0
(c)
(d)
Solution : (d) Applying conservation of mechanical energy between A and B point
 GMm =1 mv2 + æ GMm ö ; 1 mv 2 = GMm – GMm
ç ÷

r 2 è R 2 R r
v 2 = 2Gm – 2Gm
= v ^{2} – 2v ^{2} Þ v =
R r e 0
[As escape velocity ve = , orbital velocity v_{0} = ]
Problem 85. The escape velocity for a planet is ve . A tunnel is dug along a diameter of the planet and a small body is
dropped into it at the surface. When the body reaches the centre of the planet, its speed will be
(a)
ve (b)
ve
 ve
2
 Zero
Solution : (b) Gravitational potential at the surface of the earth Vs
= – GM
R
Gravitational potential at the centre of earth V
= – 3GM
^{c} 2R
By the conservation of energy
 mv ^{2} = m(V
 s
 Vc )
_{2} GM æ 3
ö GM
v 2

v = 2 ç
è
– 1÷ =

ø
= gR = ^{e}
R 2
[As ve =
2gR ]
\ v = ve
Problem 86. A small body of superdense material, whose mass is twice the mass of the earth but whose size is very small
compared to the size of the earth, starts from rest at a height the earth’s surface in time t. Then t is equal to
H << R
above the earth’s surface, and reaches
(a)
(b)
(c)
(d)
Solution : (c) As the masses of the body and the earth are comparable, they will move towards their centre of mass, which remains stationary.
H
Hence the body of mass 2m move through distance .
3
and time to reach the earth surface = = =