CHAPTER NOTES 1

 A planet moving along an elliptical orbit is

SECTION –C
MOTION of PLANETS closest to sun at nearest distance r1 and

farthest away at distance r2 .

 Kepler’s Laws:
If v1 and v2 are linear velocities at these points,
(1). Law of Orbits:
then: L = constant  mv1 r1 mv2 r2  v1 / v2  r2 / r1
All planets move around the sun in elliptical
orbits having the sun at one focus of the orbit. (2). Law of Areas:

A line joining any planet to the sun sweeps out

Note: The gravitational force is a central
equal areas in equal times, that is, the areal
force so torque on planet relative to sun is
velocity of the planet remains constant.
always zero; hence angular momentum of a
planet (or) satellite is always constant dA L mvr  r  r  r 2
    
irrespective of shape of orbit. dt 2m 2m 2 2

 A satellite revolves around the earth: PRACTICE QUESTION:

(i) Force of satellite is always towards Earth; Q.1. The figure shows elliptical orbit of a

therefore, acceleration of satellite is always planet m about the sun S. The shaded area SCD

towards centre of earth. is twice the shaded area SAB. If t1 is the time for

the planet to move from C to D, t 2 is the time to

(ii) Angular momentum (both in magnitude and
move from A to B then
direction) of satellite is constant about the
centre

(iii) Mechanical energy of satellite is constant.

(iv) Speed of satellite is: (a) t 1  4t 2 (b) t 1  2t 2 (c) t 1  t 2 (d) t 1  t 2

Maximum = Closer to earth SOLUTION: (b)

dA A1 A2
Minimum = Farthest to earth   Constant  
dt t1 t 2

 The angular momentum of the earth around Here, A1  Area under 𝐴𝐶𝐷, A2  Area under ABS
the sun:
A 
Angular momentumof a planet : L  mvo r
 t1   1  t 2  t 1  2t 2 Given : A 1  2 A2 
 A2 

 L m  GM / r r  L  r 1/2
CHAPTER NOTES 2
T  r  T  kr
2 3 2 3

ONE OR MORE CORRECT OPTIONS

Here k is a constant. T2
Q.1. The position vector of a particle in a
circular motion (i.e. uniform) about the origin
R3
sweeps out equal area in equal time. Its
Note: Kepler's laws are valid for satellites also.
(a) Velocity remains constant
 A planet revolves in an elliptical orbit
(b) Speed remains constant
around the sun. The semi-major and semi-
(c) Acceleration remains constant minor axes is a and b, then T 2  a3
(d) Tangential acceleration remains constant.
[Remember: T 2  a3 (semi-major) not depend
(i.e. Tangential acceleration is zero in uniform
on semi-minor (b)]
  2

acceleration)  at  0 & ac  v  cons tan t 
 r 
   For perigee and Apogee position of a planet:

3
SOLUTION: (b), (d) r r 
The law of periods: T  a 2 3
 T  1 2 
2

 2 
Condition: Kepler’s IInd law
Proof: From the figure: AB  AF  FB
 Speed = Constant
 Velocity = Variable Direction change   2a  r1  r2  a   r1  r2  / 2

(velocity)2
 Acceleration   variable 𝑬
r
 Tangential acceleration Perigee 𝐴𝑝𝑜𝑔𝑒𝑒
Sun F r2

d  d 𝑨 𝑩
at  | v |  (Constant)  at  0 r1 𝒄 𝒂
dt dt

 Tangential acceleration remains constant. 𝑫

(3). Law of Periods: Where a = semi-major axis

The Square of the period of revolution of any r1  Shortest distance of planet from sun
planet around the sun is directly proportional to (perigee)
the cube of its mean distance from the sun.
r2  Largest distance of planet from sun
If the period of a planet around the sun is T and (apogee)
the mean radius of its orbits is r, then
CHAPTER NOTES 3
 The distance of the earth from the sun when  
2/3 2/3
rbody  1  1
   earth    
it is at perpendicular to the major-axis of the rearth   27  9
 body 
orbit drawn from the sun is:

Smallest distance of earth from the sun  r1 Q.2. Two planets at mean distance d1 and d2

from the sun and their frequencies are n1 and n2

Largest distance of earth from the sun  r2
respectively then

(a) n12d12  n2d22 (b) n22d23  n12d13

(c) n1d12  n2d22 (d) n12d1  n22d2

SOLUTION: (b)
2 1 1
Using the property of ellipse:  
R r1 r2 T2 T2 1
 3  Cons tant  3  2 3  Constant
R d nd
2 r1  r2 2r1r2
   R
R r1r2 r1  r2  n12 d13  n22 d23

TYPE-I: Radius of the orbit of a geostationary

PRACTICE QUESTION:
satellite:
Q.1. Two heavenly bodies S1 and S2 , not far off
2 r
Time period of a planet: T 
v0 from each other are seen to revolve in orbit

  (a) Around their common centre of mass

2 r 2 r 3/2 2 gR2
T    v0  
( gR 2 / r )1/2 gR 2   r  (b) Which are arbitrary
 

(c) With S1 fixed and S2 moving round S1

gR2
 r 3/2   r  ( gR 2 /  2 )1/3

(d) With S2 fixed and S1 moving round S2
Q.1. A body revolved around the sun 27 times
faster than the earth what is the ratio of their SOLUTION: (a)

radii
 Velocity of a Planet in terms of Eccentricity
(a) 1/3 (b)1/9 (c) 1/27 (d)1/4 𝑬

Perigee 𝐴𝑝𝑜𝑔𝑒𝑒
F r2
SOLUTION: (b) Sun

𝑨 𝒂 𝑩
r1 𝒄
Given : body  27earth
𝑫
 T 2  r3  2  1 / r3  r   2/3
CHAPTER NOTES 4
The speeds of planet at apogee and perigee are

GM  1  e  GM  1  e 
Final Information:
va    and vp 
a  1e  a  1  e  (1) When a body falls from a height h to the
surface of the earth, its velocity on reaching the
TYPE-I: Speeds of planet at apogee/ perigee
surface of the earth is
and maximum/ minimum velocity of a planet:
1/2 1/2
  h    nR  
GM  1  e  GM  1  e   2 gR   (or) 2 g   Here : h  nR
va  vmin    & v p  vmax    R  h    n  1 
a 1e  a  1  e 
 When h << R, we find: v  2 gh
v p  v max  r ac 1e
and  a  
va  v min  rp a  c 1  e

(2) If the radius of earth changes, then length of

Q.1. The eccentricity of earth's orbit is 0.0167.
the day:
The ratio of its maximum speed in its orbit to its
minimum speed is  L  Constant  I11  I22

(a) 2.507 (b) 1.033 2 2  2 2  2

 MR1     MR22    T2  ?
5  24  5  T2
(c) 8.324 (d) 1.000
 2 2

SOLUTION: (b)  Inertia of earth  MR 

 5 

v p  v max  1e v 1  0.0167 Example: If the polar ice caps of earth melt,
   max   1.033
va  v min  1e vmin 1  0.0167 then affect the length of the day is

2  2 
 Geostationary satellites:  L  I  MR2     Constant  T  M
5 T 
“If the height of an artificial satellite above
‘‘Time of day would increase’’
earth’s surface be such that its periods of
revolution is equal to the period of revolution
(3)
(24 hours) of the axial motion of the earth, then
such a satellite is known as ‘geostationary
satellite’.

 Artificial satellite (i.e., geostationary satellite)

is launched on the equatorial plane.

(NOTE: New Delhi is not on the equatorial plane)

CHAPTER NOTES 5
Miscellaneous Exercise:

Q.1. A planet moves around the sun. At a given

point P, it is closest from the sun at a distance
d1 and has a speed v1 . At another point Q, when

it is farthest from the sun at a distance d2 , its

speed will be

d12 v1 d2 v1
(a) 2 (b)
d2 d1

d1 v1 d22 v1
(c) (d)
d2 d12

SOLUTION: (c)

Angular momentum remains constant

v1d1
 mv1d1  mv2d2  v2 
d2

Q.2. The mean radius of the earth is R, its

angular speed on its own axis is  and the
acceleration due to gravity at earth's surface is g.
The cube of the radius of the orbit of a
geostationary satellite will be

(a) R 2 g /  (b) R 22 / g

(c) Rg / 2 (d) R 2 g / 2

SOLUTION: (d)

GM gR 2
Orbital velocity  v0    and v0  r 
r r

R2 g
 r3 
2
CHAPTER NOTES 1

ONE OR MORE CORRECT OPTIONS

Q.1. Consider a planet moving in an elliptical

orbit round the sun. The work done on the
planet by the gravitational force of the sun

(a) Is zero in any small part of the orbit

(b) Is zero in some parts of the orbit

(c) Is zero in one complete revolution

(d) Is zero in no part of the motion

SOLUTION: (b), (c)

 In one complete revolution, the

displacement of the planet is zero; hence the
work done is zero.

 At extreme position of major axis

gravitational force is perpendicular to the
direction of motion, then work done in that
part is zero.
 DAILY ACTIVITY 1
KEPLER’S LAW of PLANETARY (c) The earth's potential energy remains

MOTION constant

(d) All are correct

1. The distance of neptune and saturn
from sun are nearly 1013 and 1012 4. Venus looks brighter than other
meters respectively. Assuming that they planets because
move in circular orbits, their periodic (a) It is heavier than other planets
times will be in the ratio
(b) It has higher density than other planets
(a) 10 (b) 100
(c) It is closer to the earth than other

(c) 10 10 (d) 1 / 10 (d) It has no atmosphere

2. The figure shows the motion of a planet 5. A planet moves around the sun. At a
around the sun in an elliptical orbit given point P, it is closest from the sun
with sun at the focus. The shaded at a distance d1 and has a speed v1 . At
areas A and B are also shown in the another point Q, when it is farthest
figure which can be assumed to be from the sun at a distance d2 , its speed
equal. If t1 and t 2 represent the time for
will be
the planet to move from a to b and d
d12v1 d2v1
to c respectively, then (a) (b)
d22 d1
(a) t1  t2 b
a d1v1 d22v1
A (c) (d)
(b) t1  t 2 d2 d12
S
(c) t1  t 2 6. The rotation period of an earth
B
d c satellite close to the surface of the earth
(d) t1  t2
is 83 minutes. The time period of
3. The earth revolves about the sun in an another earth satellite in an orbit at a
elliptical orbit with mean radius distance of three earth radii from its

9.3  107 m in a period of 1 year. surface will be

Assuming that there are no outside (a) 83 minutes (b) 83  8 minutes

influences
(c) 664 minutes (d) 249 minutes
(a) Earth's kinetic energy remains constant
7. A satellite of mass m is circulating
(b) Earth's angular momentum remains around the earth with constant angular
constant
velocity. If radius of the orbit is R0 and
 DAILY ACTIVITY 2
mass of the earth M, the angular 12. The maxi. and mini. distances of a
momentum about the centre of the comet from the sun are 8  1012 m and
earth is
1.6  1012 m . If its velocity when nearest
(a) m GMR0 (b) M GmR0 to the sun is 60 m/s, what will be its
velocity in m/s when it is farthest
GM GM
(c) m (d) M
R0 R0 (a) 12 (b) 60

(c) 112 (d) 6

8. A satellite A of mass m is at a distance of r
from the centre of the earth. Another 13. The period of moon’s rotation around
satellite B of mass 2m is at a distance of the earth is nearly 29 days. If moon’s
2r from the earth's centre. Their time mass were 2 fold, its present value
periods are in the ratio of and all other things remained
unchanged, the period of moon’s
(a) 1 : 2 (b) 1 : 16
rotation would be nearly
(c) 1 : 32 (d) 1 :2 2
(a) 29 2 days (b) 29/ 2 days
9. The earth E moves in an elliptical orbit
with the sun S at one of the foci as (c) 29 × 2 days (d) 29 days

shown in figure. Its speed of motion will 14. Two planets at mean distance d1 and
be maximum at the point
d2 from the sun and their frequencies
(a) C C are n1 and n2, then
E
(b) A (a) n12d12  n2d22 (b) n22d23  n12d13
A B
(c) B S
(c) n1d12  n2d22 (d) n12d1  n22d2
(d) D D
15. A planet is revolving around the sun
10. If mass of a satellite is doubled and
as shown in elliptical path
time period remain constant the B

ratio of orbit in the two cases will be

A C
(a) 1 : 2 (b) 1 : 1 S
D
(c) 1 : 3 (d) None of these The correct option is
11. In the solar system, which is conserved (a) The time taken in travelling DAB is less
than that for BCD
(a) Total Energy (b) K.E.
(b) The time taken in travelling DAB is
(c) Angular Velocity (d) Momentum
greater than that for BCD
 DAILY ACTIVITY 3
(c) The time taken in travelling CDA is less speed in its orbit to its minimum speed
than that for ABC is

(d) The time taken in travelling CDA is (a) 2.507 (b) 1.033
greater than that for ABC
(c) 8.324 (d) 1.000
16. In the previous question the orbital
velocity of the planet will be
minimum at

(a) A (b) B

(c) C (d) D

17. In planetary motion the areal

velocity of position vector of a planet
depends on angular velocity () and
the distance of the planet from sun (r).
If so the correct relation for areal
velocity is

dA dA
(a)  r (b)  2r
dt dt

dA dA
(c)  r 2 (d)  r
dt dt

18. Suppose the law of gravitational

attraction suddenly changes and becomes
an inverse cube law i.e. F  1/ r3 , but
still remaining a central force. Then

(a) Keplers law of areas still holds

(b) Keplers law of period still holds

(c) Keplers law of areas and period still

hold

(d) Neither the law of areas, nor the law of

period still holds

19. The eccentricity of earth's orbit is

0.0167. The ratio of its maximum
 DAILY ACTIVITY 4
SOLUTION 9. (b) Speed of the earth will be
3/2 3/2 maximum when its distance from the sun
T R   1013 
1. (c) 1   1    12   (1000) 1/2
 10 10 is minimum because mvr  constant
T2  R2   10 

2. (c) Areal velocity of the planet 10. (b) Mass of satellite does not affect on

remains constant. If the areas A and B are orbital radius.

equal, then t1  t2 .
11. (a)

3. (b) Kinetic and potential energies

12. (a)By conservation of angular
varies with position of earth w.r.t. sun.
momentum: mvr = constant
Angular momentum remains constant
 vmin  rmax  vmax  rmin
everywhere.

60  1.6  1012 60
4. (c)  vmin    12m / s
8  1012 5

5. (c)Angular momentum remains 13. (d) Time period does not depends upon
vd
constant:  mv1d1  mv2d2  v2  1 1 the mass of satellite.
d2

T2 T2 1
6. (c)For first satellite: r1  R & T1  83 min 14. (b) 3
 3  2 3  constant  n12d13  n22d23
R d nd

For second satellite: r2  4R

15. (a) During path DAB planet is nearer
3/2
r  to sun as comparison with path BCD. So
 T2  T1  2   T1 (4)3/2  8T1 = 664 min
 r1  time taken in travelling DAB is less than
that for BCD because velocity of planet will
7. (a) Angular momentum
be more in region DAB.
= Mass ×Orbital velocity ×Radius

 GM  16. (c) Because distance of point C is

 L  m   R  m GMR0
 R  0 maximum from the sun.
 0 

8. (d) Mass of the satellite does not dA L dA

17. (c)    vr  r 2
dt 2m dt
effects on time period
3/2 3/2 1/2
T r   r  1 1 18. (d)
 A  1      
TB  r2   2r  8 2 2
vmax 1  e 1  0.0167
19. (b)    1.033
vmin 1  e 1  0.0167
 DAILY ACTIVITY 1
5. A ball is dropped from a spacecraft
GEOSTATIONARY SATELLITE
revolving around the earth at a height
1. The relay satellite transmits the T.V. of 120 km. What will happen to the ball
programme continuously from one
(a) It will continue to move with velocity v
part of the world to another because:
along the original orbit of spacecraft
(a) Period is greater than the period of
(b) It will move with the same speed
rotation of the earth
tangentially to the spacecraft
(b) Period is less than the period of
(c) It will fall down to the earth gradually
rotation of the earth about its axis
(d) It will go very far in the space
(c) Period has no relation with the period
of the earth about its axis 6. An artificial satellite is placed into a
circular orbit around earth at such a
(d) Period is equal to the period of rotation
height that it always remains above a
of the earth about its axis
definite place on the surface of earth. Its
2. A geostationary satellite height from the surface of earth is
(a) Revolves about the polar axis (a) 6400 km (b) 4800 km
(b) Has a time period less than that of the (c) 32000 km (d) 36000 km
near earth satellite
7. Which of following statements is correct in
(c) Moves faster than a near earth satellite
respect of a geostationary satellite
(d) Is stationary in the space
(a) It moves in a plane containing the
3. The time period of a geostationary Greenwich meridian
satellite is (b) It moves in a plane perpendicular to
(a) 24 hours (b) 12 hours the celestial equatorial plane

(c) 365 days (d) One month (c) Its height above the earth’s surface is
about the same as the radius of the earth
4. The mean radius of the earth is R, its
angular speed on its own axis is  (d) Its height above the earth’s surface is
and the acceleration due to gravity at about six times the radius of the earth
earth's surface is g. The cube of the 8. The distance of a geo-stationary satellite
radius of the orbit of a geostationary from the centre of the earth (Radius R
satellite will be = 6400 km) is nearest to
(a) R g / 
2
(b) R  / g
2 2
(a) 5 R (b) 7 R

(c) Rg / 2 (d) R 2 g / 2 (c) 10 R (d) 18 R

 DAILY ACTIVITY 2
9. Where can a geostationary satellite be SOLUTION
installed 1. (d) Telecommunication satellites are
(a) Over any city on the equator geostationary satellite

(b) Over the north or south pole

2. (a)
(c) At height R above earth
3. (a)
(d) At the surface of earth
GM gR 2
4. (d) Orbital velocity: v0  
r r

R2 g
and v0  r   r  2
3

5. (a) Due to inertia it will continue to

move along the original path of the space
craft.

6. (d)

7. (d)

8. (b) 6R from the surface of earth and 7R

from the centre.

9. (a)
BOUNCH SERIES 168
PLANTERY MOTION L L
(a) (b)
2m m
LEVEL-1
2L 2L
1. Two planets revolve round the sun (c) (d)
m Me
with frequencies N1 and N2 revolutions
5. A satellite orbiting the circular orbit of
per year. If their average orbital radii
radius R completes 1 revolution in 3h.
be R1 and R2 respectively, then R1 / R2 is
If orbital radius of geostationary satellite
equal to is 36000 km, the orbital radius R of
(a) ( N1 / N2 )3/2 (b) ( N2 / N1 )3/2 satellite is

(a) 6000 km (b) 9000 km

(c) ( N1 / N2 )2/3 (d) ( N2 / N1 )2/3
(c) 12000 km (d) 15000 km
2. The earth moves in an elliptical orbit
6. Time period of revolution of a nearest
with the sun S at one of foci as shown in
satellite around a planet of radius R is T.
the figure. Its rotational kinetic energy
Period of revolution around another
is maximum at the point
planet, whose radius is 3R but having
same density is

(a) T (b) 3T

(a) A (b) B (c) 9T (d) 3 3T

(c) C (d) D 7. Two satellites of masses m1 and m2

3. Near earth’s surface, time period of a (where m1 is greater than m2 ) are going

satellite is 4h. Find its time period at around the earth in orbits of radius r1
height 4R from the centre of earth
and r2(r1  r2 ) . Which statement about
 1  their velocities is correct?
(a) 32 h (b)  3  h
8 2
(a) v1  v2 (b) v1  v2
(c) 8 3 2 h (d) 16 h
v1 v 2
(c) v1  v2 (d) 
4. A satellite of mass m, moving around r1 r2
the earth in a circular orbit of radius R,
8. The mean radius of the earth's orbit
has angular momentum L. the areal
round the sun is 1.5  1011 . The mean
velocity of satellite is (Me  mass of
radius of the orbit of mercury round
earth)
BOUNCH SERIES 169
the sun is 6  1010 m . The mercury will
rotate around the sun in (c) T2 (d) T2

(a) A year (b) Nearly 4 years

R3 R3
(c) Nearly 1/ 4 year (d) 2.5 years 12. A satellite is launched into a circular
orbit of radius R around the earth.
9. Imagine a light planet revolving around
A second satellite is launched into an
a very massive star in a circular orbit
orbit of radius (1.01)R. The period of
of radius R with a period of revolution T.
the second satellite is larger than that of
If the gravitational force of attraction
the first one by approximately
between planet and star is proportional
to R  5/2 , then T 2 is proportional to (a) 0.5% (b) 1.0%

(c) 1.5% (d) 3.0%

(a) R 3 (b) R 7/2
13. If the distance between the earth and
(c) R 5/2 (d) R 3/2
the sun becomes half its present value,
10. A satellite S is moving in an elliptical the number of days in a year would have
orbit around the earth. The mass of been
the satellite is very small compared to
(a) 64.5 (b) 129
the mass of earth
(c) 182.5 (d) 730
(a) The acceleration of S is always directed
towards the centre of the earth

(b) The angular momentum of S about the

centre of the earth changes in direction
but its magnitude remains constant

(c) The total mechanical energy of S varies

periodically with time

(d) The linear momentum of S remains

constant in magnitude

11. Which of the following graphs represents

the motion of a planet moving about the sun

T2 T2
(a) (b)

R3 R3
BOUNCH SERIES 170
SOLUTION  3 
2/3

 R   36000  9000 km

 24 
1. (d) According to Kepler's law, T 2  R3

If N is the frequency, then N 2  ( R )3 6. (d)

3/2 2/3
N R  R N  7. (b) Orbital velocity of a satellite around in
 2  2   1  2 
N1  R1  R2  N1 
GM 1 v r
any planet, v   v  1 2
r r v2 r1
2. (a) The earth revolves around the sun once
in a year. In order to find the rotational kinetic As r1  r2  v1  v2
energy about the sun, we may treat the earth as a
8. (d)
point mass, of moment of inertia I. If  is
angular velocity of earth, then Krot  1/2 I2
9. (b) Given : F  R 5/2 , F  1 / R 5/2 ,T  ?

In planetary motion, angular momentum

For revolution of planet, T  R  n  1/2 
remains conserved. At A, the moment of
inertia is least, so angular speed is maxi. So,  F  1 / R 
n

5 
 2 1  /2
at A, rotational kinetic energy is maxi.  T R  
 T 2  R7/2

(a) Given : T1  4h, R1  R , R2  4R ,T2  ? 10. (a)

3.
Kepler’s third law of planetary motion,
11. (c)  T 2  R3
T  kR
2 3

3/2 12. (c)Given: Orbital radius is increased by 1%.

T 2 R3  4R 
 12  13  T2     4  32 h
T2 R2  R  Time period of satellite, T  r 3/2

4. (a) Kepler’s second law of planetary  Percentage change in time period

motion area speed of planet is constant. 3 3
= (% change in orbital radius) = (1%)  1.5%
2 2
dA L
Therefore   Constant
dt 2 m
2 3
13. (b) For revolution of planet , T  r

5. (b) Given :T1  3h, R1  ?,T2  24h, R2  36000km 2 3 3

T   r   r 
  1   1    11  8
Kepler’s third law: T 2  k R3  T2   r2   2 r1 

T R 
3/2 3/2 T1 365 days
3  R   T2    129 days
 1  1   
T2  R2  24  36000  2 2 2 2
IIT MAINS 1
2
CHAPTER-6 mv Mm GM  1
  G n  v 2  n 1 Fg  n 
R R R R
PART-3
4 2R2 GM  2 R 
  n 1 v  T 
KEPLER’S LAW T2 R

1/2
4 2 n1  4 2  n1
(R) 2  T  R 
n 1 /2
1. One satellite of Earth has orbital periodic T  2
R  T 
GM  GM 
time 5 hours. If distance between them is
made four times the present distance then
3. Figure shows elliptical path abcd of a planet
new orbital periodic time of satellite would
around the sun S such that the area of triangle
become ……....
1
csa is the area of the ellipse. (See figure) With
(a) 10h (b) 80h (c) 40h (d) 20h 4
db as the semi major axis, and ca as the semi
[2003]
minor axis, If t1 is the time taken for planet to
Solution: (c)
go over path abc and t 2 for path taken over
Given: T1  5 hours, T2  ?, r1  r ( let ), r2  4r
cda then

T12 r13
 T 2  r3  
T22 r23

3
(5)2  r  1
 2     T2  5  8  40 h
T2  4r  64

(a) t1  4 t 2 (b) t1  2t 2
2. If we assume that gravitational force
exerted by sun on a satellite orbiting around Sun (c) t1  3t 2 (d) t1  t 2
with orbital radius R; depends inversely on
[2016]
nth power of R ,then orbital periodic time of
this satellite would be directly proportional to Solution: (c)

…… A
Let area of ellipse = A,  Acba  Acda 
2
(a) R  (b) R
n 1 /2
(d) R 
n 1 /2 n  2 /2
(c) R n
 A A 3A
[2004] Remember: Acsab  Acsa  Acab     
 4 2 4
Solution: (a)
 1 A
In the present case,  Acsa  4 Area of ellipse  4 

Centripetal force (Fc )  gravitational force ( Fg )

SSK PUBLICATION 2
dA A t
Kepler’s IInd law,  Constant  abcsa  1
dt Acda t 2

t1 (3 A / 4)  3A A 
  3  Acda  A  Acsab  A  4  4 
t 2 ( A / 4)

4. A particle is moving with a uniform speed in a

circular orbit of radius R in a central force
inversely proportional to the nth power of R .
If the period of rotation of the particle is T , then
n
1
(a) T  R3/2 for any n (b) T  R 2

(c) T  R( n1)/2 (d) T  R n/2

[2018]

Solution: (c)
n 1
1
If F  n
, then T  R 2
R