# Chapter 16 Gravitation Part 2 – Physics free study material by TEACHING CARE online tuition and coaching classes

Chapter 16 Gravitation Part 2 – Physics free study material by TEACHING CARE online tuition and coaching classes

File name : Chapter-16-Gravitation-Part-2.pdf

• It is a scalar quantity because it is defined as work done per unit
• Unit : Joule/kg or m2/sec2
• Dimension : [M0L2T–2]

• If the field is produced by a point mass then

V = –

I dr = –

æ- GM ö dr

[As I = – GM ]

ç     r 2  ÷                              r 2

 ò
 ò
 è
 ø

\                                                                         V = – GM + c

r

[Here c = constant of integration]

Assuming reference point at ¥ and potential to be zero there we get

0 = – GM + c Þ c = 0

¥

\                                    Gravitational potential V = – GM r

• Gravitational potential difference : It is defined as the work done to move a unit mass from one point to the other in the gravitational The gravitational potential difference in bringing

unit test mass m from point A to point B under the gravitational influence of source mass M is

WA® B

æ 1      1 ö

D V = VB   VA   =   m

= –GM ç     –     ÷

r        r

è B             A ø

• Potential due to large numbers of particle is given by scalar addition of all the

V = V1 + V2 + V3 + ……….

= – GM GM GM ……..

r1                r2                r3

 å

i=n    M

= –G            i

i=1  ri

• Point of zero potential : It is that point in the gravitational field, if the unit mass is shifted from infinity to that point then net work done will be equal to

Let m1 and m2 are two masses placed at d distance apart and P is the point of zero potential in between the two masses.

Net potential for point

P = VA  + VB   = 0

Þ             – Gm1  – Gm2  = 0

By solving

x         d x

x =     m1 d m1 – m2

# Gravitational Potential for Different Bodies.

## (1) Potential due to uniform ring

 At a point on its axis At the centre V = –       GMa 2 + r 2 V = – GMa

• Potential due to spherical shell

 Outside the surfacer > R On the surfacer = R Inside the surfacer < R V = –GMr V = –GMR V = –GMR

## (3) Potential due to uniform solid sphere

Problem 35.     In some region, the gravitational field is zero. The gravitational potential in this region                           [BVP 2003]

• Must be variable (b) Must be constant (c) Cannot be zero                  (d) Must be zero

Solution : (b)      As I = – dV , if

dx

I = 0 then V = constant.

Problem 36.     The gravitational field due to a mass distribution is E = K / x 3 in the x – direction (K is a constant). Taking the

gravitational potential to be zero at infinity, its value at a distance x is                                             [MP PET 1994]

(a)

K / x

K / 2x

K / x 2

K / 2x 2

Solution : (d)

V = -ò

E dx = –

K dx =

 ò 3

x

K   .

2x 2

Problem 37.     The intensity of gravitational field at a point situated at a distance of 8000 km from the centre of the earth is

6 N / kg . The gravitational potential at that point is – (in Joule / kg)

(a)

8 ´ 106

(b)

2.4 ´ 10 3

(c)

4.8 ´ 107

(d)

6.4 ´ 1014

Solution : (c)      Gravitational intensity at point P,

I GM

r 2

and gravitational potential V = – GM

r

\ V = I ´ r

= 6 N / kg ´ 8000 km

= 4.8 ´ 107

Joule .

kg

Problem 38.     The gravitational potential due to the earth at infinite distance from it is zero. Let the gravitational potential at

a point P be -5 J / kg . Suppose, we arbitrarily assume the gravitational potential at infinity to be + 10 then the gravitational potential at P will be

J / kg ,

(a)

-5 J / kg

(b)

+5 J / kg

(c)

-15 J / kg

(d)

+15 J / kg

Solution : (b)      Potential increases by +10 J / kg

every where so it will be +10 – 5 = +5J / kg

at P

Problem 39.     An infinite number of point masses each equal to m are placed at x =1. x = 2, x = 4, x = 8 ……… What is the total gravitational potential at x = 0

(a)

• Gm

(b)

-2Gm

(c)

• 4Gm

(d)

• 8Gm

 ê
 ú

Solution : (b)      Net potential at origin V = -é Gm + Gm + Gm +……………….. ù

é1    1    1     1 ù

ë r1               r2              r3                           û

æ           ö

ç    1    ÷

= –Gmê1 + 2 + 4 + 8 ú   = –Gmç           1 ÷ = -2Gm

 ç
 2
 ÷

ë                           û                 ç 1 –      ÷

è           ø

Problem 40.     Two bodies of masses m and M are placed a distance d apart. The gravitational potential at the position where the gravitational field due to them is zero is V, then

(a)

V = – G (m + M) d

(b)

V = – Gm

d

(c)

V = – GM

d

(d)

V = – G (

d

+    M )2

Solution : (d)     If P is the point of zero intensity, then x =                   M       .d         and      d x =            m       d

M +   m                                                       +

Now potential at point P,

V = V  + V

= – GM –  GM

1             2                      x

d x

Substituting the value of x and d x we get V = – G ( m +

d

M )2 .

# Gravitational Potential Energy.

The gravitational potential energy of a body at a point is defined as the amount of work done in bringing the body from infinity to that point against the gravitational force.

W = ò

r GMm x 2

é 1 ùr

dx = –GMm êë x úû

¥                                                                         ¥

W = – GMm

r

This work done is stored inside the body as its gravitational potential energy

\ U = – GMm

r

• Potential energy is a scalar
• Unit : Joule

• Dimension : [ML2T–2]
• Gravitational potential energy is always negative in the gravitational field because the force is always attractive in
• As the distance r increases, the gravitational potential energy becomes less negative e., it increases.
• If r = ¥ then it becomes zero (maximum)
• In case of discrete distribution of masses

Gravitational potential energy U = å u

= -é Gm1m2   + Gm2 m3

+ ……..ù

 r

i              ê

ë     12

 ú

r23                                 û

• If the body of mass m is moved from a point at a distance r1

to a point at distance

r2 (r1 > r2 )

then

change in potential energy DU =

r2  GMm

= –GMm é 1 1 ù

or DU = GMm é 1 – 1 ù

òr               2    dx

êr        r ú

êr        r ú

1         x                                 ë 2       1 û

ë 1        2 û

As r1 is greater than r2 , the change in potential energy of the body will be negative. It means that if a body is

brought closer to earth it’s potential energy decreases.

• Relation between gravitational potential energy and potential U = – GMm= mé GM ù

\                         U = mV

r               êë   r      úû

• Gravitational potential energy at the centre of earth relative to

U          = m V

mæ- 3 GM ö

= – 3 GMm

centre

centre               ç              ÷

 2
 R

è              ø        2     R

• Gravitational potential energy of a body at height h from the earth surface is given by

U   = – GMm = – gR 2m = – mgR

h           R + h

R + h

1 + h

R

# Work Done Against Gravity.

If the body of mess m is moved from the surface of earth to a point at distance h above the surface of earth, then change in potential energy or work done against gravity will be

W                              é 1     1 ù

= DU = GMmê r   r ú

ë 1            2 û

Þ                        W = GMmé 1 –   1   ù

[As r = R and r

= R + h ]

êë R

R húû                                            1                                    2

Þ                        W =

GMmh R 2 æ1 + h ö

mgh

 R

1 + h

[As

GM  = g ]

R2

 R

ç         ÷

è         ø

• When the distance h is not negligible and is comparable to radius of the earth, then we will use above formula.
 ç        ÷
• If h = nR then W = mgRæn  ö

n + 1

è        ø

• If h = R

then W = 1 mgR

2

• If h is very small as compared to radius of the earth then term h / R can be neglected

From

W =    mgh     = mgh

1 + h / R

éAs   h  ® 0ù

 ê                 ú

ë       R           û

Problem 41.     Energy required to move a body of mass m from an orbit of radius 2R to 3R is                                     [AIEEE 2002]

(a)

GMm                                     (b)

12R 2

GMm                         (c)

3R 2

GMm

8 R

(d)

GMm

6R

Solution : (d)     Work done = Change in potential energy = U  U

é   GMmù       é   GMmù            GMm       GMm

 ú
 ê
 ú

= –                – –                 = –              +

GMm .

2         1     ê

ë

r2       û     ë

r1       û

3R           2R              6R

Problem 42.     A body of mass m kg. starts falling from a point 2R above the earth’s surface. Its kinetic energy when it has fallen to a point ‘R’ above the earth’s surface [R-Radius of earth, M-Mass of earth, G-Gravitational constant]

[MP PMT 2002]

(a)

1 GMm

(b)

• GMm

(c)

• GMm

(d)

1 GMm

2    R                                             6    R                                  3    R                                         3    R

Solution : (b)      When body starts falling toward earth’s surface its potential energy decreases so kinetic energy increases.

Increase in kinetic energy = Decrease in potential energy

Final kinetic energy – Initial kinetic energy = Initial potential energy – Final potential energy

æ    GMm ö      æ    GMm ö

Final kinetic energy – 0 = ç-

r

÷ – ç-          ÷

r

è         1        ø     è          2        ø

æ – GMm ö  æ

GMm ö

æ     GMm ö   æ

GMm ö

GMm

GMm

1 GMm

\ Final kinetic energy = ç

R + h

÷ – ç-           ÷

R + h

= ç- R + 2R ÷ – ç- R + R ÷ = –   3R    +    2R        = 6     R     .

è            1 ø   è

2 ø       è

ø   è                ø

Problem 43.     A body of mass m is taken from earth surface to the height h equal to radius of earth, the increase in potential energy will be                                                                                  [CPMT 1971, 97; IIT-JEE 1983; CBSE PMT 1991;

Haryana CEE 1996; CEET Bihar 1995; MNR 1998; RPET 2000]

(a)

mgR

(b)

1 mgR

2

(c)

2mgR

(d)

1 mgR

4

Solution : (b)      Work done =  mgh    , If h = R

1 + h / R

then work done =  mgR

1 + R / R

= 1 mgR .

2

Problem 44.     If mass of earth is M, radius is R and gravitational constant is G, then work done to take 1 kg mass from earth surface to infinity will be                                                                                                                        [RPET 1997]

(a)

(b)

GM                             (c)

R

(d)  GM

2R

Solution : (b)       Work done = U        U

= U     U

= 0 – æ- GMm ö

GMm

[As m = 1kg ]

final

initial                ¥              R                     ç                ÷

 R

è                ø

 R

Problem 45.     Three particles each of mass 100 gm are brought from a very large distance to the vertices of an equilateral triangle whose side is 20 cm in length. The work done will be

(a)

0.33 ´ 10 11 Joule

(b)

– 0.33 ´ 10 11 Joule (c)

1.00 ´ 10 11 Joule

(d)

– 1.00 ´ 10 11 Joule

Solution : (d)     Potential energy of three particles system

U = – Gm1m2  – Gm2m3  – Gm1m3

r12

r23

r13

Given m1

= m2

= m3

= 100 gm

and r12

= r23

= r13

= 2cm

é- 6.67 ´ 1011 ´ (101) ´ (101

-11

\ U = 3 ê

êë

20 ´ 102

ú = -1.00 ´ 10

úû

Joule .

Problem 46.     A boy can jump to a height h on ground level. What should be the radius of a sphere of density d such that on jumping on it, he escapes out of the gravitational field of the sphere

é 4p Gd ù1/ 2

é 4p

gh ù1 / 2

é 3   gh ù1/ 2

é  3  Gd ù1/ 2

 3
 û
• ê

ë

gh ú

(b)

êë 3

Gd úû

(c)

êë 4p Gd úû

(d)

ê 4p

gh ú

 ë
 û

Solution : (c)      When a boy jumps from a ground level up to height h then its velocity of jumping v =                                            …..(i)

and for the given condition this will become equal to escape velocity v

escape =              =

…..(ii)

Equating (i) and (ii)              = R

# Escape Velocity.

Þ R = é 3

 ê

ë 4p

gh ù1 / 2

Gd úû           .

The minimum velocity with which a body must be projected up so as to enable it to just overcome the gravitational pull, is known as escape velocity.

The work done to displace a body from the surface of earth (r = R) to infinity ( r = ¥ ) is

W = ¥ GMm

= –GMm é 1 – 1 ù

òR        x 2     dx

êë ¥        R úû

Þ                        W GMm

R

This work required to project the body so as to escape the gravitational pull is performed on the body by providing an equal amount of kinetic energy to it at the surface of the earth.

If v  is the required escape velocity, then kinetic energy which should be given to the body is

1 mv2

e

\                         1 mv2  = GMm

2      e

Þ         v =

2      e            R

e

Þ         ve  =

[As GM = gR2 ]

or                        ve =

Þ         ve  = R

[As

g = 4 prGR ]

3

• Escape velocity is independent of the mass and direction of projection of the
• Escape velocity depends on the reference Greater the value of (M / R) or (gR)

will be escape velocity.

for a planet, greater

• For the earth as

\ ve =

g = 9.8m / s 2 and R = 6400 km

= 11.2km / sec

• A planet will have atmosphere if the velocity of molecule in its atmosphere

é

êvrms =

êë

ù is lesser than

 ú
 ú

û

escape velocity. This is why earth has atmosphere (as at earth moon vrms < ve )

vrms < ve ) while moon has no atmosphere (as at

• If body projected with velocity lesser than escape velocity ( v < ve ) it will reach a certain maximum height and then may either move in an orbit around the planet or may fall down back to the
• Maximum height attained by body : Let a projection velocity of body (mass m ) is v , so that is attains a maximum height h . At maximum height, the velocity of particle is zero, so kinetic energy is

By the law of conservation of energy

Total energy at surface = Total energy at height h .

Þ           – GMm + 1 mv2 = – GMm + 0

R          2               R + h

Þ             v2   = GM é 1 –    1   ù

=   GMh

2

Þ            2GM

v2 R

êë R

R + h

h

R + húû

= 1 + R

h

R(R + h)

Þ            h =            R

=  R     R é      v 2     ù

[As v =

\  2GM  = v2 ]

æ 2GM         ö      v 2

êv 2v 2 ú

e                                                      R            e

 è

ç v2 R

– 1÷

ø

e   – 1         ë e                  û

v 2

• If a body is project with velocity greater than escape velocity (v > ve ) then by conservation of Total energy at surface = Total energy at infinite

1 mv2 – GMm = 1 m(v¢)2 + 0

2               R           2

i.e.,

(v¢)2 = v2 – 2GM

Þ v2 = v 2v 2

[As

2GM  = v2 ]

R                                     e                                                                                                             R            e

\                         v¢ =

1. e, the body will move in interplanetary or inter stellar space with velocity .
• Energy to be given to a stationary object on the surface of earth so that its total energy becomes zero, is called escape

Total energy at the surface of the earth = KE + PE = 0 – GMm

R

\                        Escape energy = GMm R

• If the escape velocity of a body is equal to the velocity of light then from such bodies nothing can escape, not even Such bodies are called black holes.

The radius of a black hole is given as

R = 2GM

C 2

[As C =

, where C is the velocity of light]

Problem 47.     For a satellite escape velocity is 11km / s . If the satellite is launched at an angle of 60o with the vertical, then escape velocity will be                                                                                     [CBSE 1993; RPMT 1997; AIEEE 2003]

(a)

11km / s

• 11

km / s

km / s

33 km / s

Solution : (a)  Escape velocity does not depend upon the angle of projection.

Problem 48.   The escape velocity from the earth is about 11km / s . The escape velocity from a planet having twice the radius and the same mean density as the earth, is                         [MP PMT 1987; UPSEAT 1999; AIIMS 2001; MP PET 2001, 2003] (a) 22 km/s                                 (b) 11 km/s                     (c) 5.5 km/s                            (d) 15.5 km/s

Solution : (a)

ve   =

=     8 prGR 2

3

\ ve   µ R

if r = constant. Since the planet having double radius in comparison

to earth therefore the escape velocity becomes twice i.e. 22 km/ s .

Problem 49.     A projectile is projected with velocity kve

in vertically upward direction from the ground into the space. ( ve is

escape velocity and k < 1). If air resistance is considered to be negligible then the maximum height from the centre of earth to which it can go, will be (R = radius of earth)                                                     [Roorkee 1999; RPET 1999]

(a)

R                                        (b)

k 2 + 1

R                            (c)

k 2 – 1

R

1 – k 2

(d)

R

k + 1

Solution : (c)      From the law of conservation of energy

Difference in potential energy between ground and maximum height = Kinetic energy at the point of projection

mgh     = 1 m(kve )2 = 1 mk 2 v 2 = 1 mk 2 (

2g R)2

[As v   =              ]

1 + h / R       2

2          e          2                                                    e

By solving height from the surface of earth h =

Rk 2

1 – k 2

So height from the centre of earth r = R + h = R +

Rk 2

1 – k 2

=     R      .

1 – k 2

Problem 50.     If the radius of earth reduces by 4% and density remains same then escape velocity will

[MP PET 1991; MP PMT 1995]

• Reduce by 2% (b) Increase by 2%        (c) Reduce by 4%             (d) Increase by 4%

Solution : (c)      Escape velocity ve µ R         and if density remains constant ve µ R

So if the radius reduces by 4% then escape velocity also reduces by 4%.

Problem 51.   A rocket of mass M is launched vertically from the surface of the earth with an initial speed V. Assuming the radius of the earth to be R and negligible air resistance, the maximum height attained by the rocket above the surface of the earth is                                                                                                                                            [AMU 1995]

(a)          R

(b)

Ræ gR

– 1ö

(c)

R                                (d)

Ræ 2gR – 1ö

æ gR

 è

ç 2V 2

– 1ö

 ÷

ø

ç 2V 2        ÷

æ 2gR

 è

ç V 2

– 1ö

 ÷

ø

ç V 2         ÷

 è
 è
 ø
 ø

Solution : (c)      Kinetic energy given to rocket at the surface of earth = Change in potential energy of the rocket in reaching

from ground to highest point

Þ   1 mv2 =      mgh

Þ   v 2 =        g

Þ 1 + 1 = 2g

Þ 1 = 2g – 1 Þ 1 = 2gR v 2

Þ h =

v 2 R

2           1 + h / R

2      1 + 1

h      R      v 2

h      v 2     R         h

v 2 R

2gR v 2

Þ h =

h      R

R

 è
 ø

æ 2gR – 1ö

ç v 2               ÷

Problem 52.     A body of mass m is situated at a distance

4 Re

above the earth’s surface, where Re

How much minimum energy be given to the body so that it may escape

(a)

mgRe                                                        (b)

2mgRe

(c)

mgRe

5

(d)

mgRe

16

Solution : (c)      Potential energy of the body at a distance 4 Re

from the surface of earth

U = –

mgRe

1 + h / Re

= – mgRe

1 + 4

= – mgRe

5

[As h = 4 Re

(given)]

So minimum energy required to escape the body will be

# Kepler’s Laws of Planetary Motion.

mgRe   .

5

Planets are large natural bodies rotating around a star in definite orbits. The planetary system of the star sun called solar system consists of nine planets, viz., Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune and Pluto. Out of these planets Mercury is the smallest, closest to the sun and so hottest. Jupiter is largest and has maximum moons (12). Venus is closest to Earth and brightest. Kepler after a life time study work out three empirical laws which govern the motion of these planets and are known as Kepler’s laws of planetary motion. These are,

• The law of Orbits : Every planet moves around the sun in an elliptical orbit with sun at one of the
• The law of Area : The line joining the sun to the planet sweeps out equal areas in equal interval of i.e. areal velocity is constant. According to this law planet will move slowly when it is farthest from sun and more rapidly when it is nearest to sun. It is similar to law of conservation of angular momentum.

Areal velocity = dA = 1

r(vdt)1 rv

dt       2     dt          2

\                   dA = L

[As

L = mvr ; rv = L ]

dt       2 m                                                            m

• The law of periods : The square of period of revolution proportional to the cube of the semi-major axis of the

(T)

of any planet around sun is directly

æ r r ö3

T 2 µ a 3 or T 2 µ ç 1               2 ÷

è    2    ø

Proof : From the figure AB = AF + FB

2a = r1

• r2

a r1 + r2

2

where a = semi-major axis

r1 =

Shortest distance of planet from sun (perigee).

r2 =

Largest distance of planet from sun (apogee).

## Important data

Note:: @ Kepler’s laws are valid for satellites also.

# Velocity of a Planet in Terms of Eccentricity.

Applying the law of conservation of angular momentum at perigee and apogee

mvprp = mvara

Þ                         vp

ra

a + c1 + e

[As r

= ac,

r = a + c

and eccentricity e = c ]

va           rp

ac

1 – e                           p                                      a                                                                                            a

Applying the conservation of mechanical energy at perigee and apogee

• mv

2  – GMm1 mv 2  – GMm  Þ

v 2 – v 2 = 2GM é 1 – 1 ù

• p r

2      a                  r                   p               a

êr        r ú

p

ér 2 – r 2 ù

a

ér r ù

êë p          a úû

 r
 p

v r

 r
 a
 2

Þ                         v 2 ê a             p     ú = 2 GM ê a          p ú

[As v   =   a   a   ]

ëê    p             úû

ëê rarp     úû                                         p

Þ                         v 2 = 2 GM érp ù Þ v 2 = 2 GM æ a c ö = 2 GM æ 1 – e ö

a             r + r

ê r ú        a

a     ç a + c ÷

a     ç 1 + e ÷

a           p     ë a û

è         ø             è        ø

Thus the speeds of planet at apogee and perigee are

va   =                         ,               vp =

Note:: @ The gravitational force is a central force so torque on planet relative to sun is always zero, hence angular momentum of a planet or satellite is always constant irrespective of shape of orbit.

# Some Properties of the Planet.

 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto Mean distacne from sun, 106 km 57.9 108 150 228 778 1430 2870 4500 5900 Period of revolution, year 0.241 0.615 1.00 1.88 11.9 29.5 84.0 165 248 Orbital speed, km/s 47.9 35.0 29.8 24.1 13.1 9.64 6.81 5.43 4.74 Equatiorial diameter, km 4880 12100 12800 6790 143000 120000 51800 49500 2300 Mass (Earth =1) 0.0558 0.815 1.000 0.107 318 95.1 14.5 17.2 0.002 Density (Water =1) 5.60 5.20 5.52 3.95 1.31 0.704 1.21 1.67 2.03 Surface value of g, m/s2 3.78 8.60 9.78 3.72 22.9 9.05 7.77 11.0 0.5 Escape velocity, km/s 4.3 10.3 11.2 5.0 59.5 35.6 21.2 23.6 1.1 Known satellites 0 0 1 2 16+ring 18+rings 17+rings 8+rings 1

Problem 53.     The distance of a planet from the sun is 5 times the distance between the earth and the sun. The Time period of the planet is                                                                                                                                     [UPSEAT 2003]

(a)

5 3 / 2 years

(b)

5 2 / 3 years

(c)

51/ 3 years

(d)

51/ 2 years

Solution : (a)      According to Kepler’s law T µ R 3 / 2

Tplanet

= (5)3 / 2 Tearth

= 5(3 / 2) ´ 1 year

= 53 / 2 years .

Problem 54.     In planetary motion the areal velocity of position vector of a planet depends on angular velocity (w) and the distance of the planet from sun (r). If so the correct relation for areal velocity is                                              [EAMCET 2003]

(a)

dA µ w r dt

(b)

dA µ w 2 r dt

(c)

dA µ w r 2

dt

(d)

dA µ

dt

Solution : (c)

dA =  L

mvr

= 1 wr 2

[As Angular momentum L = mvr and v = rw ]

dt       2m         2m        2

\ dA µ wr 2 .

dt

Problem 55.     The planet is revolving around the sun as shown in elliptical path. The correct option is                     [UPSEAT 2002]

• The time taken in travelling DAB is less than that for BCD
• The time taken in travelling DAB is greater than that for BCD
• The time taken in travelling CDA is less than that for ABC
• The time taken in travelling CDA is greater than that for ABC

Solution : (a)      When the planet passes nearer to sun then it moves fast and vice-versa. Hence the time taken in travelling

DAB is less than that for BCD.

Problem 56.     The distance of Neptune and Saturn from sun are nearly 1013 and 1012 meters respectively. Assuming that they move in circular orbits, their periodic times will be in the ratio                       [NCERT 1975; CBSE PMT 1994; MP PET 2001]

(a)                                      (b) 100                      (c) 10                             (d) 1 /

T                æ R

ö3 / 2          æ 1013 ö3 / 2

Solution : (c)      Kepler’s third law T 2 µ R 3 \

Neptune   = ç   Neptune ÷        = ç

÷       = 10

10 .

TSaturn           ç RSaturn ÷

ç 1012 ÷

è               ø          è          ø

Problem 57.  The maximum and minimum distance of a comet from the sun are 8 ´ 1012 m and 1.6 ´ 1012 m. If its velocity when nearest to the sun is 60 m/s, what will be its velocity in m/s when it is farthest             [Orissa JEE 2001] (a) 12     (b) 60                                       (c) 112                      (d) 6

Solution : (a)      According to conservation of angular momentum

mvmin rmax

= mvmax rmin = constant

rmin

æ 1.6 ´ 1012 ö

vmin = vmax ´

 r

max

= 60 ´ ç

è

8 ´ 1012

÷ = 12m / s

ø

Problem 58.     A satellite A of mass m is at a distance of r from the centre of the earth. Another satellite B of mass 2m is at distance of 2r from the earth’s centre. Their time periods are in the ratio of                                                     [CBSE PMT 1993]

(a) 1 : 2                              (b) 1 : 16                   (c) 1 : 32                         (d) 1 : 2 2

Solution : (d)     Time period does not depend upon the mass of satellite, it only depends upon the orbital radius.

T        æ r

ö 3 / 2

æ r ö 3 / 2        1

According to Kepler’s law

1   = ç 1  ÷

= ç      ÷        =            .

T2          è r2 ø

è 2r ø

Problem 59.     A planet moves around the sun. At a given point P, it is closed 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                     [MP PMT 1987]

d 2v

d  v1

d  v1

d 2v

(a)

1   1

 d

2

2

•   2

d1

•   1

d 2

2   1

 d

2

1

Solution : (c)      According to law of conservation of angular momentum mv d = mv d                   \ v   = d1v1 .

 d

1   1                 2   2                      2

2

# Orbital Velocity of Satellite.

Satellites are natural or artificial bodies describing orbit around a planet under its gravitational attraction. Moon is a natural satellite while INSAT-1B is an artificial satellite of earth. Condition for establishment of artificial satellite is that the centre of orbit of satellite must coincide with centre of earth or satellite must move around great circle of earth.

Orbital velocity of a satellite is the velocity required to put the satellite into its orbit around the earth. For revolution of satellite around the earth, the gravitational pull provides the required centripetal force.

mv2

r

Þ             v =

v =

GMm r 2

= R

[As GM = gR 2 and r = R + h ]

• Orbital velocity is independent of the mass of the orbiting body and is always along the tangent of the orbit

i.e., satellites of deferent masses have same orbital velocity, if they are in the same orbit.

• Orbital velocity depends on the mass of central body and radius of
• For a given planet, greater the radius of orbit, lesser will be the orbital velocity of the satellite (v µ 1 / r ).
• Orbital velocity of the satellite when it revolves very close to the surface of the planet

v =            =

\ v =

GM  =

R

[As h = 0

and GM = gR 2 ]

For the earth v =                              = 7.9 k m / s » 8 km / sec

• Close to the surface of planet v =

[As ve   =             ]

\                         v = ve

2

i.e., v

escape   =

vorbital

It means that if the speed of a satellite orbiting close to the earth is made it will escape from the gravitational field.

times (or increased by 41%) then

• If the gravitational force of attraction of the sun on the planet varies as

F µ 1

r n

then the orbital velocity

varies as v µ       1      .

Problem 60.   Two satellites A and B go round a planet P in circular orbits having radii 4R and R respectively. If the speed of the satellite A is 3V, the speed of the satellite B will be         [MNR 1991; AIIMS 1995; UPSEAT 2000] (a) 12 V                (b) 6 V                                                  (c) 3/2 V                                                               (d) 3/2 V

Solution : (b)      Orbital velocity of satellite v =

\ v µ 1   Þ

vB     =

v A

Þ   vB    =

3V

Þ  vB

= 6V .

Problem 61.     A satellite is moving around the earth with speed v in a circular orbit of radius r. If the orbit radius is decreased by 1%, its speed will                                                                                                          [MP PET 1996, 99, 2002]

• Increase by 1% (b) Increase by 5% (c) Decrease by 1%          (d) Decrease by 0.5%

Solution : (b)      Orbital velocity v =

\ v µ 1

[If r decreases then v increases]

Percentage change in v = 1

2

(Percentage change in r) = 1

2

(1%) = 0.5% \orbital velocity increases by 0.5%.

Problem 62.     If the gravitational force between two objects were proportional to 1/R; where R is separation between them, then a particle in circular orbit under such a force would have its orbital speed v proportional to

[CBSE PMT 1994; JIPMER 2001, 02]

(a)

1 / R 2

• R 0
• R1

1 / R

Solution : (b)      If F µ 1

Rn

then v µ

1     ; here n = 1 \v µ

1     µ R 0 .

Problem 63.     The distance between centre of the earth and moon is 384000 km. If the mass of the earth is 6 ´ 1024 kg and

G = 6.67 ´ 10 11 Nm2 / kg 2 . The speed of the moon is nearly                                                 [MH CET 2002]

• 1 km / sec (b) 4 km / sec                  (c) 8 km / sec                          (d) 2 km / sec

Solution : (a)      Orbital velocity v =                  =                                                    v = 1.02 km / sec = 1km / sec (Approx.)