ALP Sol P Gravitation E

8/18/2019 ALP Sol P Gravitation E

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SOLN GRAVITATION - 1

TOPIC : GRAVITAION

PART - I1.

Tsin! = 2

2

x) –(a

Gm

Tcos! = mg

deviding we get tan ! =gx) –(a

mG2

2. Inside the shell gravitation field due to the shell will be zero but there will be some gravitational field due to theblock.

3. From modified Gauss’s theorem for gravitation

" sd.E!!

= 4 #

G %

%

&

'

((

)

* +"

,

,

rr

0r

dv

E 4 #

r2 = 4

#

G "

,

,

#rr

0r

2 drr4r

k

get E = constant

4. as E is constant, so the Potential (V = – " drE ) will be proportional to r

5. B

Applying energy conservation form P to O

Ki + U

i = K

f + U

f

O + (Mo)

- . %% &

'

(()

*

/ 22 2RR

GM –

=21 m

o V2 + (m

o) % & '() * R

GM –

0 V = %% & '((

) *

51 –1

R2GM

PHYSICS SOLUTIONS OF"ADVANCED LEVEL PROBLEMS"

Target : ISEET (IIT-JEE)




PART - II

1. The mass of the sun is same for both the cases, so we can apply

2

earth

planet

T

T%% &

'(()

* =

3

earth

planet

R

R%% &

'(()

*

2

planet

year1T %%

& '((

) * =

3

rr2 % & '(

) *

Tplanet

= 23/2 years

The life span of the man is 70 years . During 70 years , the revolution completed by that planet = 3/22

705

25 revolutions. So he will see 25 summers, 25 winters, 25 springs ... so according to that planet his age

will be 25 years.

2. (a) particle is closet to ring, when it is at it’s centre, i.e. zero.

(b) External force on “Ring + particle” is zero, hence centre of mass remain stationary. Hence distance of

centre of mass from initial position of ring = displacement of rin

=pR

0PRmm

xm0m/ 6/6

=)103.0107.2(

610399

8

6/6

66

= 0.6 m(c) According to conservation of momentum –

mR V

R = m

P v

P

where vR and v

P are speed of ring and particle in opposite direction, when particle reaches centre of ring.2.7 × 109 v

R = 3 × 108 v

P

VP = 9v

R

By conservation of energy -

– 20

2

PR

xR

mGm

/ + 0 = –

R

mGm PR + 1/2 mRV

R2 + 1/2 m

Pv

p2

GMR × M

P 7

7

8

9

::

;

<

/=

20

2 xR

1

R

1 =

2

1 v

P2 7

8

9:;

</ P

R m81

m

6.67 × 10-11 × 2.7 × 109 × 3 × 108 78

9:;

<=

10

1

8

1 =

2

1v

P2

778

9

::;

<6/

6 89

10381

107.2

vP = 9 cm /sec.

3. Total distance from apogee to perigee(a) 300 + 2(6400) + 3000 = 2a

a = 8050 kmTime period of the spacecarft

T2 =3

e

2

aGM

4#

T =2 / 3

e

aGM

2# =

2 / 3

2

2e

a

RR

GM

4#

T =

2 / 3

agR

4#=

2 / 33

3 )108050(8.9106400

4

6666

#




(b) Apply angular momentum conservation about the centre of earth, between perigee and Apogee.mv

1 r

min = m v

2 r

max.............(i)

(v1) (300 + 6400)= v

2(3000 + 6400)

2

1

v

v =

67

94

(c) Also apply energy conservation between perigee an Apogee

21 mv

12 + %%

& '((

) * =

min

e

rGmM =

21 mv

22 + %%

& '((

) * =

max

e

rGmM .............(ii)

Where rmin

= (300 + 6400)km and rmax

= (3000 + 6400)kmFrom eqn. (i) & (ii) we get

v1 = 8.35 × 103 m/sec

v2 = 5.95 × 103 m/sec.

(d) To escape, velocity at r > ? should be zero.

Applying energy conservation between perigee and r > ?.

ki + U = kf + U

f

2

1mv

12 + %%

&

'(()

*

/=

)6400300(

mGM2 = 0 + 0

v1 = 11.44 × 103 m/sec.

Increase in speed = 11.44 × 103 – 8.35 × 103

= 3.09 × 103 m/sec.

4.

at max and min. distance, Velocity will be perpendicular to the radius vector, Applying angular momentum

conservation about the sun, between initial position and the position of max or minimum distance.

MV0r

0 sin@ = MVr ..........(1)

applying energy conservation :

2

1MV

o2 + %%

&

'(()

* =

0

S

r

mGM =

2

1 mv2 + %

&

'()

* =

r

mGMS

from equ (1) and (2) get

r = A=2ro %

& '(

) * @AA==B 2sin)2(11

wheres

oo

GM

vr,A

have / sign will give rmax

and = sign will give rmin

.

5. R = Radius of earth

r = radius of orbit of geostationary satellite

T = Time period of earth about its axis

T @ r3 / 2

3 @ r – 3 / 2

331

=

2

3 – ×

r

r1




1 ! =2

3 – ×

r

r1 × 3

Vrel

= (31 – 3

2 ) R = – 13 × R

=2

3 ×

r

r1 × R × 3

Vrel = 2

3

× r

r1

× R × T

2#

= rT

rR3 #1

Alternately

3earth

earth

3s

3earth

=earthT

2# =

hours 24

2#

If the satellite were geo –stationary its T would also be 24 hours. But radius is slightly increased, so its T

will also be increased.

T2 =3

2

RGM

4%% &

'(()

* #, taking log on both the sides

2 log T = 3log(R)GM

4 log

2

/%% &

'(()

* #

Differentiating

2T

dT = 0 +

R

dR3

dT =2

3

R

dRT (here R = radius of geo –stationary satllite = 42000 km)

dT =)km42000(

)km1(

2

3(24 hours) =

420002

243

66

hours

Now, angular velocity of satellite relative to the earth

3s/earth = 3s – 3earth = T

2

dTT

2 #=/

#

3rel

=778

9

::;

<=%

&

'()

* /

# =

1T

dT1

T

21

Using binomial expansion

3rel

= % &

'()

* ==

#1

T

dT1

T

2




3rel

= dTT

22

#=

Velocity of the point directly below the satellite relative to earth's surface will be

v = (3rel

)Rearth

v = % &

'()

* #dT

T

22 (R

earth)

v = % &

'()

* 6

6#hours

420002

243

hours)24(

22 (6400 km). =

189

#m/s = 1.66 cm/sec

6. (a) Orbital speed

v =r

GMe =

km)6406400(

)106(106.6 2411

/66 =

= 7.53 km/sec.

(b) Time period

T2 = r GMe4 32

%% & '

(() * #

T2 =)106106.6(

42411

2

666

#= (6400 + 640 km)3

T = 1.63 hours

(c) Initial mechanical energy =r2

mGMe=

= Jkm)6406400(2

)220)(106)(106.6( 2411

/666

=

=

Total loss in mech. energy during 1500 rev.

= (1.4 × 105) × 1500

= 21 × 107 J.

Final mechanical energy remaining after 1500 rev.

TEf = TE

i – 1E

loss

f

m

r2

GMe= –

%

%

&

'

(

(

)

*

/6

6666= =

km)6406400(2

)220106106.6( 2311

– 21 × 107 J

Solving get rg = 6812 km

Height from earth's surface = 6812 – 6400 = 412 km

(d) Final orbital velocity

vf =

f

e

r

GM =

)km 6812(

106106.6 2311 666 =

vf = 7.67 km/sec.




(e) Time period T =v

r2# =

67.7

68122 6#6

= 1.55 hours

(f) Due to Air resistance, net torque about the earth is non –zero.

So, angular momentum about the earth will not remain conserved.

7.

from angular momentum conservation about the sun,J = m v

1 r

1 = m v

2 r

2..........(1)

from energy conservation

2

1 mv

12 + %%

&

'(()

* =

1

S

r

mGM =

2

1 mv

22 + %%

&

'(()

* =

2

S

r

mGM..........(2)

Solving eq (1) and (2) get

J = )rr(

rrGM2m

21

21s

/

8. s2m

R

GM%

&

'()

* = ms 3

2 R

M =GT

R4

G

R2

3232 #,3

M = 6 × 1041 kgM = nms

(6 × 1041) = n (2 × 1030) 11103n 6,

So x = 11.Ans. 11

9. W =R

GM

5

3 2

= MR

R

GM

5

32

= MRg5

3=

5

3× 10 × 2.5 × 1031 = 15 × 1031 J

ALP Sol P Gravitation E

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