alp sol p gravitation e
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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)
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SOLN GRAVITATION - 3
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
#
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SOLN GRAVITATION - 4
(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
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SOLN GRAVITATION - 5
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
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SOLN GRAVITATION - 6
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.
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SOLN GRAVITATION - 7
(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