SOLUTIONS TO CONCEPTS CHAPTER 11 1.
Gravitational force of attraction, GMm F= r2 =
2.
6.67 10 11 10 10 –7 = 6.67 × 10 N (0.1)2
To calculate the gravitational force on ‘m’ at unline due to other mouse.
FOD =
G m 4m 8Gm2 = (a / r 2 )2 a2
FOI =
G m 2m 6Gm2 = (a / r 2 )2 a2 G m 2m
FOB =
=
2 2
(a / r )
Gmm
FOA =
=
2 2
(a / r )
m
A
E
B
2m
m
4Gm2 a2
4m
F
D
C
3m
2Gm2 a2 2
Resultant FOF =
Gm2 Gm2 64 2 36 2 a a
Resultant FOE =
Gm2 Gm 2 64 2 4 2 a a
2
2
= 10
Gm2 a2
2
= 2 5
Gm 2 a2
The net resultant force will be, 2
2
F=
Gm2 Gm 2 Gm2 100 2 20 2 2 2 20 5 a a a 2
=
Gm2 120 40 5 = a2
2
Gm2 (120 89.6) a2
Gm2 Gm 2 40.4 = 4 2 2 2 a a a) if ‘m’ is placed at mid point of a side =
3.
then FOA =
4Gm2 in OA direction a2
A m
4Gm 2 in OB direction a2 Since equal & opposite cancel each other
m
FOB =
Foc =
Gm
2
B
2
r / 2a 3
O
2
=
4Gm in OC direction 3a 2
Net gravitational force on m =
A
4Gm 2
m
a2
b) If placed at O (centroid) the FOA =
C m
m
3Gm2 Gm 2 = (a / r3 ) a2
B
11.1
m O m
C m
Chapter 11
3Gm2 a2
FOB =
Resultant F =
2
2
2 2 3Gm 2 2 3Gm 1 = 3Gm 2 2 2 2 a a a 2
3Gm2 , equal & opposite to F, cancel a2 Net gravitational force = 0 Since FOC =
4.
2 Gm2 ˆi Gm sin 60ˆj cos 60 4a 2 4a 2
FCB =
2
M
A
B
2
Gm Gm cos 60 ˆi sin 60ˆj 4a 2 4a 2
FCA =
F = FCB + FCA
C
2 2 2Gm 2 ˆj = 2Gm r3 = r3Gm sin 60 4a 2 2 4a 2 4a 2 Force on M at C due to gravitational attraction.
=
5.
FCB =
Gm2 ˆ j 2R 2
B
A
2
GM ˆ i 4R 2
FCD =
R D
2 GM2 ˆj GM sin 45ˆj cos 45 4R 2 4R 2 So, resultant force on C,
FCA =
FC =
C
= FCA + FCB + FCD
GM2 1 ˆ GM2 i 2 4R 2 4R 2 2
1 ˆ 2 j 2
GM2 2 2 1 4R 2
FC =
mv 2 For moving along the circle, F = R or 6.
MV 2 GM2 = or V = 2 2 1 R 4R 2 GM
R h
2
=
GM 2 2 1 R 4
49.358 1011 6.67 10 11 7.4 10 22 = 2 6 (1740 1000 ) 10 2740 2740 10 6
49.358 1011 –2 2 = 65.8 × 10 = 0.65 m/s 0.75 1013 The linear momentum of 2 bodies is 0 initially. Since gravitational force is internal, final momentum is also zero. So (10 kg)v1 = (20 kg) v2 Or v1 = v2 …(1) Since P.E. is conserved =
7.
6.67 10 11 10 20 –9 = –13.34×10 J 1 When separation is 0.5 m, Initial P.E. =
11.2
Chapter 11 –13.34 × 10
–9
+0=
13.34 10 9 2 2 + (1/2) × 10 v1 + (1/2) × 20 v2 (1/ 2)
–9
–9
2
…(2)
2
– 13.34 × 10 = -26.68 ×10 + 5 v1 + 10 v2 2 –9 –9 – 13.34 × 10 = -26.68 ×10 + 30 v2
13.34 10 9 –10 = 4.44 × 10 30 –5 v2 = 2.1 × 10 m/s. –5 So, v1 = 4.2 × 10 m/s. In the semicircle, we can consider, a small element of d then R d = (M/L) R d = dM. GMRdm M F= LR 2 2GMm dF3 = 2 dF since = sin d LR d 2
v2 =
8.
/2
F =
0
2GMm 2GMm cos 0 / 2 sin d LR LR
L
d R
m
GMm 2GMm 2GMm 2GMm ( 1) = = = LR LR L L / A L2 A small section of rod is considered at ‘x’ distance mass of the element = (M/L). dx = dm G(dm) 1 = dE2 dE1 = d2 x 2 = –2
9.
Resultant dE = 2 dE1 sin G(dm) d 2 GM d dx =2× 2 = 2 2 2 d x2 d x L d x 2 d2 x 2 Total gravitational field
L/2
E=
2
x2
dE1 d
M
2Gmd dx
Ld 0
dE2
O
3/2
dx
x
x
a
Integrating the above equation it can be found that, 2GM E= d L2 4d2 10. The gravitational force on ‘m’ due to the shell of M2 is 0. R R2 M is at a distance 1 2 Then the gravitational force due to M is given by GM1m 4GM1m = = (R1 R 2 / 2 (R1 R 2 )2
R2
M1 R1 m
3
11. Man of earth M = (4/3) R Man of the imaginary sphere, having 3 Radius = x, M = (4/3)x or
M x3 = 3 M R
Gravitational force on F = or F =
m
GMm m2
x
GMmx GMx 3m = 3 2 R x R3 11.3
m2
Chapter 11 12. Let d be the distance from centre of earth to man ‘m’ then D=
R2 = (1/2) x 2 4
4x 2 R2
m
x R/2
M be the mass of the earth, M the mass of the sphere of radius d/2. 3 Then M = (4/3) R 3 M = (4/3)d
d
O
M d3 = 3 M R Gravitational force is m, or
n
Gmm GMmd Gd3Mm = = 2 d R 3 d2 R3 So, Normal force exerted by the wall = F cos. GMm GMmd R = (therefore I think normal force does not depend on x) = 2d 2R 2 R3 13. a) m is placed at a distance x from ‘O’. If r < x , 2r, Let’s consider a thin shell of man
d F
F=
dm =
m 4 3 mx 3 x = ( 4 / 3)r 2 3 r3
Thus
dm =
R/2
x
R M
mx 3 r3
m
r O
G md m Gmx Gmx 3 / r 3 = = 2 2 x x r3 b) 2r < x < 2R, then F is due to only the sphere. Gmm F= x r 2 Then gravitational force F =
c) if x > 2R, then Gravitational force is due to both sphere & shell, then due to shell, GMm F= x R 2 due to the sphere =
Gmm
x r 2
So, Resultant force =
Gm m
x r 2
+
GMm
x R 2
14. At P1, Gravitational field due to sphere M =
GM
3a a
2
=
GM 16a 2
At P2, Gravitational field is due to sphere & shell,
a 49 P1
a GM GM GM 1 1 61 GM + = = 2 2 2 a a 36 25 (a 4a a) ( 4a a ) 2 900 a P2 15. We know in the thin spherical shell of uniform density has gravitational field at its internal point is zero.
=
At A and B point, field is equal and opposite and cancel each other so Net field is zero.
A A B
Hence, EA = EB 16. Let 0.1 kg man is x m from 2kg mass and (2 – x) m from 4 kg mass. 4 0 .1 2 0 .1 =– 2 ( 2 x )2 x 11.4
B
Chapter 11 or
0 .4 0.2 =– 2 ( 2 x )2 x
or
2 1 2 2 = or (2 – x) = 2 x ( 2 x )2 x2
or 2 – x = 2 x or x(r2 + 1) = 2 2 or x = = 0.83 m from 2kg mass. 2.414 17. Initially, the ride of is a To increase it to 2a, 2
2
m a
a
2
Gm Gm 3Gm = m m a 2a a 2a 18. Work done against gravitational force to take away the particle from sphere, work done =
G 10 0.1 6.67 10 11 1 –10 = = 6.67 × 10 J 0.1 0.1 1 10 1 19. E = (5 N/kg) ˆi + (12 N/kg) ˆj a) F = E m = 2kg [(5 N/kg) ˆi + (12 N/kg) ˆj ] = (10 N) ˆi + (12 N) ˆj F = 100 576 = 26 N b) V = E r At (12 m, 0), V = – (60 J/kg) ˆi V = 60 J
100g 10cm
=
10kg
At (0, 5 m), V = – (60 J/kg) ˆj V = – 60 J c) V =
(1,2,5 )
E mdr = (10N)ˆi (24N)ˆj r
(12,5 ) ( 0,0 )
( 0,0 )
= – (120 J ˆi + 120 J ˆi ) = 240 J 0,5m d) v = – r(10N ˆi 24Nˆj ) 12m,0
= –120 ˆj + 120 ˆi = 0 20. a) V = (20 N/kg) (x + y)
MLT 2 M1L3 T 2M1 ML2 T 2 GM = L or = M R L M 0 2 –2 0 2 –2 Or M L T = M L T L.H.S = R.H.S b) E( x, y ) = – 20(N/kg) ˆi – 20(N/kg) ˆj c) F = E m = 0.5kg [– (20 N/kg) ˆi – (20 N/kg) ˆj = – 10N ˆi - 10 N ˆj | F | = 100 100 = 10 2 N 21. E = 2 ˆi + 3 ˆj The field is represented as tan 1 = 3/2 3j 5/3 2 Again the line 3y + 2x = 5 can be represented as tan 2 = – 2/3 2j 5/2 m1 m2 = –1 Since, the direction of field and the displacement are perpendicular, is done by the particle on the line. 11.5
Chapter 11 22. Let the height be h GM GM (1/2) 2 = (R h)2 R 2
Or 2R = (R + h)
2
Or 2 R = R + h Or h = (r2 – 1)R 23. Let g be the acceleration due to gravity on mount everest.
2h g = g1 R 17696 2 =9.8 1 = 9.8 (1 – 0.00276) = 9.773 m/s 6400000 24. Let g be the acceleration due to gravity in mine. d Then g= g 1 R 640 2 = 9.8 1 = 9.8 × 0.9999 = 9.799 m/s 3 6400 10 25. Let g be the acceleration due to gravity at equation & that of pole = g 2 g= g – R –5 2 3 = 9.81 – (7.3 × 10 ) × 6400 × 10 = 9.81 – 0.034 2 = 9.776 m/s 2 mg = 1 kg × 9.776 m/s = 9.776 N or 0.997 kg The body will weigh 0.997 kg at equator. 2 …(1) 26. At equator, g = g – R Let at ‘h’ height above the south pole, the acceleration due to gravity is same.
2h Then, here g = g 1 R
…(2)
2h 2 g - R = g 1 R or 1
2h 2R = 1 R g
2
2
7.3 10 5 6400 10 3 2R 2 = = 11125 N = 10Km (approximately) 2 9.81 2g 27. The apparent ‘g’ at equator becomes zero. 2 i.e. g = g – R = 0 2 or g = R or h =
or =
g = R
9 .8 = 6400 10 3
1.5 10 6 = 1.2 × 10
2 2 3.14 –6 = = 1.5 × 10 sec. = 1.41 hour 1.2 10 3 28. a) Speed of the ship due to rotation of earth v = R 2 b) T0 = mgr = mg – m R 2 T0 – mg = m R c) If the ship shifts at speed ‘v’ 2 T = mg – m R
–3
rad/s.
T=
11.6
To A
A
Chapter 11
v R 2 R = T0 - R2 v 2 2R 2 2Rv m = T0 – R T = T0 + 2v m 29. According to Kepler’s laws of planetary motion, 2 3 T R Tm
2
Te 2
3
R es 3
3
Rms R es
R ms
2
1.88 1
R ms 2/3 = (1.88) = 1.52 R es
30. T = 2
r3 GM
3.84 10
5 3
27.3 = 2 × 3.14 or 2.73 × 2.73 = or M =
6.67 10 11 M
2 3.14 3.84 10 5 6.67 10 11 M
3
2 (3.14 )2 (3.84)3 1015 24 = 6.02 × 10 kg 11 2 3.335 10 (27.3 )
mass of earth is found to be 6.02 × 10
24
kg.
3
31. T = 2
r GM
9.4 10
2
or (27540) = (6.28) or M = 32. a) V = =
3
103 11 6.67 10 M
27540 = 2 × 3.14
3
9.4 10
6 2
2
6.67 10 11 M
(6.28)2 (9.4 )3 1018 23 = 6.5 × 10 kg. 11 2 6.67 10 (27540 ) gr 2 r h
GM = r h
9.8 ( 6400 10 3 )2 6
10 (6.4 2)
3
= 6.9 × 10 m/s = 6.9 km/s
2
b) K.E. = (1/2) mv 6 10 = (1/2) 1000 × (47.6 × 10 ) = 2.38 × 10 J GMm c) P.E. = (R h) =–
40 1013 6.67 10 11 6 10 24 10 3 10 = – = – 4.76 × 10 J 8400 (6400 2000 ) 10 3
d) T =
2(r h) 2 3.14 8400 10 3 2 = = 76.6 × 10 sec = 2.1 hour 3 V 6.9 10 11.7
Chapter 11 33. Angular speed f earth & the satellite will be same 2 2 = Te Ts or
or
1 = 24 3600
1 2
or 12 I 3600 = 3.14
(R h)3 gR 2
(R h)2 (12 3600 )2 = gR 2 (3.14 )2
or
(R h)3 gR 2
(6400 h)3 109 (12 3600 )2 = 9.8 (6400 )2 10 6 (3.14 )2
(6400 h)3 10 9 4 = 432 × 10 9 6272 10 3 4 or (6400 + h) = 6272 × 432 × 10 4 1/3 or 6400 + h = (6272 × 432 × 10 ) 4 1/3 or h = (6272 × 432 × 10 ) – 6400 = 42300 cm. b) Time taken from north pole to equator = (1/2) t or
( 43200 6400)3
= (1/2) × 6.28
2
10 (6400 ) 10
6
= 3.14
( 497 )3 10 6 (64)2 1011
497 497 497 = 6 hour. 64 64 10 5 34. For geo stationary satellite, 4 r = 4.2 × 10 km h = 3.6 × 104 km Given mg = 10 N = 3.14
R2 mgh = mg R h2
2 6400 103 = 10 6400 10 3 3600 10 3
4096 = = 0.23 N 2 17980
R 23
35. T = 2
gR12
Or T = 4 2
R2
3 2
gR1 3
Or g =
42 R 2 T 2 R12
Acceleration due to gravity of the planet is =
3
42 R 2 T 2 R12
A
36. The colattitude is given by . OAB = 90° – ABO Again OBC = = OAB 6400 8 sin = = 42000 53 = sin
–1
Colatitude
O
8 –1 = sin 0.15. 53
11.8
B
C
Chapter 11 37. The particle attain maximum height = 6400 km. On earth’s surface, its P.E. & K.E.
GMm 2 Ee = (1/2) mv + R In space, its P.E. & K.E.
…(1)
GMm Es = +0 Rh GMm …(2) Es = 2R Equating (1) & (2) GMm GMm 1 mv 2 = 2R R 2
( h = R)
1 1 2 Or (1/2) mv = GMm 2 R R 2
Or v = =
GM R
6.67 10 11 6 10 24 6400 10 3
40.02 1013 6.4 10 6 7 8 = 6.2 × 10 = 0.62 × 10 =
4
Or v = 0.62 10 8 = 0.79 × 10 m/s = 7.9 km/s. 38. Initial velocity of the particle = 15km/s Let its speed be ‘v’ at interstellar space. GMm 3 2 2 (1/2) m[(15 × 10 ) – v ] = dx R x2
1 3 2 2 (1/2) m[(15 × 10 ) – v ] = GMm x R 6
2
(1/2) m[(225 × 10 ) – v ] =
GMm R
2 6.67 10 11 6 10 24 6400 10 3 40.02 2 6 8 v = 225 × 10 – × 10 32 2 6 8 8 v = 225 × 10 – 1.2 × 10 = 10 (1.05) 4 Or v = 1.01 × 10 m/s or = 10 km/s 24 39. The man of the sphere = 6 × 10 kg. 8 Escape velocity = 3 × 10 m/s 6
2
225 × 10 – v =
Vc = Or R = =
2GM R 2GM Vc 2
2 6.67 10 11 6 10 24
3 10
8 2
=
80.02 –3 –3 × 10 = 8.89× 10 m 9 mm. 9 11.9