Phy 1 (10)

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GRAVITATIONAL

Q.1. Total energy = - 0E

r 2

GMm

P.E. = - 0E2r

GMm

Q.2. Work done is independent of path traversed (by conservative force) WPOQ : WPLQ = 1 : 1

Q.3. g / 4

Q.4. The maximum height attained by the particle is

h =

Rvg2

v2

2

h =

6

23

23

104.6

)100.4(8.92

)100.4(

= 935 km.

Q.5. from COE.

- 0R

GMm =

R

GMm

2

3mv

2

1 2

R2

GMmmv

2

1 2

v =R

GM.

Q.6. Potential energy midway between 2M and M.

= -10

M

R5

)GM(

)10(R5

GMM2

Potential energy at infinity = 0For minimum speed K.E. at infinity should be zero.

By COE, - 22

v10

M

2

1

R50

GM3

= 0

v =R5

GM6.

Q.7. A/c to Kepler’s III law

3

2

1

2

2

1

R

R

T

T

RB = 443/2

104)10(1

8

km

v A = h/km1021

102

T

R2 44

A

A

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vB = h/km108

1042

T

R2 44

B

B

relative velocity of ‘B’ w.r.t. A is [vB – v A ] = - 104 km/h.

relative angular speed of ‘B’ w.r.t. A is = AB

AB

RR

vv

=3101104

1044

4

rad/hour

Q.8. As ‘m’ is a point mass, consider an element on the rod at distance r from m,

dF =2r

)dm().m(G =

2r

dr L

M)m(G

r

dr

dm

m

or, F =

dL

d2 dr r

1.L

GMm

F = dL

d2

dr r

1.

L

GMm =

)dL(d

GMm

Q.9. From conservation of mechanical energy

hR

GMmmv

2

1

R

GMmmv

2

1 21

20

. . . (i)

Also from conservation of angular momentum

mv0R sin 600

= mv1 (R + h) . . . (ii)

Solving (i) and (ii) and putting v0 =R4

GM3,

we geth 0.25 R (Approx.)

R

v1

600

h

Q.10. By conservation of ME

0 +

R

GMmmv

2

1

r

GMm 2

or v2 = 2GM

R2

1

R

1 [as r = R + h = R + R = 2R]

or v = s/km8106410gRR

GM 5

Q.11. OEOB EE

ODOA FF

2OC a

GMF


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Q.12. vr = -r

GM

du = - dr m

r

GM

u = -

R

Rr

dr GMm

u = -

R1ln

GMm

Q.13. Let r 0 is the distance between centres of two stars.Consider star of mass M,for this star

21

210

MvGM(M/ 2)

r r where r 1 = 0

Mr

2(M M /2) = 0

r

3

21

200

vGM

r / 32r P = M

0

GM

6r r 0 =

3

2

GM

6P

Since = 1

1

v

r

r 1

v1

M

M/2

v2

C F F

T =3 2

1

1

2 r 2 GM / 6P

v 3 P /M

T =

4

3

GM

9P

Q.14. 2r =nr

GM

=1nr

GM

T =GM

r 2

2 1n

T

21n

r

Q.15. F =2r

GMm

all the forces will cancel each other. So it will be inequilibrium for any value of it. m

F

F F

M

MM

r

Q.16. Let v be the escapes velocity of particle then

Total energy E = 21

mv2

-2

2

3

GMm R3R 0

22R

v2 11GM

4R

vmin =11GM

4R.

v

C


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Q.17. Mr = 5M (L - r)

r =

6

L5

now2L

GM5M =

6

L5

T

22

M

T = 2 GM6

L 2/3.

M

5M

C

r

Q.18. 1

2

2

1

g

g

T

T but g =

2)hR(

GM

2

2

1

R

hR

g

g

R/h1

T

T

1

2

h =360024

640030R

T

TT

1

12

= 2.2 km.

Q.19. xcm =5

d

m5

)d(m)0(m4

R1 = d/5R2 = 4d/5

(a) mr 2 =2

B A

d

mGm

= 20

B Ardm mGm = 10

20

2

1)r m( )r m(

5/d)m4(

)5/d4(m

2

1

= 1 : 1

T1 : T2 = 2 : 1 = 1 : 1

4n

(0, 0)

Ad

B

(d, 0)

12

(b))I(2/1

)I(2/1

KE

KE222

211

2

1

=

21

2

22

2

)T()5/d4)(m)(2/1(

)T()5/d)(m4)(2/1( =

)1(16

)1(4 2 = 1 : 4

(c)

1)5/d4(

)1)(5/d(

Tr

Tr

r

r

v

v

11

21

22

11

2

1 1 : 4

Q.20. V(r) = -GM

3

22

R2

r R3 for 0 r R

Thus V(0) =R2

GM3

V(R) = -R

GM

Thus ratio =2

3


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I(r) =3R

GMr for 0 r R

RI0I

= 0

Q.21. (a) at r = a

|E|r = a = Edue to M1 + Edue to M2) = )MM(a

G

a

GM

a

GM2122

2

2

1

(b) at r = b

|E| = 0 +22 b

GM

b

GM

(c) at r = c ; E = 0

Q.22. If we consider that a sphere of radius R is placed with centre at C1 of density 1 the force on themass at P is

F1 = G 213

)yR(mR)3/4(

towards the sphere.

If we consider that a sphere of radius R/2 is placed with centre at C2 of density 1 the force on themass at P is

F2 = 21

3

)yR2/R(

m)2/R()3/4(G

towards the sphere.If we consider that a sphere of radius R/2 is

placed with centre at C2 of density 2 the force on the mass in at P

F3 = 22

3

)yR2/R(

m)2/R()3/4(G

By the principle of superposition

F = F1 - F2 + F3 =34 R3 Gm

212

21

)y)2/R3((

8/) _(

)yR(.

Q.23. The distance X, where gravitationalpull of each stars becomes equal andopposite can be obtained

G2X

Mm = G

2)xa10(

m)M16(

x = 2a

P

O

x

so the body will reach the smaller star due to stars gravitational pull is body just crosses point P byusing COE, energy at O = Energy at P

-a2

m,M16.G - 2minmv

2

1

a8

GMm = -

a8

m)M16(G -

a2

GMm

vmin =2

3

a

GM5.


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http://www.rpmauryascienceblog.com Q.24. At nearest point P and farthest point

Q. velocities are perpendicular totheir r 1 and r 2 conservation of angularmomentum, gives

mv1r 1 = mv2r 2

r 2 s r 1

v1

PQ

v2

Conservation of energy

2

22

1

21

r

GMmmv

2

1

r

GMmmv

2

1 . . . (2)

solving (1) and (2)we getr 2 = 3r 1 so maximum distance of satellite from the centre of earth is 3R(since r 1 R) Maximum distance of satellite from surface of earth will be 3R - R = 2R.

Q.25. T

o

r

r

dtadr

o

r = r 0 – aT

At any moment I = ttanconsmr 5

2 2

r 2 = constantat t = 0, r = r o , = o at t = T, r = r 0 - at (r o - aT)

2 = or 2

= o 2o

2o

)aTr (

r

Q.26. P.E. of the system = -r

GMm

self energy of the shell = - M

0

dmr

Gm

=-r 2

GM2

mr

M

Now total initial energy (Ui) = -1

2

1 r 2

GM

r

GmM

Total final energy (Uf ) = -2

2

2 r 2

GM

r

GMm

Work done by external agent = Uf – Ui

= GMm

21 r

1

r

1 +

21

2

r

1

r

1

2

GM

= GM

2

Mm

r

1

r

1

21

Q.27. Potential energy midway between M1 and M2


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= – 2121 MM

d

G2

2d

GM

2d

GM

Potential energy at infinity = 0For minimum speed K.E. at infinity should be zero.

Conserving energy

–2 0MV2

1MM

d

GM 221

v = 2 dMMG 21

Q.28. The centre of mass C will be at a distance d/3 and2d/3 from the masses 2m and m respectively.

Both the stars rotate with same angular velocity ''around C in their individual orbits.

Ratio of angular momentum = 1

2

)3/d(m2

)3/d2(m2

2

Ratio of kinetic energies =22

22

)3/d(m22

1

)3/d2(m2

1

=

1

2

C

d/3 2d/3

Q.29. Ei = -d

GMM3

Ef = - 3

2Mv

2

13

R2

GMM

initial energy = final energy2v

2

1

d

1

R2

1GM

d

v =

dR2

R2dGM2 .

Q.30. Potential at C = Potential due to whole sphere – potential due to cutt off sphere + potential duecutt off sphere at B

vc =R4

MG

2/R

MG

R

GM

vc = -

32

1

4

11

R

GM

R32

GM

R4

MG

R

GM

vc =

32

25

R

GM

32

1832

R

GM

vc = -R32

GM25

for escape velocity2

1 2emv =

R32

GMm25 = m

R32

GM25mv

2

1 2e


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ve =R

GM

4

5.

Q.31. Applying energy conservation principle at

A and at P.

0 +

220 mv

2

1

r

GMmmv

2

1

…(1) Where M = mass of the sun

m = mass of the bodyand r = distance of closest approach.

v = velocity at P

v0

A

l

M

Sun

P

Since the angular momentum of the body about the sun will remain conserved, therefore

(mv0) = (mv)r

v0 = vr …(2)

2

020

r

vm

2

1

r

GMmmv

2

1

l

0vGMr 2r v 2022

0 2

l

r =20

40

22

v2

v4MG4GM2 2l =

220

20 GM

v11

v

GM l.

Q.32. (a) g(h) =

9

4

R

GM

)2/3(R

GM

)2/RR(

GM2222

= g

9

4

(b) g(d) = 23 R

GM

2

1

2

R

R

GM

=

2

g

(c) Required ratio =9

82x

9

4

)d(g

)h(g

.

(acceleration in part (a) will be more)

Q.33. (a) Gravitational field at B is due to larger sphere and sphere A

IB =16

MG

32

GM0

= GM16

31

16

GM

32

M64G

(b) The point y2 + z2 = 36 lies at6 m from the centre. 40 m from the smaller sphere.

Gravitational potential at the point is

v = - 240

GM

6

GM0 = 240

GM

6

)M64(G

= - 10.98 GM.(c) Similarly at y2 + z2 = 4

v = - 222

GM)r R3(

R2

GM 223

0 = -22 GM +2

GM = - 21.29 GM


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Q.34. From conservation of angular momentummv0r 0 sin = mvr sin 90

0 (when the planet is at the maximum or minimum distance from planet, = 900 ) v0r 0 sin = vr . . . (i) From conservation of mechanical energy,

2

1m 20v -

0r

GMm =

2

1mv2 -

r

GMm

Solving the equations (i) and (ii)

r =

20 sin)2(112

r

where =GM

vr 200

therefore r max = 20 sin2(11

2r

r min =

20 sin2(112

r

Q.35. M = mass of the sunr = distance between the two stars

r 1 = r 3

2r .

mm

m

21

2

r 2 =3

r r

mm

m

21

1

Centripetal force on m2

=

2

2

221

r

M9/2G

r

mGm

r

r1 r2

C.M

m1=M/3 m2=2M/3

Now 22222

2

3

r M

3

2r m

r

GM

9

2

2 =3r

GM T = Time period of revolutions.

or3

2

r

GM

T

2

; T2 =

GM

r 4 32

Since time period of earth around sun

T2 =GM

R4 32 comparing r = R


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http://www.rpmauryascienceblog.com Q.36. (a) Particle will travel in straight line AP

parallel to CO with constant acceleration

23

RG4g

AP = 2R cos 30

t =g

AP2

=g

30cosR4

O

300

C

A

300P

mg

300

(b) It will heat at point P and distance is AP which parallel to OC and AP = 2R cos 300.

Q.37. (i) We know that for satellite motion

vo =

hR

gR

r

GM

[as g =2

R

GM and r = R + h]

In this problem vo = gR22

1v

2

1e

So gR2

1

hR

gR2

, i.e. 2R = h + R or h = R = 6400 km

(ii) By conservation of ME

0 +

R

GMmmv

2

1

r

GMm 2

or v2 = 2GM

R2

1

R

1 [as r = R + h = R + R = 2R]

or v = s/km88104.610gRR

GM 5

Q.38. The force on the ring by the sphere will be equal to the force on the sphere due to ring. The massof the sphere can be assumed to be concentrated at the centre of the sphere.

a

a

a

3 a

Edue to ring = 22/322 a8

Gm3

)da(

Gmd

Therefore force on the ring will be equal to F = ME =2a8

GMm3


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http://www.rpmauryascienceblog.com Q.39.

The comet is moving along an ellipticalpath with the sun situated at one of thefoci of ellipse. Q is the farthest positionof the comet.

211

2

cRMmG

Rmv

P

R1

Q

1.2 vc

R2

1

2c

R

MGv . . . (1)

Energy and angular momentum at P

EP =1

2P

R

MmGmv

2

1

EP =1

2c

R

GMm)v2.1(m

2

1 =

11

2

R

GMm

R

GM)2.1(m

2

1

EP = - 0.28 1R

m.GM

. . . .(2)

and LP = mvPR1 . . . (3)

At Q,

EQ =2

2

QR

m.GMmv

2

1 . . . (4)

LQ = mvQR2 . . . (5)

Energy and momentum should be conservedL1 = L2 mvPR1 = mvQR2

vQ = vP2

1

RR

-0.28

2

2

12P

1 R

Rmv

2

1

R

GMm

-

2R

GMm=

2

2

2

1

1

2

R

GMm

R

R

R

GMm)2.1(

2

1

-0.28 = 0.722

1

2

2

1

R

R

R

R

0.72 X2 – X + 0.28 = 0 Where R1/R2 = X

X =2

1

R

R

44.1

44.01

Taking ve sign, we get

R2 =56.0

44.1 = 2.57 R1

Q.40. The potential energy of the mass at the bottom of the crater is2

2

3

GMm RV 3R R

1002R

where M and m are masses of the moon and the projectile respectively. If the maximum heightabove the lunar surface reached, is h, then from energy conservation,


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2e

GMm 99 1 GMm3 mv

2R 100 2 R h

where ev is the escape velocity from the lunar surface. Thus2e

2GMv

R

. This gives

2GMm 99 GMm

3 2 h 99.5R2R 100 R h

Q.41. Let v0 is the orbital velocity of satellite then20 e

2e e

mv GM m

3R (3R ) v0 =

e

GM

3R

Final momentum of satellite after the impulse I is given, is

mv1 = mv0 – I = me e

e e

GM GMm

3R 12R

v1 = e

e

GM1

2 3R . . . (i)

v0 m

Me

Let r be the minimum distance of satellite from centre of earth. If v2 is the velocity of satellite at thisdistance then by conservation of angular momentum

mv2 r = mv1 (3Re) . . . (ii)By energy conservation

2 2e1 2

e

GM m1 1mv mv

2 3R 2 - e

GM m

r

using equation (ii) and (i)

r =3

7Re is the minimum distance of satellite from earth’s centre.

Phy 1 (10)

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