Chapter # 11
Gravitation
Solved Examples 1. Sol.
Two particles of masses 1.0 kg and 2.0 kg are placed at a separation of 50 cm. Assuming that the only forces acting on the particles are their mutual gravitation, find the initial accelerations of the two particles. The force of gravitation exerted by one particle on another is F=
Gm1m2 r2 6.67 10 11
=
N m2 kg2
(1.0 kg) (2.0 kg)
(0.5m)2
= 5.3 × 10–10 N. The acceleration of 1.0 kg particle is
F 5.3 10 10 N a1 = m = 1.0 kg 1 = 5.3 × 10–10 m/s 2 This acceleration is towards the 2.0 kg particle. The acceleration of the 2.0 kg particle is a2 =
5.3 10 10 N F = 2.0 kg a2
= 2.65 × 10–10 m/s 2 This acceleration is towards the 1.0 kg particle. 2.
Find the work done in bringing three particles, each having a mass of 100 g, from large distances to the vertices of an equilateral triangle of side 20 cm.
Sol.
When the separation are large, the gravitational potential energy is zero. When the particles are brought at the vertices of the triangle ABC, three pairs AB, BC and CA are formed. The potential energy of each pair is – Gm 1m 2/r and hence the total potential energy becomes Gm1m 2 U = 3 × r 6.67 10 11N m 2 / kg2 (0.1 kg) (0.1 kg) =3× 0.20 m
= – 1.0 × 10–11 J. The work done by the gravitational forces is W = – U = 1.0 × 10–11 J. If the particles are brought by some external agency without changing the kinetic energy, the work done by the external agency is equal to the change in potential energy = – 1.0 × 10–11 J. 3.
A particle of mass M is placed at the centre of a uniform spherical shell of equal mass and radius a. Find the gravitational potential at a point P at a distance a/2 from the centre.
Sol.
The gravitational potential at the point P due to the particle at the centre is V1 = –
GM 2GM =– . a/2 a
The potential at P due to the shell is GM V2 = – a The net potential at P is V1 + V2 = –
3GM . a
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Chapter # 11
Gravitation
4.
A particle of mass 50 g experiences a gravitational force of 2.0 N when placed at a particular point. Find the gravitational field at that point.
Sol.
The gravitational field has a magnitude E=
2.0 N F = = 40 N/kg (50 10 3 kg) m
This field is along the direction of the force. 5.
The gravitational field due to a mass distribution is given by E = K/x3 in X-direction. Taking the gravitational potential to be zero at infinity, find its value at a distance x. HCV_Ch-11_Sol.Ex._5 fdlh nzO;eku forj.k ds dkj.k xq:Roh; {ks=k dh rhozrk x v{k ds vuqfn'k E = K/x3 }kjk nh tkrh gSA xq:Roh; foHko dks
vuUr ij 'kwU; ekus rks bldk eku x nwjh ij crk,aA Sol.
The potential at a distance x is x
x
E dx
V(x) = –
=–
x
K K = 2 = . 2x 2 2x
K
x3 dx
A 6.
The gravitational potential due to a mass distribution is V =
x 2 a2
. Find the gravitational field.
A Sol.
V=
x 2 a2
= A (x 2 + a2)–1/2.
If the gravitational field is E, Ex = –
1 V = – A (x 2 + a2)–3/2 (2x) x 2 Ax
=
2
( x a 2 )3 / 2
V V Ey = y = 0 and Ez = = 0. z Ax
The gravitational field is
2
( x a 2 )3 / 2
in the x-direction.
7.
Find the gravitational field due to the moon at its surface. The mass of the moon is 7.36 × 1022 kg and the radius of the moon is 1.74 × 106 m. Assume the moon to be a spherically symmetric body.
Sol.
To calculate the gravitational field at an a external point, the moon may be replaced by a single particle of equal mass placed at its centre. Then the field at the surface is E=
=
GM a2 6.67 10 11N m 2 / kg2 7.36 10 22 kg (1.74 10 6 m) 2
= 1.62 N/kg. This is about one sixth of the gravitational field due to the earth at its surface.
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Chapter # 11 8.
Gravitation
Calculate the value of acceleration due to gravity at a point (a) 5.0 km above the earth’s surface and (b) 5.0 km below the earth’s surface. Radius of earth = 6400 km and the value of g at the surface of the earth is 9.80 m/s 2.
Sol.(a) The value of g at a height h is (for h << R) 2h g = g 0 1 R
2 5.0 km = (9.80 m/s 2) 1 6400 km = 9.78 m/s 2. (b) The value at a depth h is h g = g0 1 R
5.0 km = (9.8 m/s 2) 1 6400 km = 9.79 m/s 2. 9.
A satellite is revolving round the earth at a height of 600 km. Find (a) the speed of the satellite and (b) the time period of the satellite. Radius of the earth = 6400 km and mass of the earth = 6 × 1024 kg.
Sol.
The distance of the satellite from the centre of the earth is 6400 km + 600 km = 7000 km. The speed of the satellite is v=
=
GM a
6.67 10 11N m2 / kg2 6 10 24 kg 7000 103 m
= 7.6 × 103 m/s = 7.6 km/s. The time period is 2 a T= v =
2 7000 103 m 7.6 103 m / s
= 5.8 × 103 s.
10.
Calculate the escape velocity from the moon. The mass of the moon = 7.4 × 1022 kg and radius of the moon = 1740 km.
Sol.
The escape velocity is v=
=
2GM R
2 6.67 10 11N m 2 / kg2 7.4 10 22 kg 1740 103 m
= 2.4 km/s.
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Chapter # 11
Gravitation
WORKED OUT EXAMPLES 1.
Three particles A, B and C, each of mass m, are placed in a line with AB = BC = d. Find the gravitational force on a fourth particle P of same mass, placed at a distance d from the particle B on the perpendicular bisector of the line AC.
Sol.
The force at P due to A is
FA =
G m2 ( AP )2
=
G m2 2 d2
along PA. The force at P due to C is FC =
G m2
=
(CP )2
G m2 2 d2
along PC. The force at P due to B is FB =
G m2 d2
along PB.
The resultant of FA, FB and FC will be along PB. Clearly APB = BPC = 45°. Component at FA along PB = FA cos 45° =
G m2 2 2d 2
Component at FC along PB = FC cos 45° =
.
G m2 2 2d 2
.
G m2
Component at FB along PB =
d2
Hence, the resultant of the three forces is
G m2 1 1 1 G m2 = d2 2 2 2 2 d2
1 1 along PB. 2
2.
Find the distance of a point from the earth’s centre where the resultant gravitational field due to the earth and the moon is zero. The mass of the earth is 6.0 × 1024 kg and that of the moon is 7.4 × 1022 kg. The distance between the earth and the moon is 4.0 × 105 km.
Sol.
The point must be on the line joining the centres of the earth and the moon and in between them. If the distance of the point from the earth is x, the distance from the moon is (4.0 × 105 km-x). The magnitude of the gravitatiional field due to the earth is E1 =
GMe x2
=
G 6 10 24 kg x2
and magnitude of the gravitational field due to the moon is
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Chapter # 11
Gravitation GMm
E2 =
( 4.0 10 5 km x ) 2
=
G 7.4 10 22 kg ( 4.0 10 5 km x ) 2
These fields are in opposite directions. For the resultant field to be zero E1 = E2. or,
6 10 24 kg x2
=
7.4 10 22 kg ( 4.0 10 5 km x )2
x
or,
4.0 10 km x
or,
x = 3.6 × 105 km.
5
=
6 10 24 7.4 10 22
=9
3.
Two particles of equal mass go round a circle of radius R under the action of their mutual gravitational attraction. Find the speed of each particle.
Sol.
The particles will always remain diametrically opposite so that the force on each particle will be directed along the radius. Consider the motion of one of the particles. The force on the particle is F =
Gm2 4R 2
. If
the speed is v, its acceleration is v2/R. Thus, by Newton’s law,
Gm2
= 4R 2
or,
v=
mv 2 R
GM 4R
4.
Two particles A and B of masses 1 kg and 2 kg respectively are kept 1 m apart and are released to move under mutual attraction. Find the speed of A when that of B is 3.6 cm/hour. What is the separation between the particles at this instant ?
Sol.
The linear momentum of the pair A + B is zero initially. As only mutual attraction is taken into account, which is internal when A + B is taken as the system, the linear momentum will remain zero. The particles move in opposite directions. If the speed of A is v when the speed of B is 3.6 cm/hour = 10 –5 m/s, (1 kg) v = (2 kg) (10–5 m/s) or, v = 2 × 10–5 m/s. The potential energy of the pair is –
=–
Gm A mB with usual symbols. Initial potential energy R
6.67 10 11N m 2 / kg2 2kg 1kg 1m
= – 13.34 × 10–11 J. If the separation at the given instant is d, using conservation of energy, – 13.34 × 10–11 J + 0 =–
1 13.34 10 11J m + (2kg) (10–5 m/s)2 2 d
1 (1 kg) (2 × 10–5) m/s)2 2 Solving this, d = 0.31 m. +
5.
The gravitational field in a region is given by E = (10 N/kg) ( ˆi + ˆj ). Find the work done by an external agent to slowly shift a particle of mass 2 kg from the point (0, 0) to a point (5m, 4m).
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Chapter # 11
Gravitation E = (10 N/kg) ( ˆi + ˆj ) }kjk fn;k
fdlh {ks=k esa xq:Roh; {ks=k tkrk gSA cká lzksr }kjk 2 fdxzk0 ds d.k dks fcUnq (0, 0) ls (5m, 4m) rd /khjs&/khjs ys tkus esa fd, x, dk;Z dh x.kuk dhft,A HCV_Ch-11_WOE_5 Sol.
As the particle is slowly shifted, its kinetic energy remains zero. The total work done on the particle is thus zero. The work done by the external agent should be negative of the work done by the gravitational field. The work done by the field is f
F.d r
i
Consider figure. Suppose the particle is taken from O to A and then from A to B. The force on the particle is F = m E = (2 kg) (10 N/kg) ( ˆi + ˆj ) = (20 N) ( ˆi + ˆj ).
The work done by the field during the displacement OA is 5m
W1 =
5m
Fx dx
=
0
(20N)dx
= 20 N × 5m = 100 J.
0
Similarly, the work done in displacement AB is 4m
W2 =
Fy dy =
0
4m
(20 N) dy 0
= (20 N) (4 m) = 80 J. Thus, the total work done by the field, as the particle is shifted from O to B, is 180 J. The work done by the external agent is – 180 J. Note that the work is independent of the path so that we can choose any path convenient to us from O to B. Method-II : W = F.S F m E const. Fext. mE = 20( ˆi ˆj ) W ext = 20( ˆi ˆj ) . (5 ˆi 4ˆj ) = – 100 – 80 = –180 J
6.
A uniform solid sphere of mass M and radius a is surrounded symmetrically by a uniform thin spherical shell of equal mass and radius 2 a. Find the gravitational field at a distance (a)
3 a from the centre, (b) 2
5 a from the centre. 2
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Chapter # 11
Gravitation
Sol.
Figure shows the situation. The point P1 is at a distance
3 a from the centre and P2 is at a distance 2
5 a from the centre. As P1 is inside the cavity of the thin spherical shell, the field here due to the shell 2 is zero. The field due to the solid sphere is E=
4 GM
GM 3 a 2
=
2
9 a2
This is also the resultant field. The direction is towards the centre. The point P2 is outside the sphere as well as the shell. Both may be replaced by single particles of the same mass at the centre. The field due to each of them is E =
GM 5 a 2
=
2
4 GM 25 a 2
8 GM
The resultant field is E = 2E = 7.
25 a 2
towards the centre.
The density of mass inside a solid sphere of radius a is given by = 0 a/r where 0 is the density at the surface and r denotes the distance from the centre. Find the gravitational field due to this sphere at a distance 2a from its centre. a f=kT;k okys n`<+ xksys esa fLFkr nzO;eku dk ?kuRo = 0 a/r ls fn;k tkrk gS tgk¡ 0 lrg ij ?kuRo rFkk r dsUnz ls nwjh
gSA blds dsUnz ls 2a nwjh ij bl xksys ds dkj.k xq:Roh; {ks=k dk eku D;k gksxk \ Sol.
HCV_Ch-11_WOE_7
The field is required at a point outside the sphere. Dividing the sphere in concentric shells, each shell can be replaced by a point particle at its centre having mass equal to the mass of the shell. Thus, the whole sphere can be replaced by a point particle at its centre having mass equal to the mass of the given sphere. If the mass of the sphere is M, the gravitational field at the given point is GM
E=
( 2a )
2
GM
=
4a 2
.
...........(i)
The mass M may be calculated as follows. Consider concentric shell of radius r and thickness dr. Its volume is dV = (4r2) dr and its mass is a dM = dV = 0 (4r2dr) r
= 40 ar dr.
The mass of the whole sphere is a
M=
40ar dr
= 20a3.
0
Thus, by (i) the gravitational field is E=
2G0 a 3 4a
2
=
1 G0a. 2
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Chapter # 11 8.
Gravitation
A uniform ring of mass m and radius a is placed directly above a uniform sphere of mass M and of equal radius. The centre of the ring is at a distance
3 a from the centre of the sphere. Find the gravitational
force exerted by the sphere on the ring. Sol.
The gravitational field at any point on the ring due to the sphere is equal to the field due to a single particle of mass M placed at the centre of the sphere. Thus, the force on the ring due to the sphere is also equal to the force on it by a particle of mass M placed at this point. By Newton’s third law it is equal to the force on the particle by the ring. Now the gravitational field due to the ring at a distance d =
3 a on its axis is Gmd
E=
2
(a d 2 ) 3 / 2
=
3 Gm 8a 2
.
The force on a particle of mass M placed here is F = ME =
3 GMm 8a 2
This is also the force due to the sphere on the ring. 9.
A particle is fired vertically upward with a speed of 9.8 km/s. Find the maximum height attained by the particle. Radius of earth = 6400 km and g at the surface = 9.8 m/s 2. Consider only earth’s gravitation.
Sol.
At the surface of the earth, the potential energy of the earth - particle system is –
GMm with usual R
1 mv02 where v0 = 9.8 km/s. At the maximum height the kinetic energy is 2 zero. If the maximum height reached is H, the potential energy of the earth-particle system at this symbols. The kinetic energy
instant is –
GMm . Using conservation of energy,, R H
GMm 1 GMm + mv02 = – . R 2 RH Writing GM = gR2 and dividing by m, –
– gR +
or,
v 02 gR 2 = 2 R H
v 02 R2 =R– 2g RH
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Chapter # 11
or,
Gravitation R2
R+H=
v2 R 0 2g
Putting the values of R, v0 and g on the right side,
(6400 km)2
R+H =
6400 km = or, 10. Sol.
(6400 km) 1500 km
2
(9.8 km / s)2 2 9.8m / s 2 = 27300 km
H = (27300 – 6400) km = 20900 km.
A particle hanging from a spring stretches it by 1 cm at earth’s surface. How much will the same particle stretch the spring at a place 800 km above the earth’s surface ? Radius of the earth = 6400 km. Suppose the mass of the particle is m and the spring constant of the spring is k. The acceleration due to gravity at earth’s surface is g =
GM R2
with usual symbols. The extension in the spring is mg/k.
GMm
Hence, 1cm =
k R2
.
..........(i)
At a height h = 800 km, the extension is given by GMm
x=
..........(ii)
k (R h)2
By (i) and (ii),
x R2 = 1cm (R h)2
=
(6400 km)2 (7200 km)2
= 0.79.
Hence x = 0.79 cm 11.
A simple pendulum has a time period exactly 2 s when used in a laboratory at north pole. What will be the time period if the same pendulum is used in a laboratory at equator ? Account for the earth’s rotation only. Take g =
Sol.
GM R2
= 9.8 m/s 2 and radius of earth = 6400 km.
Consider the pendulum in its mean position at the north pole. As the pole is on the axis of rotation, the bob is equilibrium. Hence in the mean position, the tension T is balanced by earth’s attraction. Thus, T=
GMm R2
= mg. The time period t is
t = 2
= 2 T /m
g .
.............(i)
2 radian At equator, the lab and the pendulum rotate with the earth at angular velocity = 24 hour in a circle of radius equal to 6400 km. Using Newton’s second law,
GMm R2
– T = m2R
or
T = m (g – 2R)
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Chapter # 11
Gravitation
where T is the tension in the string. The time period will be t = 2
= 2 . (T / m) g 2R
.............(ii)
By (i) and (ii), 2R 1 = g g 2R g
t = t
or,
12.
Sol.
1/ 2
2R t = t 1 2g
Putting the values, t = 2.004 seconds. A satellite is to revolve round the earth in a circle of radius 8000 km. With what speed should this satellite be pr ojec ted into orbit? W hat will be the tim e period? Tak e g at the surf ace = 9.8 m/s 2 and radius of the earth = 6400 km. Suppose, the speed of the satellite is v. The acceleration of the satellite is v2/r, where r is the radius of the orbit. The force on the satellite is
GMm r2
=m
GMm r2
with usual symbols. Using Newton’s second law,,
v2 r
(9.8m / s 2 )(6400 km)2 GM gR 2 = = (8000 km) r r
or,
v2 =
giving
v = 7.08 km/s.
The time period is
2(8000 km) 2r = (7.08 km / s) 118 minutes. v
13.
Two satellites S1 and S2 revolve round a planet in coplanar circular orbits in the same sence. Their periods of revolution are 1 h and 8h respectively. The radius of the orbit of S1 is 104 km. When S2 is closest to S1, find (a) the speed of S2 relative to S1 and (b) the angular speed of S2 as observed by an astronaut in S1.
Sol.
Let the mass of the planet be M, that of S1 be m 1 and of S2 be m 2. Let the radius of the orbit of S1 be R1 (= 104 km) and of S2 be R2. Let v1 and v2 be the linear speeds of S1 and S2 with respect to the planet. Figure shows the situation.
As the square of the time period is proportional to the cube of the radius,
R2 R1
3
T2 = T1
2
8h = 1h
2
= 64
manishkumarphysics.in
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Chapter # 11
or,
Gravitation
R2 R1 = 4
or, R2 = 4R1 = 4 × 104 km. Now the time period of S1 is 1 h. So,
2 R1 v1 = 1 h or,
v1 =
2 R1 4 1 h = 2 × 10 km/h.
similarly, v2 =
2 R2 4 8 h = × 10 km/h.
(a) At the closest separation, they are moving in the same direction. Hence the speed of S2 with respect to S1 is |v2 – v1| = × 104 km/h. (b) As seen from S1, the satellite S2 is at a distance R2 – R1 = 3 × 104 km at the closest s epar ation. Als o it is m oving at × 10 4 k m /h in a direc tion per pendic ular to the line joining them. Thus, the angular speed of S2 as observed by S1 is =
10 4 km / h 4
3 10 km
=
rad/h. 3
QUESTION FOR SHORT ANSWER 1.
Can two particles be in equilibrium under the action of their mutual gravitational force? Can three particles be ? Can one of the three particles be ?
2.
Is there any meaning of “Weight of the earth” ?
3.
If heavier bodies are attracted more strongly by the earth, why don’t they fall faster than the lighter bodies?
4.
Can you think of two particles which do not exert gravitational force on each other ?
5.
The earth revolves round the sun because the sun attracts the earth. The sun also attracts the moon and this force is about twice as large as the attraction of the earth on the moon. Why does the moon not revolve round the sun ? Or does it ?
6.
At noon, the sun and the earth pulls the objects on the earth’s surface in opposite directions. At midnight the sun and the earth pull these objects in same direction. Is the weight of an object, as mesured by a spring balance on the earth’s surface, more at midnight as compared to its weight at noon?
7.
An apple falls from a tree. An insect in the apple finds that the earth is falling towards it with an acceleration g. Who exerts the force needed to accelerate the earth with this acceleration g?
8.
Suppose the gravitational potential due to a small system is k/r 2 at a distance r from it. What will be the gravitational field ? Can you think of any such system? What happens if there were negative masses?
9.
The gravitational potential energy of a two-particle system is derived in this chapter as U = –
Gm1m 2 . r Does it follow from this equation that the potential energy for r = must be zero ? Can we choose the potential energy for r = to be 20 J and still use this formula ? If no, what formula should be used to calculate the gravitational potential energy at separation r ?
manishkumarphysics.in
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Chapter # 11
Gravitation
10.
The weight of an object is more at the poles than at the equator. Is it benificial to pruchase goods at equator and sell them at the pole ? Does it matter whether a spring balance is used or an equal beam balance is used ?
11.
The weight of a body at the poles is greater than the weight at the equator. Is it the actual weight or the apparent weight we are talking about ? Does your answer depend on whether only the earths rotation is taken into account or the flattening of the earth at the poles is also taken into account ?
12.
If the radius of the earth decreases by 1% without changing its mass, will the acceleration due to gravity at the surface of the earth increase or decrease? If so, by what per cent?
13.
A nut becomes loose and gets detached from a satellite revolving around the earth. Will it land on the earth ? If yes, where will it land? If no, how can an astronaut make it land on the earth ?
14.
Is is necessary for the plane of the orbit of a satellite to pass through the centre of the earth?
15.
Consider earth satellites in circular obrits. A geostationary satellite must be at a height of about 36000 km from the earth’s surface. Will any satellite moving at this height be a geostationary satellite ? Will any satellite moving at this height have a time period of 24 hours ?
16.
No part of India is situated on the equator. Is it possible to have a geostationary satellite which always remains over New Delhi?
17.
As the earth rotates about its axis, a person living in his house at the equator goes in a circular orbit of radius equal to the radius of the earth. Why does he/she not feel weightless as a satellite passenger does ?
18.
Two satellites going in equatorial plane have almost same radii. As seen from the earth one move from east to west and the other from west to east. Will they have the same time period as seen from the earth? If not, which one will have less time period ?
19.
A spacecraft consumes more fuel in going from the earth to the moon than it takes from a retun trip. Comment on this statement.
OBJECTIVE - I 1.
The acceleration of moon with respect to earth is 0.0027 m/s 2 and the acceleration of an apple falling on earth’s surface is about 10 m/s 2. Assume that the radius of the moon is one fourth of the earth’s radius. If the moon is stopped for an instant and then released, it will fall towards the earth. The initial acceleration of the moon towards the earth will be. i`Foh ds lkis{k pUnzek dk Roj.k 0.0027 m/s2 gSa rFkk i`Foh dh lrg ij fxjrs gq, lso dk Roj.k yxHkx 10 m/s2 gSaA
eku yhft;s fd pUnzek dh f=kT;k] i`Foh dh f=kT;k dk ,d pkSFkkbZ gSaA ;fn pUnzek dks ,d {k.k ds fy;s jksd fn;k tk;s rFkk fQj eqDr dj fn;k tk;s] ;g i`Foh dh vksj fxjus yxsxk ! pUnzek dk i`Foh dh vksj rkR{kf.kd Roj.k gksxk& (A) 10 m/s 2 2.
(B*) 0.0027 m/s 2
(C) 6.4 m/s 2
(D) 5.0 m/s 2
The acceleration of the moon just before it strikes the earth in the previous question is
fiNys iz'u esa i`Foh ls Vdjkus ls rqjar igys pUnzek dk Roj.k gksxk& (A) 10 m/s 2 3.
(B) 0.0027 m/s 2
(C*) 6.4 m/s 2
(D) 5.0 m/s 2
Suppose, the acceleration due to gravity at the earth’s surface is 10 m/s 2 and at the surface of Mars it is 4.0 m/s 2. A 60 kg passenger goes from the earth to the Mars in a spaceship moving with a constant velocity. Neglect all other object in the sky. Which part of fig. best represents the weight (net gravitational force) of the passenger as a function of time. eku yhft;s fd i`Foh dh lrg ij xq:Roh; Roj.k 10 m/s2 gSaA rFkk eaxy dh lrg ij bldk eku 4.0 m/s2 gSaA ,d 60 fdxzk nzO;eku okyk varfj{k ;k=kh fu;r pky ls xfr'khy varfj{k ;ku esa i`Foh ls eaxy dh vksj tkrk gSaA vkdk'k
ds vU; fi.Mksa dks ux.; eku yhft;sA fp=k dk dkSulk Hkkx varfj{k ;k=kh dk Hkkj ¼ dqy xq:Rokd"kZ.k cy ½ le; ds Qyu ds :i esa O;Dr djrk gSa&I manishkumarphysics.in
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Chapter # 11
Gravitation weight 600 N A
B
200 N C to D
(A) A 4.
(B) B
time
(C*) C
(D) D
Consider a planent in some solar system which has a mass double the mass of the earth and density equal to the average density of the earth. An object weighing W on the earth will weigh
fdlh lkSj ifjokj esa ,d xzg gSa ftldk nzO;eku i`Foh ds nzO;eku dk nqxuk rFkk ?kuRo i`Foh ds vkSlr ?kuRo ds cjkcj gSA i`Foh ij W Hkkj okyh oLrq dk ml xzg ij Hkkj gksxk& (A) W 5.
(B) 2 W
(C) W / 2
(D*) 2
1/3
W at the planet
If the acceleration due to gravity at the surface of the earth is g, the work done in slowly lifting a body of mass m from the earth’s surface to a height R (equal to the radius of the earth) is ;fn i`Foh dh lrg ij xq:Roh; Roj.k ggSa, m nzO;eku dh oLrq dh lrg ls i`Foh dh f=kT;k R ds cjkcj Å¡pkbZ rd
mBkus esa fd;k x;k dk;Z gksxk& (A*) 1/2 mgR 6.
(B) 2 mgR
(C) mgR
(D) 1/4 mgR
A person brings a mass of 1 kg from infinity to a point A. Initially the mass was at rest but it moves at a speed of 2 m/s as it reaches A. The work done by the person on the mass is -3 J. The potential at A is ,d O;fDr vuUr ls 1 fdxzk nzO;eku fcUnq A rd ykrk gSA izkjEHk esa nzO;eku fojkekoLFkk esa Fkk] fdUrq tc ;g fcUnq A ij igqaprk gSaA bldh pky 2 m/s ls gksrh gSA O;fDr }kjk nzO;eku ij fd;k x;k dk;Z -3 J twy gSA A ij foHko
gS a &
(A) –3 J/kg 7.
(B) –2 J/kg
(C*) –5 j/kg
(D) none of these
Let V and E be the gravitational potential and gravitational field at a distance r from the centre of uniform spherical shell, Consider the following two statements: (a) The plot of V against r is discontinuous (b) The plot of E against r is discontinuous (A) Both A and B are correct (B) A is correct but B is wrong (C*) B is correct but A is wrong (D) Both A and B are wrong ,d le:i xksykdkj dks'k ds dsUnz ls r nwjh ij xq:Roh; foHko rFkk xq:Roh; {ks=k Ør'k% V rFkk E gSA fuEu nks dFkuksa
ij fopkj dhft;s % (a) r ds lkis{k V dk ys[kkfp=k vlrr~ gSA (b) r ds lkis{k E dk ys[kkfp=k vlrr~ gSA (A) A o B nksu ksa lR; gSA (B) A lR; gS fdUrq B vlR; gSA (C*) B lR; gS fdUrq A vlR; gSA (D) A o B nksu ksa vlR; gSA 8.
Let V and E represent the gravitational potential and field at a distance r from the centre of uniform solid sphere, Consider the two statements: (a) The plot of V against r is discontinuous (b) The plot of E against r is discontinuous (A) Both A and B are correct (B) A is correct but B is wrong (C) B is correct but A is wrong (D*) Both A and B are wrong ekuk fd ,d le:i Bksl xksys ds dsUnz ls r nwjh ij xq:Roh; foHko rFkk {ks=k V ,oa E }kjk O;Dr fd;s tkrs gSA fuEu
nks dFkuksa ij fopkj fdft;s % (a) r ds lkis{k V dk ys[kkfp=k vlrr~ (b) r ds lkis{k E dk ys[kkfp=k vlrr~ gSA
manishkumarphysics.in
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Chapter # 11 (A) A o B (C) B lR; 9.
Gravitation
nksuksa vlrr~ gS fdUrq A vlrr~
(B) A lR; gS fdUrq B vlrr~ (D*) A o B nksu ksa vlR; gSA
Take the effect of bulging at equator of the earth and its rotation in account. Consider the following statements: HCV_Ch-11_obj I _9
(A) There are points outside the earth where the value of g is equal to its value at the equator. (B) There are points outside the earth where the value of g is equal to its value at the poles.
i`Foh dks bldh fo"kqor~ js[kk ls mHkjk gqvk rFkk blds ?kw.kZu dks ekurs gq, fuEu dFkuksa ij fopkj dhft, & (A) i`Foh ds ckgj dqN fcUnq gksaxs tgk¡ g dk eku fo"kqo r js[kk ij g ds eku ds cjkcj gksxkA (B) i`Foh ds ckgj dqN fcUnq gksxsa tgk¡ g dk eku /kzqo ij g ds eku ds cjkcj gksxkA (A) Both A and B are correct (C) B is correct but A is wrong. (A) A rFkk B nksu ksa lgh gSA (C) B lgh gS rFkk A xyr gSA. 10.
(B*) A is correct but B is wrong. (D) Both A and B are wrong (B*) A lR; gS rFkk B xyr gSA (D) A rFkk B nksu ksa xyr gSA
The time period of an earth-satellite in circular orbit is independent of
i` F oh ds pkjksa vksj o`Ù kh; d{kk es a pDdj dkVus okys mixz g dk vkorZd ky fdl ls LorU=k gks rk gs & HCV I _Ch-11_Obj.1_10
(A*) the mass of the satellite (C) none of them (A*) mixz g ds nz O ;eku ls (C) bues a ls fdlh ls Hkh ugha
(B) radius of the orbit (D) both of them (B) d{kk dh f=kT;k ls (D) nks u ks a ls
11.
The magnitude of gravitational potential energy of the moon-earth system is U with zero potential energy at infinite separation. The kinetic energy of the moon with respect to the earth is K. vuUr nqjh ij fLFkfrt ÅtkZ 'kwU; ekurs gq, i`Foh ,oa panz fudk; dh xq:Roh; fLFkfrt ÅtkZ dk ifjek.k U gSA i`Foh ds lkis{k pUnzek dh xfrt ÅtkZ K gS& (A) U < K (B*) U > K (C) U=K
12.
Fig. shows the elliptical path of a planet about the sun. The two shaded parts have equal area. If t 1 and t2 be the time taken by the planet to go from a to b and from c to d respectively lw;Z ds pkjksa vksj Hkze.k'khy xzg dh nh?kZo`Ùkkdkj d{kk fp=k esa iznf'kZr dh xbZ gSA Nk;kfdar nksuksa a ls b rd rFkk c ls d rd tkus esa yxs le; Øe'k% t1 rFkk t2 gS] rks&
13.
(A) t1 < t2 (B*) t1 = t2 (C) t1 > t2 (D) insufficient information to deduce the relation between t1 and t2. (D) nh xbZ lwpuk t1 rFkk t2.ds e/; lEcU/k LFkkfir djus ds fy;s vi;kZIr gSA A person sitting in a chair in a satellite feels weightlessness because
mixzg esa dqlhZ ij cSBk O;fDr Hkkjghurk eglwl djrk gS D;ksfa d
HCV_Ch-11_obj I _13
(A) the earth does not attract the objects in a satellite
mixzg esa j[kh oLrqvksa dks i`Foh vkdf"kZr ugha djrh gSA (B) the normal force by the chair on the person balances the earth’s attraction
dqlhZ }kjk vkneh ij yxk;k x;k vfHkyEc çfrfØ;k cy i`Foh ds vkd"kZ.k dks lUrqfyr dj nsrk gSA (C*) the normal force is zero
vfHkyEc çfrfØ;k cy 'kwU; gksrk gSA (D) the person in satellite is not accelerated
mixzg esa O;fDr Rofjr ugha gksrk gSA
manishkumarphysics.in
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Chapter # 11 14.
Gravitation
A body is suspended from a spring balance kept in a satellite. The reading of the balance is W 1 when the satellite goes in a an orbit of radius R and is W 2 when it goes in an orbit or radius 2 R. ,d mixzg esa dekuhnkj rqyk (spring balance) ls ,d oLrq dks yVdk;k tkrk gSA rqyk dk ikB;kad W 1 gS tc mixzg R f=kT;k ds d{k esa pDdj yxk jgk gS vkSj ikB~;kad W 2 gS tc ;g 2R f=kT;k dh d{kk esa pDdj yxk jgk gSA HCV_Ch-11_Obj.1_14
(A*) W 1 = W 2 15.
16.
(B) W 1 < W 2
(C) W 1 > W 2
(D) W 1 W 2
The kinetic energy needed to project a body of mass m from the earth’s surface to infinity is m nzO;eku dh ,d oLrq dks i`Foh ds lrg ls vuUr rd iz{ksfir djus ds fy;s vko';d xfrt ÅtkZ gS (A) 1/4 mgR (B) 1/2 mgR (C*) mgR (D) 2 mgR
&
A particle is kept at rest at a distance R (earth’s radius) above the earth’s surface. The minimum speed with which it should be projected so that it does not return is i`Foh dh lrg ls R (i`Foh dh f=kT;k) Å¡pkbZ ij ,d d.k fLFkj voLFkk esa j[kk x;k gSA bldks fdruh U;wure pky
ls iz{ksfir fd;k tk;s fd ;g iqu% ykSVdj ugha vk;sA (A)
17.
GM 4R
(B)
GM 2R
(C*)
GM R
(D)
2GM R
A satellite is orbiting the earth close to its surface. A particle is to be projected from the satellite to just escape from the earth. The escape speed from the earth is e. Its speed with respect to the satellite
,d mixzg i`Foh dh lrg ds utnhd pDdj yxk jgk gSA ,d d.k dks mixzg ls bl rjg ç{ksfir fd;k tkrk gS fd ;g i`Foh ls iyk;u gh dj ik,aA i`Foh ls iyk;u pky e gSA bldh pky mixzg ds lkis{k (A) will be less than e. (C) will be equal to e. (A) e ls de gksxhA (C) e ds cjkcj gksxhA
(B) will be more than e. (D*) will depend on direction of projection (B) e ls vf/kd gksxhA (D*) ç{ksi.k dh fn'kk ij fuHkZj djsxhA
HCV_11_Obj.I_17
OBJECTIVE - II 1.
Let V and E denote the gravitational potential and gravitational field at a point. It is possible to have ekukfd fdlh fcUnq ij V rFkk E xq:Roh; foHko ,oa xq:Roh; {ks=k dks O;Dr djrs gSA ;g gksuk lEHko gS& (A*)V = 0 and E = 0 (B*)V = 0 and E 0 (C*) V 0 and E = 0 (D*) V 0 and E 0
2.
Inside a uniform spherical shell (A) the gravitational potential is zero (B*) the gravitational field is zero (C*) the gravitational potential is same everywhere (D*) the gravitational field is same everywhere
le:i xksyh; dks'k esa& (A) xq:Roh; foHko 'kwU; gksrk gSA (B*) xq:Roh; {ks=k 'kwU; gksrk gSA (C*) izR;sd LFkku ij xq:Roh; foHko ,d leku gksrk gSA (D*) izR;sd LFkku ij xq:Roh; {ks=k ,d leku gksrk gSA 3.
A uniform spherical shell gradually shrinks maintaining its shape. The gravitational potential at the centre HCV_Ch-11_Obj.II_3
,d leku xksyh; dks'k viuh vkd`fÙk dks cuk, j[krs gq, /khjs&/khjs fldqM+rk gSA blds dsUnz ij xq:Roh; foHko (A) increase (A) c<+rk gSA 4.
(B*) decreases (B*) ?kVrk gS
(C) remains constant (C) ogh jgrk gSS
(D) oscillates (D) nksyu djr
gS
Consider a planet moving in an elliptical orbit round the sun. The work done on the planet by the gravitational force of the sum manishkumarphysics.in
15
Chapter # 11
Gravitation
(A) is zero in any small part of the orbit (C*) is zero in one complete revolution
(B*) is zero in some parts of the orbit (D) is zero in no part of the motion
lw;Z ds pkjksa vksj nh?kZo`Ùkkdkj d{kk esa ifjØek djrs gq, xzg ij fopkj dhft;sA lw;Z ds xq:Rokd"kZ.k cy }kjk xzg ij fd;k x;k dk;Z& (A) d{kk ds fdlh Hkh vYi Hkkx esa 'kwU; gksrk gSA (B*) d{kk ds dqN Hkkxksa esa 'kwU; gksrk gSA (C*) ,d lEiw.kZ ifjØek esa 'kwU; gksrk gSA (D) fdlh Hkh Hkkx esa 'kwU; ugha gksrk gSA 5.
Two satellites A and B move round the earth in the some orbit. The mass of B is twice the same of A. i`Foh ds pkjksa vksj fdlh d{kk esa nks mixzg A rFkk B pDdj yxk jgs gSaA B dk nzO;eku A ls nqxquk gS & HCV_11_Obj II_5
(A*) Speeds of A and B are equal A rFkk B dh pky cjkcj gksxhA (B) the potential energy of earth + A is same as that of earth + B. i`Foh + A fudk; dh fLFkfrt ÅtkZ i`Foh + B dh fLFkfrt ÅtkZ ds cjkcj gksxhA (C) The kinetic energy of A and B are equal A rFkk B dh xfrt ÅtkZ cjkcj gksxhA (D) The total energy of earth + A is same as that of earth + B i`Foh + A fudk; dh dqy ÅtkZ] i`Foh + B dh fudk; dh dqy ÅtkZ ds cjkcj gksxhA 6.
Which of the following quantities remain constant in a planetary motion (consider elliptical orbits) as seen from the sun? (A) speed (B) Angular speed (C) Kinetic energy (D*) Angular momentum
xzgksa dh xfr esa ¼ nh?kZo~Ùkkdkj d{kkvksa esa ½ lw;Z ls ns[kus ij fuEu esa ls dkSulh jkf'k;k¡ fu;r jgrh gS& (A) pky (B)sdks.kh; pky (C) xfrt ÅtkZ (D*) dks.kh; lao sx
EXERCISE
1.
Two spherical balls of mass 10 kg each are placed 10 cm apart. Find the gravitational force of attraction between them. 10 fdxzk nzO;eku dh nks xksykdkj xsnas 10 cm lseh nwj j[kh gqbZ gSA buds e/; yxus okyk xq:Roh; cy Kkr dhft;sA 1. 6.67 × 10–7 N
2.
Four particles having masses m, 2m, 3m and 4m are placed at the four corners of a square of edge a. Find the gravitational force acting on a particle of mass m placed at the centre. a Hkqtk okys oxZ ds pkjksa dksuksa ij m 2m, 3m rFkk 4m nzO;eku okys pkj d.k j[ks gq, gSA blds dsUnz ij fLFkr m
nzO;eku ds d.k ij yxus okyk xq:Roh; cy Kkr dhft;sA 2. 3.
4 2Gm2 a2
Three equal masses m are placed at the three corners of an equilateral triangle of side a. Find the force exerted by this system on another particle of mass m placed at (a) the mid-point of a side, (b) at the centre ofthe triangle. a Hkqtk okys lecgq f=kHkqt ds rhuks dksuksa ij rhu d.k j[ks x;s gS] izR;sd dk nzO;eku m gSA bl fudk; ds dkj.k m
nzO;eku ds ,d vU; d.k ij yxus okyk cy Kkr fdft;s tks& (a) ,d Hkqtk ds e/; fcUnq ij j[kk x;k gSA (b) f=kHkqt ds dsUnz ij j[kk x;k gSA 3.
4.
(a)
4Gm 2 3a 2
, (b) zero
Three uniform spheres each having a mass M and radius a are kept in such a way that each touches the other two. Find the magnitude of the gravitational force on any of the spheres due to the other two. rhu ,d leku xksyksa esa izR;sd dk nzO;eku M rFkk f=kT;k R gS] buds bl izdkj j[kk x;k gSA fd izR;sd vU; nks Li'kZ manishkumarphysics.in
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Chapter # 11
Gravitation
djrk gSA fdlh Hkh ,d xksys ij vU; nks xksyksa ds dkj.k yxus okyk xq:Roh; cy Kkr dhft;sA 4.
5.
3 Gm2 4a 2
Four particles of equal mass M move along a circle of radius R under the action of their mutual gravitational attraction. Find the speed of each particle. leku nzO;eku M okys pkj d.k ikjLifjd xq:Rokd"kZd cy ds izHkko esa R f=kT;k ds o`Ùkkdkj iFk ij xfr'khy gSA izR;sd
d.k dh pky Kkr dhft;sA 5.
6.
GM 2 2 1 R 4
Find the acceleration due to gravity of the moon at a point 1000 km above the moon’s surface. The mass of the moon is 7.4 × 1022 kg and its rdius is 1740 km. pUnzek dh lrg ls 1000 km Åij pUnzek ds dkj.k xq:Roh; Roj.k Kkr dhft;sA pUnzek dk nzO;eku 7.4 × 1022 fdxzk
rFkk bldh f=kT;k 1740 km fdeh gSA 6. 7.
0.65 m/s 2
Two small bodies of masses 10 kg and 20 kg are kept distance 1.0 m apart and released. Assuming that only mutual gravitational froces are acting, find the speeds of the particles when the separation decrease to 0.5 m. 1ehVj nwjh ij j[kh gqbZ 10 fdxzk rFkk 20 fdxzk nzO;ekuksa okyh nks NksVh oLrq,¡ eqDr dj nh tkrh gSA ;g ekurs gq,
fd dsoy ikjLifjd xq:Rokd"kZ.k gh yx jgk gS] tc d.kksa ds e/; 0.5 m nwjh jg tkrh gS] budh pkyksa ds eku Kkr dhft;s 7. 8.
4.2 × 10–5 m/s and 2.1 × 10–5 m/s
A semicircular wire has a length L and mass M. A particle of mass m is placed at the centre of the circle. Find the gravitational attraction on the particle due to the wire. ,d v/kZ o`Ùkkdkj rkj dh yEckbZ L rFkk nzO;eku M gSA o`Ùk ds dsUnz ij m nzO;eku dk ,d d.k j[kk x;k gSA rkj ds
dkj.k d.k ij yxus okyk xq:Roh; vkd"kZ.k Kkr dhft;sA 8.
9.
2 GMm L2
Derive an expression for the gravitational field due to a uniform rod of length L and mass M at a point on its perpendicular bisector at a distance d from the centre. L yEckbZ ,oa M nzO;eku okyh ,d le:i NM+ dss dkj.k blds yEc v/kZd ij NM+ ds dsUnz ls d nqjh ij xq:Roh; {ks=k
ds fy;s O;atu O;qRiUu fdft;sA 2 Gm 9.
10.
d L2 4d2
Two concentric spherical shells have masses M1, M2 and radii R1, R2 (R1 < R2). What is the force exerted by this system on a particle of mass m 1 if it is placed at a distance (R1 + R2)/2 from the centre? HCV_Ch-11_Ex._10
nks ldsUnzh; xksyh; dks'k dk nzO;eku M1, M2 rFkk f=kT;k R1, R2 (R1 < R2) gSA bl fudk; }kjk m 1 nzO;eku ds d.k ij fdruk cy yxk;k tk,xk vxj ;g dsUnz ls (R1 + R2)/2 dh nwjh ij gSA 10. 11.
4 GM1m (R1 R 2 )2
A tunnel is dug along a diameter of the earth. Find the force on a particle of mass m placed in the tunnel at a distance x from the centre. manishkumarphysics.in
17
Chapter # 11
Gravitation
i`Foh ds O;kl ds vuqfn'k ,d lqjxa [kksnh x;h gS m nzO;eku dk ,d d.k lqjax ds dsUnz ls x nwjh ij j[ks gq, m nzO;eku ds d.k ij cy Kkr dhft;sA GMem
11. 12.
R3
x
A tunnel is dug along a chord of the earth at a perpendicular distance R/2 from the earth’s centre. The wall of the tunnel may be assumed to be frictionless. Find the force exerted by the wall on a particle of mas m when it is at a distance x from the centre of the tunnel. i`Foh dh thok ds vuqfn'k ,d lqjax x;h gS] ftldh i`Foh ds dsUnz ls yEcor~ nwjh R/2 gSA ftldh dh nhokj ?k"kZ.k
jfgr ekuh tk ldrh gSA m nzO;eku dk ,d d.k lqjax ds dsUnz ls x nqjh ij gS] ij nhokj }kjk yxk;k x;k cy Kkr dhft;s 12. 13.
GMem 2R 2
A solid sphere of mass m and radius r is placed inside a hollow thin spherical shell of mass M and radius R as shown in figure. A particle of mass mis placed on the line joining the two centres at a distance x from the point of contact of the sphere and the shell. Find the magnitude of the resultant gravitational force on this particle due to the sphere and the shell if (a) r < x < 2 r, (b) 2r < x < 2R and (c) x > 2R. fp=k es iznf'kZr fd;k x;k gS fd m nzO;eku o r f=kT;k okyk ,d Bksl] M nzO;eku o R f=kT;k okys ,d irys xksyh;
dks'k esa j[kk gqvk gSA nksuksa ds dsUnzksa dks feykus okyh js[kk ij ,oa xksys rFkk dks'k ds lEidZ fcUnq ls x nwjh ij m nzO;eku dk ,d d.k j[kk x;k gSA bl d.k ij xksys rFkk dks'k ds dkj.k yxus okys ifj.kkeh xq:Roh; cy dk ifjek.k Kkr fdft;s] ;fn (a) r < x < 2 r, (b) 2r < x < 2R and (c) x > 2R.
13.
14.
(a)
Gmm( x r ) r
3
Gmm
(b)
(x r)
2
Gmm
(c)
( x R)
2
Gmm
+
( x r )2
A uniform metal sphere of radius a and mass M is surrounded by a thin uniform spherical shell of equal mass and radius 4a (figure). The centre of the shell falls on the surface of the inner sphere. Find the gravitational field at the points P1 and P2 shown in the figure. M nzO;eku rFkk a f=kT;k okyk /kkrq dk le:i xksyk] leku nzO;eku o 4a f=kT;k okys irys ,oa le:i xksyh; dks'k
esa ifjc) gS (fp=k ) dks'k dk dsUnz vkarfjd xksys dh lrg ij fLFkr gSA fp=k esa iznf'kZr fcUnqvksa P1 rFkk P2 ij xq:Roh; {ks=k Kkr dhft;sA
14.
15.
GM 16a
2
61GM
,
900 a 2
A thin spherical shell having uniform density is cut in two parts by a plane and kept separated as shown in figure. The point A is the centre of the plane section of the first part and B is the centre of the plane manishkumarphysics.in
18
Chapter # 11
Gravitation
section of the second part. Show that the gravitational feld at A due to the first part is equal in magnitude ot the gravitational field at B due to the second part.
,d iryk ,oa le:i ?kuRo okyk xksyh; dks'k] ,d lery }kjk esa dksV dj fp=kkuqlkj j[kk x;k gSA izFke Hkkx ds lery dkV dk dsUnz A rFkk f}rh; Hkkx ds lery dkV dk dsUnz B gSA O;Dr dhft;s fd izFke Hkkx ds dkj.k A ij xq:Roh; {ks+=k dk fijek.k B ij f}rh; Hkkx ds dkj.k {ks=k ds ifjek.k ds rqY; gksxkA
16.
Two small bodies of masses 2.00 kg and 4.00 kg are kept at rest at a separation of 2.0 m. Where should a particle of mass 0.10 kg be placed to experience no net gravitational force from these bodies? The particle is placed at this point. What is the gravitational potential energy of the system of three particles with usual reference level ? 16.0.83 m from the 2.00 kg body towards the other body, – 3.06× 10–10 J 2.00 kg rFkk 4.00 kg nzO;eku okyh nks NksVh oLrq,¡ 2.0 m nqjh ij fLFkr voLFkk esa j[kh gqbZ gSA 0.10 kg nzO;eku
dk ,d NksVk d.k dgk¡ ij j[kk tk;s fd bl ij bu oLrqvksa ds dkj.k dqy xq:Roh; cy 'kwU; gks\ d.k bl fcUnq ij j[k fn;k x;k gSA lkekU; lanHkZ Lrj ds lkis{k bu rhuksa d.kkss ds fudk; dh xq:Roh; fLFkfrt ÅtkZ fdruh gksxh ? 16.0.83 m from the 2.00 kg body towards the other body, – 3.06× 10–10 J
17.
Three particles of mass m each are placed at the three corners of an equilateral triangle of side a. Find the work which should be done on this system to increase the sides of the triangle to 2a. a Hkqtk okys leckgq f=kHkqt ds rhukss dksuksa ij] m nzO;eku ¼ izR;sd ½ okys rhu d.k j[ks x;s gSA f=kHkqt dh Hkqtk,¡ c<+kdj 2a djus 17.
18.
ds fy;s x;s dk;Z dk eku Kkr dhft;sA
3 Gm 2 2a
A particle of mass 100 g is kept on the surface of a uniform sphere of mass 10 kg and radius 10 cm. Find the work to be done against the gravitational force between them to take the particle away from the sphere. 100 xzke nzO;eku dk ,d d.k 10 kg nzO;eku rFkk 10 cm.f=kT;k okys xksys dh lrg ij j[kk gqvkk gSA d.k dks bl
xksys ls nwj ys tkus ds fy;s buds e/; xq:Rokd"kZ.k cy ds fo:) fd;k x;k dk;Z Kkr dhft;sA 18. 19.
6.67 × 10–10 J
The gravitational field in a region is given by E = (5 N/kg) i + (12 N/kg) j . (a) Find the magnitude of the gravitational force acting on a particle of mass 2 kg placed at the origion. (b) Find the potential at the points (12 m, 0) and (0, 5 m) if the potential at the origin is taken to be zero. (c) Find the change in gravitational potential energy if a particle of mass 2 kg is taken from the origin to the point (12 m, 5m). (d) Find the change in potential energy if the particle is taken from (12m, 0) to (0, 5m). fdlh {ks=k esa xq:Rokd"kZ.k {ks+=k E = (5 N/kg) i + (12 N/kg) j }kjk O;Dr fd;k tkrk gSA (a) ewy fcUnq ij j[ks 2
fdxzk nzO;eku ds d.k ij yxus okys xq:Roh; cy dk ifjek.k Kkr dhft;sA (b) ;fn ewy fcUnq ij foHko 'kwU; ekuk tk;s rks fcUnqvksa(12 m, 0) rFkk (0, 5 m) ij foHko Kkr dhft;sA (c) ;fn 2 kg fdxzk nzO;eku okyk d.k ewy fcUnq (12 m, 5m) rd ys tk;k tk;s rks xq:Roh; fLFkfrt ÅtkZ esa ifjorZu Kkr dhft;sA (d) ;fn d.k dks (12m, 0) ls (0, 5m)rd ys tk;k tk;s rks fLFkfrt ÅtkZ esa ifjorZu Kkr dhft;sA 19. 20.
(a) 26 N
(b) – 60 J/kg, – 60 J/kg
(c) – 240 J
(d) zero
The gravitational potential in a region is given by V = (20 N/kg) (x + y). (a) Show that the equation is dimensionally correct. (b) Find the gravitational field at the point (x, y). Leave your answer in terms of manishkumarphysics.in
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Chapter # 11
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the unit vectors i , j , k . (c) Calculate the magnitude of the gravitational force on a particle of mass 500 g placed at the origin. 20 U;wVu
fdlh {ks=k esa xq:Roh; foHko V = fdxzk (x + y) }kjk O;Dr fd;k tkrk gSA (a) P;O;Dr dhft;s fd lehdj.k
foeh; :i ls lgh gSA (b) fcUnq (x, y)ij xq:Roh; {ks=k Kkr dhft;sA viuk mÙkj ,dkad lfn'kksa i , j , k ds inks esa fyf[k;sA (c) ewy fcUnq ij j[ks gq, 500 xzke nzO;eku okys d.k ij xq:Roh; cy ds ifjek.k dh x.kuk dhft;sA 20. 21.
(b) – 20 ( i j ) N/kg
(c) 10 2 N
The gravitational field in a region is given by E = (2 i + 3 j )N/kg. Show that no work is done by the gravitational field when a particle is moved on the line 3y + 2x = 5. [Hint : If a line y = mx + c makes angle with the X-axis, m = tan]
U;wVu }kjk fdxzk
fdlh {ks=k esa xq:Roh; {ks=k E = (2 i + 3 j )
O;Dr fd;k tkrk gSA O;Dr dhft;s fd tc dksbZ d.k js[kk 3y
+ 2x = 5 ds
vuqfn'k foLFkkfir fd;k tkrk gSA rks xq:Rohl {ks=k }kjk dksbZ dk;Z ugh fd;k tkrk gSA [ladsr : ;fn js[kk y = mx + c, X v'k ls dks.k cukrh gS, m = tan] 22.
dFind the height over the earth’s surface at which the weight of a body becomes half of its value at the surface.
i`Foh dh lrg ls og Å¡pkbZ Kkr dhft;s] tgk¡ oLrq dk Hkkj] lrg ij blds eku dk vk/kk jg tk;sA 22. 23.
( 2 – 1) times the radius of the earth
What is the acceleration due to gravity on the top of Mount Everest ? Mount Everst is the height mountain peak of the world at the height of 8848 m. The value at sea level is 9.80 m/s 2.
ekm.M ,ojsLV ds 'kh"kZ ij xq:Roh; Roj.k dk eku fdruk gS\ ekm.V ,ojsLV lalkj dh mPpre pksVh gaS ftldh Å¡pkgZ 8848 m gSA leqnz dh lrg ij eku 9.80 m/s 2. 23. 24.
9.77 m/s 2
Find the acceleration due to the gravity in a mine of depth 640 m if the value at the surface is 9.800 m/ s 2. The radius of the earth is 6400 km. 640 m xgjh [kku esa xq:Roh; Roj.k Kkr dhft;s i`Foh dh lrg ij bldk eku 9.800 m/s 2. gSA i`Foh dh f=kT;k 6400 km gSA 24. 9.799 m/s 2
25.
A body is weighed by a spring balance to be 1.000 kg at the north pole. How much will it weigh at the equator? Account for the earth’s rotation only. mÙkjh /kzqoa ij fLizax rqyk ls fdlh oLrq dk Hkkj 1.000 fdxzk ekik tkrk gSA fo"kqor~ js[kk ij bldk Hkkj fdruk ekik
tk;sxk\ dsoy i`Foh dk ?kw.kZu izHkkoh gS\ 25. 26.
0.997 kg
A body stretches a spring by a particular length at the earth’s surface at equator. At what height above the south pole will it stretch the same spring by the same length ? Assume the earth to be spherical.
i`Foh dh lrg ij] Hkqe/; js[kk ij oLrq fdlh fLizax dks fdlh fof'k"V yEckbZ rd [khprh gSA nf{k.kh /kzqo ls fdruh Å¡pkbZ ij ;g blh fLizax dks bruh gh yEckbZ rd [khpsxh\ eku yhft;s fd i`Foh xksykdkj gSA 26. 27.
10 km approx
At what rate should the earth rotate so that the apparent g at the equator becomes zero ? What will be the length of the day in this situation?
i`Foh fdl nj ls ?kw.kZu djs fd] Hkwe/; js[kk ij fLFkr oLrq dk izsf{kr Hkkj 'kwU; gks tk;s\ bl ifjfLFkfr esa fnu dh vof/k fdruh gks tk;sxh\ 27.
1.237 × 10–3 rad/sec, 1.41 h manishkumarphysics.in
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Chapter # 11 28.
Gravitation
A pendulum having a bob of mass m is hanging in a ship sailing along the equator from east to west. When the ship is stationary with respect to water the tension in the string is T 0. (a) Find the speed of the ship due to rotation of the earth about its axis. (b) Find the difference between T 0 and the earth’s attraction on the bob. (c) If the ship sails at speed v, what is the tension in the string ? Angular speed of earth’s rotation is and radius of the earth is R.
fo"kqor js[kk ds vuqfn'k iwoZ ls ifjpe dh vksj xfr'khy tgkt esa] ,d yksyd yVd jgk gS] ftls ckWc dk nzO;eku m gSA tc tgkt ikuh ds lkis{k fLFkj gS] Mksjh esa ruko T 0.gSA (a) i`Foh ds viuh v{k ij ?kw.kZu ds dkj.k tgkt dh pky Kkr dhft;s (b) ckWc ij i`Foh ds vkd"kZ.k ,os T0 ds ek/; vUrj Kkr dhft;sA (c) ;fn tgkt v pky ls py jgk gS] Mksjh esa ruko fdruk gS\ i`Foh ds ?kw.kZu dh dks.kh; pky gS rFkk i`Foh dh f=kT;k R gSA 28. 29.
(a) R
(b) m2R
(c) T 0 + 2 mv approx.
The time taken by Mars to revolve round the sun is 1.88 years. Find the ratio of average distance between Mars and thesun to that between the earth and the sun. eaxy xzg dks lw;Z ds pkjksa vksj ifjØek esa 1.88 o"kZ yxrs gSA eaxy ls lw;Z dh vkSlr nwjh rFkk i`Foh dh lw;Z ls vkSlr
nwjh dk vuqikr Kkr dhft;sA 29. 30.
1.52
The moon takes about 27.3 days to revolve round the earth in a nearly circular orbit of radius 3.84 × 105 km. Calculate the mass of the earth from these data. i`Foh ds pkjksa vksj 3.84 × 105 fdeh f=kT;k dh yxHkx 27.3 fnu yxrs gSA bu vkadMks ds vk/kkj ij i`Foh ds nzO;eku
dh x.kuk dhft;sA 30. 31.
6.02 × 1024 kg
A Mars satellite moving in an orbit of radius 9.4 × 103 km takes 27540 s to complete one revolution. Calculate the mass of Mars. eaxy dk ,d mixzg 9.4 × 103 fdeh f=kT;k dh d{kk esa ,d ifjØek iwjh djus esa 27540 lsd.M yxkrk gSA eaxy
ds nzO;eku dh x.kuk dhft;sA 31. 32.
6.5 × 1023 kg
A satelite of mass 1000 kg is supposed to orbit the earth at a height of 2000 km above the earth’s surface. Find (a) its speed in the orbit, (b) its kinetic energy, (c) the potential energy of the earthsatellite system and (d) its time period. Mass of the earth = 6 × 1024 kg. 1000 fdxzk nzO;eku dk ,d mixzg] i`Foh dh lrg ls 2000 fdeh mij fLFkr d{kk esa ifjØe.k dj jgk gSA Kkr
dhft;sA (a) d{kk esa bldh pky (b) bldh xfrt Å¡tkZ (c) i`Foh mixzg fudk; dh fLFkfrt ÅtkZ rFkk (d) bldk vkorZdky i`Foh dk nzO;eku = 6 × 1024 fdxzk 32. (a) 6.90 km/s (b) 2.38 × 1010 J 33.
(c) – 4.76 × 1010 J with usual reference (d) 2.01 hours
(a) Find the radius of the circular orbit of a satellite moving with an angular speed equal to the angular speed of earth’s rotation. (b) If the satellite is directly above the north pole at some instant, find the time it takes to come over the equatorial plane. Mass of the earth = 6 × 1024 kg. (a) i`Foh dk ?kw.kZu pky ds cjkcj ?kw.kZu pky ls ?kqe jgs mixzg dh d{kk dh f=kT;k Kkr fdft;sA (b) ;fn fdlh {k.k
ij mixzg Bhd mÙkjh /kzqo ds Åij gS] bldks Hkwe/; js[kh; ry esa vkus esa yxk le; Kkr dhft;sA i`Foh dk nzO;eku = 6 × 1024 fdxzk 34.
33. (a) 42300 km (b) 6 hours What is the true weight of an object in a geostationary satellite that weighed excatly 10.0 N at the north pole? fdlhj oLrq dk mÙkjh /kzqo ij Hkkj 10.0 U;wVu gS] Hkw fLFkj mixzg esa bldk okLrfod Hkkj fdruk gksxk\ 34.
35.
0.23 N
The radius of a planet is R1 and a satellite revolves round it in a circle of radius R2. The time periof of revolution is T. Find the acceleration due to the gravitation of the planet at its surface. ,d xzg dh f=kT;k R1 gS rFkk ,d mixzg blds pkjksa vksj R2 f=kT;k dh o`Ùkkdkj d{kk esa ifjØe.k dj jgk gSA ifjØe.k
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Chapter # 11
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dk vkorZdky T gSA bldh lrg ij xzg ds dkj.k xq:Roh; Roj.k Kkr dhft;sA 35. 36.
T 2R12
Find the minimum colatitude which can directly receive a signal from a geostationary satellite. ,d Hkw LFkklh mixzg ls ladsr izkIr djus ds fy;s fdlh LFkku ij U;wure vkufr dks.k (colatitude) Kkr dhftlsA 36.
37.
4 2R 32
sin–1 (0.15)
A particle is fired vertically upward from earth’s surface and it goes upto a maximum height of 6400 km. Find the initial speed of the particle. ,d d.k i`Foh dh lrg ls Bhd m/okZ/kj fn'kk esa iz{ksfir fd;k tkrk gSA d. ;g 6400 fdeh vf/kdre Å¡pkbZ rd
tkrk gSA d.k dh izkjfEHkd pky Kkr dhft;sA 37. 38.
7.9 km/s
A particle is fired vertically upward with a speed of 15 km/s. With what speed will it move in intersteller space. Assume only earth’s gravitational field. ,d d.k m/okZ/kj Åij dh vksj 15 fdeh@ ls] pky ls iz{ksfir fd;k tkrk gSA vUrj rkjdh; varfj{k esa ;g fdl pky
ls xfr djsxk\ dsoy i`Foh dk xq:Roh; {ks=k izHkkoh gSA 38. 39.
10.0 km/s
A mass of 6 × 1024 kg (equal to the mass of the earth) is to be compressed in a sphere in such a way that the escape velocity from its surface is 3 × 108 m/s. What should be the radius of the sphere? 6 × 1024 fdxzkk nzO;eku ¼ i`Foh ds nzO;eku ds cjkcj ½ xksykdkj vkd`fr esa bruk laihfM+r fd;k tkrk gS fd bldh
lrg ls iyk;u osx 3 × 108 eh@ ls gks tkrk gSA bl xksys dh f=kT;k fdruh gksxh\ 39.
9 mm
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