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GRAVITATION

8.2 Kepler’s Laws

Q. Write down the statement of Kepler’s Laws of planetary motion ?

Solution : There are three laws of Kepler’s :

(i) Law of orbits or Kepler’s first law : All planets move in elliptical orbits with the sun situated at one ofthe foci (S or S’) of the ellipse as shown in figure.

(ii) Laws of areas or Kepler’s second law : The line that joins any planet to the sun sweeps equal areas inequal intervals of time i.e., the areal velocity of any planet around the sun remains constant.

(iii) Laws of periods or Kepler’s third law : The square of the time period (T) of revolution of a planet isproportional to the cube of the semi-major axis of the ellipse traced out by the planet i.e., T2 a3, where ais the length of semi majour axis.

Q. Kepler’s second law is the consequence of which law ? OR On which law Kepler’s second law isbased ?

Solution : Kepler’s second law is the consequence of conservation of angular momentum.

Q. Deduce Kepler’s second law ? OR How Kepler’s second law can be understood from theconservation of angular momentum ?

Solution : The law of areas can be understood as a consequence of conservation of angular momentum. Let

the sun be at the origin and let the position and momentum of the planet be denoted by r

and p

respectively. Then the area swept out by the planet of mass m in time interval t is as shown in figure, A

given by )tvr½(A

.

Hence m/)pr½(t/A , (since m/pv

) = )m2/(L

where v

is the velocity, L

is the angular momentum equal to )pr( . For a central force, which is directed

along L,r

is a constant as the planet goes around. Hence, t/A

is a constant according to the last

equation. This is the law of areas.

Q. What is central force ? Give one example.

Solution : A central force is always directed towards or away from a fixed point, i.e., along the positionvector of the point of application of the force with respect to the fixed point. Further, the magnitude of acentral force F depends on r. A central force is such that the force on the planet is along the vector joiningthe sun and the planet. Gravitation force is a central force.

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Q. Which quantity is always conserved in the motion under a central force ?

Solution : In the motion under a central force the angular momentum is always conserved.

Q. Write down two important results which will arise in the motion under a central force ?

Solution : Two important results which arises in the motion under the central force :

(i) The motion of a particle under the central force is always confined to a plane.

(ii) The position vector of the particle with respect to the centre of the force (i.e., the fixed point) has aconstant areal velocity.

Q. Let the speed of the planet at the perihelion P in figure be vP and the sun-planet distance SP be r

P,

Relate {rP, v

P} to the corresponding quantities at the aphelion {r

A, v

A}. Will the planet take equal

times to transverse BAC and CPB ? [NCERT Solved Example 8.1]

Solution : The magnitude of the angular momentum at P is Lp = m

pr

pv

p, since inspection tells us that pr

and

pv

are mutually perpendicular. Similarly, LA = m

pr

Av

A. From angular momentum conservation

mpr

pv

p = m

pr

Av

A

orp

A

A

p

r

r

v

v . Since r

A > r

p, v

p > v

A.

The area SBAC bounded by the ellipse and the radius vectors SB and SC is larger than SBPC. FromKepler’ second law, equal areas are swept in equal times. Hence the planet will take a longer time totransverse BAC than CPB.

8.3 Universal Law of Gravitation :

Q. Explain Newton’s law of gravitation ? OR Write down the force between two point massesseparated by some distance in vector form.

Solution : The force between two point masses separated by some distance is given by Newton’s law of

gravitation which will be stated as : the force F

on a point mass m2 due to another point mass m

1 has the

magnitude 2

21

r

mmG|F|

.

Equation can be expressed in vector form as rr

mmG)r(

r

mmGF

221

221

= r

|r|

mmG

321

, where G is

the universal gravitational constant, r is the unit vector from m1 to m

2 and 12 rrr

as shown in above

figure.

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Q. Write down the unit and dimensional formula of gravitational constant G.

Solution : The unit of G is Nm2/kg2 and dimensional formula [M–1L3T–2].

Q. What is the nature of the gravitational force ?

Solution : The gravitational force is attractive central force.

Q. Find the acceleration of the moon revolving in an orbit of radius Rm ?

Solution : The moon revolving in an orbit of radius Rm was subject to a centripetal acceleration due to

earth’s gravity of magnitude.

2m

2

m

2

mT

R4

R

Va

where V is the speed of the moon related to the time period T by the relation V = 2Rm/T.

Q. Write down the expression of total force on m1 due to other point masses m

2, m

3 and m

4 ?

Solution : If we have a collection of point masses, the force on any one of them is the vector sum of thegravitational forces exerted by the other point masses as shown.

The total force on m1 is 412

41

14312

31

13212

21

121 r

r

mGmr

r

mGmr

r

mGmF

Q. Three equal masses of m kg each are fixed at the vertices of an equilateral triangle ABC. (a) Whatis the force acting on a mass 2m placed at the centroid G of the triangle ? (b) What is the force if themass at the vertex A is doubled ? [NCERT Solved Example 8.2]

Take AG = BG = CG = 1 m

Solution : (a) zero (b) jGm2 2

Q. How will you determine the force between a hollow spherical shell of uniform density and a pointmass situated outside ?

Solution : The force of attraction between a hollow spherical shell of uniform density and a point masssituated outside is just as if the entire mass of the shell is concentrated at the centre of the shell.

Q. What is the force on point mass which is situated inside a hollow spherical shell of uniformdensity ?

Solution : The force of attraction due to a hollow spherical shell of uniform density, on a point masssituated inside it is zero.

8.4 The Gravitational Constant :

Q. Describe Cavendish’s experiment which is used to determine the value of G ?

Solution : The value of the gravitational constant G entering the Universal law of gravitation can bedetermined experimentally and this was first done by English scientist Henry Cavandish. The apparatusused by him is shown in figure.

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The bar AB has two small lead spheres attached at its ends. The bar is suspended from a rigid support by afine wire. Two large lead spheres are brought close to the small ones but on opposite sides as shown. Thebig spheres attract the nearby small ones by equal and opposite force as shown. There is no net force on thebar but only a torque which is clearly equal to F times the length of the bar, where F is the force of attractionbetween a big sphere and its neighbouring small spheres. Due to this torque, the suspended wire getstwisted till such time as the restoring torque of the wire equals the gravitational torque. If is the angle oftwist of the suspended wire, the restoring torque is proportional to , equal to , where is the restoringcouple per unit angle of twist. can be measured independently e.g., by applying a known torque andmeasuring the angle of twist. The gravitation force between the spherical balls is the same as if their massesare concentrated at their centres. Thus if d is the separation between the centres of the big and its neighbouringsmall ball, M and m their masses, the gravitational force between the big sphere and its neighbouring small

ball is, 2d

MmGF . If L is the length of the bar AB, then the torque arising out of F is F multiplied by L. At

equilibrium, this is equal to the restoring torque and hence Ld

MmG

2.

Observation of this enables one to calculate G from this equation.

Since Cavandish’s experiment, the measurement of G has been refined and the currently accepted value is

G = 6.67 × 10–11 N m2/kg2

8.5 Acceleration due to Gravity of the Earth :

Q. A particle mass ‘m’ is situated inside the earth of mass ME and radius R

E at the distance r from the

centre of earth. Find the force on this particle due to earth.

Solution :

Consider the earth to be made up of concentric shells and the point mass m situated at a distance r from thecentre. For the shells of radius greater than r, the point P lies inside, they exert no gravitational force onmass m kept at P. The shells with radius less than or equal to r make up a sphere of radius r for which thepoint P lies on the surface. This smaller sphere therefore exerts a force on a mass m at P as if its mass M

r is

concentrated at the centre. Thus the force on the mass m at P has a magnitude 2

r

r

)M(GmF . We assume

that the entire earth is of uniform density and hence its mass is

3EE R

3

4M where M

E is the mass of the

earth RE is its radius and is the density. On the other hand the mass of the sphere M

r of radius r is

3r3

4

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and hence rR

GmM

r

r

R

MGm

r

r

3

4GmF

3E

E2

3

3E

E2

3

If the mass m is situated on the surface of earth, then r = RE and the gravitational force on it is given by

2E

E

R

mMGF .

Q. Derive the expression of acceleration due to gravity due to earth on its surface ?

Solution : The acceleration experienced by the mass m, which is usually denoted by the symbol g is related

to F by Newton’s 2nd law by relation F = mg. Thus 2E

E

R

GM

m

Fg

8.6 Acceleration Due to Gravity below and above the Surface of Earth :

Q. Discuss the variation of g with height and depth ?

Solution : Variation of g with height above the surface of earth :

Consider a point mass m at a height h above the surface of the earth as shown in figure. The radius of theearth is denoted by R

E. Since this point is outside the earth, its distance from the centre of the earth is

(RE + h). If F(h) denoted the magnitude of the force on the point mass m, we get

2E

E

)hR(

mGM)h(F

.

The acceleration experienced by the point mass is F(h)/m g(h) and we get

2E

E

)hR(

GM

m

)h(F)h(g

=

2E2

E2E

E )R/h1(g)R/h1(R

GM

, where 2E

E

R

GMg .

For 1R

h

E

(near the surface of earth), using binomial expression,

ER

h21g)h(g .

Variation of g with depth :

Now, consider a point mass m at a depth d below the surface of the earth, so that its distance from the centreof the earth is (R

E – d) as shown in figure.

The earth can be thought of as being composed of a smaller sphere of radius (RE – d) and a spherical shell

of thickness d. The force on m due to the outer shell of thickness d is zero. As far as the smaller sphere ofradius (R

E – d) is concerned, the point mass is outside it and hence the force due to this smaller sphere is

F(d) = G MS m/(R

E – d)2, here M

s is the mass of the smaller sphere.

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The value of Ms is given by, 3Es dR

3

4M =

3E

3E

E )dR(3

4

R3

4

M

= 3

E

3EE

R

)dR(M .

Subtituting for Ms in F(d) = G M

S m/(R

E – d)2, we get F(d) = G M

E m(R

E – d)/R

E3.

Hence the acceleration due to gravity at a depth d is given by

m

)d(F)d(g )dR(

R

GME3

E

E = )R/d1(gR

dRg E

E

E

2E

E

R

GMg

Hence acceleration due to earth’s gravity is maximum on its surface and decreases whether you go up ordown from the surface.

8.7 Gravitational Potential Energy :

Q. What is gravitational potential energy ? Derive the expression of gravitational potential energyfor two point masses m

1 and m

2 separated by distance r ?

Solution : The force of gravity is a conservative force and we can calculate the potential energy of a bodyarising out of this force, called the gravitational potential energy. The gravitational potential energy at apoint is the amount of work done in displacing the particle from infinity to that point.

We know that a point outside the earth, the force of gravitation on a particle directed towards the center of

the earth is 2E

r

mGMF , where M

E = mass of earth, m = mass of the particle and r its distance from the

center of the earth. Now we calculate the work done in lifting a particle from to r, we get

W = V =

r

Fdr

r

2E dr

r

mGM =

r

mGM1

r

1mGM

r

1mGM E

E

r

E

.

Hence the gravitational potential energy associated with two particles of masses m1 and m

2 separated by

distance by a distance r is given by V = r

mGm 21 (if we choose V = 0 as r ).

Q. Find the potential energy of a system of four particles placed at the vertices of a square of side l.Also obtain the potential at the centre of the square. [NCERT Solved Example 8.3]

Solution : Consider four masses each of mass m at the corners of a square of side l. We have four masspairs at distance l and two diagonal pairs at distance 2 l.

Hence, ll 2

Gm2

Gm4)r(W

22

ll

22 Gm41.5

2

12

Gm2

The gravitational potential at the center of the square (r = 2 l/2) is l

Gm24)r(U .

8.8 Escape Speed :

Q. Define escape speed ? Derive the expression of escape speed from the surface of earth ?

Solution : Escape speed on earth (or any other planet) is defined as the minimum speed with which a bodyhas to be projected from the surface of earth (or any other planet) so that it just crosses the gravitationalfield of earth (or of that planet) and never returns.

Derivation of escape speed :

Let Ve is the escape speed of an object of mass m from the surface of spherical astronomical body of mass

M and radius R then from conservation of mechanical energy principle,

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0R

GMmmV

2

1 2e

R

GM2Ve

Using the relation g = GME/R

E2, we get Ee gR2V .

Q. Express the escape speed in terms of the density of planet.

Solution : Let the astronomical body is of uniform density then M = 3R3

4 and hence 2

e GR3

8V .

Q. Does the escape speed depend on the mass of projected particle ?

Solution : The escape speed of an object at a given point in the field is independent of its mass.

Q. Does the escape speed depend on angle of projection ?

Solution : The escape speed does not depend on angle of projection.

Q. What is value of escape speed from the surface of earth ?

Solution : The value of escape speed is 11.2 km/s from the surface of earth.

Q. Moon has no atmosphere. Give reason ?

Solution : The escape speed for the moon turns out to be 2.3 km/s. This is the reason that moon has noatmosphere. Gas molecules if formed on the surface of the moon has no atmosphere. Gas molecules ifformed on the surface of the moon having velocities larger than this will escape the gravitational pull of themoon.

Q. Two uniform solid spheres of equal radii R, but mass M and 4 M have a centre to centreseparation 6 R, as shown in figure. The two spheres are held fixed. A projectile of mass m is projectedfrom the surface of the sphere of mass M directly towards the centre of the second sphere. Obtain anexpression for the minimum speed v of the projectile so that it reaches the surface of the secondsphere. [NCERT Solved Example 8.4]

Solution :

2/1

R5

GM3v

8.9 Earth Satellites :

Q. What is earth satellite ?

Solution : Earth satellites are objects which revolve around the earth. In particular, their orbits around theearth are circular or elliptic.

Q. What is the use of artificial earth satellites ?

Solution : Artificial earth satellites for practical use in fields like telecommunication, geophysics andmeterology.

Q. Derive the expression of time period of a earth satellite ? OR Derive Kepler’s third law forcircular orbit?

Solution : We will consider a satellite in a circular orbit of a distance (RE + h) from the centre of the earth,

where RE = radius of the earth. If m is the mass of the satellite and V its speed, the centripetal force required

for this orbit is )hR(

mV)lcentripeta(F

E

2

directed towards the centre. The centripetal force is provided

by the gravitational force, which is 2

E

E

)hR(

GmM)ngravitatio(F

where M

E is the mass of the earth.

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Equating R.H.S. of both equation, we get )hR(

GMV

E

E2

. Its time period T therefore is

E

2/3EE

GM

)hR(2

V

)hR(2T

. Squaring both sides of this equation, we get TT2 = k(R

E + h)3

(where k = 42/GME) which is Kepler’s law of periods, as applied to motion of satellites around the earth.

Q. Find the time period of a satellite which is very near to the surface of earth ?

Solution : For a satellite very close to the surface of earth h can be neglected in comparison to RE in

equation. Hence, for such satellites, T is T0, where g/R2T E0 . If we substitute the numerical values

g = 9.8 m s–2 and RE = 6400 km., we get s

8.9

104.62T

6

0

which is approximately 85 minutes.

Q. The planet Mars has two moons, phobos and delmos. (i) phobos has a period 7 hours, 39 minutesand an orbital radius of 9.4 × 103 km. Calculate the mass of mars. (ii) Assume that earth and marsmove in circular orbits around the sun, with the martian orbit being 1.52 times the orbital radius ofthe earth. What is the length of the martian year in days ? [NCERT Solved Example 8.5]

Solution : (i) 6.48 × 1023 kg (ii) 0.99986

Q. Weighing the Earth : You are given the following data : g = 9.81 ms–2, RE = 6.37 × 106 m, the

distance to the moon R = 3.84 × 108 m and the time period of the moon’s revolution is 27.3 days.Obtain the mass of the Earth M

E in two different ways. [NCERT Solved Example 8.6]

Solution : 6.02 × 1024 kg

Q. Express the constant k (k = 42/GME) in days and kilometres. Give k = 10–13 s2 m–3. The moon is at

a distance of 3.84 × 105km from the earth. Obtain its time period of revolution in days.[NCERT Solved Example 8.7]

Solution : K = 1.33 × 10–14d2 km–3, T = 27.3 d

8.10 Energy of an Orbiting Satellite :

Q. Derive the expression of kinetic energy, potential energy and total energy of an orbitingsatellite ?

Solution : The kinetic energy of the satellite in a circular orbit with speed v is )hR(2

GmMmv

2

1KE

E

E2

,

the potential energy at distance (R + h) from the centre of the earth is )hR(

GmME.P

E

E

. The total E is

)hR(2

GmME.PE.KE

E

E

.

Q. If the total energy is E, potential energy is U and kinetic energy is K, then find the ratioE : U : K ?

Solution : As K = )hR(2

GmM

E

E

, U =

)hR(

GmM

E

E

, E =

)hR(2

GmM

E

E

, hence E : U : K = –1 : –2 : 1.

Q. Does the kinetic and potential energy of a satellite in elliptic orbit change from point to point inthe orbit ? What will happen to total energy of satellite in this orbit ?

Solution : When the orbit of a satellite becomes elliptic, both the K.E. and P.E. vary from point to point.The total energy which remains constant is negative

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8.11 Geostationary and Polar Satellites :

Q. What is Geostationary satellite ? Find the height above the surface of earth of such satellites ?

Solution : Geostationary satellite are those satellite for which the circular orbit is in the equatorial plane ofthe earth, having the same period as the period of rotation of the earth about its own axis and hence it wouldappear stationary viewed from a point on earth.

Height of geostationary satellite above the surface of earth :

From E

2/3EE

GM

)hR(2

V

)hR(2T

, we get

3/1

2E

2

E4

GMThR

=

3/1

2

22

4

gRT

2E

R

GMg

Put T = 24 hours, R = 6400 km then we get h = 35800 km.

Q. Give one example of Geostationary satellite and write down one use of this ?

Solution : The INSAT group of satellites sent up by India are one such group of Geostationary satelliteswhich are widely used for telecommunications in India.

Q. What is polar satellites ? What are the uses of such satellite ?

Solution : Polar satellites are low altitude ( h 500 to 800 km) satellites, they go around the poles of theearth in a north-south direction.

Use of polar satellites : in remote sensing, meterology and as well as for environmental studies of the earth.

8.12 Weightlessness :

Q. What is weightlessness ?

Solution : When an object is in free fall, it is weightless and this phenomenon is usually called thephenomenon of weightlessness.

Q. Explain the phenomenon of weightlessness in satellites around the earth ?

Solution : In a satellite around the earth, every part and parcel of the satellite has an acceleration towardsthe center of the earth which is exactly the value of earth’s acceleration due to gravity at that position. Thusin the satellite everything inside it is in a state of free fall. This is just as if we were falling towards the earthfrom a height. Thus, in a manned satellite, people inside experience no gravity.

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NCERT EXERCISE

8.1 Answer the following :

(a) You can shield a charge from electrical forces by putting it inside a hollow conductor.Can you shield a body from the gravitational influence of nearby matter by putting itinside a hollow sphere or by some other means ?

(b) As astronaut inside a small space ship orbiting around the earth cannot detect gravity. Ifthe space station orbiting around the earth has a large size, can he hope to detectgravity ?

(c) If you compare the gravitational force on the earth due to the sun to that due to the moon,you would find that the Sun’s pull is greater than the moon’s pull. (you can check thisyourself using the data avail able in the succeeding exercise). However, the tidal effect ofthe moon’s pull is greater than the tidal effect of sun. Why ?

8.2 Choose the correct alternative :

(a) Acceleration due to gravity increases/decreases with increasing altitude.

(b) Acceleration due to gravity increases/decreases with increasing depth (assume the earth tobe a sphere of uniform density).

(c) Acceleration due to gravity is independent of mass of the earth/mass of the body.

(d) The formula – G Mm(1/r2 – 1/r

1) is more/less accurate than the formula mg(r

2 – r

1) for the

difference of potential energy between two points r2 and r

1distance away from the centre

of the earth.

8.3 Suppose there existed a planet that went around the sun twice as fast as the earth. What would be itsorbital size as compared to that of the earth ?

8.4 Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is4.22 × 108 m. Show that the mass of Jupiter is about one-thousandth that of the sun.

8.5 Let us assume that our galaxy consists of 2.5 × 1011 stars each of one solar mass. How long will a starat a distance of 50,000 ly from the galactic centre take to complete one revolution ? Take thediameter of the Milky Way to be 105 ly.

8.6 Choose the correct alternative :

(a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite isnegative of its kinetic/potential energy.

(b) The energy required to launch an orbiting satellite out of earth’s gravitational influenceis more/less than the energy required to project a stationary object at the same height(as the satellite) out of earth’s influence.

8.7 Does the escape speed of a body from the earth depend on (a) the mass of the body, (b) the locationfrom where it is projected, (c) the direction of projection, (d) the height of the location from wherethe body is launched ?

8.8 A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed,(b) angular speed, (c) angular momentum, (d) kinetic energy, (e) potential energy, (f) total energythroughout its orbit ? Neglect any mass loss of the comet when it comes very close to the Sun.

8.9 Which of the following symptoms is likely to affilict an astronaut in space (a) swollen feet,(b) swollen face, (c) headache, (d) orientational problem.

8.10 In the following two exercises, choose the correct answer from among the given ones :

The gravitational intensity at the centre of a hemispherical shell of uniform mass density has thedirection indicated by the arrow (see in the figure) (i) a, (ii) b, (iii) c, (iv) 0.

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8.11 The direction of the gravitational intensity at an arbitrary point P is indicated by the a r r o w(i) d, (ii) e, (iii) f, (iv) g.

8.12 A rocket is fired from the earth towards the sun. At what distance from the earth’s centre is thegravitational force on the rocket zero ? Mass of the sun = 2 × 1030 kg, mass of the earth = 6 × 1024 kg.Neglect the effect of other planets etc. (orbital radius = 1.5 × 1011 m).

8.13 How will you ‘weigh the sun’, that is estimate its mass ? The mean orbital radius of the earth aroundthe sun is 1.5 × 108 km.

8.14 A saturn year is 26.5 times the earth year. How far is the saturn from the sun if the earth is1.50 × 108 km away from the sun ?

8.15 A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earthat a height equal to half the radius of the earth ?

8.16 Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half waydown to the centre of the earth if it weighed 250 N on the surface ?

8.17 A rocket is fired vertically with a speed of 5 km s–1 from the earth’s surface. How far from the earthdoes the rocket go before returning to the earth ? Mass of the earth = 6.0 × 1024 kg; mean radius ofthe earth = 6.4 × 106 m; G = 6.67 × 10–11 N m2 kg–2.

8.18 The escape speed of a projectile on the earth’s surface is 11.2 km s–1. A body is projected out withthrice this speed. What is the speed of the body far away from the earth ? Ignore the presence of thesun and other planets.

8.19 A satellite orbits the earth at the height of 400 km above the surface. How much energy must beexpended to rocket the satellite out of the earth’s gravitational influence ? Mass of thesatellite = 200 kg; mass of the earth = 6.0 × 1024 kg; radius of the earth = 6.4 × 106 m;G = 6.67 × 10–11 N m2 kg–2.

8.20 Two stars each of one solar mass (= 2 × 1030 kg) are approaching each other for a head on collision.When they are a distance 109 km, their speeds are negligible. What is the speed with which theycollide ? The radius of each star is 104 km. Assume the stars to remain undistorted until they collide.(Use the known value of G).

8.21 Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontaltable. What is the gravitational force and potential at the mid point of the line joining the centres ofthe spheres ? Is an object placed at that point in equilibrium ? If so, is the equilibrium stable orunstable ?

8.22 As you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly36,000 km from the surface of the earth. What is the potential due to earth’s gravity at the site of thissatellite ? (Take the potential energy at infinity to be zero). Mass of the earth = 6.0 × 1024 kg,radius = 6400 km.

8.23 A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a speed of 1.2 rev.per second. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objectscalled pulsars belong to this category). Will an object placed on its equator remain stuck to thissurface due to gravity ? (mass of the sun = 2 × 1030 kg).

8.24 A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launchedit out of the solar system ? Mass of the space ship = 1000 kg; mass of the sun = 2 × 1030 kg; mass ofmars = 6.4 × 1023 kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28 × 108 km;G = 6.67 × 10–11 N m2 kg–2.

8.25 A rocket is fired vertically with a speed of 5 km s–1 from the earth’s surface. How far from the earthdoes the rocket go before returning to the earth ? Mass of the earth = 6.0 × 1024 kg; mean radius ofthe earth = 6.4 × 106 m; G = 6.67 × 10–11 N m2 kg–2.

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8.1 (a) No.

(b) Yes, if the size of the sphere ship is large enough for him to detect the variation in g.

(c) Tidal effect depends inversely on the cube of the distance unlike force, which depends inverselyon the square of the distance.

8.2 (a) decreases; (b) decreases; (c) mass of the body; (d) more

8.3 Smaller by a factor of 0.63

8.5 3.54 × 108 years

8.6 (a) Kinetic energy, (b) less

8.7 (a) No, (b) No, (c) No, (d) Yes

8.8 All quantities vary over an orbit except angular momentum and total energy.

8.9 (b), (c) and (d)

8.10 8.10 and 8.11 for these two problems, complete the hemisphere to sphere. At both P, and C potentialis constant and hence intensity = 0. Therefore, for the hemisphere, (c) and (e) are correct.

8.12 2.6 × 108 m

8.13 2.0 × 1030 kg

8.14 1.43 × 1012 m

8.15 28 N

8.16 125 N

8.17 8.0 × 106 m from the earth’s centre

8.18 31.7 km/s

8.19 5.9 × 109 J

8.20 2.6 × 106 m/s

8.21 0, 2.7 × 10–8 J/kg; an object placed at the mid point is in an unstable equilibrium

8.22 –9.4 × 106 J/kg

8.23 GM/R2 = 2.3 × 1012ms–2, 2R = 1.1 × 106ms–2; here is the angular speed of rotation. Thus in therotating frame of the star, the inward force is much greater than the outward centrifugal force as itsequator. The object will remain stuck (and not fly off due to centrifugal force). Note, if angular speedof rotation increases say by a factor of 2000, the object will fly off.

8.24 3 × 1011 J

8.25 8.0 × 106 m from the earth’s centre

ANSWERS

PG – 13


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ADDITIONAL QUESTIONS AND PROBLEMS

Q. Draw graphs showing the variation of acceleration due to gravity with (a) height above the Earth’ssurface, (b) depth below the Earth’s surface.

A.

Q. Why are space rockets usually launched from west to east in the equatorial line ?

A. We know that earth revolves from west to east about its polar axis. Therefore, all the particles on the earthhave velocity from the west to east. This velocity is maximum in the equatorial line, as v = R , where R isthe radius of earth and is the angular velocity of revolution of earth about its polar axis. When a rocket islaunched from west to east in equatorinal plane, the maximum linear velocity is added to the launchingvelocity of the rocket, due to it, launching becomes easier.

Q. The escape speed of the projectile on the earth’s surface is ve. A body is projected out with thrice of

this speed. What is the speed of the body after crossing the gravitational field of the earth ? Ignorethe presence of Sun and the another planets.

A. 22 ve

Q. What is the potential energy of a body of mass m relative to the surface of Earth at a (a) heighth = R above its surface (b) depth d = R below its surface ?

Q. The mass and the diameter of a planet are twice those of Earth. What will be the time period of thatpendulum on this planet, which is a second pendulum on the Earth.

Q. Although gravitational pull of sun on earth is more than that of moon, yet moon’s contributiontowards tide formation on earth is greater than that due to sun. Why ?

A. The distance between moon and earth is very small as compared to the distance between earth and sun.Since, the tidal effect on oscen water of earth is inversely proportional to the cube of the distance, therefore,tidal effect on oscen water due to moon is larger than that due to sun.

Q. Why does a body lose weight at the centre of the earth ?

Q. Explain, why a tennis ball bounces higher on hills than in plains.

Q. Does the force of attraction between two bodies depend upon the presence of other bodies andproperties of intervening medium ?

A. The gravitational force of attraction between the two bodies is independent of the presence of other bodiesand properties of intervening medium.

Q. Why is Newton’s law of gravitation called a universal law ?

A. Newton’s law of gravitation holds good irrespective of the nature of two bodies, i.e., big or small, at alltimes, at all locations and for all distances in the universe.

Q. On earth value of G = 6.67 × 10–11 Nm2kg–2. What is its value on moon, where g is nearly one-sixththan that of earth ?

A. The value of G is same on moon as on the surface of earth because G is a universal gravitational constant.

Q. The value of g on the moon is 1/6th of that of earth. If a body is taken from the earth to the moon,then what will be the change in its (i) weight, (ii) inertial mass and (iii) gravitational mass ?

A. Weight of body on moon will become 1/6th of that on the earth, but there will be no change in inertial andgravitational masses.

Q. Moon travellers tie heavy weight at their back before landing on the moon. Why ?

A. The value of g on moon is small, therefore, the weight of moon travellers will also be small.

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Q. Is it possible to put an artificial satellite on an orbit in such a way that it always remains visibledirectly over chandigarh ?

A. No, because to put an artificial satellite in an orbit such that it always remains directly over a particularplace, its time period should be the same as that of the earth in the equatorial plane. As Chandigarh does notlie on the equatorial plane, a geostationary satellite cannot be seen over Chandigarh.

Q. An astronaut, while revolving in a circular orbit happens to throw a ball outside. Will the ball reachthe surface of earth ?

A. The ball will never reach the surface of earth as it will continue to move in the same circular orbit and willchase the astronaut.

Q. Distinguish between g and G.

A. The value of g changes with height depth and rotation of the earth and is zero at the centre of the earth.

The value of G is constant at all locations. The value of G is never zero any where.

Q. Generally the path of a projectile from the surface of earth is parabolic but it is elliptical forprojectiles going to a very great height. Why ?

A. Upto ordinary heights, the change in the distance of a projectile from the centre of the earth is negligiblesmall, as compared to the radius of the earth. Due to which the projectile moves under a uniformgravitational force and its path is parabolic. But if projectile is going to a very great height, the gravitationalforce decreases, which is inversely proportional to the square of the distance of the projectile from thecentre of the earth. Under such a variable force, the path of projectile is elliptical.

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