Gravitation.pdf

Vidyamandir Classes

Vidyamandir Classes Gravitation

Gravitation NEWTON’S UNIVERSAL LAW OF GRAVITATION

Section - 1

Every body attracts in this universe every other body with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Let m1 and m2 be the masses of two bodies and r be the separation between them.



m m F 1 2 r2 G m1 m2 F r2

Where G = 6.67 × 10–11 Nm2/kg2

ACCELERATION DUE TO GRAVITY (of Earth) :

Section - 2

If M is the mass of earth and R is the radius, the earth attracts a mass m on its surface by a force F given as :

F

GM m R2

This force imparts an acceleration to the mass m which is known as

Acceleration Due to Gravity(g) By Newton’s IInd Law, we have : Acceleration =

force mass

GM m F  m

R2 m

GM



g



On the surface of earth, g =



R2

GM

R2 Substituting the values of G, M, R we get :

g = 9.81 m/s2.

1

Section 1 & 2

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Variation in acceleration due to gravity : (a)

Above the Surface of Earth : At a point which is located at a height h above the earth’s surface, acceleration due to gravity (g) is given as :

GM m force  R  h  g   mass m

2



GM

 R  h 2

2

 (b)

 R  g  g   where g = acceleration due to gravity at the surface.  R  h

Below the Surface of Earth : Let us place a mass m at a depth h below the surface of earth. This mass m is attracted towards the centre of earth by gravitational force due to shaded sphere of radius (R – h). The net gravitational force on m due to the remaining portion of earth (outside the shaded portion) is zero. Let the mass of the shaded portion  M  Assuming earth to be a sphere of uniform density d, d



M 4  R2 3

M   (volume of shaded part) (density) 3



M 

4 M R  h 3   R  h  M   4 3  R   R3 3

The net force on m = force exerted by shaded portion

Force 

G Mm

 R  h 

2



GM m  R  h 

 R  h

2

R3

3



GM m R3

 R  h

h  g   g 1   R 

Where g = acceleration due to gravity at the surface. Note : It should be noted that value of g decreases if we move above the surface or below the surface of the earth.

2

Section 2


Vidyamandir Classes (a)

Gravitation

With Latitudes : Consider a point A on the earth’s surface where the line joining centre O to A makes an angle  with the equatorial plane as shown. The angle  is said to be the latitude of the point A.

Let us place a small mass m at A. Analysing the forces on m, from the reference frame of (or relative to) the earth, we find two forces - gravitational force (mg) and pseudo force (m 2r). Here r is the radius of the circular path followed by m.  r = R cos The resultant force on masses m is given by





m2 g 2  m 2 4 r 2  2  mg  m 2 r cos     

F 

The effective gravity at point A is g 

F  m

g 2   4 r 2  2 2 g r cos  2



 2 R   2R  2 g  g 1    cos   2   cos   g   g     

Where g is the acceleration due to gravity at the surface of the earth neglecting the effect of the rotation. In Particular 1.

At Poles :

 = 90°  g = g Hence the rotation of the earth has no effect on the gravity at poles. 2.

At Equator :

 = 0°  2 R  g   g 1     g  


Section 2

3


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ESCAPE VELOCITY

Section - 3

It is the minimum velocity with which a body must be projected from the surface of earth so that it permanently overcomes and escapes the gravitational field of the earth. We can also say that a body projected with escape velocity will be able to go to a point which is at infinite distance from the earth. Let us imagine what happens to a body of mass m if it is thrown from the earth with a velocity Ve (escape velocity). As the body moves away from the earth, it slows down (due to gravitational pull of the earth) and hence its kinetic energy is converted into gravitational potential energy of the mass-earth system. Let us imagine that is just able to reach upto infinity (where G.P.E is zero) K.E. lost by mass m = gain in G.P.E. of mass-earth system 1 mVe2   G.P.E. f   G.P.E.i 2

1  GM m mVe2  0     2 R  



Ve 

2G M R

or

Ve 

2g R

Substituting the values of g = 9.81 m/s2 and R = 6400 Km, we get : Ve = 11.3 km/s Hence any object thrown with a velocity of 11.3 km/s or more will escape the gravitational field of the earth and will never come back to the earth.

MOTION OF A SATELLITE AROUND EARTH

Section - 4

Consider a satellite of mass m revolving in a circle around the earth which is located at the centre of its orbit. If the satellite is at a height h above the earth’s surface, the radius of its orbit is r = R + h, where R is the radius of the earth. The gravitational force between m and M provides the centripetal force necessary for circular motion.

Orbital Velocity : The velocity of a satellite in its orbit is called orbital velocity. Let v be the orbital velocity of satellite, then G Mm r2

4



mv 2 r

Section 3 & 4


Vidyamandir Classes



v

Gravitation

GM r

GM Rh

or v 

Hence the orbital velocity of a satellite is decided by the radius of its orbit or its height above the earth’s surface. For a satellite very close to the earth’s surface,

v

GM  r

GM  R

gR

Time Period : The time taken to complete one revolution is called the time period. It is given by :

T



2 r r  2 r v GM 4 2 3  T  r GM

2 r r T GM

2

Note : For a satellite whose time period is 24 hrs (same as the time period of the earth’s rotation) Radius of orbit r = 3

GMT 2 4 2

(T = 24 hrs.)

This satellite seems stationary if observed from the surface of earth. This is also known as Geostationary Satellite.

Total Energy of the Satellite TE

=K+U

(K : kinetic energy and U : potential energy)

TE 

1  G M m mv 2     2 r  

TE 

1 G M  GM m m  2  r  r

G M   GM m TE      r   2r   

TE  

GM m 2r

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Section 4

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Vidyamandir Gravitation Classes

GMm 2r The energy that must be given to the orbiting satellite to make it escape to infinity is known as binding energy.

Binding Energy = – (Total Energy) 

Trajectory of a Satellite for different speeds : Let v be the velocity given to a satellite. Let Vc represent the velocity for a circular orbit and Ve be the escape velocity.

Vc 

GM r

and

Ve 

2G M r

Where r is the distance of the satellite from centre of the earth. (i)

v < Vc

The satellite follows an elliptical path with centre of earth as the farther focus. In this case, if satellite is projected from near the surface of earth, it will hit the earth’s surface without completing the orbit.

(ii)

v = Vc

The satellite follows a circular orbit with the centre of earth as the centre of orbit.

(iii) Vc < v < Ve

The satellite follows an elliptical orbit with the centre of earth as the focus nearer to the point of projection.

(iv) v = Ve

The satellite escapes from the field of earth along a parabolic trajectory.

(v)

The satellite escapes the field of earth along a hyperbolic trajectory.

v > Ve

KEPLER’S LAWS

1.

Section - 5

Law of Orbits : Each planet revolves around the sun in an elliptical orbit with the sun at one focus of the ellipse.

2.

Law of Areas : This law states that the radius vector from the sun to the planet sweeps out equal areas in equal time intervals. Both shaded areas are equal if t he t ime from A t o B is equal to the time from to Q.

3.

Law of Periods : It states that the square of the time taken by the planet about the sun is proportional to the cube of the planet’s mean distance from the sun. If T be the time period of the planet and r be the mean distance of planet from the sun (average of maximum and minimum distances from sun) r  rmax r  min 2

 6

T2/r3 is same for all planets. Section 5


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Gravitation

Illustration - 1 Two planets have masses in the ratio 1 : 10 and radii in the ratio 2 : 5. Compare (i) their densities, (ii) the acceleration due to gravity on their surface, (iii) escape velocities from their surfaces and (iv) the periods of revolutions of satellites near to their surfaces. SOLUTION : Let M1, M2 be the masses and R1, R2 be radii of planets.  (i)

M1 1  M 2 10

and

2

(iii)

4 3  3  R2     M2   

Ratio of g1 and g2

g1 M R  1  1  2  = g 2 M 2  R1  10

R1 2  R2 3

d1 Ratio of densities = d l 2

   M1     4  R13  3 



Escape velocity = Ve1 Ve2

(iv)



M1 M2





d1  1   5  25      d2  10   2  16

(ii)

Acceleration due to gravity at surface  g 

GM



5 8

2G M R

Time period of a satellite near surface (orbit radius = R) =

3

2

R2 1 5 1  .  R1 10 2 2

3

d1 M R   1  2 d 2 M 2  R1 

5   2



T1  T2



10  2  1  5 

2 R R GM

M 2  R1  M1  R 2 

R1 R2

2 4  5 5

R2

Illustration - 2 Consider an earth satellite so positioned that it appears stationary to an observer on the earth. What would be the height at which the satellite should be positioned ? SOLUTION : Time period of the satellite = Time period of earth’s rotation = 24 hrs. If r is the radius of orbit, then, T  2  r GM 

24  3600 

Mearth = 6 × 1024 kg ; G = 6.67 × 10–11 Nm2/kg2.

 r ~ 42330 km

r 2 r GM

Substituting :

r

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If h be the height of satellite, then h=r–R  h = (42330 – 6400) km 

h = 35930 km

Section 5

7

Vidyamandir Gravitation Classes

Vidyamandir Classes

Illustration - 3 A body is thrown from the surface of earth with the velocity 1 km/s. Find the maximum height above the surface of the earth upto which it will go ? SOLUTION : Let u = 1km/s, h = height attained by the body and R = radius of the earth We have, GM = gR2 As the body goes away, it loses K.E. and gains P.E. Loss in K.E. = Gain in G.P.E. 1/2 mu2 – 0 = (G.P.E.)f – (G.P.E.)i



1 2 GM GM g R2 g R2 u    2 R Rh R Rh u2





2g R h

2



h R  R  h

u2R 2 g R  u2 2

 GM m   GM m  1 mu 2      2 R  h R    

103    6400  103    2 2  9.8  6400  103  103  

h = 51430.408 m = 51.43 km.

Illustration - 4 (a) Find the height from the earth’ surface where g will be 25% of its value on the surface of earth.(R = 6400 km) (b) Find the % age change in the value of g when we go 4 km below the surface of earth. SOLUTION : (a)

At a height h above the earth’s surface, we have  R  g  g    R  h 



R 1  Rh 2

At a depth h below earth’ surface, we have  h g   g 1    R

2

 R  g / 4 g   R  h

(b)

2





g  g 4  g 6400



% age =

g 4  1 g 6400 

g 1  g 1600

g 1  100  % decrease. g 16

 h = R = 6400 km.

8

Section 5


Vidyamandir Classes

Gravitation

Illustration - 5 Calculate the orbital velocity of a satellite revolving at a height h above the earth’ s surface if h = R. Also calculate the time period of this satellite. (g = 9.8 m/s2, R = 6400 km) SOLUTION : For orbital velocity in a circular orbit, we have

v

 v 

GM  r

GM Rh



(r = R + h)

gR 2 gR  (GM = gR2 and h = R) 2R 2

v

6400  103  9.8  5.6 km / s 2

Time period = T =



T

2   2R  2 r  v 4 2  103

4  103  9.8  14212 s  3.95 hrs 2

Illustration - 6 The mean distance of Mars from the sun is 1.524 times that of earth from the sun. Find the number of years required for Mars to make one revolution about the sun. SOLUTION : 3/ 2

For planets revolving around the sun T2  r3 

T12 T22





r13

r  T1   1 T2  r2 

3/ 2

 r1   T1 = T2    r2  = (1 yr) (1.524)3/2 = 1.88 yrs.

r23

(T1 : time period of mass) (T2 : time period earth)

Illustration - 7 Two particles of mass m and M are initially at rest at an infinite distance apart. They move towards each other and gain speeds due to gravitational attraction. Find their speeds when the separation between the masses becomes equal to d. SOLUTION : Let v1 and v2 be the speeds of two masses m and M respectively when they are at a separation d. As they approach each other, the kinetic energy increases and G.P.E. decreases. Hence for the system : Loss in G.P.E. = Gain in K.E.

GM m 1 1  mv12  Mv22 d 2 2 As there is no external force on this system, its total momentum remains conserved. Pi = Pf



0 = mv1 – Mv2 Combining the two equations, we have

(G.P.E.)i – (G.P.E.)f = K.Ef – K.E.i  GM m 1 2 1 2 0     mv1  Mv2   0 d  2 2  


v1 

2G M 2 and v2  d m  M 

2 G m2 d m  M 

Section 5

9

Vidyamandir Classes

Gravitation

Illustration - 8 What should be the time period of rotation of earth so that the bodies at the equator would weigh 40 % of their actual weight ? SOLUTION : Gravity at the equator is given by :  2 R  g   g 1     g    Apparent weight at the equator





2  2 

 2 R   W   mg 1     g   

 2

 40 2 R  mg  mg  1    100 g  

0.6 g R



R 0.6 g

6400  103  6489.245 sec 0.6  10

Illustration - 9 A space-ship into a circular orbit close to the earth’s surface. What additional velocity must be imparted to the ship so that it is able to escape the gravitational pull of the earth ? (R = 6400 km, g = 9.8 m/s2) SOLUTION : The orbital velocity in a circular orbit close to

ve  v 

the earth is v 

gR The velocity required to escape





2 1

gR

ve  v  0.414  9.8  6400  103

 ve  2 g R

 3278.71 m / s  3.278 km / s

Hence additional velocity required is :

Illustration - 10 Calculate the time period of revolution and orbital speed of a satellite describing an equatorial orbit at 1400 km above the earth’s surface. If the satellite is travelling in the same direction as the rotation of the earth, calculate the interval between two successive appearances of the satellite for an observer on a fixed point on the earth just below the satellite. SOLUTION : Radius of the orbit = r = R + h

 6

= 6400 + 1400 = 7800 km = 7.8 × 10 m

3

   6830 sec 2 9.8   6.4  106  4 2 7.8  106

Time period of the satellite (Ts) is given by : 2

Ts 

2

4 4 r3 r3  2 GM g R

Orbital velocity of satellite  v 

 v  10

Section 5



9.8  6.4  106 7.8  106



GM r

2

 7.17 km / s


Vidyamandir Classes

Gravitation 2 2 2   TSE TS TE

If the observer and satellite are rotating in the same sense, the relative time period of satellite is given by :

SE  S  E



 TSE 

TS TE TE  TS

6830   3600  24   7416.26 sec 3600  24  6830

TSE 

Illustration - 11 Two satellites S and S revolve around a planet in coplanar circular orbit in the same 1 2 sense. Their period of revolutions are 1 hr and 8 hrs respectively. The radius of the orbit of S1 is 104 km. When S2 is closest to S1, find : (i) the instantaneous speed of S2 relative to S1, (ii) the instantaneous angular speed of S2 actually observed by an astronaut in S1. SOLUTION : Let r1, r2 be the radii of orbits and T1, T2 be the periods for the satellites. 2

T 2  r3

 

r23



r13

T22 T12

3

 

 10

4

8   1

2



Relative speed = | V2 – V1 | = | 3.14  104 – 6.28  104 |

2  r2 Speed of S 2  v2  T2



1

vs2 s1  vs2  vs1

(i)

r2  4  10 km

 v2 

   6.28  104 km / hr

v1 

4

2  4  104

2  r1 T1

2  104

3

 T2   r2       T1   r1 



Speed of S1  v1 



= 3.14  104 km/hr

21 

(ii)

8 4



= 3.14  10 km.hr

v2  v1 r2  r1

3.14  104 3  104



1 rad / s 3600

Illustration - 12 A satellite of mass 1000 kg is rotating around the earth in a circular orbit of radius 3R. What extra energy should be given to this satellite if it is to be lifted into an orbit of radius 4R. SOLUTION : Energy required = (TE)f – (TE)i  GMm   GMm  GMm GMm            8R 6R  2  4 R    2  3R  



GM m GM m  4  3  24 R 24 R



9.8  6400  103 





2

 1000

24  6400  103



 2.614  199 J

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Section 5

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NOW ATTEMPT OBJECTIVE WORKSHEET BEFORE PROCEEDING AHEAD IN THIS EBOOK Vidyamandir Classes

Gravitation

THINGS TO REMEMBER

1.

Newton’s Universal Law of Gravitation Let m1 and m2 be the masses of two bodies and r be the separation between them. m m F 1 2 r2 G m1 m2 F r2 Where G = 6.67 × 10–11 Nm2/kg2

2.

Acceleration due to gravity (of Earth) (a)

On the surface of earth, g =

GM R2

Variation in acceleration due to gravity : (b)

Above the Surface of Earth : 2

 R  g  g   where g = acceleration due to gravity at the surface.  R  h (c)

Below the Surface of Earth : h  g   g 1   R 

Where g = acceleration due to gravity at the surface. (d)

At a Latitude  2

 2 R   2R  g  g 1    cos 2   2   cos   g   g      Where g is the acceleration due to gravity at the surface of the earth neglecting the effect of the rotation. In Particular 1. At Poles :

 = 90°  g = g Hence the rotation of the earth has no effect on the gravity at poles. 12

Things to Remember

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At Equator :  = 0°  2 R  g   g 1     g  

3.

Escape Velocity : It is the minimum velocity with which a body must be projected from the surface of earth so that it permanently overcomes and escapes the gravitational field of the earth. We can also say that a body projected with escape velocity will be able to go to a point which is at infinite distance from the earth.

Ve 

4.

2G M R

or

Ve 

2 g R = 11.3 km/s

Motion of a Satellite of mass M around earth (M) : (a)

Orbital Velocity :

v

GM r

or v 

GM Rh

For a satellite very close to the earth’s surface,

v (b)

GM  r

GM  R

gR

Time Period :

2 r r T GM



4 2 3 T  r GM 2

Note : For a satellite whose time period is 24 hrs (same as the time period of the earth’s rotation) Radius of orbit r = 3

GMT 2 4 2

(T = 24 hrs.)

This satellite seems stationary if observed from the surface of earth. This is also known as Geostationary Satellite. (c)

Total Energy of the Satellite GMm ) 2r The energy that must be given to the orbiting satellite to make it escape to infinity is known as binding energy.

Binding Energy = – (Total Energy –


Things to Remember

13


5.

Vidyamandir Classes

Kepler’s Laws : 1.

Law of Orbits : Each planet revolves around the sun in an elliptical orbit with the sun at one focus of the ellipse.

2.

Law of Areas : This law states that the radius vector from the sun to the planet sweeps out equal areas in equal time intervals. Both shaded areas are equal if the time from A to B is equal to the time from to Q.

14

Things to Remember


Vidyamandir Classes

My Chapter Notes


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Illustration - 1


Vidyamandir Classes


Vidyamandir Classes


Gravitation.pdf

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