##### Document Text Contents

Page 1

CBSE-i

CLASS

XI

UNIT-6

CENTRAL BOARD OF SECONDARY EDUCATION

Shiksha Kendra, 2, Community Centre, Preet Vihar, Delhi-110 092 India

PHYSICS

GRAVITATION

Page 33

CBSE-i

Unit 6 : Gravitation 13

this theorem in all our calculations.

5.7 vEctor form of nEwton’s lAw of GrAvItAtIon

Since force is a vector quantity, it must be

expressed in a vector form. The gravitational

force can also be expressed in a vector form

by attaching a unit vector to the expression for

this force.

By convention, the direction of unit vector is

always taken as directed from the body experiencing the force (body 1) towards

the body exerting the force (body 2). Therefore,

F

= G 1 2

2

ˆm m

r

r

If we are calculating the force on body A due to body B, then r̂ will be the unit

vector drawn from A towards B.

It will be imperative to mention here that this vector notation is consistent with

the basic fact that the gravitational force is always an attractive force. This implies

that the gravitational force is a central force, and hence the direction of this force

has to be along the line joining the center of the two bodies.

5.8 prIncIplE of supErposItIon

Newton‘s law has been stated for two point bodies. How do we calculate the

force on a body if there are more than two bodies interacting with one other?

Figure 12: Diagram showing direction of

the gravitational force acting along the

line joining the center of two bodies.

1

A

2

B

r̂

Page 34

CBSE-i

Unit 6 : Gravitation14

The solution to this situation lies in what is called the principle of superposition.

In a group of objects, the net gravitational force, on any one of the objects, is

the vector sum of the forces due to all the other objects. The principle implies

that we first calculate the gravitational force that acts on an object due to each

of the other objects as if all other objects are absent. After doing this for all

possible pairs, the net force on the object under consideration is calculated by

the vector sum of all the forces acting on it.

→

1F =

→ → → →

+ + + +12 13 14 15 ...F F F F

here

→

1F is the net force on object 1 due to all the other objects 2, 3, 4, 5, …

The Principle of superposition is based on the fact that the gravitational interaction between

two bodies is independent of the presence of other bodies in the neighborhood.

The same concept is applicable to electrostatic force which will be studied in class XII.

IllustratIon. three objects of masses 5 kg, 3 kg and 3 kg are placed at the corners

of an equilateral triangle of side 20 cm. Calculate the net gravitational

force on the object of 5 kg.

solutIon. The magnitude of the force on object A

due to object B is

FAB = G 2

A Bm m

r

It is in the direction of AB

Putting the values we get,

FAB 5 × 10

–9 N

Similarly, the magnitude of the force on

A due to body C is

Fac 5 × 10

–9 N

It is the direction of AC.

As per the principle of superposition, the net force on the body A is the

vector sum of forces

→

ABF and

→

ACF . Applying the laws of vector addition,

A

B C

Page 65

CBSE-i

45Unit 6 : Gravitation

multiPle CHoiCe QueStionS

Q.1 If the acceleration due to gravity on the surface of the earth (of radius R) is g,

the gain in potential energy of a body, of mass m, when raised from the surface

to a height R would be

(a) mg

4

R (b) mg

2

R (c) mgR (d) 2 mgR

(e) 4 mgR

Q.2 Acceleration due to gravity ‘g’ at the center of the earth is

(a) 9.8 m/s2 (b) 4.9 m/s2 (c) infinite (d) zero

Q.3 If the ratio of the radii of the orbits of mars and earth around the Sun is 1.526,

the time period of mars (in earth years) would be (nearly)

(a) 1.89 years (b) 32 years (c) 45 years (d) 48 years

Q.4 If a satellite of mass m is revolving around the earth at a distance r from its

centre, its total energy is

(a) 2GMm

r

− (b) GMm

r

− (c)

2

GMm

r

− (d)

2

GMm

r

Q.5 A satellite is orbiting around the earth (and close to it) with a total energy E. If

the satellite’s kinetic energy is made 2E,

(a) its period of revolution would get doubled

(b) radius of its orbit would get halved

(c) radius of its orbit would get doubled

(d) the satellite would escape away from the earth.

Q.6 Kepler’s second law is based upon the

(a) Newton’s second law

(b) Law of conservation of energy

(c) Law of conservation of momentum

(d) law of conservation of angular momentum

post content

student worksheet 5

Page 66

CBSE-i

46 Unit 6 : Gravitation

Q.7 The average density of a hypothetical planet is twice that of the earth. If the

acceleration due to gravity on the surface of the planet is equal to that of the

earth and if the radius of the earth is R, the radius of the planet must be

(a)

4

R (b)

2

R (c) 2R (d) 4R

Q.8 For a satellite moving in an orbit around the earth, the magnitude of the ratio

of its kinetic energy to its potential energy is

(a) 1

2

(b) 2 (c) 1

2

(d) 2

CBSE-i

CLASS

XI

UNIT-6

CENTRAL BOARD OF SECONDARY EDUCATION

Shiksha Kendra, 2, Community Centre, Preet Vihar, Delhi-110 092 India

PHYSICS

GRAVITATION

Page 33

CBSE-i

Unit 6 : Gravitation 13

this theorem in all our calculations.

5.7 vEctor form of nEwton’s lAw of GrAvItAtIon

Since force is a vector quantity, it must be

expressed in a vector form. The gravitational

force can also be expressed in a vector form

by attaching a unit vector to the expression for

this force.

By convention, the direction of unit vector is

always taken as directed from the body experiencing the force (body 1) towards

the body exerting the force (body 2). Therefore,

F

= G 1 2

2

ˆm m

r

r

If we are calculating the force on body A due to body B, then r̂ will be the unit

vector drawn from A towards B.

It will be imperative to mention here that this vector notation is consistent with

the basic fact that the gravitational force is always an attractive force. This implies

that the gravitational force is a central force, and hence the direction of this force

has to be along the line joining the center of the two bodies.

5.8 prIncIplE of supErposItIon

Newton‘s law has been stated for two point bodies. How do we calculate the

force on a body if there are more than two bodies interacting with one other?

Figure 12: Diagram showing direction of

the gravitational force acting along the

line joining the center of two bodies.

1

A

2

B

r̂

Page 34

CBSE-i

Unit 6 : Gravitation14

The solution to this situation lies in what is called the principle of superposition.

In a group of objects, the net gravitational force, on any one of the objects, is

the vector sum of the forces due to all the other objects. The principle implies

that we first calculate the gravitational force that acts on an object due to each

of the other objects as if all other objects are absent. After doing this for all

possible pairs, the net force on the object under consideration is calculated by

the vector sum of all the forces acting on it.

→

1F =

→ → → →

+ + + +12 13 14 15 ...F F F F

here

→

1F is the net force on object 1 due to all the other objects 2, 3, 4, 5, …

The Principle of superposition is based on the fact that the gravitational interaction between

two bodies is independent of the presence of other bodies in the neighborhood.

The same concept is applicable to electrostatic force which will be studied in class XII.

IllustratIon. three objects of masses 5 kg, 3 kg and 3 kg are placed at the corners

of an equilateral triangle of side 20 cm. Calculate the net gravitational

force on the object of 5 kg.

solutIon. The magnitude of the force on object A

due to object B is

FAB = G 2

A Bm m

r

It is in the direction of AB

Putting the values we get,

FAB 5 × 10

–9 N

Similarly, the magnitude of the force on

A due to body C is

Fac 5 × 10

–9 N

It is the direction of AC.

As per the principle of superposition, the net force on the body A is the

vector sum of forces

→

ABF and

→

ACF . Applying the laws of vector addition,

A

B C

Page 65

CBSE-i

45Unit 6 : Gravitation

multiPle CHoiCe QueStionS

Q.1 If the acceleration due to gravity on the surface of the earth (of radius R) is g,

the gain in potential energy of a body, of mass m, when raised from the surface

to a height R would be

(a) mg

4

R (b) mg

2

R (c) mgR (d) 2 mgR

(e) 4 mgR

Q.2 Acceleration due to gravity ‘g’ at the center of the earth is

(a) 9.8 m/s2 (b) 4.9 m/s2 (c) infinite (d) zero

Q.3 If the ratio of the radii of the orbits of mars and earth around the Sun is 1.526,

the time period of mars (in earth years) would be (nearly)

(a) 1.89 years (b) 32 years (c) 45 years (d) 48 years

Q.4 If a satellite of mass m is revolving around the earth at a distance r from its

centre, its total energy is

(a) 2GMm

r

− (b) GMm

r

− (c)

2

GMm

r

− (d)

2

GMm

r

Q.5 A satellite is orbiting around the earth (and close to it) with a total energy E. If

the satellite’s kinetic energy is made 2E,

(a) its period of revolution would get doubled

(b) radius of its orbit would get halved

(c) radius of its orbit would get doubled

(d) the satellite would escape away from the earth.

Q.6 Kepler’s second law is based upon the

(a) Newton’s second law

(b) Law of conservation of energy

(c) Law of conservation of momentum

(d) law of conservation of angular momentum

post content

student worksheet 5

Page 66

CBSE-i

46 Unit 6 : Gravitation

Q.7 The average density of a hypothetical planet is twice that of the earth. If the

acceleration due to gravity on the surface of the planet is equal to that of the

earth and if the radius of the earth is R, the radius of the planet must be

(a)

4

R (b)

2

R (c) 2R (d) 4R

Q.8 For a satellite moving in an orbit around the earth, the magnitude of the ratio

of its kinetic energy to its potential energy is

(a) 1

2

(b) 2 (c) 1

2

(d) 2