Table of Contents

## Physics Notes for Class 9 Chapter 5 Gravitation

Physics Notes for Class 9 Chapter 5 Gravitation.

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**Chapter: 05**

**Gravitation**

**COMPREHENSIVE QUESTIONS:**

### 1. **State and explain the law of universal Gravitation. Also show that the law obeys Newtonβs third law of motion.**

**Ans. Law of universal Gravitation: **

**Statement:**

Everybody in the universe attracts every other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

**Derivation:**

Consider two spherical bodies of masses βm1β and βm2β separated by distance βrβ as shown in fig.

According to the Newtonβs law of universal gravitation, the force of gravity βFgβ between them is:

Fg Ξ± m1 m2

———(i)

And Fg Ξ±

12 ———-(ii)

Ξ±

1 2

Combining eq(i) and (ii), we get

= 2

Where βGβ is the constant of proportionality and is known as universal gravitational constant. The value of βGβ is 6.67 Γ 10-11 Nm2 kg-2. It does not depend on the medium between the two bodies.

**Newtonβs third law of motion and universal gravitation:**

The law of universal gravitation also obeys the Newtonβs third law of motion. We can see that the force acting on mass βm2β due to mass m1 is F12. Also the force acting on mass m1 due to mass βm2β is same force βF21β, Both these forces are equal but opposite in direction. Therefore, we can say that the forces acting on two bodies due to gravitation force is the example of action and reaction i-e

F12 = – F21

### 2. **Determine the mass of earth by applying the law of gravitation.**

**Ans. Determination of Mass of Earth:**

The mass of the earth can be determined with the help of law of universal gravitation. Let an object of mass βm0β be placed on the surface of the earth. The distance between the centre of the body and the earth is equal to the radius of earth βrEβ. If the mass of earth is βmEβ then the force βFgβ with which the earth attracts the body towards its centre is given by law of gravitation.

Fg = G

Also, we know that force of gravity (Fg) is equal to the weight of the body (W). i-e

Fg = W

So,

W = G

We know that W = mog2

mog = G

0

g = G

2

By cross Multiplication, we get

2 G= g

2

Divide βGβ on both sides,

=

g 2

= —— (i)

By putting the values of g, and G in eq (i) we will obtain value of.

-11

Nm kg

-2

Gravitational constant = G= 6.67 Γ 10

Acceleration due to gravity = g = 9.8 ms

And, radius of earth rE = 6.4 Γ 106m

Now,

9.8 Γ (6.4 Γ 106)2

=

6.67 Γ 10

9.8 Γ 40.96 Γβ11

0

=12

6.67 Γ 10

401.40 Γ 10β11

= 12

6.67 Γ 10

401.40 Γ 10

β11Γ10

= 12

11

6.67

= 60.18 Γ 10

12+11

= 60.18 Γ 10 23

= 6.018 Γ 10

= Γ 23+1

Thus, the mass of earth is approximately 6 Γ 1024kg.

### 3. **What is gravitational field and gravitational field strength Show that weight of an object changes with location.**

**Ans. Gravitational Field (Gravity as a field force):**

Gravitational field is region surrounding the earth in which another object feels force of attraction toward its centre According to the field theory; every mass creates a gravitational field composed of field lines that permeates outward into space.

The earth creates a gravitational field that pulls objects towards its centre by force of gravity. At any point, earthβs gravitational Field can be described by the gravitational field strength (g).

The gravitational field or gravitational field strength is defined as a measure of gravitational force (Fg) exerted on a per unit mass (m) of an object.

**Mathematical Form:**

Gravitational Field =

Or g =

**Quantity and Unit:**

The gravitational field (g) is a vector quantity and its nit is Newton per kilogram i-e Nkg-1

**Gravitational Field strength:**

Β The gravitational field strength tells us how strong a gravitational field is. The gravitational field is represented by field lines as shown in figure, which shows the strength of gravitational field decreases, as the distance from the ear h increase. The gravitational field strength of earth near its surface is equal to the acceleration of free fall (g) at its surface i-e g = 9.8 ms-2 According to Newtonβs second Fg law, a = F/m.

As gravitational field strength (g) is defined as m, so the value of βgβat any given point is equal to the acceleration due to gravity. For this reason, the gravitational field strength is 9.8 Nkg-1 means 9.8N force is exerted on every 1 kg mass of an object. Therefore, the acceleration is same for any object, regardless of mass.

**Weight changes with location:**

As we know that weight is the magnitude of force due to gravity i-e W = mog. Now we can refine our definition of weight as mass time gravitational field strength this new definition helps to explain why our weight changes with our location in the universe by calculating the value of g

i-e.

=

4

This equation shows that gravitational field strength depends only on mass of earth βmEβ and radius of earth βrEβ. Greater will be the value of βgβ if lesser the distance from earthβs centre. Therefore, on the surface of any planet, the value of earthβs centre varies as we change location. βgβ as well as our weight will depends on the planetβs mass and radius.

### 4. **How is the Value of βgβ changing by going to higher altitude? Write the relevant formula.**

**Ans. Variation of βgβ with altitude:**

The value of βgβ doesnβt depend upon the mass of the body. It means that light and heavy bodies should fall towards the centre of earth with constant acceleration. However, the value of βgβ depends upon the distance of the body from the centre of the earth. Greater the distance from the centre of the earth, smaller will be the value of βgβ. That is why the value of βgβ at the poles is greater than at equator because earth is not a perfect square; its equatorial radius is greater than the radius at the poles.

**Derivation:**

=

Consider a body of mass βmoβ placed on earthβs surface, as shown in the figure, now we know that law of universal gravitation is given by.

As, 2

Fg = W = 0g

0

So,

g =

β¦ β¦ β¦ β¦ (i)

2

g =

If we lift the body to a height βhβ, then the value of βgβ at the height is given by

= ( )

From equ(i), GmE = gr2E +β 2

2

β

=

β¦ β¦ β¦ β¦ eq (ii) ( + )

Both eq (i) and (ii) shows that βgβ is inversely proportional to the square of distance of the body from the earth centre. Thus, the value of βgβ decreases with altitude (height).

### 5. **Derive the formula for the orbital speed of an artificial satellite.**

**Ans: Satellites:**

Satellites are the objects revolving around the planet in fixed orbits. Artificial satellites are man-made objects that revolve around the earth or other planets in different orbit with uniform speed due to gravitational force.

A satellite requires a centripetal force (Fc) to revolve around the earth and this necessary centripetal force is provided by the gravitational force of attraction between the earth and satellite.

**Derivation: **

Consider a satellite of mass βmsβ revolving in a circular orbit with uniform velocity βvβ from earth of mass βmEβ Let βrβ be the distance between the centre of earth and centre of satellite as shown in figure.

Now, the gravitational force by Newtonβs law of universal gravitation is:

Fg= β¦β¦β¦.eq (i)

The necessary centripetal force βFcβ required for uniform circular motion is

**Given by:**

2

Fc=

β¦β¦β¦.eq (ii)

As Centripetal force is provided by gravitational force, therefore

Fc = Fg β¦β¦β¦.eq (iii)

Now, putting the value of Fg and Fc in eq (iii),

We get 2

=

2

v =

Taking square root on both sides

=

=

Where r = rE + h, so

=

This equation represents the orbital speed of satellite where h is the height of satellite from surface of earth and β βis the radius of earth. So, the orbital of satellite speed depends upon the mass of earth and the distance from the centre of the earth to the centre of mass of the satellite and doesnβt depend upon mass of satellite.

**6. Derive a formula to calculate the value of g.**

**Value of g:**

The Newtonβs law of Universal gravitation shows that value of g depends on mass of all reacting bodies and distance to it. So the value of βgβ can be determined by using law of gravitation. Consider an object of mass βmoβ placed on the surface of earth and rE is the distance between their centers as shown in figure The gravitational force between the object and earth is as follow. 2

Fg = G ———- (i)

As, we know that

Fg = W = mog ———– (ii)

Comparing eq (i) and (ii)

mog = G

2

g = G

By putting the values of G, mE & rE, we will obtain the value of g

As,

mE = 6 Γ 1024 kg

rE= 6.4 Γ 106 m

g =

6.67 Γ 10β11 Γ 6 Γ 1024

G = 6.67 Γ 10-11 Nm2

kg-2

g = 40.02 Γ 10β11+24

(6.4 Γ 106)2

g

= 40.02 Γ 1013

40.96 Γ 1012

g = 40.02

Γ 1013 Γ 10β12

40.96 Γ 1012

40.96

g = 0.977 Γ 101

g = 9.77 Γ 101-1

g = 9.77 Γ 100 (..100=1)

g = 9.77

or

g = 9.8ms-2

This is value of βgβ at the surface of earth.

**CONCEPTUAL QUESTIONS:**

**Q1. If there is an attractive force between all objects, why donβt we feel ourselves gravitating toward nearby massive buildings?**

Ans. Gravitational force pulls us to massive buildings but the forces are relatively small due to the small masses of us and buildings when compared to the mass of the earth. Therefore, the attractive force would be almost unnoticeable.

**Q2. Does the sun exert a larger force on the Earth than that exerted on the sun by the earth? Explain.**

Ans. By Newtonβs third law, the force exerted on the Sun by the Earth is exactly equal to the force the Sun exerts on the Earth but in opposite direction.

**Q3. What is the importance of gravitational constant βGβ? Why is it difficult to calculate?**

Ans. Constant always play an important in bringing the equality in the dimensions on both sides of an equation and its value depends upon the factors relating the interaction between the two bodies similar is the case with gravitational constant βGβ here it maintains the equality on both sides of the universal gravitational lawβs equation. Itβs difficult to calculate it because thereβs no theoretical derivation to it is just an experimentally measured value.

**Q4. If Earth somehow expanded to a larger radius, with no charge in mass, how would your weight is affected? How would it be affected if earth instead shrunk?**

Ans: According to law of universal Gravitation

F= ———— (i)

As we know that

F=W=mg

mg =

So eq (i) becomes 2

g = ———— (ii)

From eq (ii), we say that force of gravity is inversely proportional to the radius of earth. So if radius of Earth gets larger weight would get smaller. If the earth shrunk, the radius of earth decreases and as a result weight gets increases.

**Q5. What would happen to your weight on earth if the mass of the earth doubled but its radius stayed the same?**

Ans: As we know that

g= ———— (i)

Above eq (i) shows that value of βgβ depends on mass of earth ββ and radius of earth β β.

If we only double the mass of earth the value of βgβ becomes double and weight depends on the value of βgβ. i.e.

W= mg ———— (ii)

Eq (ii) shows that if value of βgβ is doubled, the weight will also double.

**Q6. Why lighter and heavier objects fall at the same rate toward the earth?**

Ans: Because value of βgβ does not depend upon the mass of the body, it depends only on mass of earth and radius of earth. Therefore, lighter and heavier bodies fall towards earth with same acceleration.

**Q7. The value of βgβ changes with location on earth, however we take the same value of βgβ as 9.8ms-2 for ordinary calculations why?**

Ans: The value of βgβ depends upon the distance from the centre of earth. Greater the distance from the centre of earth, smaller will be the value of βgβ and vice versa. The change in the value of βgβ is significant only at very large distance. Therefore, we take same value of βgβ for ordinary calculation.

**Q8. Moon is attracted by the earth, why it does not fall on earth?**

Ans: The moon is natural satellite of the earth. It revolves around the earth in a specific orbit. The earth attracts the moon towards itself with gravitational force. The gravitational force of earth provides necessarily centripetal force which compels the moon to move in the circular path. Because of this orbital motion moon does not fall on earth.

**Q9. Why for some height larger and smaller satellites must have same orbital speeds?**

The orbital speed depends upon the mass of ear h and distance from the centre of earth to the centre of mass of satellite and does not depend upon the mass of satellite. Therefore, for some particular distance from the centre of earth, all the satellites have the same orbital speed irrespective of the size of satellite.

**Assignments**

**5.1 The mass of earth is 6 Γ 1kg and that of the moon is 7.4 Γ kg. If the distance between the earth and the moon is 3.841 km, calculate the force exerted by the earth on the moon, Γ**

**Data: **

= 6Γ1024 kg

Mass of earth =

Mass of moon = 1

= 7.4 Γ1022 kg

Distance = r= 3.84Γ10 5

2

= 3.84 Γ10 Γ 10 m

3

5

= 3.84 Γ 108cm 2 β2

Gravitational constant = G = 6.67 Γ 10β11N

Gravitational force = Fg = ?

**Solution:**

By using formula

Fg = G

24Γ . Γ 22

Putting Values

= . Γ β11Γ Γ

6 67 10 ( . 6 Γ10 8)2 7 4 10. Γ

Γ10

=Γ .3 84 Γ β11+24+22

6 67 6 .7 4 10 16

14 7456. Γ

10

= 296. 148

Γ1035

14 7456

16

10

= 20.083 Γ 1035β16

= 20.083 Γ 1019

N

Fg= 2.008 Γ 1020N

Fg = 2 Γ 1Nm and have acceleration due to gravity on its

**5.2 If the radius of the moon is 1.74 Γ1 surface as 1.6β . Calculate the mass of moon.**

**Data:**

Radius of Moon = r = 1.74 Γ 106 m β2

Acceleration due to gravity on moon = g = 1.6

Gravitational constant = G = 6.67 Γ 10

β11

N

Find: Mass of moon = m = ?

2 β2

Solution:

By using formula m =

Putting Values

=. Γ( . Γ 6)2 . Γ 10

1 74

β11

Γ . Γ 12

=.10

. Γ 10

3 027

. Γ 10β11

= 12

Γ

4. 843 10

6 67 β11

10

= 0.726 Γ1012+11

23

= 0.73 Γ 10 kg

m = 7.3 Γ 1 kg

10

**5.3 An astronaut of mass 65.0 kg (weighting 637N on earth) is walking on the surface of the moon, which has a mean radius of 1.74 Γ 1 m and a mass of 7.35 Γ1 kg. What is the weight of the astronaut on moon? What is the free β fall acceleration at the surface of the moon?**

Data: mass of astronaut = m = 65kg

Radius of moon = rM = 1.74Γ106m Γ 0β11 2 β2

**Find:**

= 7.35 Γ 1022 kg

Mass of moon =

Gravitational constant = G = 6.67 1 Nk

Weight of astronaut = W= ?

Free β fall acceleration on moon = gM= ?

**Solution:**

g =

By Using Formula

6 67 7 35

10

10 22

= . Γ(.β11ΓΓ. )Γ

Values Putting 1.74 Γ 10

. Γ

= 6 67

- 35
- 6 2 Γ 10β11+22

. 3 027

10

= 49. 02 1011

12

Γ

3 027

10 12

Γ

= 16.19 Γ 1011β12

= 16.2 Γ 10β1 m/ 2

= 16.19 Γ10β1 m/ 2

g = 1.62 m/ Now we find weight of Astronaut

As we know that

W = mg

Putting values

= 65 Γ 1.62 N

= 105.3 N

W = 105 N