Gravitation Physics G9 Cha 5 Bise Rawalpindi

Physics Class 9th Notes gravitation Chapter 5 Board of Intermediate and Secondary Education Rawalpindi Board Pdf Download Short questions, Long questions, and Numerical Problems.

Gravitation Chapter 5 Bise Rawalpindi

i) Earth’s gravitational force of attraction vanishes at

  • A. 6400 km 
  • B. infinity
  • C. 42300 km
  • D. 1000 km

ii) Value of g increases with the

  • A.  increase in mass of the body
  • B. an increase in altitude
  • C. a decrease in altitude
  • D. none of the above

iii) The value of g at a height one Earth’s radius above the surface of the Earth is:

  • A. 2 g  
  • B. ½ g
  • C. 1/3 g
  • D. ¼ g

iv) The value of g on the moon’s surface is 1.6 ms 2. What will be the weight of a 100 kg body on the surface of the moon?

  • A. 100 N    
  • B. 160 N
  • C. 1000 N  
  • D. 1600 N

v) The altitude of geostationary orbits in which communication satellites are launched above the surface of the Earth is:

  • A. 850 km  
  • B. 1000 km
  • C. 6400 km
  • D. 42,300 km

vi) The orbital speed of a low orbit satellite is:

  • A. zero 
  • B. 8 ms-1
  • C. 800 ms-1 
  • D. 8000 ms-1

Questions Answer

Q.2) What is meant by the force of gravitation?

THE FORCE GRAVITATION

The force of attraction between two masses is called the force of gravitation. It is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. Everybody in the universe attracts every other body. Mathematically,

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Where m1m2 are the two masses, d is the distance between their centers and G is the universal constant of gravitation.

Q.3) Do you attract the Earth or does the Earth attract you? Which one is attracting a larger force?  You or the Earth

Answer:

Both attract each other with the same magnitude of the force and we can understand it in two ways. First, we can use Newton’s third law which states “to every action, there is always an equal but opposite reaction”.  So, the forces are equal.

Second, according to Gravitation law,

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As the force is directly proportional to the product of two masses. This means that if we calculate the force that the earth exerts on us, we multiply the mass of the earth and our mass. And if we calculate the force that we exert on the earth, we again multiply the mass of the earth and our mass. In both cases, the forces have the same magnitude. But the mass of the earth is so great compared to a person’s mass that this force is not large enough to cause any effect on the earth.

Q.4) What is a field force?

Answer:

FIELD FORCE

A field force is a non-contact force acting on a body at a distance.

For example, the gravitational force is a field force that attracts the bodies which are not in contact. Magnets and charged particles also have force fields around them.

Q.5) Why earlier scientists could not guess about the gravitational force?

Answer:

Issac Newton was the one who was trying to solve the mystery of why planets revolve around the sun. Thus when an apple fell on the ground while he was sitting under the tree, the idea of gravity flashed in his mind. He related this concept with the concept of how planets revolve around the sun by explaining the cause of falling apples and also the cause that makes the planets revolve around the sun.

Q.6) How can you say that gravitational force is a field force?

Answer:

The gravitational force is a field force because it can attract the bodies from a distance without contact.

For example, the gravitational force between earth and the moon.

Similarly, planets of the solar system are revolving around the sun under the influence of this field force.

Similarly, planets of the solar system are revolving around the sun under the influence of this field force.

Q.7) Explain, what is meant by gravitational field strength?

Answer:

The force of attraction experienced by a unit mass in a gravitational field is a measure of the field strength at that point. The field strength is governed by Inverse Square Law and grows weaker as the distance between two masses increases. Theoretically, the gravitational field of every massive body extends to infinity.

Q.8) Why is the law of gravitation important to us?

Answer:

The gravitational law is very important because of the release of satellites from the earth, the motion of the earth around the sun, the motion of the moon around the earth, and the gravitational attraction of the sun to other planets. All these are explained by gravitational law. The mass of the earth can also be calculated by gravitational law. While designing huge or high-rise buildings, the placement of the center of gravity is critical. Similarly, in the design of massive machinery; oil rigs, high-speed vehicles, etc. the placement of the center of gravity is equally important. The Law of gravitation is the basic tool to work such details out.

Q.9) Explain the law of gravitation.

Answer:

LAW OF GRAVITATION

Everybody in the universe attracts 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 their centers.

Explanation:        

Consider two bodies of masses m1 and m2. The distance between the centers of masses is d.

According to the law of gravitational force, 

According to the law of gravitational force, 

By combining (i) and (ii) we get

By combining (i) and (ii) we get

Here G is the universal constant of gravitation. In SI units the value of G is 6.673 10-11 Nm2 kg-2.

Here G is the universal constant of gravitation. In SI units the value of G is 6.673 10-11 Nm2 kg-2.

Q.10) How can the mass of Earth be determined?

Answer:

MASS OF THE EARTH

Consider a body of mass m on the surface of the Earth. Let the mass of the Earth Me and radius of the Earth R.

According to the law of gravitation, the gravitational force F of the Earth acting on a body is given by

But the force with which Earth attracts a body towards its centre is equal to its weight W.Therefore,

But the force with which Earth attracts a body towards its center is equal to its weight W.

Therefore,

By putting the values the mass Me can be determined

By putting the values the mass Me can be determined

Thus, mass of the Earth is 6.0 x 1024 kg. 

Thus, the mass of the Earth is 6.0 x 1024 kg. 

Q.11) Can you determine the mass of our moon? If yes, then what do you need to know?

Answer:

We can determine the mass of the moon by using the gravitational law.

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If we know the value of acceleration due to gravity on the surface of the moon and the radius of the moon, then we can calculate the mass of the moon.

Q.12) Why does the value of g vary from place to place?

Answer:

As we know that

where R = R'+h i.e. R' = Radius of earth and h is the altitude of an object from the surface of the earth. So, the value of g is inversely proportional to the square of the distance from the center of the earth.  Thus, the value of g varies place to place.

where R = R’+h

i.e. R’ = Radius of earth and the altitude of an object from the surface of the earth. So, the value of g is inversely proportional to the square of the distance from the center of the earth.  Thus, the value of g varies from place to place.

Q.13) Explain how the value of g varies with altitude.

Answer:

As we know that

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According to the above equation, the value of g is inversely proportional to the square of the distance from the center of the earth. As altitude increases or decreases, the distance (R+h) also increases or decreases. Thus the value of g also varies accordingly.

Q.14) What are artificial satellites?

Answer:

Artificial satellites are human-built objects which revolve around a planet. Most of the artificial satellites revolving around the earth are used for communication purposes. Some of them carry instruments to perform experiments in space.

Q.15) How does Newton’s law of gravitation help in understanding the motion of satellites?

Answer:

The motion of satellites is due to the gravitational force between the earth and the satellite. The satellites are in a state of freely falling and have an acceleration equal to gravity due to the earth.

Q.16) On what factors does the orbital speed of a satellite depend?

Answer:

Since orbital speed v0 is

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The formula shows that the orbital speed of a satellite depends on the value of g, radius of the earth, and altitude. The satellites are in a state of freely falling and have an acceleration equal to gravity due to the earth

Q.17) Why are communication satellites stationed in geostationary orbits?

Answer:

Communication satellites take 24 hours to complete their one revolution around the earth and the earth also completes its one rotation about its axis in 24 hours. Due to this reason, these communication satellites appear to be stationary on earth. The communication satellites need to communicate with fixed earth stations round the clock. For this purpose, the satellite has to be in direct line of sight of the earth stations. Only a geostationary satellite can meet this requirement. Otherwise more satellites would be needed to take turns.    

Q1. Who gave the idea of gravity?

Ans: The first man who came up with the idea of gravity was Isaac Newton. It was an evening in 1665 when he was trying to solve the mystery of why planets revolve around the Sun. Suddenly an apple fell from the tree under which he was sitting. The idea of gravity flashed in his mind. He discovered not only the cause of the falling apple but also the cause that makes the planets revolve around the Sun and the moon around the Earth. This unit deals with the concepts related to gravitation.

Q2. What is meant by the force of gravitation?

Ans:   The force of gravitation:

There exists a force due to which everybody in the universe attracts every other body. This force is called the force of gravitation.

Q3. Explain Newton’s law of gravitation?

Ans: See Q # 5.9 from Exercise.

Q4. Explain that the gravitational forces are consistent with Newton’s third law of motion?

Ans:   Law of gravitation and Newton’s third law of motion:

It is to be noted that mass m1 attracts m2 towards it with a force of F while mass m2 attracts m1 towards it with a force of the same magnitude F but in the opposite direction. If the force acting on m1 is considered as action then the force acting on m2 will be the reaction. The action and reaction due to the force of gravitation are equal in magnitude but opposite in direction. This is consistent with Newton’s third law of motion which states that to every action there is always an equal but opposite reaction.

Q5. Explain gravitational field as an example of a field of force?

Ans:   Gravitational field:

The field in a region of space in which a particle would experience a gravitational force is called a gravitational field.

It is assumed that a gravitational field exists all around the Earth due to the gravitational force of attraction of the

Earth.

The weight of a body is due to the gravitational force with which the Earth attracts a body. Gravitational force is a non-contact force.

For example, the velocity of a body, thrown up, goes on decreasing while on return its velocity goes on increasing. This is due to the gravitational pull of the Earth acting on the body whether the body is in contact with the Earth or not. Such a force is called the field force. It is assumed that a gravitational field exists all around the Earth. This field I directed towards the center of the Earth.

Q6. Explain, what is meant by gravitational field strength?

Ans:   Gravitational field strength:

In the gravitational field of the Earth, the gravitational force per unit mass is called the gravitational field strength of the Earth. It is 10 N kg-1 near the surface of the Earth.

The gravitational field becomes weaker and weaker as we get farther and farther away from the Earth. At any place, its value is equal to the value of g at that point.

Q7. How can the mass of the Earth be determined?

Ans: See Q # 5.10 from Exercise.

Q8. Why does the value of g vary from place to place? Explain how the value of g varies with altitude.

Ans: See Q # 5.13 from Exercise.

Q9. Explain the variation of ‘g’ with altitude.

OR

       What is the effect of the following on the gravitational acceleration?

  1. Mass of a freely falling body.
  2. Distance of freely falling body from the center of the Earth.
  3. Is there any difference between the values of g at the equator and the poles? Explain.

Ans:   (a)  Since g = GMe / R2        …….. (i)

Equation (i) shows that the value of g does not depend upon the mass of the body. This means that light and heavy bodies should fall toward the center of the Earth with the same acceleration.

(b)  The value of g varies inversely as the square of the distance i.e. g  1 / R2   if the distance from the center of the earth is increased then the value of g will decrease. That is why the value of g at hills (Murree) is less than its value of the seashore (Karachi).

(c)   Earth is not a perfect sphere. It is flattened at the poles, for this reason, the value of g at the pole is more than at the equator. Because the polar radius is less than equal to the oral radius. (g  1 / R2 )


Problems 

5.1 Find the gravitational force of attraction between two spheres each of mass 1000 kg. The distance between the centers of the spheres is 0.5 m.

                                                                                   (2.67 × 10-4 N)

Solution:    Mass = m1 = m2 = 1000 kg

Distance between the centers = d = 0.5 m

Gravitational constant = G = 6.673 × 10-11 Nm2kg-2

                                Gravitational force = F = ?

F = Gm1m2/d2

                                            F = 6.673 * (10)-11 * 1000*1000/(0.5)2

                                               = 6.673 * (10)-11 * (10)6 /0.25

= 6.673 * (10)-5 /0.25

= 26.692 * (10)-5

= 2.67 × 10-4 N

5.2 The gravitational force between two identical lead spheres kept at 1 m apart is 0.006673 N. Find their masses.    (10, 000 kg each)

Solution:    Gravitational force = F = 0.006673 N

Gravitational constant = G = 6.673 × 10-11 Nm2kg-2

                                Distance between the masses = d = 1m

Mass = m1 = m2 =?

F = Gm1m2/d2

                                F = m2/d2                     (Let m1 = m2 = m)

m2 = F*d2/G

m2 = 0.006673 * (1)-3 / 6.673*(10)-11

Taking square root on both side

                                   = 108

m = 104 = 10000 kg each

Therefore, the mass of each lead sphere is 10000 kg.

5.3 Find the acceleration due to gravity on the surface of Mars. The mass of Mars is 6.42 × 1023 kg and its radius is 3370 km.

Solution:    Mass of Mars = Mm = 6.42 × 1023 kg

Radius of Mars = Rm = 3370 km = 3370 × 1000 m = 3.37 × 108 m

Acceleration due to gravity of the surface of Mars = gm =?

gm = G Mm/R2m

                or          gm = 6.673 × 10-11 × 6.42*1023 / (3.37*106)2

                                                 = 6.673*10-11 * 6.42*1023 / 11.357

= 42.84/11.357 = 3.77 m-2

5.4 The acceleration due to gravity on the surface of the moon is 1.62 ms-2. The radius of the Moon is 1740 km. Find the mass of the moon.

Solution:    Acceleration due to gravity = gm = 1.62 ms-2

Radius of the moon = Rm = 1740 km = 1740 × 1000 m = 1.74 × 106 m

Mass of moon = Mm = ?

gm = GMm/R2m

                or           Mm =gm *R2m/ G  

Mm = 1.62*(1.74*106)2/6.673*10-11

= 4.86*1012*1011/6.673

Mm = 7.35 × 1022 kg

5.5 Calculate the value of g at a height of 3600 km above the surface of the Earth. (4.0 ms-2)

Solution:    Height = h = 3600 km = 3600 × 1000 m = 3.6 × 106 m

Mass of Earth = Me = 6.0 × 1024 kg

Gravitational acceleration = gh =?

gh = GMe/(R+h)2

gh = 6.673 × 10-11 × 6.0*1024/(6.4*106+3.6*106)2

                               = 6.673 × 10-11 × 6.0*1024/(10.0*106)2

= 6.673 × 10-11 × 6.0*1024/(100*1012)

                                     = 6.673 × 10-11 × 6.0 × 1010 = 40 × 10-1 = 4.0 ms-2

5.6 Find the value of g due to the Earth at a geostationary satellite. The radius of the geostationary orbit is 48700 km.         (0.17 ms-2)

Solution:    Radius = R = 48700 × 1000 m = 4.87 × 107 m

Gravitational acceleration = g =?

g = GMe/R

                                g = 6.673 × 10-11 × 6.0*1024/(4.87*107)2

= 6.673 × 10-11 × 6.0*1024/(23.17*1014)2

= 4.0038/23.717

= 0.17 ms-2

5.7 The value of g is 4.0 ms-2 at a distance of 10000 km from the center of the Earth. Find the mass of the Earth.   (5.99 × 1024 kg)

Solution:      Gravitational acceleration = g = 4.0 ms-2

                                Radius of Earth = Re = 10000 km = 10000 × 1000 m = 107 m

Mass of Earth = Me =?

Me = gR2/G

Me = 4.0*(107)2 / 6.673*10-11

= 0.599 × 1025 kg = 5.99 × 1024 kg

5.8 At what altitude the value of g would become one-fourth that on the surface of the Earth?    (One Earth’s radius)

Solution:    Mass of Earth = Me = 6.0 × 1024 kg

Radius of Earth = Re = 6.4 × 106 m

Gravitational acceleration = gh =  g = × 10 ms-2 = 2.5 ms-2

                                Altitude above Earth’s surface = h =?

gh = GMe/(R+h)2

                or    (R + h)2 = GMe/gh

                Taking square root on both sides

or √(R+h)2 = √GMe/gh

                or    R + h = √G GMe/gh

                or    h = √GMe/gh – R

or    h = √6.673*10-11*6.0*1024/2.5 – 6.4*106  – 6.4 × 106

= √40.038*1013*/2.5 – 6.4*106

                          = √16*1013m2 – 6.4*106

                              = -6.0 × 106 m

As height is always taken as positive, therefore

h = 6.0 × 106 m = One Earth’s radius

5.9 A polar satellite is launched at 850 km above Earth. Find its orbital speed. (7431 ms-1)

Solution:    Height = h = 850 km = 850 × 1000 m = 0.85 × 106 m

Orbital velocity = vo =?

vo = √GMe/R+h

vo = √6.673*10-11 * 6.0*1024/6.4*106 + 0.85*106 =

                           = √40.038*1013 /7.25*106

= √5.55*107

                         = 7.431 × 103 = 7431 ms-1

5.10 A communication is launched at 42000 km above Earth. Find its orbital speed.  (2876 ms-1)

Solution:    Height = h = 42000 km = 42000 × 1000 m = 42 × 106 m

Orbital velocity = vo =?

vo = √GMe/R+h

vo = √6.673*10-11 * 6.0*1024/ 6.4*106+42*106 

= √40.038*1013/48.4*106

                          = √400.38*1012/48.4*106

                          = √8.27*106

                          = 2.876 × 103 = 2876 ms-1