Gravitation

Gravitation

Class: IX

Gravitation:

• Gravitation

• Universal Law of Gravitation

• Free Fall

• Acceleration Due to Gravity (g)

• Difference between G and g

• Mass and Weight

• Weight of an Object on the Surface of Moon

Gravitation:

Earth attracts everything towards it by an unseen force of attraction. This force of attraction is known as gravitation or gravitational pull.

What is the Centripetal Force? 

.    We know that an object in circular motion keeps on changing its direction.

·     Due to this, the velocity of the object also changes.

·     A force called Centripetal Force acts upon the object that keeps it moving in a circular path.

·     The centripetal force is exerted from the centre of the path.

·     Without the Centripetal Force objects cannot move in circular paths, they would always travel straight.

·     For Example, The rotation of Moon around the Earth is possible because of the centripetal force exerted by Earth.

Newton's Observations:

·     Why does Apple fall on Earth from a tree? – Because the earth attracts it towards itself.

·     Can Apple attract the earth? - Yes. It also attracts the earth as per Newton's third law (every action has an equal and opposite reaction). But the mass of the earth is much larger than Apple's mass thus the force applied by Apple appears negligible and Earth never moves towards it.

·     Newton thus suggested that all objects in this universe attract each other. This force of attraction is called Gravitational Force.

Universal Law of Gravitation:

Every object in the universe attracts other object by a force of attraction, called gravitation, which is directly proportional to the product of masses of the objects and inversely proportional to the square of distance between them. This is called Law of Gravitation or Universal Law of Gravitation.

Let masses (M) and (m) of two objects are distance (d) apart. Let F  be the attraction force between two masses.

Why we study the universal law of gravitation?

It explains many important phenomena of the universe –

=> Earth’s gravitational force

=> Why the moon always moves in a circular motion around the earth and the sun

=> Why all planets revolve around the sun

=> How the sun and moon can cause tides

Kepler’s first law (law of elliptical orbit):

 A planet moves round the sun in an elliptical orbit with sun situated at one of its foci.

Kepler’s second law (law of areal velocities):

 A planet moves round the sun in such a way that its areal velocity is constant.

Kepler’s third law (law of time period):

 A planet moves round the sun in such a way that the square of its period is proportional to the cube of semi major axis of its elliptical orbit.

T² ∝ R³

Here R is the radius of orbit.

T² = (4π²/GM)R ³

Kepler’s third law (law of time period):

A planet moves round the sun in such a way that the square of its period is proportional to the cube of semi major axis of its elliptical orbit.

T² ∝ R³

Here R is the radius of orbit.

T² = (4π²/GM)R ³

Q1. Two spheres of masses 100 kg and 900 kg each of radius 10 m and 20 m respectively are in touch. Find the gravitational force acting between them.

Solution:             F= GMm /d²

F= (6.673 x 10^-11)(100)(900)/(30)²

F= (6.673 x 10^-11)(100)(900)/(900)

F= (6.673 x 10^-11)(100)

F= 6.673 x 10^-9 N

Q2. The gravitational force acting between point objects M and m separated by a distance d is F. What will be the force of gravitation between them if each mass is quadrupled and distance between them is halved?

Solution:             F= GMm /d²

                               F’= G(4M)(4m) / (d/2)²

                               F’= 64 GMm /d²

                               F’= 64 F

Q3. What will be the gravitational force between two bodies if the distance between them is doubled?

Solution:             F= GMm /d²

F’= GMm /(2d)²

F’= GMm /4d²

F’= ¼ [GMm /d²]

F’=F/4

Q4. What will be the gravitational force between two bodies if the distance between them is halved?

Solution:             F= GMm /d²

F’= GMm /(d/2)²

F’= 4 [GMm /d²]

F’= 4F

Q5. Find the force with which two friends having same mass of 50 kg and sitting on benches separated by 3 meters, attracts each other.

Solution:              F= GMm /d²

F= (6.673 x 10^-11)(50)(50)/(3)²

F= (6.673 x 10^-11)(2500)/(9)

F= (6.673 x 10^-9)(2.778)

F= 1.8536 x 10^-8 N

Q6. Calculate the force of gravity between two bodies weighing 100 kg and 50 kg kept at a distance 5 m apart.

Solution:              F= GMm /d²

F= (6.673 x 10^-11)(100)(50)/(5)²

F= (6.673 x 10^-11)(100)(50)/(25)

F= (6.673 x 10^-11)(200)

F= 1.3346x 10^-8 N

Q7. If earth describes an orbit around the sun of double its present radius, the year on earth will be of ______________.

Solution:             T² α R³

                                (T₂ / T₁)² = (R₂/R₁)³

                               (T₂ / 1Y)² = (2R/R)³

                                               (T₂)² = (2)³

                                      (T₂)² = 8

                                          T₂ = 2√2 Y

                                          T₂ = 2(1.414) Y

    T₂ = 2.828 years.

Q8. Calculate the force of gravity between two bodies weighing 10 kg and 10 kg kept at a distance 10 m apart.

Solution:              F= GMm /d²

F= (6.673 x 10^-11)(10)(10)/(10)²

F= (6.673 x 10^-11)(100)/(100)

F= (6.673 x 10^-11)(1)

F= 6.673 x 10^-11 N

Q9. The gravitational force acting between point objects M and m separated by a distance d is F. What will be the force of gravitation between them if each mass is halved and distance between them is doubled?

Solution:             F= GMm /d²

                               F’= G(M/2)(m/2) / (2d)²

                               F’=  GMm /16d²

                               F’=  F/16

Q10. The gravitational force acting between point objects M and m separated by a distance d is F. What will be the force of gravitation between them if each mass is doubled and distance between them is halved?

Solution:             F= GMm /d²

                               F’= G(2M)(2m) / (d/2)²

                               F’= 16 GMm /d²

                               F’= 16 F

Free Fall:

When an object falls from any height under the influence of gravitational force only, it is known as free fall.

Acceleration Due to Gravity:

When an object falls towards the earth there is a change in its acceleration due to the gravitational force of the earth. So this acceleration is called acceleration due to gravity.

The acceleration due to gravity is denoted by g.

The unit of g is same as the unit of acceleration, i.e., ms⁻²

Mathematical Expression for g:

From the second law of motion, force is the product of mass and acceleration.    

           F = ma

For free fall, acceleration is replaced by acceleration due to gravity. 

Therefore, force becomes:                      

            F = mg           ….(i)

But from Universal Law of Gravitation,

Value of ‘g' may vary at different parts of the earth :

·     From the equation g = GM/ r² it is clear that the value of ‘g' depends upon the distance of the object from the earth's centre.

·     This is because the shape of the earth is not a perfect sphere. It is rather flattened at poles and bulged out at the equator.

·     Hence, the value of ‘g' is greater at the poles and lesser at the equator. However, for our convenience, we take a constant value of ‘g' throughout.

Factors Affecting the Value of g:

·        As the radius of the earth increases from the poles to the equator, the value of g becomes greater at the poles than at the equator.

·        As we go at large heights, value of g decreases.

To Calculate the Value of g:

Value of universal gravitational constant, G = 6.7 × 10^–11 N m²/ kg²,

Mass of the earth, M = 6 × 10²⁴ kg, and

Radius of the earth, R = 6.4 × 10⁶ m

Putting all these values in equation (iii), we get:

 

Thus, the value of acceleration due to gravity of the earth, g = 9.8 m/s².

Difference between Gravitation Constant (G) and Gravitational Acceleration (g)