Class 9 Science Chapter 10 Important Questions of Gravitation. All the extra question answers are taken from intext part of NCERT Textbook issued for academic session 2022-2023. After reading the NCERT Book, these important practice questions provide a perfect way to revise the entire chapter for exams or tests.

## Class 9 Science Chapter 10 Important Questions 2022-2023

### Class 9 Science Chapter 10 Important Extra Questions Set – 1

### What will happen to the gravitational force between two bodies if the masses of one body is doubled?

If the mass of one body is doubled, force is also doubled.

### Why is “G” called the universal gravitational constant?

The constant G is universal because it is independent of the nature and sizes of bodies, the space where they are kept and at the time at which the force is considered.

### Who formulated the universal law of gravitation?

Isaac Newton

### How is gravitation different from gravity?

Gravitation is the force of attraction between any two bodies while gravity refers to attraction between any body and the earth.

### What does a small value of G indicate?

A small value of G indicates that the force of gravitational attraction between two ordinary sized objects is a very weak force.

#### Phenomena based on the Universal Law of Gravitational

The universal law of gravitation was able to explain successfully

- the force that binds us to the earth.
- the motion of the moon around the earth.
- the motion of planets around the sun.
- the tides due to the moon and the sun.

### Class 9 Science Chapter 10 Important Extra Questions Set – 2

### At what place on the earth’s surface is the weight of a body maximum?

At the poles.

### A what place on the earth’s surface is the weight of a body minimum?

At the equator.

### If the mass of a body is 9.8 kg on the earth what would its mass on the moon?

It will remain the same on the moon, i.e. 9.8kg.

### Do fluids possess weight?

Yes, fluids have weight.

### If the earth attracts two objects with equal force can we say that their masses must be equal?

No

#### The Standard Kilogram

The standard kilogram is the mass of a block of a platinum alloy kept at the international bureau of weights and measures near Paris in France.

### Class 9 Science Chapter 10 Important Extra Questions Set – 3

### Is weight a force?

Yes.

### What keeps the moon in uniform circular motion around the earth?

Gravitational force between moon and the earth keeps moon in uniform circular motion around the earth.

### When a body is dropped from a height, what is its initial velocity?

Zero.

### When a body is thrown vertically upwards, what is its final velocity?

Zero.

### Is the time taken by a body to rise to the highest point equal to the time taken to fall from the same height?

Yes.

##### When does an object show weightlessness?

Weightlessness is a state when an object does not weigh anything. It occurs only when a body is in a state of free fall under the effect of gravity alone.

### Class 9 Science Chapter 10 Important Extra Questions Set – 4

### Is the acceleration due to gravity acting on a freely falling body directly proportional to the (a) mass of the body? (b) time of fall of the body?

(a) No

(b) No

### Suppose gravity of earth suddenly becomes zero, then which direction will the moon begin to move if no other celestial body affects it?

The moon will begin to move in a straight line in the direction in which it was moving at that instant because the circular motion is due to centripetal force provided by the gravitational force of the earth.

### Why does a body reach the ground quicker at poles than at the equator when dropped from the same height?

The acceleration due to gravity is more at the equator. The time taken for a body’s less if the acceleration due to gravity is more when the initial velocities and the distance travelled are same. So, when dropped from the same height a body reaches the ground quicker at the poles than at the equator.

### What is the source of centripetal force that a planet requires to revolve around the sun? On what factors does that force depend?

Gravitational force. This force depends on the product of the masses of the planet and sun and the distance between them.

### Suppose that the radius of the earth becomes twice of its original radius without any change in its mass. Then what will happen to you weight?

We know that F = GMm/r² as weight of a body is the force with which a body is attracted towards the earth.

W = GMm/r²

If the radius of the earth becomes twice of its original radius, then

W = GMm/((2r)²)

= GMm/ 4r² = W/4

i.e. weight will be reduced to one-fourth of the original.

##### The earth is acted upon by gravitational of sun, even though it does not fall into the sun. Why?

The gravitational force is responsible for providing the necessary centripetal force which allows the earth to move around the sun at the defined path of orbit. So, the earth does not fail into the sun.

### Class 9 Science Chapter 10 Important Extra Questions Set – 5

### Prove that if the earth attracts two bodies placed at same distance from the centre of the earth with the same force, then their masses are equal.

Let P and Q be the two bodies,

the mass of body P = m₁

And the mass body Q = m₂

As per the universal law of gravitation the force of attraction between the earth and the body P is given by,

Fp = (G × Mₑ ×m₁)/R² ….(1)

Where R is the distance of the body from the center of the earth.

Similarly, the force of attraction between the earth and the body Q is given by

FQ = (G × Mₑ × m₂)/R² ….(2)

Since the two forces, i.e. FP and FQ are equal, thus from (1) and (2)

(G × Mₑ × m₁)/R² = (G × Mₑ ×m₂)/R²

### On the earth, a stone is thrown from a height in a direction parallel to the earth’s surface while another stone is simultaneously dropped from the same height. Which stone would reach the ground first and Why?

For both the stones

Initial velocity, u = 0

Acceleration in downward direction = g

Now, h = ut + 1/2gt²

h = 0 + 1/2gt²

h = 1/2gt²

t = √(2h/g)

Both stones will take the same time to reach the ground because the two stones fall from the same height.

### Calculate the average density of the earth in term of g, G and R

We know that g = GM/R² or M = (gR²)/G

Average density of the earth, D = (Mass )/(Volume ) = (gR²)/(G × V_e )

(Where Ve is the volume of the earth)

D = (gR²)/(G 4/3) πR³ = 3g/4πGR

### Prove that if a body is thrown vertically upwards, the time of ascent is equal to the time of descent.

Upward motion v = u – gt₁ 0 = u – gt₁, t₁ = u/g ….(1) Downward motion v = u + gt₂ v = 0 + gt₂

As the body falls back to the earth with the same velocity it was thrown vertically upwards.

v = u

u = 0 + gt₂

t₂ = u/g ….(2)

From (1) and (2) we get t₁ = t₂

Time of ascent = Time of desert

##### The earth attracts an apple. Does the apple also attract the earth? If it does why, does the earth not move towards the apple?

According to Newton’s third law of motion, action and reaction are equal and opposite. It means that the force on the apple due to earth’s attraction is equal to that on the earth due to apple’s attraction. But we know acceleration ∝ 1/m.

As the mass of the earth is very large as compared to that of the apple, the accelerated experienced by the earth will be so small that it will not be noticeable.

### Class 9 Science Chapter 10 Important Extra Questions Set – 6

### Two object of masses m1 and m2 having the same size are dropped simultaneously from heights h₁ and h₂, respectively. Find out the ratio of time they would take in reaching the ground. Will this ratio remain the same if (i) one of the objects is hollow and the other one is solid and (ii) both of them are hollow, size remaining the same in each case? Give reasons.

As u =, h₁ = 1/(2 ) gt₂/1

h₂ = 1/(2 ) gt₂/2

t₁/t₂ = √(h₁/h₂ )

Ratio will not change in either case because acceleration remains the same. In case of free fall acceleration does not depend upon mass and size.

### Derive expression for force attraction between two bodies and then define gravitational constant.

Every body 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 them.

Let us considered two bodies A and B of masses m₁ and m₂ which are separated by a diseases r. Then the force of gravitation (F) acting on the two bodies is given by

F ∝ m₁ × m₂ …..(1)

and F ∝1/(r² ) …..(2)

Combining (1) and (2) we get

F ∝ m₁ × m₂ )/r²

Or F = G × m₁ m₂ )/r² …..(3)

Where G is a constant known as universal gravitational constant.

Here is the masses m₁ and m2 of the two bodies are of 1kg and the distance (r) between them is 1m then putting m₁ = 1kg, m₂ = 1kg and r = 1m in the above formula we get

G = F

Thus, the gravitational constant G is numerically equal to the force of gravitation which exists between two bodies of unit masses kept at a unit distance from each other.

### Define acceleration due to gravity. Derive an expression for acceleration due to gravity in terms of mass of the earth (M) and universal gravitational constant (G).

The acceleration produced in the motion of a body falling under the force of gravity is called acceleration due to gravity. It denoted by g.

The force (F) of gravitational attraction on a body of mass m due to earth of mass M and Radius R is given by

F = G Mm/R² …..(1)

We know from Newton’s second law of motion that the force is the product of mass and acceleration.

F = ma

But the acceleration due to gravity is represented by the symbol g. Therefore, we can write

F = ma …..(2)

From the equation (1) and(2) we get

mg = G Mm/R² or g = GM/R² ……(3)

When body is at a distance r from the center of the earth then g =GM/r²

### Show that the weight of an object on the moon is 1/6 th of its weight on the earth.

Suppose the mass of the moon is Mm and its radius is Rm. If a body of mass m is placed on the surface of the moon, then weight of the body on the moon is

Wm = (GMₘ m)/Rₘ² …..(1)

Weight of the same body on the earth’s surface will be

We = (GMₑ m)/(Rₑ)² ……(2)

Where Me = mass of earth and Re = radius of earth.

Dividing the equation (1) and (2) we get

Wm/(We ) = Mₘ/Mₑ × Rₑ² )/(Rₘ)² …….(3)

Now, mass of the earth Me = 6 × 10²⁴ kg

Mass of the moon Mm = 7.4 × 10²² kg

Radius of the earth Re = 6400 km

and radius of the moon, Rm = 1740 km

Thus the equation (3) becomes

Wm/(We ) =(6 × 10²⁴ kg)/( 7.4 × 10²² kg) × ( 6400km/(1740km ))2

Or Wm/(We ) = 1/(6 ) or Wm ≈ Wₑ/6

The weight of the body on the moon is about one sixth of its weight on the earth.

### How does the weight of an object vary with respect to mass and radius of the earth? In a hypothetical case, if the diameter of the earth becomes half of its present value and its mass becomes four times of its present value, then how would the weight of any object on the surface of the earth be affected?

Weight of an object is directly proportional to the mass of the earth and inversely proportional to the square of the radius of the earth. i.e.,

Weight of a body ∝ M/R²

Original weight, W0 = mg = mGM/R²

When hypothetically M becomes 4 M and R becomes R/2.

Then weight becomes Wn = mG4M/(R/2)² = (16 mG) M/R² 16 ×W0

The weight will become 16 times.

##### Why can one jump higher on the surface of the moon than on the earth?

Because the value of acceleration due to gravity (g) on the moon’s surface is nearly 1/6th to that of the surface of the earth.

### Class 9 Science Chapter 10 Important Extra Questions Set – 7

### From a cliff of 49 m high a man drops a stone. One second later, he throws another stone. They both hit the ground at the same time. Find out the speed with which he threw the second stone.

For the first stone

u = 0 ms⁻¹, h = 49 m,

As we know

s = ut + 1/2 gt²

49 = 0 × t+ 1/2 × 9.8 × t²

t2 = 98/9.8 = 10

t = √10 = 3.16 s

i.e. first stone would take 3.16s to reach the ground.

For the second stone,

the time taken by the second stone to reach the ground is one second less than the that taken by the first stone as both the stones reach the ground at the same time.

That is for the second stone, t = (3.16 – 1) s = 2.16 s

For the second stone,

g = 9.8 ms⁻², h = 49 m, t = 2.16 s, u = ?

s = ut + 1/2gt²

49 = u × 2.16 + 1/2 × 9.8 x (2.16)²

49 – 22.86 = 2.16 u or 26.14 = 2.16u

u = 26.14/2.16 = 12.1 ms⁻¹

i.e. the second stone was thrown downward with a speed of 12.1 ms⁻¹.

### A stone is dropped from the top of a 40 m high tower. Calculate its speed after 2s. Also find the speed with which the stone strikes the ground.

(i) As v = u + gt

v = 0 + (-10) × 2 = -20 ms⁻¹

(ii) As v = u² + 2gs

Or v² – 0² = 2(-10) × (-40)

Or v = √800

20√(2ms⁻¹)

### Calculate the value of acceleration due to gravity g using the relationship between g and G.

We know that G = 6.67 × 10⁻¹¹) Nm² kg⁻²)

Mass of the earth, Me = 6 × 10²⁴ kg

And Radius of the earth Re = 6.4 × 10⁶ m

As g = (G × Mₑ)/(Rₑ)²

g = ( 6.67 × 10⁻¹¹) × 6 × 10²⁴ )/(6.4 × 10⁶ )²

g = ( 6.67 × 6 ×10)/(6.4 ×6.4) m/s² = 9.8 m/s²

### Class 9 Science Chapter 10 Important Extra Questions Set – 7

### From a cliff of 49 m high a man drops a stone. One second later, he throws another stone. They both hit the ground at the same time. Find out the speed with which he threw the second stone.

For the first stone

u = 0ms-1, h = 49m,

As we know

s = ut + 1/2gt2

49 = 0 ×t+ 1/2×9.8 ×t^2

t2 = 98/9.8 = 10

t = √10 = 3.16 s

i.e. first stone would take 3.16s to reach the ground.

For the second stone,

the time taken by the second stone to reach the ground is one second less than the that taken by the first stone as both the stones reach the ground at the same time.

That is for the second stone, t = (3.16 – 1) s = 2.16 s

For the second stone,

g = 9.8 ms-2, h = 49 m, t = 2.16 s, u = ?

s = ut + 1/2gt2

49 = u ×2.16+ 1/2×9.8 (2.16)^2

49 – 22.86 = 2.16 u or 26.14 = 2.16u

u = 26.14/2.16 = 12.1 ms-1

i.e. the second stone was thrown downward with a speed of 12.1 ms-1.

### A stone is dropped from the top of a 40 m high tower. Calculate its speed after 2s. Also find the speed with which the stone strikes the ground.

(i) As v = u + gt

v = 0 + (-10) × 2 = -20ms-1

(ii) As v = u2 + 2gs

Or v2 – 02 = 2(-10) × (-40)

Or v = √800

20√(2ms^(-1) )

### Calculate the value of acceleration due to gravity g using the relationship between g and G.

We know that G = 6.67 ×〖10〗^(-11) Nm^2 kg^(-2)

Mass of the earth, Me = 6 ×〖10〗^24 kg

And Radius of the earth Re = 6.4 ×〖10〗^6 m

As g = (G × M_e)/R_(e^2 )

g = ( 6.67 ×〖10〗^(-11)×6 ×〖10〗^24 )/(6.4 ×〖10〗^6 )^2

g = ( 6.67 ×6 ×10)/(6.4 ×6.4) m/s2 = 9.8m/s2

##### If the small and big stones are dropped from the roof of a house simultaneously, they will reach the ground at the same time. Why?

The acceleration due to gravity does not depend upon the mass of the stone or body. Both the bodies fall with the same acceleration towards the surface of the earth. Thus a big stone will fall with the same acceleration as a small stone. So, both the stones will reach the ground at the same time when dropped simultaneously.

###### Difference between g and G

Acceleration due to gravity (g) | Universal gravitational constant (G) |
---|---|

1. Acceleration due to gravity is the acceleration acquired by a body due to the earth’s gravitational pull on it. | 1. Gravitational constant is numerically equal to the force of attraction between two masses of 1kg that are separated by a distance of 1 m. |

2. g is a vector quantity. | 2. G is a scaler quantity. |

3. It is different at different places on the surface of the earth. Its value also value varies from one celestial body to another. | 3. The G is universal constant i.e. its value is the same ( i.e. 6.7 × 10^-11 Nm^2 kg^-2) everywhere in the universe. |