NCERT Solutions for Class 9 Science Chapter 10 Gravitation

1. How does the force of gravitation between two objects change when the distance between them is reduced to half ?

Answer: By universal law of gravitation, gravitational force F is given by the following expression,

F = Gm₁m₂/ r²

Where, m₁ and m₂ are the masses of two objects and r is distance between them, and G is gravitational constant.

Now, if distance is reduced to half, then distance r = r/2

∴ Gravitational force will be,

F = Gm₁m₂ / (r/2)²

= 4 Gm₁m₂ / r²

I.e F = 4F

Therefore, gravitation force becomes four times when distance between the objects is reduced to half.

2. Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object?

Answer: Heavy object will not fall faster than light object because all objects are attracted towards ground with constant acceleration in absence of air resistance called acceleration due to gravity. Acceleration due to gravity is constant and does not depend on mass of an object.

3. What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 10²⁴kg and radius of the earth is 6.4 × 10⁶ m.)

Answer: Using universal law of gravitation,

Gravitational force F = Gm₁m₂ / r²

Mass of earth = m₁ = 6 x 10²⁴ kg

Mass of an object = m₂ = 1 kg

Radius of earth = 6.4 x 10⁶ m

Gravitational constant = 6.7 x 10^-11 Nm² / kg²

As object is on the surface of earth therefore, r = R

∴ F = Gm₁m₂ / R²

F = 6.7 x 10^-11 x 6 x 10²⁴ x 1 / (6.4 x 10⁶)²

F = 6.7 x 6 x 1 x 10⁷/ 6.4 x 6.4 x 10⁶ x 10⁶

F = 6.28 x 10⁷/ 6.4 x 10⁶ N

F = 9.8 N

4. The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth? Why?

Answer: As per law of gravitation, two objects attracts each other with equal and opposite force. Thus, earth attracts moon with equal force as the moon attracts the earth.

5. If the moon attracts the earth, why does the earth not move towards the moon?

Answer: Earth and moon both attract each other. Mass of earth very large as compared to moon. Therefore, acceleration experienced by earth from moon is very small in comparison to acceleration experienced by moon from earth. Thus, earth does not move towards moon.

6. What happens to the force between two objects, if (i) the mass of one object is doubled? (ii) the distance between the objects is doubled and tripled? (iii) the masses of both objects are doubled?

Answer: Expression of universal law of gravitation states that :

F = Gm₁m₂ / r²

(i) The mass of one object is doubled? –

Force of gravitation (F) is directly proportional to product of masses. Therefore, if the mass of the object is doubled, gravitational force also gets doubled.

(ii) The distance between the objects is doubled and tripled? –

Gravitational force F is inversely proportional to distance between the objects. Therefore if distance between the object is doubled gravitational force becomes one fourth, and if distance between objects is tripled then gravitational force becomes one ninth of its original value.

(iii) The masses of both objects are doubled? –

If mass of both the objects are doubled then the gravitational force gets four times of its original value. This is because gravitational force is directly proportional to product of masses of the objects.

7. What is the importance of universal law of gravitation?

Answer: Universal law of gravitation explains that each and every body in the universe attracts other body. It is important as it explains force the binds us to earth, provide reasons behind the motion of planets around the sun, moon around the earth, etc.

8. What is the acceleration of free fall?

Answer: Object falling towards the earth, is said to be under the influence of acceleration of free fall. It is denoted by g and has value 9.8 m/s². This value is constant for all object irrespective of their masses.

9. What do we call the gravitational force between the earth and an object?

Answer: Gravitational force between the earth and object is known as weight.

10. Amit buys few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why? [Hint: The value of g is greater at the poles than at the equator.]

Answer: Weight of object on earth is measured by W = mg. Here m is mass of object and g is the acceleration due to gravity. Value of g is greater at poles in comparison to equator. Therefore, gold at poles will weigh more and same gold on equator will weigh less. So, friend will not agree with weight of gold Amit bought.

11. Why will a sheet of paper fall slower than one that is crumpled into a ball?

Answer: Crumpled paper ball has less surface area and therefore, resistance to its motion decreases through air. Thus, sheet of paper fall slower than crumpled paper ball.

12. Gravitational force on the surface of the moon is only 1 /6 as strong as gravitational force on the earth. What is the weight in newtons of a 10 kg object on the moon and on the earth?

Answer: Given : Mass of object = 10 kg

Weight of object on moon = ( ⅙ ) Weight of object on earth

Weight = mass x acceleration due to gravity

∴ Weight of object on earth = 10 x 9.8 = 98 N

So Weight of object on moon = (1/6) 98 = 16.33 N

13. A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate (i) the maximum height to which it rises, (ii) the total time it takes to return to the surface of the earth.

Answer: 1) Using equation of motion under gravity

v² – u² = 2gh

It is given that initial velocity u of ball is 49 m/s.

Final velocity of ball v = 0 m/s

Acceleration due to gravity in upward direction = – 9.8 m/s²

Let maximum height attained by the ball be h

∴ 0² – (49)² = 2 (-9.8)h

∴ 49² = 19.6h

∴ h = 49² / 19.6 = 122.5 m

2) Using equation of motion

v = u + gt

0 = 49 + (-9.8) t

9.8 t = 49

∴ t = 5 s

Also, it is known that time of ascent = time of descent

∴ Total time taken by the ball to return on the earth surface = 5 + 5 = 10 s.

14. A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity.

Answer: By using equation of motion under gravity

v² = u² + 2gs

V = √u² + 2 gs

= √0 + 2 (9.8)19.6

= √384.16

v = 19.6 m/s

Therefore final velocity of stone when released from the height of 19.6 m is 19.6 m/s

15. A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s², find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

Answer: Given initial velocity of stone u = 40 m/s.

Acceleration due to gravity g = -10 m/s²[ as stone is moving upward ]

Let h be maximum height attained by stone.

Also, final velocity = 0

Using equation of motion under gravity,

v² = u² + 2gs

∴ 0 = 40² + 2 (-10)s

∴ 20 s = 1600

∴ s = 80 m = h

Total distance covered by object is sum of upward and downward distance = 80 + 80 = 160 m

Net displacement of stone = 80 + (-80) = 0 m

16. Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6 × 10²⁴kg and of the Sun = 2 × 10³⁰ kg. The average distance between the two is 1.5 × 10¹¹ m.

Answer: By universal law of gravitation,

F = Gm₁m₂/ r²

Given : m₁ = mass of earth = 6 × 10²⁴kg

m₂ = mass of sun = 2 × 10³⁰kg

R = average distance between earth and sun = 1.5 × 10¹¹m.

G = gravitational constant = 6.7 × 10^-11 Nm² kg^-2

F = ( 6.7 × 10^-11 x 6 × 10²⁴ x 2 × 10³⁰ ) / ( 1.5 × 10¹¹)²

∴ F = 80.4 x 10²¹ / (1.5)²

∴ F = 80.4 x 10²¹ / 2.25

= 35.73 x 10²¹N

17. A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.

Answer: Let t be the time after which stone meet

∴ For stone dropped from the tower initial velocity u = 0

Let s be displacement of stone stop dropped from the tower at time t

Acceleration due to gravity = 9.8 m/s2

Consider following equation of motion,

s = ut + ½ gt²

= 0 x t + ½ (9.8)t²

s = 4.9 t²

Now, consider for the stone thrown upward.

Initial velocity u = 25 m/s

Let s’ be the displace of stone from the ground at time t

∴ Acceleration due to gravity = – 9.8 m/s²

By using equation of motion,

s’ = ut + ½ gt²

= 25t – ½ (9.8) t²

= 25 t – 4.9 t²

∴ The combined distance of stones meeting each other at time t is same as the height of the tower which is 100 m.

∴ s + s’ = 100

4.9 t² + 25 t – 4.9 t² = 100

25 t = 100

∴ t = 4 s

In 4 seconds distance covered by falling stone can be represented as –

s = ut + ½ gt²

= 0(t) + ½ (9.8)(4)²

= 4.9 x 16

= 78.4m

Therefore, in 4 seconds stones will meet at the height of (100 – 78.4)m from the ground i.e. 21.6 m

18. A ball thrown up vertically returns to the thrower after 6 s. Find (a) the velocity with which it was thrown up, (b) the maximum height it reaches, and (c) its position after 4 s.

Answer: (a) The velocity with which it was thrown up –

Ball takes 6 s to return to thrower. Also, time of ascent = time of descent.

∴ Time t = ( ½ ) x 6

t = 3 s

Therefore, time taken for upward journey is 3 s.

Let v be the final velocity at the ball at maximum height.

∴ v = 0

Also, acceleration due to gravity g = – 9.8 m/s²

Using equation of motion

v = u + gt

0 = u + (-9.8) x 3

u = 29.4 m/s

Ball is thrown upward with the velocity 29.4 m/s

(b) The maximum height it reaches –

Using equation of motion :

s = ut + ½ gt²

= 29.4 x 3 + ½ (-9.8) 3²

= 88.2 – 4.9(3)²

= 88.2 – 44.1

= 44.1 m

Maximum height reached by the ball is 44.1 m

(c) Its position after 4 s –

Ball attains maximum height after 3 seconds. After attaining this height it starts falling downward. Thus initial velocity u = 0. Therefore position after 4s of throw can be given as 4s- 3s = 1s

By using equation of motion,

s = ut + ½ gt²

= 0 + ½ (9.8) (1)²

= 4.9 x 12

= 4.9 m

We know that, maximum height reached by the ball = 44.1 m

∴ At 4 s ball is at height 44.1 – 4.9 m = 39.2 m

19. In what direction does the buoyant force on an object immersed in a liquid act?

Answer: Object immersed in liquid is experience buoyant force in vertically upward direction.

20. Why does a block of plastic released under water come up to the surface of water?

Answer: When object is immersed in water gravitational force and buoyant force acts on the object. Gravitational force pulls the object down, whereas, buoyant force pushes the object at the top. When buoyant force is more than gravitational force, in such situation object under water comes up on the surface. Same is the reason, why block of plastic released underwater comes up on the surface.

21. The volume of 50 g of a substance is 20 cm³. If the density of water is 1 g cm³, will the substance float or sink?

Answer: If density of substance is greater than density of liquid, then substance sinks. In contrast to this, if density of substance is less than density of liquid, then substance floats.

In this case, density of substance = mass / volume = 50 / 20 = 2.5 g/cm³

Therefore, density of object is more than density of water so, the substance will sink in water.

22. The volume of a 500g sealed packet is 350 cm³. Will the packet float or sink in water if the density of water is 1 g/ cm³? What will be the mass of the water displaced by this packet?

Answer: Density of 500g sealed packet = mass / volume = 500 / 350 = 1.42 g / cm³. Here density of water is less than the density of packet, therefore, sealed packet will sink. Moreover, mass of water displaced by packet will be same as volume of the packet i.e 350 cm³.

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