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## Chapter 10 Gravitation

Intext Questions
Question 1: State the universal law of gravitation.
Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them.

Question 2: Write the formula to find the magnitude of the gravitational force between the Earth and an object on the surface of the Earth.
Let the two objects of masses M and m lie at a distance d from each other. Let the force of attraction between two objects be F. According to the Law of Gravitation, the force between two objects is directly proportional to the product of their masses. That is,
F ∝ M × m → eq 1
And force between two objects is inversely proportional to the square of the distance between them. That is,
F ∝ 1/d² → eq 2
Combining eq 1 and 2
F M × m/d²
F = G M × m/d²
G is the constant of proportionality and is called universal gravitational constant.

Question 3: What do you mean by free fall?
When a body moves exclusively under the influence of the Earth’s gravity, it is said to be in freefall.

Question 4: What do you mean by acceleration due to gravity?
When an object falls freely towards the surface of Earth from a certain height, then its velocity changes and this change in velocity produces acceleration in the object which is known as acceleration due to gravity. It is represented by the letter ‘g’ and has a value of 9.8 m s⁻².

Question 5: What are the differences between the mass of an object and its weight?

• Mass
→ Mass is the quantity of matter contained in the body.
→ It is the measure of its inertia.
→ It has only the magnitude hence it is a scalar quantity.
→ It is constant everywhere.
→ SI unit is kg.
→ Mass is measured by using beam balance.
• Weight
→ Weight is the force of gravity acting on the body.
→ It is the measure of gravity.
→ It has both magnitude and direction hence it is a vector quantity.
→ It is not constant.
→ SI unit is Newton (N).
→ Weight is measured by using spring balance.

Question 6: Why is the weight of an object on the moon 1/6th its weight on the Earth?
The mass of the moon is 1/100 times and its radius is 1/4 times that of Earth. As a result, the gravitational attraction on the moon is about 1/6 when compared to the Earth. Hence, the weight of an object on the moon is 1/6th its weight on the Earth.

Question 7: Why is it difficult to hold a school bag having a strap made up of a thin and strong string?
It is difficult to hold a school bag having a strap made of a thin and strong string. We know that the smaller the surface area, larger is the pressure exerted. The pressure on the shoulders is relatively large, so it is difficult to hold a school bag with a thin strap.

Question 8: What do you mean by buoyancy?
The upward force exerted by fluids on an object immersed in it known as buoyancy.

Question 9: Why does an object float or sink when placed on the surface of water?
An object floats or sinks when placed on the surface of water because of two reasons:
• If the density of an object is greater than that of water, the object sinks in water.
• If the density of an object is lesser than that of water, the object floats.

Question 10: You have a bag of cotton and an iron bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality, one is heavier than other. Can you say which one is heavier and why?
In reality, the cotton bag is heavier than iron bar. This is because the cotton bag has more volume than that of an iron bar so it displaces more air and hence the buoyant force acting on the cotton bag due to the air is more than that acting on the iron bar (by Archimedes principle) and
True weight = Apparent weight + Buoyant force
Apparent weight = 100 kg for both but the buoyant forces are different.

Exercise Questions
Question 1: How does the force of gravitation between two objects change when the distance between them is reduced to half?
The Universal Law of Gravitation states that the force of attraction between two objects is
F = G M × m/d²
where,
M and m are the masses of two objects, and d is the distance between two objects.
When the distance is reduced to half,
F = G M × m/d²
F = G M × m/(d/2)²
F = 4 G M × m/d²
F = 4F
Therefore, when the distance between the two objects is reduced to half, the force of gravitation will increase 4 times.

Question 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?
Gravitational force acts on all object in proportional to their masses. But a heavy object does not fall faster than a light object. This is because Acceleration = Force/Mass. As the force is directly proportional to mass, acceleration is constant for a body of any mass.

Question 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 × 10²⁴ kg and radius of the earth is 6.4 × 10⁶ m.)

Given
Mass of the object = 1 kg
Mass of the Earth = 6 × 10²⁴ kg
Radius of the Earth = 6.4 × 10⁶ m
Magnitude of gravitational force = ?
We know that, F = G M × m/d²
Substituting the values, we get
F = 6.67 × 10⁻¹¹ × 6 × 10²⁴ × 1/(6.4 × 10⁶)²
F = 40.02 × 10¹³/40.96 × 10¹²
F = 9.8 N

Question 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?
The Earth attracts the moon with a force same as the force with which the moon attracts the Earth. However, these forces are opposite in direction.

Question 5: If the moon attracts the earth, why does the earth not move towards the moon?
According to the universal law of gravitation and Newton’s third law of motion, the force of attraction between two objects is the same, but opposite in the direction. So, the Earth attracts the moon with a force as the force as the force with which the moon attracts Earth but is opposite in direction. We know that, a larger mass has less acceleration and lesser mass has more acceleration. So the Earth accelerates at a rate lesser than the acceleration rate of the moon. Hence, the Earth does not move towards the moon.

Question 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?
The force between two objects is given by G M × m/d²
i) If mass of one object is doubled,
F = G M × m/r²
= G M × 2m/r²
F = 2F
Therefore, if the mass of the one object is doubled, then the force between two objects will get doubled.

ii) If the distance between the objects is doubled,
F = G M × m/d²
= G M × m/(2d)²
= G M × m/4d²
F = F/4
Therefore, if the distance between the objects is doubled, then the force between two objects will get 1/4th.

If the distance between the objects is tripled,
F = G M × m/d²
= G M × m/(3d)²
= G M × m/9d²
F = F/9
Therefore, if the distance between the objects is tripled, then the force between two objects will get 1/9th.

iii) If the masses of both objects are doubled,
F = G M × m/d²
= G M² × m²/d²
F = 4F

Question 7: What is the importance of universal law of gravitation?
The universal law of gravitation successfully explained several phenomena which were believed to be unconnected:
• the force that binds us to the Earth
• the motion of the moon around the Earth
• the motion of the planets around the Sun
• the tides due to the moon and the Sun

Question 8: What is the acceleration of free fall?
Acceleration of free fall is the acceleration produced when a body falls under the influence gravitational force of the earth alone. It is represented by the letter ‘g’ and has a value of 9.8 m s⁻² on Earth.

Question 9: What do we call the gravitational force between the earth and an object?
The gravitational force between the Earth and an object is called weight.

Question 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.]
The weight of body on Earth’s surface is mg. The value of weight is greater on poles and lesser at the equator. So, his friend won’t agree with the weight of gold brought at the poles.

Question 11: Why will a sheet of paper fall slower than one that is crumpled into a ball?
A sheet of paper has more surface area when compared to a crumpled paper ball and air resistance will be more. Hence, a sheet of paper falls slower than crumpled ball.

Question 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 newton of a 10 kg object on the moon and on the earth?

Given
Acceleration due to Earth’s gravity = 9.8 m s⁻²
Mass of the object = 10 kg
Acceleration due to Moon’s gravity = gₘ
Weight on Earth = Wₑ
Weight on Moon = Wₘ
We know that, Weight = mass × acceleration due to gravity
gₘ = (1/6)gₑ
Wₘ = m × gₘ = m × 1/6gₑ
Wₘ = 10 × 1/6 × 9.8 = 16.33 N
Wₑ = m × gₑ = 10 × 9.8 = 98 N

Question 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.

Given
Initial velocity = 49 m/s
Final velocity at maximum height = 0 m/s
Acceleration due to gravity = -9.8 m s⁻² (acceleration due to gravity is in opposite direction)
We know that, 2gs = v² - u²
Substituting the values, we get
2 × (-9.8) × s = 0² - 49²
-19.6s = -2401
s = -2401/-19.6
s = 122.5 m
Total time = Time taken to reach maximum height + time taken to return to the surface of the Earth
Time taken to reach maximum height = ?
We know that, v = u + gt
Substituting the values, we get
0 = 49 + (-9.8)t
0 = 49 -9.8t
-49 = -9.8t
t = -49/-9.8
t = 5 s
Therefore, it takes 5 s to reach maximum height.
Time taken to reach the surface of Earth = ?
We know that, v = u + gt
Substituting the values, we get
0 = 49 + (9.8)t
0 = 49 + 9.8t
-49 = 9.8t
t = 49/9.8
t = 5 s
Therefore, it takes 5 s to reach Earth’s surface.
Total time = 5 + 5 = 10 s

Question 14: A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.

Given
Acceleration due to gravity = 9.8 m s⁻²
Distance = 19.6 m
Initial velocity = 0 m/s
We know that, 2gs = v² - u²
Substituting the values, we get
2 × 9.8 × 19.6 = v²
384.16 = v²
v = 19.6 m/s

Question 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?

Given
Acceleration due to Earth’s gravity = 10 m/s²
Initial velocity = 40 m/s
Final velocity = 0 m/s
Distance travelled = ?
We know that, 2gs = v² - u²
Substituting the values, we get
2 × 10 × s = 0² - 40²
20s = -1600
s = -1600/20
s = 80 m
Total distance travelled = s + s = 80 + 80 = 160 m
Displacement = 0 m as the initial and final points are the same.

Question 16: Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6 × 10 ²⁴ kg and of the Sun = 2 × 10³⁰ kg. The average distance between the two is 1.5 × 10¹¹ m.

Given
Mass of the Earth = 6 × 10 ²⁴ kg
Mass of the Sun = 2 × 10³⁰ kg
Distance between Sun and Earth = 1.5 × 10¹¹ m
Force of gravitation = ?
We know that, F = GMm/d²
Substituting the values, we get
F = 6.67 × 10⁻¹¹ × 6 × 10²⁴ × 2 × 10³⁰/(1.5 × 10¹¹)²
F = 80.04 × 10⁴³/2.25 × 10²²
F = 35.57 × 10²¹ N

Question 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.

Given
i) When the stone is thrown from the top of a tower.
Initial velocity = 0 m/s
Distance = x m
s = ut + 1/2 gt²
s = 1/2 × 9.8 × t²
s = 4.9t² → eq 1
ii) When another stone is thrown upwards.
Initial velocity = 25 m/s
100 - x = ut - 1/2 gt²
100 - x = 25t - 1/2 × 9.8 × t²
100 - x = 25t - 4.9t² → eq 2

Adding eq 1 and eq 2, we get
x + 100 - x = 25t - 4.9t² + 4.9t²
100 = 25t
t = 4 s
From eq 1
x = 4.9 × 4² = 78.4 m
Therefore, the two stones will meet each other at 4 s at the height of 21.6 m (100 - 78.4 = 21.6 m).

Question 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.

Given
Total time taken = 6 s
Final velocity = 0 m/s
Acceleration due to gravity = 9.8 m/s²
Time to ascend = Time to descend
3 s = 3 s
a) We know that, v = u + gt
Substituting the values, we get
0 = u + -9.8 × 3
0 = u - 29.4
u = 29.4 m/s

b) We know that, 2gs = v² - u²
Substituting the values, we get
2 × 9.8 × s = 0² - (29.4)²
19.6s = -864.36
s = 864.36/19.6
s = 44.1 m

c) Distance travelled in 3 s is the maximum height. Distance travelled in another 1 s.
s = utₐ - 1/2 × g × (tₐ)²
s = 1/2 × 9.8 × (1)²
s = 4.9 m
Its height above ground = 44.1 - 4.9 = 39.2 m

Question 19: In what direction does the buoyant force on an object immersed in a liquid act?
The buoyant force n an object that is immersed in a liquid will be vertically upward direction.

Question 20: Why does a block of plastic released under water come up to the surface of water?
The density of plastic is less than that of water. Therefore, the force of buoyancy on plastic block will be greater than the weight of the plastic block. So, the acceleration of the plastic block is going to be in upward direction. Hence, the plastic block comes up to the surface of water.

Question 21: The volume of 50 g of a substance is 20 cm³. If the density of water is 1 g cm⁻³, will the substance float or sink?

Given
Volume of 50 g substance = 20 cm³
We know that, density = Mass/Volume
= 50/20 = 2.5 g cm⁻³
Density of water is 1 g cm⁻³. Since, the density of the substance is greater than the density of water, so the substance will sink.

Question 22: The volume of a 500 g sealed packet is 350 cm³. Will the packet float or sink in water if the density of water is 1 g cm⁻³? What will be the mass of the water displaced by this packet?

Given
Volume of 500 g sealed packet = 350 cm³
We know that, Density = Mass/Volume
= 500/350 = 1.42 g cm⁻³
Density of water = 1 g cm⁻³
Since the density of sealed packet is greater than the density of water, the packet will sink.
Mass of water displaced by the packet is equal to the volume of the packet times the density of water.
= ρ × V
= 1 × 350
= 350 g
Therefore, 350 g of water will be displaced.