Class 9 - Physics - Gravitation

**Question1.**

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

Answer:

According to Universal law of gravitation, the force of attraction between 2 bodies is:-

F = (G m_{1}m_{2})/r^{2}

Where m_{1}m_{2 }= the masses of the two bodies,

r = distance between them

G = gravitational Constant

When the distance is reduced to half i.e. r’ = (1/2) r then,

F’ = (G m_{1}m_{2})/r^{2}

= (G m_{1}m_{2})/(r/2)^{2}

= (4 G m_{1}m_{2})/ (r^{2}) = 4F

Therefore when the distance between the objects is reduced to half, then the force of gravitation increases by four times the original force.

**Question2.**

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:

All objects fall on ground with constant acceleration known as acceleration due to gravity (g).

It is constant and the value of ‘g’ does not depend upon the mass of an object.

Therefore heavy objects do not fall faster than light objects provided there is no air resistance.

**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 × 10^{24} kg and radius of the earth is 6.4 × 10^{6 }m.)

Answer:

Given:

Mass of earth m_{1} =6 × 10^{24} kg

Mass of the object m_{2} = 1 kg

Radius of the earth =6.4 × 10^{6 }m

Universal gravitational constant G = 6.67 x 10^{-11} Nm^{2}kg^{-2}

By applying universal law of gravitation: - F = (G m_{1} m_{2}/r^{2})

= (6.67 x 10^{-11} x 6 × 10^{24} x 1)/ (6.4 × 10^{6})^{2}

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?

Answer:

The earth attracts the moon with an equal force with which the moon attracts the earth but these forces are in opposite directions.

By universal law of gravitation, the force between moon and the sun will be:

F = (G m_{1} m_{2}/d^{2})

Where m_{1} m_{2} = masses of earth and moon respectively.

d = distance between the earth and moon.

**Question 5.**

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

Answer:

According to universal law of gravitation and Newton third law, we know that the force of attraction between two objects is same but in opposite direction.

Thus the earth attracts the moon with the same force as the moon exerts on earth but in opposite directions.

Since earth is much larger in size than moon, so the acceleration cannot be noticed on earth surface.

**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?

Answer:

According to universal law of gravitation, the force between two objects (m_{1} and m_{2}) is proportional to their masses and inversely

proportional to the square of the distance(R) between them.

F = (G m_{1} m_{2})/ (R^{2})

- If the mass of one object is doubled,

Then, F = (G x 2 m_{1} x m_{2})/ R^{2}

=> F= 2F, hence force is doubled.

(ii) If the distance between the objects is doubled and tripled,

Then, F = (G x m_{1} x m_{2})/ (2R)^{ 2}

=> F = 4F, hence force become one – fourth of its initial force.

And F = ((G x m_{1} x m_{2})/ (3R)^{ 2}

=> F = 9F, hence force becomes one-ninth of its initial force.

(iii) If the masses of both objects are doubled,

Then F= (G x2m_{1} x 2m_{2} /R^{2})

=> F = 4F, hence force will be four times more than its actual value.

**Question 7.**

What is the importance of universal law of gravitation?

Answer:

The universal law of gravitation explains several phenomena which were believed to be unconnected:

(i) The force that binds us to the earth;

(ii) The motion of the moon around the earth;

(iii) The motion of planets around the Sun; and

(iv) The tides due to the moon and the Sun.

**Question 8.**

What is the acceleration of free fall?

Answer:

When anybody is in free fall, the only force acting on the object is the earth’s gravitational field.

By Newton’s second law of motion all the forces produce acceleration so all the objects accelerate toward the earth’s surface due to gravitational attraction of the earth.

This acceleration is known as acceleration due to gravity near earth’s surface. It is denoted by ‘g’ and its value is 9.8m/s^{2}

and it is constant for all objects near earth’s surface (irrespective of their masses).

**Question 9. **

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

Answer:

Gravitational force between the earth and an object is known as the weight of the object.

** 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.]

Answer:

Weight of a body on the surface of earth;

W = mg where (m = mass of the body and g= acceleration due to gravity)

The value of g is more at poles as compared to equator. Therefore gold will weigh less at the equator as compared to poles.

Therefore Amit’s friend will not agree with the weight of the gold bought.

**Question 11.**

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

Answer:

A sheet of paper has more surface area as compared to a crumpled paper ball. A sheet of paper has to face more air resistance.

As result a sheet of paper falls slower than the 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?

Answer:

Given:-

Mass of the object m = 10kg

Acceleration due to gravity on earth = g_{e} or g = 9.8 m/s^{2}

Acceleration due to the gravity on moon = g_{m}

Weight on the moon = W_{m}

Weight on the earth= W_{e}

Weight = m x g

g_{m} = (1/6) g_{e} (given)

Therefore W_{m} = m g_{m} = m x (1/6) g_{e}

W_{m} = 10 x (1/6) x 9.8 = 16.34 N

W_{e} = mxg_{e} = 10 x 9.8 = 98N

**Question 13.**

A ball is thrown vertically upwards with a velocity of 49 m/s.

Calculate

(i) Maximum height to which it rises,

(ii) Total time it takes to return to the surface of the earth.

Answer:

Given:-

Initial velocity u = 49m/s

Final velocity v at max height = 0

- Acceleration due to gravity on earth g = (-) 9.8 m/s
^{2}(As ball is thrown up therefore negative).

Using third equation of motion

v^{2} = u^{2} - 2gs

- 0 = (49)
^{2}– 2 x 9.8 x s - s = (49)
^{2}/ (2 x 9.8) - s = 122.5m
- Total time T = Time to ascend(T
_{a}) + Time to descend (T_{d})

v= u – gt

- 0 =49 – 9.8 x T
_{a} - T
_{a}= (49/9.8) = 5s.

Also T_{d} = 5s

Therefore T = T_{a} + T_{d}

T =5 + 5 = 10s

**Question 14.**

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

Answer:

Given:-

Initial velocity u =0

Height of the tower = total distance = 19.6m

g = 9.8 m/s^{2}

Using third equation of motion:-

v^{2} = u^{2} + 2gs

= 0 + 2 x 9.8 x 19.6

v^{2 }= 384.16

- v = √384.16
- v = 19.6m/s

**Question 15.**

A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s^{2}, find the maximum height reached by the stone.

What are the net displacement and the total distance covered by the stone?

Answer:

Given:-

Initial velocity u =40m/s

g = 10 m/s^{2}

Max height final velocity =0

Using third equation of motion:-

v^{2} = u^{2} - 2gs (-ive as object is going upwards)

=> 0 = (40)^{2} – 2 x 10 x s

=> s = (40 x 40) / 20

Maximum height s = 80m

Total Distance = s + s = 80 + 80 = 160m

Total displacement = 0 (As initial point is same as the final point)

**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 × 10^{30} kg.

The average distance between the two is 1.5 × 10^{11} m.

Answer:

Given-

Mass of the earth m_{e} = 6 × 10^{24} kg

Mass of the sun m_{s} =2 × 10^{30} kg

Average distance r = 1.5 × 10^{11} m

Gravitation constant G = 6.67 x 10^{-11} N m^{2}/ kg^{2}

Using Universal law of Gravitation:-

F = (G m_{e} m_{s})/r^{2}

= (6.67 x 10^{-11 }x 6 × 10^{24} x 2 × 10^{30})/ (1.5 × 10^{11})^{2}

F = 3.56 x 10^{23} 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.

Answer:

Given:-

For S_{1}:- When the stone is thrown from the top of the tower,

Distance travelled = x

Initial velocity u =0

Time taken = t

s= ut + (1/2) gt^{2}

=> x = 0 + (1/2) gt^{2}

=> x = 5t^{2} (1)

For S_{2}:- When the stone is thrown upwards,

Distance travelled = (100 – x)

Initial velocity u = 25 m/s

Time taken = t

s= ut + (1/2) gt^{2}

=> (100 – x) = 25t + (1/2) x 10 x t^{2}

x = 100 -25t + 5t^{2} (2)

From (1) and (2)

5t^{2} = 100 -25t + 5t^{2}

=> t = (100/25) =4sec.

Two stones will meet after 4sec.

From (1) => x = 5t^{2} = 5 x 4 x 4= 80m.

Putting the value of x in (100-x)

= (100-80) = 20m.

This means 2 stones will meet a distance of 20m from the ground after 4sec.

**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.

Answer:

Given:-

g= 10m/s^{2}

Total time T = Time to ascend (T_{a}) + Time to descend (T_{d}) = 6sec

=> T_{a} =T_{d} = 3sec

(a) Final velocity at maximum height v = 0

From first equation of motion:-

v = u - gt_{a} ((-ive) as stone is going up).

=> u = v +gt_{a}

=0 + 10 x 3

= 30m/s

The velocity with which stone was thrown up is 30m/s.

- From second equation of motion:-

s=ut_{a} - (1/2) g (t_{a})^{ 2}

= 30 x 3 – (1/2) x 10 x (3)2

=90-45=45m

The maximum height stone reaches is 45m.

- In 3sec, it reaches the maximum height.

Distance travelled in another 1sec = s’

s' = ut + (1/2) gt^{ 2}

= 0 + 10 x 1 x 1

= 5m.

Distance travelled in another 1sec = 5m.

In 4sec: - Position of point p (45 – 5) = 40m from the ground.

Question 19.

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

Answer:

The buoyant force will be in vertically upward direction on an object which is immersed in a liquid.

**Question 20.**

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

Answer:

The density of plastic is less than that of water, so the force of buoyancy on plastic block will be greater than the weight of plastic block displaced.

Hence, the acceleration of plastic block will be in upward direction, and comes up to the surface of water.

**Question 21.**

The volume of 50 g of a substance is 20 cm^{3}. If the density of water is 1 g cm^{–3}, will the substance float or sink?

Answer:

Density of the substance= (Mass/Volume)

= (50/20) = 2.5g/cm^{3}

Density of water = 1g/cm^{3}

Density of the substance > Density of water

The substance will float.

**Question 22.**

The volume of a 500 g sealed packet is 350 cm^{3}. 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?

Answer:

Density of sealed packet = (500g/350) = 1.42 g/cm^{3}

Density of sealed packet > Density of water

Therefore the packet will sink.

By Archimedes Principle,

Volume of water displaced = Force exerted on the sealed packet.

Volume of water displaced = 350cm^{3}

Therefore mass of water displaced =ρ x V

= 1 x 350

Mass of water displaced = 350g.

.