Class 09 : Gravitation : NCERT Solution

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

Answer: The force of gravitation varies inversely as square of the distance between two objects.

$F\propto \frac{1}{d^2}$

When all other factors remain constant, and distance is 1 unit then;
$F=\frac{1}{1^2}=1$

When distance is ½ unit then;

$F=\frac{1}{{\left(\frac{1}{2}\right)}^2}$
Or, $F=4$

This means that the force of gravitation between two objects becomes four times when the distance between them is reduced to half.

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?

Answer: All objects fall on the earth due to acceleration due to gravity. The value of g is constant at a given place. Hence, irrespective of their masses, all objects fall at the same rate.

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.)

Answer: Given; mass of object = 1 kg

Mass of earth $=6\times {10}^{24}$
Radius of earth $=6.4\times {10}^6$
$G=6.67\times {10}^{-11}$

Gravitational force between the earth and the object can be calculated as follows:

$F=G\frac{Mm}{R^2}$

$=6.67\times {10}^{-11}\times \frac{6\times {10}^{24}\times 1}{{(6.4\times {10}^6)}^2}$

$=\frac{6.67\times 6\times {10}^{13}}{6.4\times 6.4\times {10}^{12}}$

$=0.97\times 10=9.8N$

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 force of attraction between two objects depends on product of masses of the two objects. Hence, the force of attraction between two objects is constant for a given distance between them. So, both the earth and the moon would attract each other with equal force.

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

Answer: The force of attraction between the earth and the moon is constant. Mass of the earth is greater than the mass of the moon. Due to greater mass, the earth accelerates at a faster rate than moon. Due to this, the earth does not move towards the moon.

Question 6: What happens to the force between two objects, if

(a) the mass of one object is doubled?

Answer: F ∝ Mm

If mass of one object is doubled then;

$F\propto 2Mm$

This means the force will become double.

(b) the distance between the objects is doubled and tripled?

Answer: $F\propto \frac{1}{d^2}$

If distance is doubled then;

$F\propto \frac{1}{4}d$

This means the force becomes four times.

If distance is tripled then;

$F\propto \frac{1}{9}d$

This means the force becomes nine times.

(c) the masses of both objects are doubled?

Answer: $F\propto Mm$

If masses of both objects are doubled then;

$F\propto 2M2m$
Or, $F\propto 4Mm$

This means the force becomes four times.

Question 7: What is the importance of universal law of gravitation?

Answer: The universal law of gravitation helps in explaining many phenomena. Some of them are as follows:

• We keep on remaining on the earth’s surface.
• The earth moves around the sun.
• The planets and their satellites remain in their respective orbits.
• The tides happen at regular intervals.

Question 8: What is the acceleration of free fall?

Answer: When a body is in free fall, the acceleration it undergoes is called acceleration of free fall. It is same as acceleration due to gravity. The value of g is 9.8 m/s2.

---

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

Answer: 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: Since the value of g is greater at the poles than at the equator, the weight of gold at the poles is more than the weight at the equator. So, Amit’s friend will not agree with the weight of 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 will experience greater air resistance than the crumpled ball of paper; while falling. So, the sheet of paper will fall slower than the crumpled ball of paper.

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 newtons of a 10 kg object on the moon and on the earth?

Answer: Weight of an object on earth can be calculated as follows:

$W=m\times g$
$=10\times 9.8=98N$

Weight of the same object on moon is 1/6th its weight on earth:

Weight on moon $=\frac{98}{6}=16.33N$

Question 13: A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate

(a) the maximum height to which it rises,
(b) the total time it takes to return to the surface of the earth.

Answer: This can be calculated by using the following equation of motion:

$v^2-u^2=2gs$

where; u = initial velocity, v = final velocity, g = acceleration due to gravity and s = distance.

In this case, final velocity v = 0 because the object will come to rest on reaching the maximum height.

Or, $0-{49}^2=2\times -9.8\times s$
Or, $2401=19.6\times s$
Or, $s=\frac{2401}{19.6}=122.5m$

The time taken to reach the maximum height can be calculated as follows:

$v=u+g\times t$
Or, $49=0+9.8\times t$
Or, $t=\frac{49}{9.8}=5s$

Now, time of ascent = time of descent

So, total time to return to the surface of the earth $=5+5=10s$

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

Answer: This can be calculated by following formula:

$v^2-u^2=2gs$
Or, $v^2-0=2\times 9.8\times 19.6$
Or, $v^2=19.6\times 19.6$
Or, $v=19.6m/s$

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?

Answer: This can be calculated by following formula:

$v^2-u^2=2gs$

Here, final velocity will be zero;

Or, $0–{40}^2=2\times –10\times s$
Or, $1600=20\times s$
Or, $s=\frac{1600}{20}=80m$

The stone will finally return on the starting point, i.e. on the ground. So, net displacement is zero.

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

Answer: The force of gravitation between the earth and the sun can be calculated as follows:

$F=G\frac{Mm}{R^2}$

$=6.67\times {10}^{-11}\times \frac{6\times {10}^{24}\times {10}^{30}}{{(1.5\times {10}^{11})}^2}$

$=\frac{6.67\times 6\times 2\times {10}^{43}}{2.25\times {10}^{22}}$

$=3.58\times {10}^{22}N$

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: Let us assume that the stones meet at a height of x from the ground. The first stone will travel 100 – x to reach this point. The time taken to reach this distance by the first stone can be calculated as follows:

$s=ut+\frac{1}{2}g{t}^2$

Or, $100-x=0+\frac{1}{2}\times 9.8\times t^2$

Or, $100-x=4.9\times t^2$
Or, $x=100-4.9\times t^2$

Let this be the equation (1).

Now, time taken by second stone to reach the distance x can be calculated as follows:

$x=ut+\frac{1}{2}g\times t^2$

Or, $x=25t-4.9\times t^2$

Let it be the equation (2)

From equation (1) and (2), we get;

25t = 100
Or, t = 4 s

Now, putting the value of t, in equation (2) we get;

$x=25\times 4–4.9\times 4^2$
Or, $x=100–4.9\times 16$
Or, $x=100–78.4=21.6m$

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; initial velocity u = ?, final velocity v = 0, t = 3s, s = ?

Initial velocity can be calculated as follows:

$v=u+g\times t$
Or, $0=u-9.8\times 3$
Or, $u=9.8\times 3=29.4m/s$

Now, maximum height can be calculated as follows:

$s=ut+\frac{1}{2}g\times t^2$

$=29.4\times 3-\frac{1}{2}\times 9.8\times 3^2$

$=88.2-44.1=44.1m$

Distance after 4 s can be calculated as follows:

$s=ut+\frac{1}{2}g\times t^2$

$=29.4\times 4-\frac{1}{2}\times 9.8\times 16$

$=117.6-78.4=39.2m$

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

Answer: Upward direction

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

Answer: Because of buoyancy

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?

Answer: Density = mass/volume

Hence, density of object = 50g/20 cm3
= 2.5 g cm-3

Here; density of object is more than the density of water.

Hence, the object will sink in water.

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?

Answer: Density = mass/volume

Hence, density of object = 500g/350 cm3
= 1.43 g cm-3

Here; density of object is more than the density of water.

Hence, the object will sink in water.

Thank you very much for reading carefully, if you have any other questions, you can share it with us through comments, if this information was important to you, please let us know through comments.

Please do comment and share.
Thank You.