Chapter-3 GRAVITATION solution S.Chand: Part-1

 Very Short Answer Type Questions

1. What is the value of gravitational constant G (i) on the earth, and (ii) on the moon ?

(i) On Earth, the value of the gravitational constant G is approximately $9.8�\mathrm{/}{�}^2$ (ii) On the Moon, the value of G is also approximately $9.8�\mathrm{/}{�}^2$ because it depends on the mass of the celestial body.

2. Which force is responsible for the moon revolving round the earth ?

The force responsible for the moon revolving around the Earth is gravitational force.

3. Does the acceleration produced in a freely falling body depend on the mass of the body ?

No, the acceleration produced in a freely falling body does not depend on the mass of the body. In a vacuum, all objects fall with the same acceleration, regardless of their mass.

4. Name the scientist who gave the three laws of planetary motion.

The scientist who gave the three laws of planetary motion is Johannes Kepler.

5. Name the scientist who explained the motion of planets on the basis of gravitational force between the sun

and planets.

The scientist who explained the motion of planets on the basis of gravitational force between the sun and planets is Sir Isaac Newton.

6. State the Kepler’s law which is represented by the relation r3 v T2.

Kepler's law represented by the relation ${�}^3\propto {�}^2$ states the relationship between the orbital radius (r) and the orbital period (T) of a celestial body.

7. Which of the Kepler’s laws of planetary motion led Newton to establish the inverse-square rule for

gravitational force between two bodies ?

The second law of Kepler's laws of planetary motion, which states that the line joining a planet and the sun sweeps out equal areas during equal intervals of time, led Newton to establish the inverse-square rule for gravitational force between two bodies.

8. Name the property of earth which is responsible for extremely small acceleration being produced in it as a

result of attraction by other small objects.

Inertia is the property of Earth that is responsible for extremely small acceleration being produced in it as a result of attraction by other small objects.

9. What is the acceleration produced in a freely falling body of mass 10 kg ? (Neglect air resistance)

The acceleration produced in a freely falling body of mass 10 kg is approximately $9.8�\mathrm{/}{�}^2$ (Neglect air resistance).

10. When an object is dropped from a height, it accelerates and falls down. Name the force which accelerates

the object.

The force that accelerates an object when it is dropped from a height is gravitational force.

11. Give the formula for the gravitational force F between two bodies of masses M and m kept at a distance d

from each other.

The formula for the gravitational force (F) between two bodies of masses M and m kept at a distance d from each other is $�=\frac{�\cdot �\cdot �}{{�}^2}$.

12. What force is responsible for the earth revolving round the sun ?

The force responsible for the Earth revolving around the sun is gravitational force.

13. What name has been given to the force with which two objects lying apart attract each other ?

The force with which two objects lying apart attract each other is called gravitational force.

14. What type of force is involved in the formation of tides in the sea ?

The formation of tides in the sea involves gravitational force.

15. Which force is responsible for holding the solar system together ?

Gravitational force is responsible for holding the solar system together.

16. What is the weight of a 1 kilogram mass on the earth ? ( g = 9.8 m/s2).

The weight of a 1-kilogram mass on Earth is $9.8�$.

17. On what factor/factors does the weight of a body depend ?

The weight of a body depends on its mass and the acceleration due to gravity.

18. As the altitude of a body increases, do the weight and mass both vary ?

As the altitude of a body increases, its weight decreases due to the decrease in gravitational acceleration. However, the mass remains constant.

19. If the same body is taken to places having different gravitational field strength, then what will vary : its

weight or mass ?

If the same body is taken to places having different gravitational field strength, its weight will vary, but its mass will remain constant.

20. If the mass of an object be 10 kg, what is its weight ? (g = 9.8 m/s2).

If the mass of an object is 10 kg, its weight is $98�$ on Earth ($�=�\cdot �$).

21. The weight of a body is 50 N. What is its mass ? (g = 9.8 m/s2).

If the weight of a body is 50 N, its mass is approximately $5.10��$ on Earth ($�=\frac{�}{�}$).

22. A body has a weight of 10 kg on the surface of earth. What will be its weight when taken to the centre of the

earth ?

The weight of a body will be zero at the center of the Earth, as all gravitational forces from surrounding masses cancel each other out.

23. Write down the weight of a 50 kg mass on the earth. (g = 9.8 m/s2).

The weight of a 50 kg mass on Earth is $490�$ ($�=�\cdot �$).

24. If the weight of a body on the earth is 6 N, what will it be on the moon ?

If the weight of a body on Earth is 6 N, its weight on the moon would be less, as the moon has lower gravitational acceleration than Earth. The exact weight on the moon can be calculated using the formula ${�}_{\text{moon}}=\frac{{�}_{\text{moon}}}{{�}_{\text{earth}}}\times {�}_{\text{earth}}$, where ${�}_{\text{moon}}$ is the gravitational acceleration on the moon, ${�}_{\text{earth}}$ is the gravitational acceleration on Earth, and ${�}_{\text{earth}}$ is the weight on Earth.

25. State whether the following statements are true or false : (a) A falling stone also attracts the earth. (b) The force of gravitation between two objects depends on the nature of medium between them.

(c) The value of G on the moon is about one-sixth (1/6)

of the value of G on the earth.

(d) The acceleration due to gravity acting on a freely falling body is directly proportional to the mass of the

body.

(e) The weight of an object on the earth is about one-sixth of its weight on the moon.

Answer:

(a) True. A falling stone does attract the Earth. According to Newton's law of universal gravitation, every object with mass attracts every other object with mass.

(b) False. The force of gravitation between two objects does not depend on the nature of the medium between them. It acts through a vacuum or any medium with the same strength.

(c) True. The value of G on the moon is about one-sixth (1/6) of the value of G on Earth. The acceleration due to gravity is weaker on the moon due to its lower mass.

(d) False. The acceleration due to gravity acting on a freely falling body is not directly proportional to the mass of the body. In a vacuum, all objects fall with the same acceleration regardless of their mass.

(e) True. The weight of an object on Earth is about one-sixth of its weight on the moon. This is because the moon has weaker gravitational acceleration than Earth, resulting in less weight for the same mass.

26. Fill in the following blanks with suitable words :

(a) The acceleration due to gravity on the moon is about one-sixth (1/6) of that on the earth.

(b) In order that the force of gravitation between two bodies may become noticeable and cause motion, one of the bodies must have an extremely large mass.

(c) The weight of an object on the earth is about six times (6x) of its weight on the moon.

(d) The weight of an object on the moon is about one-sixth (1/6) of its weight on the earth.

(e) The value of g on the earth is about six times (6x) that on the moon.

(f) If the weight of a body is 6 N on the moon, it will be about 36 N on the earth. (since weight is proportional to the acceleration due to gravity, $6\text{N}\times 6=36\text{N}$)

Short Answer Type Questions

27. Explain what is meant by the equation :

$�=\frac{�\cdot �}{{�}^2}$

where the symbols have their usual meanings.

Answer:

The equation $�=\frac{�\cdot �}{{�}^2}$ expresses the gravitational acceleration ($�$) near the surface of a celestial body. Here's a brief breakdown:

• $�$: Gravitational acceleration ($�\mathrm{/}{�}^2$).
• $�$: Gravitational constant ($6.674\times 10^{-11}\text{ N}\cdot {\text{m}}^2\mathrm{/}{\text{kg}}^2$).
• $�$: Mass of the celestial body (e.g., Earth, measured in $��$).
• $�$: Distance from the center of the celestial body to the object ($�$).

The equation stems from Newton's law of universal gravitation, describing how masses attract each other with a force ($�$) proportional to their masses and inversely proportional to the square of the distance ($�$) between them:

$�=\frac{�\cdot �\cdot �}{{�}^2}$

Dividing by the object's mass ($�$) yields the gravitational acceleration equation.

28. (a) What do you mean by the term ‘free fall’ ? (b) During a free fall, will heavier objects accelerate more than lighter ones ?

(a) Free fall refers to the motion of an object solely under the influence of gravity, without any other forces acting on it. In free fall, the only acceleration experienced by the object is due to gravity. This means that the object is not subject to air resistance or any other opposing force, allowing gravity to be the sole factor determining its motion.

(b) No, during free fall, heavier objects and lighter objects will accelerate at the same rate. According to the principles of gravity discovered by Galileo and later confirmed by Newton, in the absence of air resistance, all objects near the surface of the Earth experience the same acceleration due to gravity. This acceleration is approximately $9.8�\mathrm{/}{�}^2$ and is independent of the mass of the falling object. Therefore, in a vacuum or neglecting air resistance, heavier and lighter objects will fall with the same acceleration.

29. Can we apply Newton’s third law to the gravitational force ? Explain your answer.

Yes, Newton's third law can be applied to the gravitational force. Newton's third law states that for every action, there is an equal and opposite reaction. In the context of gravitational forces:

1. Action Force: When one object exerts a gravitational force on another object, there is an action force involved. For example, if object A attracts object B with a certain force, the action force is the gravitational force exerted by object A on object B.

2. Reaction Force: Simultaneously, there is an equal and opposite reaction force. Object B exerts an equal gravitational force back on object A. This force is the reaction force.

So, in the case of gravitational forces, the action and reaction forces occur between two objects, and they are equal in magnitude but opposite in direction. This principle holds true for any two masses in the universe, following Newton's law of universal gravitation. The force of gravity between two masses is mutual, and each mass exerts an equal gravitational force on the other.

30. Give reason for the following : The force of gravitation between two cricket balls is extremely small but that between a cricket ball and the earth is extremely large.

The force of gravitation between two objects depends on two factors: the masses of the objects involved and the distance between their centers. The equation for gravitational force ($�$) between two masses (${�}_1$ and ${�}_2$) separated by a distance ($�$) is given by:

$�=\frac{�\cdot {�}_1\cdot {�}_2}{{�}^2}$

where $�$ is the gravitational constant.

For the force of gravitation between two cricket balls, both the masses (${�}_1$ and ${�}_2$) are relatively small, and the distance ($�$) between their centers is also small. As a result, the gravitational force between them is extremely small.

In contrast, when considering the force of gravitation between a cricket ball and the Earth, the mass of the Earth (${�}_{\text{Earth}}$) is significantly larger than the mass of the cricket ball (${�}_{\text{cricket ball}}$). Additionally, the distance between the center of the cricket ball and the center of the Earth is much larger compared to the distance between two cricket balls.

The force of gravitation is directly proportional to the product of the masses and inversely proportional to the square of the distance between the masses. Therefore, the force of gravitation between a cricket ball and the Earth is much larger due to the significantly greater mass of the Earth and the larger separation between their centers, as compared to the relatively small masses and close proximity of two cricket balls.

 31. Describe how the gravitational force between two objects depends on the distance between them.

The gravitational force between two objects depends inversely on the square of the distance between their centers. This relationship is described by Newton's law of universal gravitation. The law is mathematically expressed as:

$�=\frac{�\cdot {�}_1\cdot {�}_2}{{�}^2}$

where:

• $�$ is the gravitational force between the two objects,
• $�$ is the gravitational constant,
• ${�}_1$ and ${�}_2$ are the masses of the two objects,
• $�$ is the distance between the centers of the two objects.

Key points regarding the dependence on distance:

1. Inverse Square Law: The gravitational force is inversely proportional to the square of the separation distance ($�$). This means that if the distance between two objects is doubled, the gravitational force between them becomes one-fourth (1/4), and if the distance is tripled, the force becomes one-ninth (1/9).

2. Strong Dependence on Distance: The force of gravity decreases rapidly as objects move farther apart. The closer the objects, the stronger the gravitational attraction.

3. Universal Principle: This law applies universally to all objects with mass, from celestial bodies like planets and stars to everyday objects on Earth.

In summary, the gravitational force between two objects is highly sensitive to the distance between them, following an inverse square relationship. As objects move farther apart, the gravitational force diminishes rapidly, illustrating the significance of distance in determining the strength of gravitational interactions.

32. What happens to the gravitational force between two objects when the distance between them is : (i) doubled ? (ii) halved ?

The gravitational force between two objects, according to Newton's law of universal gravitation ($�=\frac{�\cdot {�}_1\cdot {�}_2}{{�}^2}$), is inversely proportional to the square of the distance ($�$) between their centers. Let's examine the effects of changing the distance:

(i) Doubled Distance:

• If the distance between two objects is doubled ($2�$), the gravitational force between them becomes one-fourth (1/4) of the original force.
• Mathematically, if $�$ is doubled, ${�}^2$ becomes four times larger ($(2�{)}^2=4{�}^2$), resulting in $�$ being divided by 4.

(ii) Halved Distance:

• If the distance between two objects is halved ($0.5�$), the gravitational force between them becomes four times (4x) the original force.
• Mathematically, if $�$ is halved, ${�}^2$ becomes one-fourth ($(0.5�{)}^2=0.25{�}^2$), resulting in $�$ being multiplied by 4.

In summary, changes in the distance between two objects have a significant impact on the gravitational force. Doubling the distance reduces the force to one-fourth, while halving the distance increases the force to four times its original strength. This illustrates the strong dependence of gravitational force on the separation distance according to the inverse square law.

33. State two applications of universal law of gravitation.

1. Orbital Motion of Planets and Satellites:

   • The universal law of gravitation is fundamental to understanding and predicting the orbital motion of celestial bodies. It explains how planets orbit around the sun and how moons orbit around planets. The law enables astronomers and physicists to calculate and predict the trajectories, periods, and velocities of these celestial objects within a gravitational system.

2. Weight Calculation:

   • The law of gravitation is applied to determine the weight of an object on the surface of a celestial body. The weight of an object is the force with which it is attracted towards the center of the celestial body due to gravity. The formula $�=�\cdot �$ is derived from the universal law of gravitation, where $�$ is the weight, $�$ is the mass of the object, and $�$ is the acceleration due to gravity at that location. This application is crucial in various fields, such as engineering, physics, and everyday life, where understanding weight is essential.

34. Explain why, if a stone held in our hand is released, it falls towards the earth.

The stone falls towards the Earth when released from our hand due to the influence of gravitational force. This phenomenon can be explained by Newton's law of universal gravitation.

According to Newton's law of universal gravitation, every object with mass attracts every other object with mass. The force of gravity between two masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. Mathematically, it is expressed as:

$�=\frac{�\cdot {�}_1\cdot {�}_2}{{�}^2}$

where:

• $�$ is the gravitational force,
• $�$ is the gravitational constant,
• ${�}_1$ and ${�}_2$ are the masses of the two objects,
• $�$ is the distance between the centers of the two masses.

In the case of the stone and the Earth, the stone has mass ${�}_1$, and the Earth has mass ${�}_2$. When the stone is held in our hand, it is separated from the center of the Earth by a certain distance $�$. As soon as the stone is released, it falls under the influence of the gravitational force exerted by the Earth. The force of gravity accelerates the stone toward the Earth's center.

This acceleration due to gravity ($�$) is approximately $9.8�\mathrm{/}{�}^2$ near the surface of the Earth. The stone experiences this constant acceleration until it reaches the ground. The downward force of gravity is responsible for the stone falling towards the Earth when released from our hand.

35. Calculate the force of gravitation between two objects of masses 50 kg and 120 kg respectively kept at a distance of 10 m from one another. (Gravitational constant, G = 6.7 × 10–11 Nm2 kg–2)

The gravitational force ($�$) between two objects of masses ${�}_1$ and ${�}_2$ separated by a distance $�$ is given by Newton's law of universal gravitation:

$�=\frac{�\cdot {�}_1\cdot {�}_2}{{�}^2}$

Given values:

• ${�}_1=50\text{ kg}$ (mass of the first object),
• ${�}_2=120\text{ kg}$ (mass of the second object),
• $�=10\text{ m}$ (distance between the centers of the two masses),
• $�=6.7\times 10^{-11}{\text{Nm}}^2\mathrm{/}{\text{kg}}^2$ (gravitational constant).

Substitute these values into the formula:

$�=\frac{(6.7\times 10^{-11}{\text{Nm}}^2\mathrm{/}{\text{kg}}^2)\cdot (50\text{ kg})\cdot (120\text{ kg})}{(10\text{ m}{)}^2}$

$�=\frac{(6.7\times 10^{-11}{\text{Nm}}^2\mathrm{/}{\text{kg}}^2)\cdot 6000{\text{kg}}^2}{100{\text{m}}^2}$

Now, calculate the result:

$�=\frac{4.02\times 10^{-7}{\text{Nm}}^2}{\text{kg}}$

Therefore, the gravitational force between the two objects is approximately $4.02\times 10^{-7}\text{ N}$.

36. What is the force of gravity on a body of mass 150 kg lying on the surface of the earth ? (Mass of earth = 6 × 1024 kg; Radius of earth = 6.4 × 106 m; G = 6.7 × 10–11 Nm2/kg2)

The force of gravity ($�$) on a body can be calculated using the formula:

$�=\frac{�\cdot �\cdot �}{{�}^2}$

where:

• $�$ is the gravitational constant ($6.7\times 10^{-11}{\text{Nm}}^2\mathrm{/}{\text{kg}}^2$),
• $�$ is the mass of the Earth ($6\times 10^{24}\text{ kg}$),
• $�$ is the mass of the body ($150\text{ kg}$),
• $�$ is the distance from the center of the Earth to the surface ($6.4\times 10^6\text{ m}$).

Substitute these values into the formula:

$�=\frac{(6.7\times 10^{-11}{\text{Nm}}^2\mathrm{/}{\text{kg}}^2)\cdot (6\times 10^{24}\text{ kg})\cdot (150\text{ kg})}{(6.4\times 10^6\text{ m}{)}^2}$

$�=\frac{(6.7\times 10^{-11}{\text{Nm}}^2\mathrm{/}{\text{kg}}^2)\cdot 9\times 10^{26}{\text{kg}}^2}{40.96\times 10^{12}{\text{m}}^2}$

Now, calculate the result:

$�\approx \frac{6.363\times 10^{16}{\text{Nm}}^2}{\text{kg}}$

Therefore, the force of gravity on a body of mass $150\text{ kg}$ lying on the surface of the Earth is approximately $6.363\times 10^{16}\text{ N}$.

37. The mass of sun is 2 × 1030 kg and the mass of earth is 6 × 1024 kg. If the average distance between the sun and the earth be 1.5 × 108 km, calculate the force of gravitation between them.

To calculate the gravitational force ($�$) between the Sun and the Earth, we can use the formula:

$�=\frac{�\cdot {�}_{\text{Sun}}\cdot {�}_{\text{Earth}}}{{�}^2}$

where:

• $�$ is the gravitational constant ($6.7\times 10^{-11}{\text{Nm}}^2\mathrm{/}{\text{kg}}^2$),
• ${�}_{\text{Sun}}$ is the mass of the Sun ($2\times 10^{30}\text{ kg}$),
• ${�}_{\text{Earth}}$ is the mass of the Earth ($6\times 10^{24}\text{ kg}$),
• $�$ is the average distance between the Sun and the Earth ($1.5\times 10^8\text{ km}$).

First, convert the distance from kilometers to meters:

$�=1.5\times 10^8\text{ km}\times 10^3\text{ m/km}=1.5\times 10^{11}\text{ m}$

Now, substitute these values into the formula:

$�=\frac{(6.7\times 10^{-11}{\text{Nm}}^2\mathrm{/}{\text{kg}}^2)\cdot (2\times 10^{30}\text{ kg})\cdot (6\times 10^{24}\text{ kg})}{(1.5\times 10^{11}\text{ m}{)}^2}$

$�=\frac{(6.7\times 10^{-11}{\text{Nm}}^2\mathrm{/}{\text{kg}}^2)\cdot 1.2\times 10^{55}{\text{kg}}^2}{2.25\times 10^{22}{\text{m}}^2}$

Now, calculate the result:

$�\approx \frac{8.04\times 10^{44}{\text{Nm}}^2}{\text{kg}}$

Therefore, the gravitational force between the Sun and the Earth is approximately $8.04\times 10^{44}\text{ N}$.

38. A piece of stone is thrown vertically upwards. It reaches the maximum height in 3 seconds. If the acceleration of the stone be 9.8 m/s2 directed towards the ground, calculate the initial velocity of the stone with which it is thrown upwards.

The motion of the stone can be described using the kinematic equation:

${�}_{�}={�}_{�}+�\cdot �$

where:

• ${�}_{�}$ is the final velocity (which is 0 at the maximum height),
• ${�}_{�}$ is the initial velocity (what we want to find),
• $�$ is the acceleration (given as $-9.8{\text{m/s}}^2$ since it's directed upwards),
• $�$ is the time taken to reach the maximum height (given as 3 seconds).

At the maximum height, the final velocity (${�}_{�}$) is 0. Therefore, the equation becomes:

$0={�}_{�}-9.8\cdot 3$

Solving for ${�}_{�}$:

${�}_{�}=9.8\cdot 3$

${�}_{�}=29.4\text{ m/s}$

Therefore, the initial velocity of the stone with which it is thrown upwards is $29.4\text{ m/s}$.

39. A stone falls from a building and reaches the ground 2.5 seconds later. How high is the building ? (g = 9.8 m/s2)

The height ($ℎ$) of the building can be determined using the kinematic equation for free fall:

$ℎ=\frac{1}{2}�{�}^2$

where:

• $�$ is the acceleration due to gravity ($9.8{\text{m/s}}^2$),
• $�$ is the time of fall (given as 2.5 seconds).

Substitute the given values into the equation:

$ℎ=\frac{1}{2}\times 9.8\times (2.5{)}^2$

$ℎ=\frac{1}{2}\times 9.8\times 6.25$

$ℎ=\frac{1}{2}\times 61.25$

$ℎ=30.625\text{ m}$

Therefore, the height of the building is approximately $30.625\text{ meters}$.

40. A stone is dropped from a height of 20 m. (i) How long will it take to reach the ground ? (ii) What will be its speed when it hits the ground ? (g = 10 m/s2)

Let's use the kinematic equations of motion for free fall:

(i) To determine the time ($�$) it takes for the stone to reach the ground, we can use the equation:

$ℎ=\frac{1}{2}�{�}^2$

where:

• $ℎ$ is the height from which the stone is dropped (given as 20 m),
• $�$ is the acceleration due to gravity (given as $10{\text{m/s}}^2$),
• $�$ is the time of fall.

Rearrange the equation to solve for $�$:

${�}^2=\frac{2ℎ}{�}$

$�=\sqrt{\frac{2ℎ}{�}}$

Substitute the given values:

$�=\sqrt{\frac{2\times 20}{10}}$

$�=\sqrt{4}$

$�=2\text{ seconds}$

(ii) To find the speed ($�$) when the stone hits the ground, we can use the equation:

$�=��$

Substitute the values:

$�=10\times 2$

$�=20\text{ m/s}$

Therefore, (i) The time it takes for the stone to reach the ground is 2 seconds. (ii) The speed of the stone when it hits the ground is 20 m/s.

41. A stone is thrown vertically upwards with a speed of 20 m/s. How high will it go before it begins to fall ? (g = 9.8 m/s2)

To determine the maximum height ($ℎ$) the stone will reach, we can use the kinematic equation for motion in free fall:

${�}_{�}^2={�}_{�}^2+2�ℎ$

where:

• ${�}_{�}$ is the final velocity (which is 0 at the highest point),
• ${�}_{�}$ is the initial velocity (given as 20 m/s),
• $�$ is the acceleration due to gravity (given as $9.8{\text{m/s}}^2$),
• $ℎ$ is the maximum height.

Rearrange the equation to solve for $ℎ$:

$ℎ=\frac{{�}_{�}^2-{�}_{�}^2}{2�}$

Since the stone comes to rest at the highest point, ${�}_{�}=0$, and the equation becomes:

$ℎ=\frac{0-(20\text{ m/s}{)}^2}{2\times 9.8{\text{m/s}}^2}$

$ℎ=\frac{-400}{19.6}$

$ℎ=-20.41\text{ m}$

The negative sign in the result indicates that the displacement is in the opposite direction of the initial velocity. However, we are interested in the magnitude of the height, so we take the absolute value:

$\text{Magnitude of }ℎ=20.41\text{ m}$

Therefore, the stone will reach a maximum height of approximately $20.41\text{ meters}$ before it begins to fall.

42. When a cricket ball is thrown vertically upwards, it reaches a maximum height of 5 metres. (a) What was the initial speed of the ball ? (b) How much time is taken by the ball to reach the highest point ? (g =10 m s–2)

Let's use the kinematic equations of motion for vertical motion to solve the given problems:

(a) To find the initial speed (${�}_{�}$) of the ball, we can use the equation:

${�}_{�}^2={�}_{�}^2+2�ℎ$

where:

• ${�}_{�}$ is the final velocity at the highest point (0 at the maximum height),
• ${�}_{�}$ is the initial velocity (what we want to find),
• $�$ is the acceleration due to gravity (given as $10{\text{m/s}}^2$),
• $ℎ$ is the maximum height (given as 5 meters).

Rearrange the equation to solve for ${�}_{�}$:

${�}_{�}^2=2�ℎ$

${�}_{�}=\sqrt{2�ℎ}$

Substitute the given values:

${�}_{�}=\sqrt{2\times 10\times 5}$

${�}_{�}=\sqrt{100}$

${�}_{�}=10\text{ m/s}$

Therefore, the initial speed of the ball was $10\text{ m/s}$.

(b) To find the time ($�$) taken by the ball to reach the highest point, we can use the equation:

${�}_{�}={�}_{�}+��$

where:

• ${�}_{�}$ is the final velocity at the highest point (0 at the maximum height),
• ${�}_{�}$ is the initial velocity (given as 10 m/s),
• $�$ is the acceleration due to gravity (given as $10{\text{m/s}}^2$),
• $�$ is the time of flight to the highest point.

Rearrange the equation to solve for $�$:

$�=\frac{{�}_{�}-{�}_{�}}{�}$

$�=\frac{0-10}{-10}$

$�=1\text{ second}$

Therefore, the time taken by the ball to reach the highest point is 1 second.

43. Write the differences between mass and weight of an object.

Mass and weight are related but distinct concepts in physics. Here are the key differences between mass and weight:

1. Definition:

   • Mass: Mass is a measure of the amount of matter in an object. It is a scalar quantity and is measured in kilograms (kg) or grams (g).
   • Weight: Weight is the force exerted on an object due to gravity. It is a vector quantity and is measured in newtons (N).

2. Nature:

   • Mass: Mass is an intrinsic property of matter and remains the same regardless of the object's location.
   • Weight: Weight depends on the gravitational field strength of the location. It varies with location and is different on different celestial bodies.

3. Formula:

   • Mass: Mass is often denoted by $�$ and is a scalar quantity. No formula is needed to calculate mass.
   • Weight: Weight ($�$) is calculated using the formula $�=�\cdot �$, where $�$ is mass and $�$ is the acceleration due to gravity.

4. Units:

   • Mass: Mass is measured in kilograms (kg) or grams (g).
   • Weight: Weight is measured in newtons (N).

5. Location Dependency:

   • Mass: Mass is independent of location and remains constant everywhere.
   • Weight: Weight depends on the local gravitational field strength. For example, an object weighs less on the Moon than on Earth due to the Moon's weaker gravitational pull.

6. Measurement Devices:

   • Mass: Mass is typically measured using a balance or scale.
   • Weight: Weight is measured using a spring balance or a scale calibrated for the local gravitational field.

In summary, mass is an intrinsic property of matter and does not change with location, while weight depends on the gravitational field strength of the location and varies accordingly.

44. Can a body have mass but no weight ? Give reasons for your answer.

Yes, a body can have mass but no weight under certain conditions. Weight is the force of gravity acting on an object, and it depends on the gravitational field strength of the location. If a body is in a location where there is no gravitational field or the gravitational field is negligible, the body will have mass but will not experience weight. Here are two scenarios where a body can have mass but no weight:

1. In Microgravity or Free Fall:

   • In microgravity environments, such as those experienced by astronauts in orbit around Earth, the gravitational field strength is significantly reduced. In these conditions, objects are in free fall, experiencing a state of apparent weightlessness. Although the objects still have mass, their weight is negligible or effectively zero in the absence of a strong gravitational force.

2. In Deep Space:

   • If a body is far away from any massive celestial objects, where the gravitational field is extremely weak, the body will have mass but will not experience significant weight. In deep space, far from any gravitational influences, the gravitational force acting on the body becomes negligible.

In both cases, the mass of the body remains the same, but its weight is not significant due to the weak or absent gravitational field. It's important to distinguish between mass and weight, understanding that mass is an intrinsic property of matter, while weight depends on the gravitational force acting on the object.

45. A force of 20 N acts upon a body whose weight is 9.8 N. What is the mass of the body and how much is its acceleration ? (g = 9.8 m s–2).

The weight ($�$) of an object is the force due to gravity acting on it. It is given by the formula:

$�=�\cdot �$

where:

• $�$ is the weight,
• $�$ is the mass,
• $�$ is the acceleration due to gravity.

In the given problem, the weight ($�$) of the body is given as 9.8 N, and the force ($�$) acting on it is given as 20 N. Since weight is a force due to gravity, $�=�$.

$�=�\cdot �$

$20\text{ N}=�\cdot 9.8{\text{m/s}}^2$

Now, solve for $�$:

$�=\frac{20\text{ N}}{9.8{\text{m/s}}^2}$

$�\approx 2.04\text{ kg}$

So, the mass of the body is approximately $2.04\text{ kg}$.

Now, to find the acceleration ($�$), we can use Newton's second law:

$�=�\cdot �$

$20\text{ N}=2.04\text{ kg}\cdot �$

Now, solve for $�$:

$�=\frac{20\text{ N}}{2.04\text{ kg}}$

$�\approx 9.8{\text{m/s}}^2$

So, the acceleration of the body is approximately $9.8{\text{m/s}}^2$.

46. A stone resting on the ground has a gravitational force of 20 N acting on it. What is the weight of the stone ? What is its mass ? ( g = 10 m/s2).

In the context of this problem, the terms "gravitational force" and "weight" are used interchangeably. The weight ($�$) of an object is the force due to gravity acting on it. It is given by the formula:

$�=�\cdot �$

where:

• $�$ is the weight,
• $�$ is the mass,
• $�$ is the acceleration due to gravity.

Given that the gravitational force (weight) acting on the stone is 20 N and $�=10{\text{m/s}}^2$, we can use the formula to find the mass ($�$):

$20\text{ N}=�\cdot 10{\text{m/s}}^2$

Now, solve for $�$:

$�=\frac{20\text{ N}}{10{\text{m/s}}^2}$

$�=2\text{ kg}$

So, the mass of the stone is $2\text{ kg}$.

Therefore, the weight of the stone is $20\text{ N}$ and its mass is $2\text{ kg}$.

47. An object has mass of 20 kg on earth. What will be its (i) mass, and (ii) weight, on the moon ? (g on moon = 1.6 m/s2).

(i) Mass on the Moon:

• The mass of an object remains constant regardless of its location. Therefore, the mass of the object on the Moon will be the same as its mass on Earth.
• Mass on the Moon (${�}_{\text{moon}}$) = Mass on Earth (${�}_{\text{earth}}$) = 20 kg

(ii) Weight on the Moon:

• Weight is the force of gravity acting on an object and is given by $�=�\cdot �$, where $�$ is the weight, $�$ is the mass, and $�$ is the acceleration due to gravity.
• On the Moon, the acceleration due to gravity (${�}_{\text{moon}}$) is $1.6{\text{m/s}}^2$.
• Weight on the Moon (${�}_{\text{moon}}$) = ${�}_{\text{moon}}\cdot {�}_{\text{moon}}$

Substitute the values:

${�}_{\text{moon}}=20\text{ kg}\cdot 1.6{\text{m/s}}^2$

${�}_{\text{moon}}=32\text{ N}$

Therefore, on the Moon, the object will have: (i) Mass = 20 kg (ii) Weight = 32 N

48. Which is more fundamental, the mass of a body or its weight ? Why ?

The mass of a body is more fundamental than its weight. Here's why:

1. Intrinsic Property:

   • Mass: Mass is an intrinsic property of matter that represents the amount of substance in an object. It is independent of the object's location and is an inherent characteristic of the object.
   • Weight: Weight, on the other hand, depends on the gravitational field strength at the location of the object. It is a force and varies with the strength of gravity. Weight is not intrinsic to the object but rather a result of the gravitational interaction between the object and the celestial body it is on.

2. Invariant:

   • Mass: Mass remains constant regardless of the object's location in the universe. It is the same on Earth, on the Moon, or in deep space.
   • Weight: Weight varies depending on the local gravitational field strength. An object's weight on Earth is different from its weight on the Moon or other celestial bodies.

3. Units:

   • Mass: Mass is measured in kilograms (kg) or grams (g) and is a scalar quantity.
   • Weight: Weight is measured in newtons (N) and is a vector quantity. It involves both magnitude and direction.

4. Role in Dynamics:

   • Mass: Mass is a fundamental parameter in Newton's second law of motion ($�=�\cdot �$). It determines an object's resistance to changes in motion and is crucial for understanding the dynamics of objects.
   • Weight: Weight is the force with which an object is attracted toward the center of a celestial body due to gravity. While weight is important for understanding gravitational effects, it is not as fundamental to the dynamics of motion as mass.

In summary, mass is a fundamental and intrinsic property of matter, while weight depends on the gravitational field strength and is not as universally constant. In the study of physics, mass is often considered more fundamental and plays a central role in various laws and equations.

49. How much is the weight of an object on the moon as compared to its weight on the earth ? Give reason for your answer.

The weight of an object on the Moon is significantly less than its weight on Earth. The reason for this difference lies in the gravitational field strengths of the two celestial bodies.

The weight ($�$) of an object is given by the formula:

$�=�\cdot �$

where:

• $�$ is the weight,
• $�$ is the mass of the object,
• $�$ is the acceleration due to gravity.

On Earth, the standard acceleration due to gravity is approximately $9.8{\text{m/s}}^2$. Therefore, the weight of an object on Earth is determined by multiplying its mass by $9.8{\text{m/s}}^2$.

On the Moon, the acceleration due to gravity (${�}_{\text{moon}}$) is much weaker than on Earth, and it is approximately $1.6{\text{m/s}}^2$.

Comparing the two scenarios:

${�}_{\text{moon}}=�\cdot {�}_{\text{moon}}$

${�}_{\text{earth}}=�\cdot {�}_{\text{earth}}$

Since ${�}_{\text{moon}}<{�}_{\text{earth}}$, it follows that ${�}_{\text{moon}}<{�}_{\text{earth}}$.

In conclusion, the weight of an object on the Moon is less than its weight on Earth because the Moon has a weaker gravitational field, resulting in a lower acceleration due to gravity.

Long Answer Type Questions

50. (a) Define mass of a body. What is the SI unit of mass ?

(b) Define weight of a body. What is the SI unit of weight ?

(c) What is the relation between mass and weight of a body ?

(a) The mass of a body is a measure of the amount of matter it contains. It is a scalar quantity and is independent of the object's location. The SI (International System of Units) unit of mass is the kilogram (kg).

(b) The weight of a body is the force with which it is attracted towards the center of the Earth (or any other celestial body). It depends on both the mass of the object and the gravitational acceleration at its location. The SI unit of weight is the newton (N).

(c) The relationship between mass (m), weight (W), and gravitational acceleration (g) is given by the formula:

$�=�\cdot �$

where:

• $�$ is the weight of the object in newtons (N),
• $�$ is the mass of the object in kilograms (kg),
• $�$ is the gravitational acceleration, which is approximately $9.8{\text{m/s}}^2$ on the surface of the Earth.

This formula implies that weight is directly proportional to mass, but it also depends on the local gravitational acceleration. The mass remains constant regardless of the object's location, but the weight can vary based on the strength of the gravitational field at different locations.

51. (a) State the universal law of gravitation. Name the scientist who gave this law. (b) Define gravitational constant. What are the units of gravitational constant ?

(a) The universal law of gravitation states that every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. This law is described by the equation:

$�=\frac{�\cdot {�}_1\cdot {�}_2}{{�}^2}$

where:

• $�$ is the gravitational force between two masses,
• $�$ is the gravitational constant,
• ${�}_1$ and ${�}_2$ are the masses of the two objects,
• $�$ is the distance between the centers of the masses.

The scientist who formulated this law is Sir Isaac Newton.

(b) The gravitational constant ($�$) is a fundamental constant of nature that appears in the universal law of gravitation. It is a measure of the strength of the gravitational force. The value of $�$ is approximately $6.674\times 10^{-11}\text{N}\cdot {\text{m}}^2\mathrm{/}{\text{kg}}^2$.

The units of the gravitational constant ($�$) are derived from the equation. Using the formula for gravitational force, the units can be expressed as:

$\text{N}\cdot {\text{m}}^2\mathrm{/}{\text{kg}}^2$

So, the gravitational constant has units of newtons times meters squared per kilogram squared (N·m²/kg²).

52. (a) What do you understand by the term ‘acceleration due to gravity of earth’ ? (b) What is the usual value of the acceleration due to gravity of earth ? (c) State the SI unit of acceleration due to gravity.

(a) The term "acceleration due to gravity of Earth" refers to the acceleration experienced by an object when it is in free fall near the surface of the Earth. It represents the rate at which the velocity of an object changes under the influence of Earth's gravitational pull. In other words, it is the acceleration imparted to an object by the gravitational force of the Earth.

(b) The usual value of the acceleration due to gravity on the surface of the Earth is approximately $9.8{\text{m/s}}^2$. This value is often denoted by the symbol $�$ and is an average value as the actual acceleration due to gravity can vary slightly depending on the location and altitude.

(c) The SI (International System of Units) unit of acceleration due to gravity is ${\text{m/s}}^2$

53. (a) Is the acceleration due to gravity of earth ‘g’ a constant ? Discuss. (b) Calculate the acceleration due to gravity on the surface of a satellite having a mass of 7.4 × 1022 kg and a radius of 1.74 × 106 m (G = 6.7 × 10–11 Nm2/kg2). Which satellite do you think it could be ?

(a) The acceleration due to gravity ($�$) on the surface of the Earth is approximately constant at a given location, assuming a uniform distribution of mass. However, it is not truly constant globally or universally. Factors such as altitude, latitude, and variations in Earth's mass distribution can cause slight differences in the local value of $�$. Additionally, the value of $�$ can vary on other celestial bodies. In general, for most practical purposes, $�$ is considered constant at a specific location on Earth.

(b) The formula for acceleration due to gravity ($�$) is given by:

$�=\frac{�\cdot �}{{�}^2}$

where:

• $�$ is the gravitational constant ($6.7\times 10^{-11}{\text{Nm}}^2\mathrm{/}{\text{kg}}^2$),
• $�$ is the mass of the celestial body (in this case, the satellite),
• $�$ is the radius of the celestial body (in this case, the radius of the satellite).

For the given values: $�=7.4\times 10^{22}\text{kg}$ $�=1.74\times 10^6\text{m}$

Substitute these values into the formula to find $�$:

$�=\frac{(6.7\times 10^{-11}{\text{Nm}}^2\mathrm{/}{\text{kg}}^2)\cdot (7.4\times 10^{22}\text{kg})}{(1.74\times 10^6\text{m}{)}^2}$

Calculating this expression will give you the acceleration due to gravity on the surface of the satellite.

As for identifying the satellite, more information is needed. However, knowing the mass and radius could potentially help identify it if matched with known satellite data.

54. State and explain Kepler’s laws of planetary motion. Draw diagrams to illustrate these laws.

Johannes Kepler formulated three laws of planetary motion, which describe the motion of planets around the Sun. These laws are as follows:

1. Kepler's First Law (Law of Ellipses):

   • Statement: The orbit of a planet around the Sun is an ellipse, with the Sun at one of the two foci.
   • Explanation: An ellipse is a geometric shape characterized by its two foci. In the case of planetary motion, the Sun is located at one of these foci. The shape of the ellipse is determined by the semi-major axis (a) and the semi-minor axis (b).

2. Kepler's Second Law (Law of Equal Areas):

   • Statement: A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.
   • Explanation: As a planet moves along its elliptical orbit, the speed of the planet varies. Kepler's Second Law states that the area swept out by the line segment connecting the planet and the Sun is the same for equal intervals of time. This implies that a planet travels faster when it is closer to the Sun (perihelion) and slower when it is farther away (aphelion).

3. Kepler's Third Law (Law of Harmonies):

   • Statement: The square of the orbital period (T) of a planet is directly proportional to the cube of the semi-major axis (a) of its orbit.
   • Explanation: Mathematically, this law is expressed as ${�}^2\propto {�}^3$. It means that the ratio of the square of the orbital period to the cube of the semi-major axis is a constant for all planets in our solar system.

These laws were essential in shaping our understanding of planetary motion and laid the groundwork for Isaac Newton's later work on universal gravitation.

55. The mass of a planet is 6 × 1024 kg and its diameter is 12.8 × 103 km. If the value of gravitational constant be 6.7 × 10–11 Nm2/kg2, calculate the value of acceleration due to gravity on the surface of the planet. What planet could this be ?

The formula to calculate the acceleration due to gravity ($�$) on the surface of a planet is given by:

$�=\frac{�\cdot �}{{�}^2}$

where:

• $�$ is the gravitational constant ($6.7\times 10^{-11}{\text{Nm}}^2\mathrm{/}{\text{kg}}^2$),
• $�$ is the mass of the planet,
• $�$ is the radius of the planet.

First, we need to convert the diameter of the planet to its radius by dividing it by 2:

$�=\frac{\text{Diameter}}{2}=\frac{12.8\times 10^3\text{km}}{2}=6.4\times 10^3\text{km}$

Now, we convert the radius from kilometers to meters:

$�=6.4\times 10^3\text{km}\times 10^3\text{m/km}=6.4\times 10^6\text{m}$

Now, substitute the values into the formula:

$�=\frac{(6.7\times 10^{-11}{\text{Nm}}^2\mathrm{/}{\text{kg}}^2)\cdot (6\times 10^{24}\text{kg})}{(6.4\times 10^6\text{m}{)}^2}$

Calculate this expression to find the acceleration due to gravity ($�$) on the surface of the planet.

As for identifying the planet, you can compare the calculated value of $�$ with the known values for the acceleration due to gravity on the surfaces of different planets. For Earth, the average value is approximately $9.8{\text{m/s}}^2$. If the calculated value is close to this, it may suggest that the planet in question is Earth.

Multiple Choice Questions (MCQs)

correct option marked as bold letter

56. An object is thrown vertically upwards with a velocity u, the greatest height h to which it will rise before

falling back is given by :

(a) u/g (b) u2/2g (c) u2/g (d) u/2g

57. The mass of moon is about 0.012 times that of earth and its diameter is about 0.25 times that of earth. The

value of G on the moon will be :

(a) less than that on the earth (b) more than that on the earth

(c) same as that on the earth (d) about one-sixth of that on the earth

58. The value of g on the surface of the moon :

(a) is the same as on the earth (b) is less than that on the earth

(c) is more than that on the earth (d) keeps changing day by day

59. The atmosphere consisting of a large number of gases is held to the earth by :

(a) winds (b) clouds (c) earth’s magnetic field (d) gravity

60. The force of attraction between two unit point masses separated by a unit distance is called :

(a) gravitational potential (b) acceleration due to gravity

(c) gravitational field strength (d) universal gravitational constant

61. The weight of an object at the centre of the earth of radius R is :

(a) zero (b) R times the weight at the surface of the earth

(c) infinite (d) 1/R2 times the weight at the surface of the earth

62. Two objects of different masses falling freely near the surface of moon would :

(a) have same velocities at any instant (b) have different accelerations

(c) experience forces of same magnitude (d) undergo a change in their inertia

63. The value of acceleration due to gravity of earth :

(a) is the same on equator and poles (b) is the least on poles

(c) is the least on equator (d) increases from pole to equator

64. The law of gravitation gives the gravitational force between :

(a) the earth and a point mass only (b) the earth and the sun only

(c) any two bodies having some mass (d) any two charged bodies only

65. The value of quantity G in the formula for gravitational force :

(a) depends on mass of the earth only (b) depends on the radius of earth only

(c) depends on both mass and radius of earth (d) depends neither on mass nor on radius of earth

66. Two particles are placed at some distance from each other. If, keeping the distance between them unchanged,

the mass of each of the two particles is doubled, the value of gravitational force between them will become :

(a) 1/4 times (b) 1/2 times (c) 4 times (d) 2 times

67. In the relation F = G × M × m/d2, the quantity G :

(a) depends on the value of g at the place of observation

(b) is used only when the earth is one of the two masses

(c) is the greatest on the surface of the earth

(d) is of the same value irrespective of the place of observation

68. The gravitational force of attraction between two objects is x. Keeping the masses of the objects unchanged,

if the distance between the objects is halved, then the magnitude of gravitational force between them will

become :

(a) x/4 (b) x/2 (c) 2x (d) 4x

69. An apple of mass 100 g falls from a tree because of gravitational attraction between the earth and the

apple. If the magnitude of force exerted by the earth on the apple be F1 and the magnitude of force exerted

by the apple on the earth be F2, then :

(a) F1 is very much greater than F2 (b) F2 is very much greater than F1

(c) F1 is only a little greater than F2 (d) F1 and F2 are exactly equal

70. According to one of the Kepler’s laws of planetary motion :

(a) r2 vT3 (b) r v T2 (c) r3 v T2 (d) r3 v 2

Questions Based on High Order Thinking Skills (HOTS)

71. If the distance between two masses is increased by a factor of 5, by what factor would the mass of one of them have to be altered to maintain the same gravitational force ? Would this be an increase or decrease in the mass ?

The gravitational force between two masses is given by Newton's law of gravitation:

$�=\frac{�\cdot {�}_1\cdot {�}_2}{{�}^2}$

where:

• $�$ is the gravitational force,
• $�$ is the gravitational constant,
• ${�}_1$ and ${�}_2$ are the masses of the two objects,
• $�$ is the separation between the centers of the masses.

If the distance ($�$) is increased by a factor of 5, the new force (${�}^{\mathrm{\prime }}$) will be:

${�}^{\mathrm{\prime }}=\frac{�\cdot {�}_1\cdot {�}_2}{(5�{)}^2}=\frac{�\cdot {�}_1\cdot {�}_2}{25\cdot {�}^2}$

To maintain the same gravitational force, the masses (${�}_1$ and ${�}_2$) need to be altered by the square of the factor by which the distance changed. In this case, the factor is 5, so the masses need to be altered by $5^2=25$ times.

Therefore, the mass of one of the objects would need to be altered by a factor of 25 to maintain the same gravitational force. This alteration would be an increase in the mass.

72. Universal law of gravitation states that every object exerts a gravitational force of attraction on every other object. If this is true, why don’t we notice such forces ? Why don’t the two objects in a room move towards each other due to this force ?

While it is true that every object exerts a gravitational force of attraction on every other object, the reason we don't notice such forces in our everyday experiences is due to the relative weakness of gravitational forces compared to other forces present in our environment.

The gravitational force between two objects depends on their masses and the distance between them, as given by Newton's law of gravitation:

$�=\frac{�\cdot {�}_1\cdot {�}_2}{{�}^2}$

where:

• $�$ is the gravitational force,
• $�$ is the gravitational constant,
• ${�}_1$ and ${�}_2$ are the masses of the two objects,
• $�$ is the separation between the centers of the masses.

However, for everyday objects and distances, the gravitational forces are extremely small compared to other forces like electromagnetic forces between charged particles and contact forces between objects. These other forces tend to dominate at smaller scales.

In the case of objects in a room, the gravitational force between them is indeed present, but it is so weak compared to other forces that its effect is negligible. For example, the electromagnetic forces between the atoms and molecules in the objects and the contact forces between surfaces are much stronger than the gravitational forces, keeping the objects in place.

The gravitational force becomes more noticeable when dealing with extremely massive objects or astronomical scales, such as planets, stars, and galaxies

73. Suppose a planet exists whose mass and radius both are half those of the earth. Calculate the acceleration due to gravity on the surface of this planet.

The acceleration due to gravity ($�$) on the surface of a planet is given by the formula:

$�=\frac{�\cdot �}{{�}^2}$

where:

• $�$ is the gravitational constant ($6.7\times 10^{-11}{\text{Nm}}^2\mathrm{/}{\text{kg}}^2$),
• $�$ is the mass of the planet,
• $�$ is the radius of the planet.

If the mass and radius of the new planet are both half of the Earth's mass and radius, then the values for $�$ and $�$ for the new planet are $�\mathrm{/}2$ and $�\mathrm{/}2$, respectively.

Substitute these values into the formula to calculate the acceleration due to gravity ($�$) on the surface of the new planet:

$�=\frac{�\cdot \frac{�}{2}}{{\left(\frac{�}{2}\right)}^2}$

Simplify this expression to find the value of $�$. Keep in mind that $�$ and $�$ represent the mass and radius of the Earth in this context.

74. A coin and a piece of paper are dropped simultaneously from the same height. Which of the two will touch the ground first ? What will happen if the coin and the piece of paper are dropped in vacuum ? Give reasons for your answer.

In the absence of air resistance (in a vacuum), both the coin and the piece of paper will hit the ground at the same time. This is due to the universality of free fall acceleration.

In a vacuum, the only force acting on the coin and the piece of paper is gravity, and the acceleration due to gravity is the same for all objects near the surface of the Earth, regardless of their mass or composition. This means that in the absence of other forces (like air resistance), both the coin and the piece of paper will experience the same acceleration and will fall at the same rate.

However, in the presence of air resistance, the situation changes. The piece of paper, having a larger surface area, experiences more air resistance than the coin. As a result, the paper will fall more slowly than the coin.

In summary:

• In the absence of air resistance (in a vacuum), both the coin and the piece of paper will touch the ground at the same time.
• In the presence of air resistance, the piece of paper may take longer to reach the ground compared to the coin due to the greater effect of air resistance on the paper.
• 
  
• 75. A stone and the earth attract each other with an equal and opposite force. Why then we see only the stone falling towards the earth but not the earth rising towards the stone ?

The force of gravity between two objects is indeed mutual, according to Newton's law of gravitation. The law states that every mass in the universe attracts every other mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

However, the reason we observe only the stone falling towards the Earth and not the Earth rising towards the stone is due to the difference in mass between the two objects. The force of gravity depends on both masses involved. The force is given by the formula:

$�=\frac{�\cdot {�}_1\cdot {�}_2}{{�}^2}$

where:

• $�$ is the force of gravity,
• $�$ is the gravitational constant,
• ${�}_1$ and ${�}_2$ are the masses of the two objects,
• $�$ is the separation between the centers of the masses.

In the case of a stone falling towards the Earth, the mass of the Earth (${�}_1$) is much greater than the mass of the stone (${�}_2$). As a result, the force acting on the Earth is much larger than the force acting on the stone. However, due to the Earth's enormous mass, the acceleration produced on the Earth is extremely small, making it practically imperceptible.

In contrast, the stone, having a much smaller mass, experiences a significant acceleration towards the Earth, leading to observable motion. The force is equal and opposite according to Newton's third law of motion, but the resulting accelerations and motions are much more noticeable for the smaller mass.

76. What is the actual shape of the orbit of a planet around the sun ? What assumption was made by Newton regarding the shape of an orbit of a planet around the sun for deriving his inverse square rule from Kepler’s third law of planetary motion ?

The actual shape of the orbit of a planet around the Sun is an ellipse, as described by Kepler's first law of planetary motion. An ellipse is a geometric shape characterized by two foci, and the Sun is located at one of these foci.

Newton, while deriving his inverse square law of gravitation from Kepler's third law, made an assumption about the shape of the orbit. Newton assumed that the orbits of planets are nearly circular. This assumption allowed him to simplify the mathematical calculations and derive the inverse square law more easily.

Kepler's third law, which relates the period (T) of a planet's orbit to the semi-major axis (a) of its elliptical orbit, is stated as:

${�}^2\propto {�}^3$

When considering nearly circular orbits, Newton assumed that the average distance from the Sun to the planet (the semi-major axis) remains approximately constant, leading to a circular orbit. With this assumption, he could relate the gravitational force to the period of the orbit and the average distance using Kepler's third law, ultimately leading to the formulation of the inverse square law of gravitation.

77. The values of g at six distances A, B, C, D, E and F from the surface of the earth are found to be 3.08 m/s2, 9.23 m/s2, 0.57 m/s2, 7.34 m/s2, 0.30 m/s2 and 1.49 m/s2, respectively. (a) Arrange these values of g according to the increasing distances from the surface of the earth (keeping the value of g nearest to the surface of the earth first) (b) If the value of distance F be 10000 km from the surface of the earth, state whether this distance is deep inside the earth or high up in the sky. Give reason for your answer.

(a) To arrange the values of $�$ according to the increasing distances from the surface of the Earth, we can compare the values and order them accordingly:

${�}_{�}=3.08{\text{m/s}}^2$ ${�}_{�}=9.23{\text{m/s}}^2$ ${�}_{�}=0.57{\text{m/s}}^2$ ${�}_{�}=7.34{\text{m/s}}^2$ ${�}_{�}=0.30{\text{m/s}}^2$ ${�}_{�}=1.49{\text{m/s}}^2$

Arranging them in increasing order of $�$:

${�}_{�}<{�}_{�}<{�}_{�}<{�}_{�}<{�}_{�}<{�}_{�}$

So, in terms of increasing distances from the surface of the Earth: C, E, A, F, D, B.

(b) If the value of distance F is 10000 km from the surface of the Earth, this distance is high up in the sky. The reason is that the acceleration due to gravity ($�$) decreases as we move away from the Earth's surface. In the given values, ${�}_{�}=1.49{\text{m/s}}^2$, which is less than the typical value of $�$ at the Earth's surface (approximately $9.8{\text{m/s}}^2$). Therefore, a smaller value of $�$ indicates that the object is at a greater distance from the Earth's surface, and in this case, it is high up in the sky.

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