Chapter Notes

Gravitation

We have learned about motion and force as the cause of motion. A force is needed to change the speed or direction of motion of an object. We observe that an object dropped from a height falls towards the Earth. All the planets go around the Sun, and the Moon goes around the Earth. In each of these cases, a force must be acting on the objects. Isaac Newton realized that the same force is responsible for all of these phenomena. This force is called gravitational force.

In this chapter, we will explore gravitation and the universal law of gravitation. We will discuss the motion of objects under the influence of gravitational force on the Earth, how the weight of a body varies from place to place, and the conditions for objects to float in liquids.

Gravitation

The Moon goes around the Earth, and an object thrown upwards reaches a certain height and then falls downwards. Newton thought that if the Earth can attract an apple, can it not attract the Moon? He believed the same force is responsible in both cases. He reasoned that at each point in its orbit, the Moon falls towards the Earth instead of going off in a straight line, so it must be attracted by the Earth. However, we don't see the Moon actually falling towards the Earth.

To understand this, consider a stone tied to a thread being whirled around. The stone moves in a circular path with a certain speed, constantly changing direction. This change in direction involves acceleration. The force causing this acceleration and keeping the stone moving along the circular path acts towards the center. This force is called centripetal force, which means 'center-seeking' force.

In the absence of centripetal force, the stone would fly off along a straight line, tangent to the circular path.

The motion of the Moon around the Earth is due to centripetal force, which is provided by the Earth's gravitational attraction. Without this force, the Moon would move in a uniform straight line.

A falling apple is attracted towards the Earth. Does the apple attract the Earth? According to Newton's third law of motion, it does. However, according to the second law of motion, acceleration is inversely proportional to mass for a given force. The mass of an apple is tiny compared to the Earth's mass, so we don't see the Earth moving towards the apple. The same applies to why the Earth doesn't visibly move towards the Moon.

All the planets in our solar system go around the Sun. This suggests a force exists between the Sun and the planets. Newton concluded that all objects in the universe attract each other. This force of attraction is the gravitational force.

Universal law of gravitation

Every object in the universe attracts every other object with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. The force acts along the line joining the centers of the two objects.

Consider two objects, A and B, with masses $M$ and $m$, respectively, separated by a distance $d$. Let $F$ be the force of attraction between them. According to the universal law of gravitation:

The force is directly proportional to the product of their masses: $F \propto M \times m$

The force is inversely proportional to the square of the distance between them: $F \propto \frac{1}{d^2}$

Combining these proportionalities: $F \propto \frac{M \times m}{d^2}$

This can be written as an equation: $F = G \frac{M \times m}{d^2}$

where $G$ is the universal gravitation constant. $G = \frac{F d^2}{M \times m}$

The SI unit of $G$ is $\text{N m}^2 \text{ kg}^{-2}$, obtained by substituting the units of force, distance, and mass.

The value of $G$ was determined by Henry Cavendish and is approximately $6.673 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$.

Saying that $F$ is inversely proportional to the square of $d$ means that if $d$ increases by a factor of 6, $F$ becomes $\frac{1}{36}$ times smaller.

Given

• Mass of the Earth, $M = 6 \times 10^{24} \text{ kg}$
• Mass of the Moon, $m = 7.4 \times 10^{22} \text{ kg}$
• Distance between the Earth and the Moon, $d = 3.84 \times 10^{5} \text{ km} = 3.84 \times 10^{8} \text{ m}$
• Gravitational constant, $G = 6.7 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$

To Find

Force exerted by the Earth on the Moon, $F$

Formula

$F = G \frac{M \times m}{d^2}$

Solution

Substitute the given values into the formula:

$F = \frac{6.7 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2} \times 6 \times 10^{24} \text{ kg} \times 7.4 \times 10^{22} \text{ kg}}{(3.84 \times 10^{8} \text{ m})^2}$

$F = 2.02 \times 10^{20} \text{ N}$

Final Answer The force exerted by the Earth on the Moon is $2.02 \times 10^{20} \text{ N}$.

Importance of the universal LAW OF GRAVITATION

The universal law of gravitation explained several phenomena previously thought to be unrelated:

• The force that binds us to the Earth.
• The motion of the Moon around the Earth.
• The motion of planets around the Sun.
• The tides caused by the Moon and the Sun.

Free Fall

Free fall is the motion of an object falling towards the Earth under the influence of gravity alone. When an object falls towards the Earth, its velocity changes due to the Earth's attraction. This change in velocity is called acceleration. The acceleration due to the Earth's gravitational force is called acceleration due to gravity, denoted by $g$. The unit of $g$ is the same as that of acceleration, $\text{m s}^{-2}$.

The magnitude of the gravitational force $F$ on an object of mass $m$ is: $F = mg$

From the universal law of gravitation: $mg = G \frac{M \times m}{d^2}$

Therefore, the acceleration due to gravity is: $g = G \frac{M}{d^2}$

where $M$ is the mass of the Earth, and $d$ is the distance between the object and the Earth.

For objects on or near the surface of the Earth, $d$ is equal to $R$, the radius of the Earth: $g = G \frac{M}{R^2}$

The Earth is not a perfect sphere, so the radius varies from the poles to the equator. As a result, the value of $g$ is greater at the poles than at the equator. However, for most calculations near the Earth's surface, $g$ can be considered constant.

To calculate the value of $g$

To calculate the value of $g$, substitute the values of $G$, $M$, and $R$:

• $G = 6.7 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$
• $M = 6 \times 10^{24} \text{ kg}$
• $R = 6.4 \times 10^{6} \text{ m}$

$g = \frac{6.7 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2} \times 6 \times 10^{24} \text{ kg}}{(6.4 \times 10^{6} \text{ m})^2} = 9.8 \text{ m s}^{-2}$

Thus, the acceleration due to gravity on Earth is approximately $9.8 \text{ m s}^{-2}$.

Motion of objects under the INFLUENCE OF GRAVITATIONAL FORCE OF THE EARTH

During free fall, an object experiences acceleration independent of its mass. This means all objects, regardless of size or composition, fall at the same rate, neglecting air resistance.

Since $g$ is constant near the Earth, the equations of uniformly accelerated motion apply, with $a$ replaced by $g$:

$v = u + gt$

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

$v^2 = u^2 + 2gs$

where $u$ is the initial velocity, $v$ is the final velocity, $s$ is the distance covered, and $t$ is the time.

In these equations, $a$ is positive when in the direction of motion and negative when opposing the motion.

Given

• Time, $t = 0.5 \text{ s}$
• Initial velocity, $u = 0 \text{ m s}^{-1}$
• Acceleration due to gravity, $g = 10 \text{ m s}^{-2}$
• Acceleration of the car, $a = +10 \text{ m s}^{-2}$ (downward)

To Find

(i) Speed on striking the ground, $v$ (ii) Average speed, $v_{avg}$ (iii) Height of the ledge from the ground, $s$

Formula

$v = at$

$v_{avg} = \frac{u+v}{2}$

$s = \frac{1}{2}at^2$

Solution

(i) Calculate the speed on striking the ground

$v = at = (10 \text{ m s}^{-2})(0.5 \text{ s}) = 5 \text{ m s}^{-1}$

Answer for part (i) = $5 \text{ m s}^{-1}$

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(ii) Calculate the average speed

$v_{avg} = \frac{u+v}{2} = \frac{0 \text{ m s}^{-1} + 5 \text{ m s}^{-1}}{2} = 2.5 \text{ m s}^{-1}$

Answer for part (ii) = $2.5 \text{ m s}^{-1}$

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(iii) Calculate the height of the ledge

$s = \frac{1}{2}at^2 = \frac{1}{2}(10 \text{ m s}^{-2})(0.5 \text{ s})^2 = \frac{1}{2}(10 \text{ m s}^{-2})(0.25 \text{ s}^2) = 1.25 \text{ m}$

Answer for part (iii) = $1.25 \text{ m}$

---

Given

• Distance traveled, $s = 10 \text{ m}$
• Final velocity, $v = 0 \text{ m s}^{-1}$
• Acceleration due to gravity, $g = 9.8 \text{ m s}^{-2}$
• Acceleration of the object, $a = -9.8 \text{ m s}^{-2}$ (upward motion)

To Find

(i) Initial velocity, $u$ (ii) Time taken to reach the highest point, $t$

Formula

$v^2 = u^2 + 2as$

$v = u + at$

Solution

(i) Calculate the initial velocity

$v^2 = u^2 + 2as$

$0 = u^2 + 2(-9.8 \text{ m s}^{-2})(10 \text{ m})$

$-u^2 = -2 \times 9.8 \times 10 \text{ m}^2 \text{ s}^{-2}$

$u = \sqrt{196} \text{ m s}^{-1} = 14 \text{ m s}^{-1}$

Answer for part (i) = $14 \text{ m s}^{-1}$

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(ii) Calculate the time taken to reach the highest point

$v = u + at$

$0 = 14 \text{ m s}^{-1} - 9.8 \text{ m s}^{-2} \times t$

$t = \frac{14 \text{ m s}^{-1}}{9.8 \text{ m s}^{-2}} = 1.43 \text{ s}$

Answer for part (ii) = $1.43 \text{ s}$

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Mass

Mass is the measure of an object's inertia. The greater the mass, the greater the inertia. Mass remains constant whether the object is on Earth, the Moon, or in space. Mass does not change from place to place.

Weight

The weight of an object is the force with which it is attracted towards the Earth. This force depends on the object's mass ($m$) and the acceleration due to gravity ($g$).

Since $F = ma$, the weight $W$ is: $W = mg$

The SI unit of weight is the same as that of force, the newton (N). Weight is a force acting vertically downwards, having both magnitude and direction.

At a given location, the weight of an object is directly proportional to its mass: $W \propto m$. This is why weight can be used as a measure of mass at a specific place. However, mass remains constant everywhere, while weight depends on location because $g$ varies.

Weight of an object on THE MOON

The weight of an object on the Moon is the force with which the Moon attracts it. The Moon's mass is less than the Earth's, so it exerts a smaller gravitational force.

Let $m$ be the mass of the object, $W_m$ be its weight on the Moon, $M_m$ be the mass of the Moon, and $R_m$ be the Moon's radius. According to the universal law of gravitation:

$W_m = G \frac{M_m \times m}{R_m^2}$

Let $W_e$ be the weight of the same object on Earth, $M$ be the mass of the Earth, and $R$ be the Earth's radius:

$W_e = G \frac{M \times m}{R^2}$

Celestial Body	Mass (kg)	Radius (m)
Earth	$5.98 \times 10^{24}$	$6.37 \times 10^{6}$
Moon	$7.36 \times 10^{22}$	$1.74 \times 10^{6}$

Substituting the values:

$W_m = G \frac{7.36 \times 10^{22} \text{ kg} \times m}{(1.74 \times 10^{6} \text{ m})^2} = 2.431 \times 10^{10} G \times m$

$W_e = G \frac{5.98 \times 10^{24} \text{ kg} \times m}{(6.37 \times 10^{6} \text{ m})^2} = 1.474 \times 10^{11} G \times m$

Dividing the equations:

$\frac{W_m}{W_e} = \frac{2.431 \times 10^{10}}{1.474 \times 10^{11}} \approx \frac{1}{6}$

Therefore, the weight of an object on the Moon is approximately $\frac{1}{6}$ of its weight on Earth.

$W_m = \frac{1}{6} \times W_e$

Given

• Mass, $m = 10 \text{ kg}$
• Acceleration due to gravity, $g = 9.8 \text{ m s}^{-2}$

To Find

Weight, $W$

Formula

$W = m \times g$

Solution

Substitute the given values into the formula:

$W = 10 \text{ kg} \times 9.8 \text{ m s}^{-2} = 98 \text{ N}$

Final Answer The weight of the object is 98 N.

Given

• Weight of object on Earth, $W_e = 10 \text{ N}$

To Find

Weight of object on Moon, $W_m$

Formula

$W_m = \frac{1}{6} \times W_e$

Solution

Substitute the given values into the formula:

$W_m = \frac{1}{6} \times 10 \text{ N} = 1.67 \text{ N}$

Final Answer The weight of the object on the surface of the moon would be $1.67 \text{ N}$.

Thrust and Pressure

Thrust is the force acting on an object perpendicular to a surface.

Pressure is the thrust acting on a unit area. $\text{Pressure} = \frac{\text{thrust}}{\text{area}}$

The SI unit of pressure is newton per square meter ($\text{N/m}^2$) or $\text{N m}^{-2}$, also known as pascal (Pa).

Given

• Mass of the wooden block, $m = 5 \text{ kg}$
• Dimensions $= 40 \text{ cm} \times 20 \text{ cm} \times 10 \text{ cm}$

To Find

Pressure exerted by the wooden block on the table top if it is made to lie on the table top with its sides of dimensions (a) $20 \text{ cm} \times 10 \text{ cm}$ and (b) $40 \text{ cm} \times 20 \text{ cm}$.

Formula

$\text{Thrust} = F = m \times g$

$\text{Pressure} = \frac{\text{thrust}}{\text{area}}$

Solution

(a) Calculate the pressure when the block lies on its side of dimensions $20 \text{ cm} \times 10 \text{ cm}$

Thrust is equal to the weight of the wooden block:

$F = 5 \text{ kg} \times 9.8 \text{ m s}^{-2} = 49 \text{ N}$

Area of the side:

$A = 20 \text{ cm} \times 10 \text{ cm} = 200 \text{ cm}^2 = 0.02 \text{ m}^2$

Pressure:

$\text{Pressure} = \frac{49 \text{ N}}{0.02 \text{ m}^2} = 2450 \text{ N m}^{-2}$

Answer for part (a) = $2450 \text{ N m}^{-2}$

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(b) Calculate the pressure when the block lies on its side of dimensions $40 \text{ cm} \times 20 \text{ cm}$

Thrust remains the same: $F = 49 \text{ N}$

Area of the side:

$A = 40 \text{ cm} \times 20 \text{ cm} = 800 \text{ cm}^2 = 0.08 \text{ m}^2$

Pressure:

$\text{Pressure} = \frac{49 \text{ N}}{0.08 \text{ m}^2} = 612.5 \text{ N m}^{-2}$

Answer for part (b) = $612.5 \text{ N m}^{-2}$

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The same force acting on a smaller area exerts a larger pressure, and a smaller pressure on a larger area. This is why nails have pointed tips, knives have sharp edges, and buildings have wide foundations.

Pressure in fluids

Liquids and gases are fluids. Fluids exert pressure on the base and walls of their container. Pressure in a confined fluid is transmitted equally in all directions.

Buoyancy

Buoyancy is the upward force exerted by a fluid on an object immersed in it. This force explains why objects feel lighter in water and why some objects float. The magnitude of the buoyant force depends on the density of the fluid.

Why objects float or sink WHEN PLACED ON THE SURFACE OF WATER?

Whether an object floats or sinks depends on its density compared to the density of the liquid. Density is defined as mass per unit volume.

• If the object's density is less than the liquid's density, it floats. The upthrust of the water on the object is greater than the object's weight.
• If the object's density is greater than the liquid's density, it sinks. The upthrust of the water on the object is less than the object's weight.

Archimedes' Principle

Archimedes' principle states that when a body is immersed fully or partially in a fluid, it experiences an upward force (buoyant force) equal to the weight of the fluid displaced by it.

Archimedes' principle is used in designing ships and submarines. Lactometers (for testing milk purity) and hydrometers (for measuring liquid density) are based on this principle.

Gravitation

We have learned that a force is needed to change the speed or direction of motion of an object. Objects dropped from a height fall towards the Earth, planets orbit the Sun, and the Moon orbits the Earth. In all these cases, a force is acting on these objects. This force is called gravitational force.

In this chapter, we will explore gravitation, the universal law of gravitation, the motion of objects under the influence of Earth's gravitational force, how the weight of a body varies from place to place, and the conditions for objects to float in liquids.

Gravitation

The Moon orbits the Earth, and objects thrown upwards eventually fall back down. Isaac Newton realized that the same force attracting an apple to the Earth might also be responsible for the Moon's orbit. He figured that, instead of traveling in a straight line, the Moon is constantly falling towards the Earth due to Earth's attraction. However, we don't see the moon crashing on Earth.

Materials Required

• Thread
• Small stone

Procedure

1. Take a piece of thread and tie a small stone at one end.
2. Hold the other end of the thread and whirl it around.
3. Note the motion of the stone.
4. Release the thread.
5. Note the direction of motion of the stone.

Observation

Before the thread is released, the stone moves in a circular path, changing direction at every point. When the thread is released, the stone flies off in a straight line.

Conclusion

The stone's circular motion is due to a force pulling it towards the center. When this force is removed, the stone moves in a straight line tangent to the circle.

Before the thread is released, the stone moves in a circular path with a certain speed and changes direction at every point. The change in direction involves a change in velocity or acceleration. The force that causes this acceleration and keeps the body moving along the circular path is acting towards the centre. This force is called the centripetal force.

In the absence of this force, the stone flies off along a straight line. This straight line will be a tangent to the circular path.

The motion of the Moon around the Earth is due to centripetal force. The Earth's gravitational attraction provides this centripetal force. Without this force, the Moon would move in a straight line.

The Earth attracts a falling apple. Does the apple also attract the Earth? Yes, it does! According to Newton's third law of motion, for every action, there is an equal and opposite reaction. The apple attracts the Earth with the same force that the Earth attracts the apple. However, we don't see the Earth moving towards the apple because, according to Newton's second law of motion, acceleration is inversely proportional to mass. The Earth's mass is so much larger than the apple's mass that the Earth's acceleration is negligible.

Similarly, the Sun and the planets exert gravitational forces on each other, causing the planets to orbit the Sun.

Newton concluded that every object in the universe attracts every other object. This attractive force is called gravitational force.

Universal law of gravitation

The universal law of gravitation states that every object in the universe attracts every other object with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. The force acts along the line joining the centers of the two objects.

Consider two objects, A and B, with masses $M$ and $m$, respectively, separated by a distance $d$. Let $F$ be the gravitational force between them.

The universal law of gravitation can be expressed mathematically as:

$F \propto M \times m$

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

Combining these proportionalities, we get:

$F \propto \frac{M \times m}{d^{2}}$

Introducing a constant of proportionality, G, known as the universal gravitation constant, we have:

$F = \text{G} \frac{M \times m}{d^{2}}$

Solving for G gives

$\text{G} = \frac{F d^{2}}{M \times m}$

The SI unit of G is $\text{N m}^2 \text{ kg}^{-2}$.

The value of G was experimentally determined by Henry Cavendish and is approximately $6.673 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$.

The universal law of gravitation applies to all bodies, regardless of their size or location in the universe.

Saying that $F$ is inversely proportional to the square of $d$ means, for example, that if $d$ gets bigger by a factor of 6, $F$ becomes $\frac{1}{36}$ times smaller.

Given

• Mass of the Earth, $M = 6 \times 10^{24} \text{ kg}$
• Mass of the Moon, $m = 7.4 \times 10^{22} \text{ kg}$
• Distance between Earth and Moon, $d = 3.84 \times 10^{5} \text{ km} = 3.84 \times 10^{8} \text{ m}$
• Gravitational constant, $\text{G} = 6.7 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$

To Find

Force exerted by the Earth on the Moon, $F$

Formula

$F = \text{G} \frac{M \times m}{d^{2}}$

Solution

Substitute the given values into the formula:

$F = \frac{6.7 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2} \times 6 \times 10^{24} \text{ kg} \times 7.4 \times 10^{22} \text{ kg}}{(3.84 \times 10^{8} \text{ m})^{2}}$

$F = \frac{6.7 \times 6 \times 7.4 \times 10^{-11+24+22}}{3.84^2 \times 10^{16}} \text{ N} = \frac{297.48 \times 10^{35}}{14.7456 \times 10^{16}} \text{ N}$

$F = 20.17 \times 10^{19} \text{ N} \approx 2.02 \times 10^{20} \text{ N}$

Final Answer The force exerted by the Earth on the Moon is approximately $2.02 \times 10^{20} \text{ N}$.

Importance of the universal law of gravitation

The universal law of gravitation explains:

• The force that binds us to the Earth.
• The motion of the Moon around the Earth.
• The motion of planets around the Sun.
• The tides caused by the Moon and the Sun.

Free Fall

When an object falls towards the Earth due to the Earth's gravitational force alone, it is said to be in free fall.

Materials Required

• Stone

Procedure

1. Take a stone.
2. Throw it upwards.
3. Observe that it reaches a certain height and then falls down.

Observation

The stone falls back to the earth.

Conclusion

The Earth attracts objects towards it due to gravitational force.

During free fall, the direction of motion remains constant, but the magnitude of the velocity changes. This change in velocity means the object is accelerating. This acceleration is due to the Earth's gravitational force and is called acceleration due to gravity, denoted by $g$. The unit of $g$ is the same as that of acceleration, which is $\text{m s}^{-2}$.

According to Newton's second law of motion, $F = ma$. In the case of free fall, the force is the gravitational force, and the acceleration is $g$. Therefore, the magnitude of the gravitational force $F$ on an object of mass $m$ is:

$F = m g$

From the universal law of gravitation, we also know that:

$F = \text{G} \frac{M \times m}{d^{2}}$

where $M$ is the mass of the Earth and $d$ is the distance between the object and the center of the Earth.

Combining these two equations, we get:

$m g = \text{G} \frac{M \times m}{d^{2}}$

$g = \text{G} \frac{M}{d^{2}}$

For objects on or near the surface of the Earth, the distance $d$ is approximately equal to the radius of the Earth, $R$. Thus:

$g = \text{G} \frac{M}{R^{2}}$

The Earth is not a perfect sphere; its radius is larger at the equator than at the poles. Therefore, the value of $g$ is slightly greater at the poles than at the equator. However, for most calculations, we can consider $g$ to be approximately constant near the Earth's surface.

To calculate the value of $g$

To calculate the value of $g$, we substitute the values of G, $M$ (mass of the Earth), and $R$ (radius of the Earth) into the equation:

$G = 6.7 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$

$M = 6 \times 10^{24} \text{ kg}$

$R = 6.4 \times 10^{6} \text{ m}$

$g = \text{G} \frac{M}{R^{2}} = \frac{6.7 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2} \times 6 \times 10^{24} \text{ kg}}{(6.4 \times 10^{6} \text{ m})^{2}}$

$g = \frac{6.7 \times 6 \times 10^{13}}{6.4^2 \times 10^{12}} \text{ m s}^{-2} = \frac{40.2 \times 10^{13}}{40.96 \times 10^{12}} \text{ m s}^{-2} \approx 9.8 \text{ m s}^{-2}$

Thus, the value of acceleration due to gravity of the Earth, $g = 9.8 \text{ m s}^{-2}$.

Motion of objects under the INFLUENCE OF GRAVITATIONAL FORCE OF THE EARTH

Materials Required

• A sheet of paper
• A stone

Procedure

1. Take a sheet of paper and a stone.
2. Drop them simultaneously from the first floor of a building.
3. Observe whether both reach the ground simultaneously.

Observation

The stone reaches the ground before the paper.

Conclusion

The paper experiences more air resistance than the stone, slowing its descent.

In a vacuum (where there is no air resistance), all objects, regardless of their mass or size, fall at the same rate due to gravity.

Since $g$ is constant near the Earth, the equations of motion for uniformly accelerated objects can be used with $a$ replaced by $g$:

$v = u + gt$

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

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

where:

• $u$ is the initial velocity
• $v$ is the final velocity
• $s$ is the distance covered
• $t$ is the time

In these equations, $g$ is positive if it is in the direction of motion and negative if it opposes the motion.

Given

• Time, $t = 0.5 \text{ s}$
• Initial velocity, $u = 0 \text{ m s}^{-1}$
• Acceleration due to gravity, $g = 10 \text{ m s}^{-2}$
• Acceleration of the car, $a = +10 \text{ m s}^{-2}$ (downward)

To Find

(i) Speed on striking the ground, $v$ (ii) Average speed, $v_{avg}$ (iii) Height of the ledge, $s$

Formula

$v = u + at$

$v_{avg} = \frac{u+v}{2}$

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

Solution

(i) Calculate the final velocity

$v = u + at = 0 + (10 \text{ m s}^{-2})(0.5 \text{ s}) = 5 \text{ m s}^{-1}$

Answer for part (i) = $5 \text{ m s}^{-1}$

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(ii) Calculate the average speed

$v_{avg} = \frac{u+v}{2} = \frac{0 + 5 \text{ m s}^{-1}}{2} = 2.5 \text{ m s}^{-1}$

Answer for part (ii) = $2.5 \text{ m s}^{-1}$

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(iii) Calculate the height of the ledge

$s = ut + \frac{1}{2}at^2 = (0 \text{ m s}^{-1})(0.5 \text{ s}) + \frac{1}{2}(10 \text{ m s}^{-2})(0.5 \text{ s})^2$

$s = 0 + \frac{1}{2} \times 10 \text{ m s}^{-2} \times 0.25 \text{ s}^2 = 1.25 \text{ m}$

Answer for part (iii) = $1.25 \text{ m}$

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Given

• Distance travelled, $s = 10 \text{ m}$
• Final velocity, $v = 0 \text{ m s}^{-1}$
• Acceleration due to gravity, $g = 9.8 \text{ m s}^{-2}$
• Acceleration of the object, $a = -9.8 \text{ m s}^{-2}$ (upward motion)

To Find

(i) Initial velocity, $u$ (ii) Time taken to reach the highest point, $t$

Formula

$v^2 = u^2 + 2as$

$v = u + at$

Solution

(i) Calculate the initial velocity

$v^2 = u^2 + 2as \implies 0 = u^2 + 2(-9.8 \text{ m s}^{-2})(10 \text{ m})$

$-u^2 = -2 \times 9.8 \times 10 \text{ m}^2 \text{ s}^{-2} = -196 \text{ m}^2 \text{ s}^{-2}$

$u = \sqrt{196} \text{ m s}^{-1} = 14 \text{ m s}^{-1}$

Answer for part (i) = $14 \text{ m s}^{-1}$

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(ii) Calculate the time taken

$v = u + at \implies 0 = 14 \text{ m s}^{-1} - 9.8 \text{ m s}^{-2} \times t$

$t = \frac{14 \text{ m s}^{-1}}{9.8 \text{ m s}^{-2}} = 1.43 \text{ s}$

Answer for part (ii) = $1.43 \text{ s}$

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Mass

Mass of an object is the measure of its inertia. The greater the mass, the greater the inertia. Mass remains constant whether the object is on Earth, the Moon, or in outer space. Mass does not change from place to place.

Weight

The weight of an object is the force with which it is attracted towards the Earth. This force depends on the mass ($m$) of the object and the acceleration due to gravity ($g$).

Since $F = ma$, the force of attraction (weight) is:

$W = m \times g$

The SI unit of weight is the same as that of force, which is the newton (N). Weight is a force acting vertically downwards; it has both magnitude and direction.

The value of $g$ is constant at a given place. Therefore, at a given place, the weight of an object is directly proportional to its mass: $W \propto m$. This is why we can use weight as a measure of mass at a given location. Mass remains the same everywhere, while weight depends on location because $g$ depends on location.

Weight of an object on THE MOON

The weight of an object on the Moon is the force with which the Moon attracts the object. The Moon's mass is less than the Earth's mass, so it exerts a smaller gravitational force.

Let:

• $m$ be the mass of the object
• $W_m$ be the weight of the object on the Moon
• $M_m$ be the mass of the Moon
• $R_m$ be the radius of the Moon

Using the universal law of gravitation, the weight of the object on the Moon is:

$W_{m} = \text{G} \frac{M_{m} \times m}{R_{m}^{2}}$

Let $W_e$ be the weight of the same object on Earth, where $M$ is the mass of the Earth and $R$ is its radius.

$W_{e} = \text{G} \frac{M \times m}{R^{2}}$

Celestial body	Mass (kg)	Radius (m)
Earth	$5.98 \times 10^{24}$	$6.37 \times 10^{6}$
Moon	$7.36 \times 10^{22}$	$1.74 \times 10^{6}$

Substituting the values from the table, we get:

$W_{m} = \text{G} \frac{7.36 \times 10^{22} \text{ kg} \times m}{(1.74 \times 10^{6} \text{ m})^{2}} = 2.431 \times 10^{10} \text{G} \times m$

$W_{e} = \text{G} \frac{5.98 \times 10^{24} \text{ kg} \times m}{(6.37 \times 10^{6} \text{ m})^{2}} = 1.474 \times 10^{11} \text{G} \times m$

Dividing the two equations:

$\frac{W_{m}}{W_{e}} = \frac{2.431 \times 10^{10}}{1.474 \times 10^{11}} \approx 0.165 \approx \frac{1}{6}$

Therefore:

$\frac{\text{Weight of the object on the moon}}{\text{Weight of the object on the earth}} = \frac{1}{6}$

Weight of the object on the moon $= (1/6) \times \text{its weight on the earth}$

Given

• Mass, $m = 10 \text{ kg}$
• Acceleration due to gravity, $g = 9.8 \text{ m s}^{-2}$

To Find

Weight, $W$

Formula

$W = m \times g$

Solution

Substitute the given values into the formula

$W = 10 \text{ kg} \times 9.8 \text{ m s}^{-2} = 98 \text{ N}$

Final Answer The weight of the object is 98 N.

Given

• Weight of object on the Earth, $W_e = 10 \text{ N}$

To Find

Weight of object on the Moon, $W_m$

Formula

$W_m = \frac{1}{6} \times W_e$

Solution

Substitute the given values into the formula

$W_m = \frac{1}{6} \times 10 \text{ N} = 1.67 \text{ N}$

Final Answer The weight of the object on the surface of the moon would be 1.67 N.

Thrust and Pressure

Thrust is the net force acting on an object in a particular direction, especially a force acting perpendicularly on a surface. Pressure is the force per unit area acting on an object.

To understand thrust and pressure, consider these situations:

Situation 1: Pushing a drawing pin into a bulletin board. You apply a force (thrust) on the head of the pin, which is directed perpendicular to the board's surface. This force is concentrated on the small area of the pin's tip.

Situation 2: Standing vs. lying on loose sand. When you stand, your weight (thrust) acts on the small area of your feet, causing them to sink deeper. When you lie down, the same weight is distributed over a much larger area, so you don't sink as far.

In both situations, the effect of the force depends on the area over which it is applied.

The pressure is defined as:

$\text{Pressure} = \frac{\text{thrust}}{\text{area}}$

The SI unit of pressure is newtons per square meter ($\text{N m}^{-2}$), also known as pascal (Pa).

$1 \text{ Pa} = 1 \text{ N m}^{-2}$

Given

• Mass of the wooden block $= 5 \text{ kg}$
• Dimensions $= 40 \text{ cm} \times 20 \text{ cm} \times 10 \text{ cm}$

To Find

Pressure exerted by the wooden block on the tabletop when it lies on its side with dimensions: (a) $20 \text{ cm} \times 10 \text{ cm}$ (b) $40 \text{ cm} \times 20 \text{ cm}$

Formula

$\text{Thrust} = F = m \times g$

$\text{Pressure} = \frac{\text{Thrust}}{\text{Area}}$

Solution

First, calculate the thrust (force) exerted by the wooden block:

$\text{Thrust} = F = m \times g = 5 \text{ kg} \times 9.8 \text{ m s}^{-2} = 49 \text{ N}$

(a) Calculate the pressure when the block lies on its side with dimensions $20 \text{ cm} \times 10 \text{ cm}$

Area $= \text{length} \times \text{breadth} = 20 \text{ cm} \times 10 \text{ cm} = 200 \text{ cm}^2 = 0.02 \text{ m}^2$

$\text{Pressure} = \frac{49 \text{ N}}{0.02 \text{ m}^2} = 2450 \text{ N m}^{-2}$

Answer for part (a) = $2450 \text{ N m}^{-2}$

---

(b) Calculate the pressure when the block lies on its side with dimensions $40 \text{ cm} \times 20 \text{ cm}$

Area $= \text{length} \times \text{breadth} = 40 \text{ cm} \times 20 \text{ cm} = 800 \text{ cm}^2 = 0.08 \text{ m}^2$

$\text{Pressure} = \frac{49 \text{ N}}{0.08 \text{ m}^2} = 612.5 \text{ N m}^{-2}$

Answer for part (b) = $612.5 \text{ N m}^{-2}$

---

The same force acting on a smaller area exerts a larger pressure, and a smaller pressure on a larger area. This is why nails have pointed tips, knives have sharp edges, and buildings have wide foundations.

Pressure in fluids

Fluids include liquids and gases. Like solids, fluids also exert pressure. They exert pressure on the base and walls of the container in which they are enclosed. Pressure exerted in any confined mass of fluid is transmitted undiminished in all directions.

Buoyancy

Buoyancy is the upward force exerted by a fluid on an object immersed in it.

Materials Required

• Empty plastic bottle
• Airtight stopper
• Bucket filled with water

Procedure

1. Take an empty plastic bottle and close its mouth with an airtight stopper.
2. Put the bottle in a bucket filled with water. Observe that the bottle floats.
3. Push the bottle into the water. Notice the upward push.
4. Try to push it further down. You will find it difficult to push deeper.
5. Release the bottle. Observe that it bounces back to the surface.

Observation

The bottle floats. When pushed into the water, there is an upward push. The upward force increases as the bottle is pushed deeper. When released, the bottle bounces back to the surface.

Conclusion

Water exerts an upward force on the bottle, opposing the downward force of gravity. This upward force is called buoyancy.

The Earth's gravitational force pulls the bottle downwards, but the water exerts an upward force on the bottle, causing it to be pushed upwards. The upthrust or buoyant force is the upward force exerted by the water on the bottle. All objects experience a force of buoyancy when they are immersed in a fluid. The magnitude of this buoyant force depends on the density of the fluid.

Why objects float or sink WHEN PLACED ON THE SURFACE OF WATER?

Materials Required

• Beaker filled with water
• Iron nail

Procedure

1. Take a beaker filled with water.
2. Take an iron nail and place it on the surface of the water.
3. Observe what happens.

Observation

The nail sinks.

Conclusion

The downward force on the nail is greater than the upward force of the water.

The force due to the gravitational attraction of the Earth on the iron nail pulls it downwards. There is an upthrust of water on the nail, which pushes it upwards. But the downward force acting on the nail is greater than the upthrust of water on the nail. So it sinks.

Materials Required

• Beaker filled with water
• Piece of cork
• Iron nail of equal mass

Procedure

1. Take a beaker filled with water.
2. Take a piece of cork and an iron nail of equal mass.
3. Place them on the surface of the water.
4. Observe what happens.

Observation

The cork floats, while the nail sinks.

Conclusion

The cork is less dense than water, so it floats. The nail is more dense than water, so it sinks.

The density of a substance is defined as the mass per unit volume. The density of cork is less than the density of water, which means that the upthrust of water on the cork is greater than the weight of the cork. So it floats. The density of the iron nail is more than the density of water, which means that the upthrust of water on the iron nail is less than the weight of the nail. So it sinks.

Objects with a density less than that of the liquid float on the liquid. Objects with a density greater than that of the liquid sink in the liquid.

Archimedes' Principle

Materials Required

• Piece of stone
• Rubber string or spring balance
• Container of water

Procedure

1. Take a piece of stone and tie it to one end of a rubber string or a spring balance.
2. Suspend the stone by holding the balance or the string.
3. Note the elongation of the string or the reading on the spring balance due to the weight of the stone.
4. Slowly dip the stone in the water in a container.
5. Observe what happens to the elongation of the string or the reading on the balance.

Observation

The elongation of the string or the reading of the balance decreases as the stone is gradually lowered in the water.

Conclusion

The decrease in elongation or balance reading indicates an upward force acting on the stone, which is the buoyant force.

The elongation produced in the string or the spring balance is due to the weight of the stone. Since the extension decreases once the stone is lowered in water, it means that some force acts on the stone in upward direction. As a result, the net force on the string decreases and hence the elongation also decreases. This upward force exerted by water is known as the force of buoyancy.

Archimedes' principle states that when a body is immersed fully or partially in a fluid, it experiences an upward force (buoyant force) that is equal to the weight of the fluid displaced by it.

Archimedes' principle has many applications. It is used in designing ships and submarines. Lactometers, which are used to determine the purity of a sample of milk, and hydrometers, used for determining the density of liquids, are based on this principle.

Gravitation

We have learned about motion and force as the cause of motion. A force is needed to change the speed or direction of an object. We see objects dropped from a height fall towards the Earth, planets go around the Sun, and the Moon goes around the Earth. In each case, a force must be acting. Isaac Newton realized that the same force is responsible for all of these motions. This force is called gravitational force.

In this chapter, we will explore gravitation and the universal law of gravitation, the motion of objects influenced by Earth’s gravitational force, how weight varies, and conditions for objects to float in liquids.

Gravitation

The Moon orbits the Earth, and objects thrown upwards eventually fall back down. Newton realized that if Earth can attract an apple, it might also attract the Moon. He believed the same force was responsible for both. He reasoned that the Moon is constantly falling towards Earth instead of moving in a straight line, implying it's attracted by Earth.

To understand this, consider the following: Imagine whirling a stone tied to a thread in a circle. The stone moves in a circular path with a certain speed, constantly changing direction. This change in direction means there is a change in velocity, and thus acceleration. The force causing this acceleration, keeping the stone moving in a circle, is directed towards the center and is called centripetal force.

Without this force, the stone would fly off in a straight line, tangent to the circular path.

The Moon's motion around Earth is due to centripetal force, provided by Earth's gravitational attraction. Without this force, the Moon would move in a straight line.

A falling apple is attracted to the Earth. Does the apple also attract the Earth? Yes, it does, according to Newton's third law of motion. However, according to the second law of motion, for a given force, acceleration is inversely proportional to mass. The apple's mass is tiny compared to Earth's, so we don't see Earth move towards the apple. The same reasoning applies to why Earth doesn't visibly move towards the Moon.

In the solar system, planets orbit the Sun. This suggests a force exists between the Sun and the planets. Newton concluded that all objects in the universe attract each other. This attraction is gravitational force.

Universal law of gravitation

Every object in the universe attracts every other object with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. The force acts along the line joining the centers of the two objects.

Imagine two objects, A and B, with masses $M$ and $m$, separated by a distance $d$. Let the gravitational force between them be $F$.

The force is directly proportional to the product of their masses:

$F \propto M \times m$

The force is inversely proportional to the square of the distance between them:

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

Combining these proportionalities:

$\begin{aligned} F & \propto \frac{M \times m}{d^{2}} \\ \text { or, } F & =\mathrm{G} \frac{M \times m}{d^{2}} \end{aligned}$

Here, G is the universal gravitation constant, the constant of proportionality. Rearranging the equation to solve for G:

$F \times d^{2}=\mathrm{G} M \times m$ or $\quad \mathrm{G}=\frac{F d^{2}}{M \times m}$

The SI unit of G is found by substituting the units of force, distance, and mass: $\text{N m}^{2} \text{ kg}^{-2}$.

Henry Cavendish determined the value of G using a sensitive balance. The accepted value is $6.673 \times 10^{-11} \text{ N m}^{2} \text{ kg}^{-2}$.

The law is universal because it applies to all objects, regardless of size or location.

Inverse-square

The term "inversely proportional to the square" means that if the distance $d$ increases by a factor of 6, the force $F$ decreases by a factor of $1/36$.

Given

• Mass of Earth, $M=6 \times 10^{24} \text{ kg}$
• Mass of Moon, $m=7.4 \times 10^{22} \text{ kg}$
• Distance between Earth and Moon, $d = 3.84 \times 10^{5} \text{ km} = 3.84 \times 10^{8} \text{ m}$
• Gravitational constant, $\mathrm{G} = 6.7 \times 10^{-11} \text{ N m}^{2} \text{ kg}^{-2}$

To Find

Force exerted by the Earth on the Moon, $F$

Formula

$F=\mathrm{G} \frac{M \times m}{d^{2}}$

Solution

Substitute the given values into the formula

$F = \frac{6.7 \times 10^{-11} \text{ N m}^{2} \text{ kg}^{-2} \times 6 \times 10^{24} \text{ kg} \times 7.4 \times 10^{22} \text{ kg}}{(3.84 \times 10^{8} \text{ m})^{2}}$

$F = \frac{6.7 \times 6 \times 7.4 \times 10^{-11+24+22}}{(3.84)^2 \times 10^{16}} \text{ N}$

$F = \frac{297.48 \times 10^{35}}{14.7456 \times 10^{16}} \text{ N} = 2.02 \times 10^{19} \text{ N}$

Final Answer The force exerted by the earth on the moon is $2.02 \times 10^{20} \text{ N}$.

Importance of the universal LAW OF GRAVITATION

The universal law of gravitation explained phenomena previously thought to be unrelated:

• The force that binds us to the Earth.
• The Moon's orbit around the Earth.
• The planets' orbits around the Sun.
• The tides caused by the Moon and the Sun.

Free Fall

Objects fall towards Earth due to gravitational force. When objects fall under the influence of gravity alone, it's called free fall.

During free fall, the direction of motion doesn't change, but the magnitude of the velocity changes due to Earth's attraction. This change in velocity means there is acceleration. This acceleration, caused by Earth's gravitational force, is called acceleration due to gravity, denoted by $g$. Its unit is the same as acceleration: $\text{m s}^{-2}$.

From Newton's second law, force equals mass times acceleration. For a falling object with mass $m$, the gravitational force $F$ is:

$F=m g$

Combining this with the universal law of gravitation:

$\begin{aligned} & \quad m g=\mathrm{G} \frac{M \times m}{d^{2}} \\ & \text { or } g=\mathrm{G} \frac{M}{d^{2}} \end{aligned}$

Where $M$ is the mass of the Earth, and $d$ is the distance between the object and the Earth.

For objects on or near Earth's surface, $d$ equals $R$, the Earth's radius:

$\begin{aligned} m g & =\mathrm{G} \frac{M \times m}{R^{2}} \\ g & =\mathrm{G} \frac{M}{R^{2}} \end{aligned}$

Earth isn't a perfect sphere. Its radius increases from the poles to the equator, so $g$ is greater at the poles than at the equator. For most calculations near Earth, $g$ is considered constant. However, for objects far from Earth, the acceleration due to gravity is given by the earlier equation: $g=\mathrm{G} \frac{M}{d^{2}}$.

To calculate the value of $g$

To calculate $g$, use the values of G, $M$, and $R$:

• Universal gravitational constant, $\mathrm{G}=6.7 \times 10^{-11} \text{ N m}^{2} \text{ kg}^{-2}$
• Mass of the Earth, $M=6 \times 10^{24} \text{ kg}$
• Radius of the Earth, $R=6.4 \times 10^{6} \text{ m}$

$g = \mathrm{G} \frac{M}{R^{2}}$

$g = \frac{6.7 \times 10^{-11} \text{ N m}^{2} \text{ kg}^{-2} \times 6 \times 10^{24} \text{ kg}}{(6.4 \times 10^{6} \text{ m})^{2}}$

$g = \frac{6.7 \times 6 \times 10^{13}}{6.4^2 \times 10^{12}} \text{ m s}^{-2} = \frac{40.2}{40.96} \times 10 \text{ m s}^{-2}$

$g \approx 9.8 \text{ m s}^{-2}$

The acceleration due to gravity on Earth is approximately $9.8 \text{ m s}^{-2}$.

Motion of objects under the INFLUENCE OF GRAVITATIONAL FORCE OF THE EARTH

During free fall, an object experiences acceleration independent of its mass. This means all objects, regardless of size or composition, should fall at the same rate, neglecting air resistance.

Since $g$ is constant near Earth, equations of uniformly accelerated motion apply, with $a$ replaced by $g$:

$v=u+g t$

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

$v^{2}=u^{2}+2 g s$

$u$ and $v$ are the initial and final velocities, and $s$ is the distance covered in time $t$.

When applying these equations, $g$ is positive if it's in the direction of motion and negative if it opposes the motion.

Given

• Time, $t=0.5 \text{ s}$
• Initial velocity, $u=0 \text{ m s}^{-1}$
• Acceleration due to gravity, $g=10 \text{ m s}^{-2}$
• Acceleration of the car, $a=+10 \text{ m s}^{-2}$ (downward)

To Find

(i) Speed on striking the ground, $v$ (ii) Average speed during the 0.5 s, $v_{avg}$ (iii) Height of the ledge from the ground, $s$

Formula

$v = a t$

$v_{avg} = \frac{u+v}{2}$

$s = \frac{1}{2} a t^{2}$

Solution

(i) Calculate the speed on striking the ground

$v = a t = 10 \text{ m s}^{-2} \times 0.5 \text{ s} = 5 \text{ m s}^{-1}$

Answer for part (i) = $5 \text{ m s}^{-1}$

---

(ii) Calculate the average speed

$v_{avg} = \frac{u+v}{2} = \frac{0 \text{ m s}^{-1} + 5 \text{ m s}^{-1}}{2} = 2.5 \text{ m s}^{-1}$

Answer for part (ii) = $2.5 \text{ m s}^{-1}$

---

(iii) Calculate the height of the ledge

$s = \frac{1}{2} a t^{2} = \frac{1}{2} \times 10 \text{ m s}^{-2} \times (0.5 \text{ s})^{2} = \frac{1}{2} \times 10 \text{ m s}^{-2} \times 0.25 \text{ s}^{2} = 1.25 \text{ m}$

Answer for part (iii) = $1.25 \text{ m}$

---

Given

• Distance travelled, $s=10 \text{ m}$
• Final velocity, $v=0 \text{ m s}^{-1}$
• Acceleration due to gravity, $g=9.8 \text{ m s}^{-2}$
• Acceleration of the object, $a=-9.8 \text{ m s}^{-2}$ (upward motion)

To Find

(i) The velocity with which the object was thrown upwards, $u$ (ii) The time taken by the object to reach the highest point, $t$

Formula

$v^{2}=u^{2}+2 a s$

$v=u+a t$

Solution

(i) Calculate the initial velocity

$v^{2}=u^{2}+2 a s$

$0 = u^{2} + 2 \times (-9.8 \text{ m s}^{-2}) \times 10 \text{ m}$

$-u^{2} = -2 \times 9.8 \times 10 \text{ m}^{2} \text{ s}^{-2}$

$u = \sqrt{196} \text{ m s}^{-1} = 14 \text{ m s}^{-1}$

Answer for part (i) = $14 \text{ m s}^{-1}$

---

(ii) Calculate the time taken to reach the highest point

$v=u+a t$

$0 = 14 \text{ m s}^{-1} - 9.8 \text{ m s}^{-2} \times t$

$t = \frac{14 \text{ m s}^{-1}}{9.8 \text{ m s}^{-2}} = 1.43 \text{ s}$

Answer for part (ii) = $1.43 \text{ s}$

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Mass

Mass is the measure of an object's inertia. The greater the mass, the greater the inertia. Mass remains constant whether the object is on Earth, the Moon, or in space. It does not change from place to place.

Weight

Earth attracts every object with a force that depends on the object's mass ($m$) and the acceleration due to gravity ($g$). Weight is the force with which an object is attracted towards Earth.

Since $F = m \times a$, weight ($W$) is:

$W=m \times g$

Weight is a force, so its SI unit is the newton (N). It acts vertically downwards and has both magnitude and direction.

The value of $g$ is constant at a given place. Therefore, at a given place, weight is directly proportional to mass: $W \propto m$. This is why we can use weight to measure mass at a specific location. Mass remains constant everywhere, but weight depends on location because $g$ varies.

Weight of an object on THE MOON

The weight of an object on the Moon is the force with which the Moon attracts it. The Moon's mass is less than Earth's, so it exerts a smaller gravitational force.

Let an object have mass $m$. Its weight on the Moon is $W_m$. The Moon has mass $M_m$ and radius $R_m$.

Using the universal law of gravitation, the weight on the Moon is:

$W_{m}=\mathrm{G} \frac{M_{m} \times m}{R_{m}^{2}}$

The weight of the same object on Earth ($W_e$) is:

$W_{e}=\mathrm{G} \frac{M \times m}{R^{2}}$

Where $M$ is the mass of Earth and $R$ is the radius of Earth.

Celestial body	Mass (kg)	Radius (m)
Earth	$5.98 \times 10^{24}$	$6.37 \times 10^{6}$
Moon	$7.36 \times 10^{22}$	$1.74 \times 10^{6}$

Substituting the values into the equations:

$\begin{aligned} W_{m} & =\mathrm{G} \frac{7.36 \times 10^{22} \text{ kg} \times m}{\left(1.74 \times 10^{6} \text{ m}\right)^{2}} \\ W_{m} & =2.431 \times 10^{10} \mathrm{G} \times m \\ \text { and } W_{e} & =1.474 \times 10^{11} \mathrm{G} \times m \end{aligned}$

Dividing the two equations:

$\begin{aligned} \frac{W_{m}}{W_{e}} & =\frac{2.431 \times 10^{10}}{1.474 \times 10^{11}} \\ \text { or } \frac{W_{m}}{W_{e}} & =0.165 \approx \frac{1}{6} \end{aligned}$

$\frac{\text { Weight of the object on the moon }}{\text { Weight of the object on the earth }}=\frac{1}{6}$

Weight of the object on the moon $=(1 / 6) \times \text { its weight on the earth. }$

Given

• Mass, $m=10 \text{ kg}$
• Acceleration due to gravity, $g=9.8 \text{ m s}^{-2}$

To Find

Weight, $W$

Formula

$W=m \times g$

Solution

Substitute the given values into the formula

$W = 10 \text{ kg} \times 9.8 \text{ m s}^{-2} = 98 \text{ N}$

Final Answer The weight of the object is 98 N .

Given

• Weight of object on Earth, $W_e = 10$ N

To Find

Weight of object on Moon, $W_m$

Formula

$W_{m} = \frac{1}{6} \times W_{e}$

Solution

Substitute the given values into the formula

$W_{m} = \frac{1}{6} \times 10 \text{ N} = 1.67 \text{ N}$

Final Answer The weight of object on the surface of the moon would be 1.67 N .

Thrust and Pressure

Why can a camel run easily in the desert? Why does an army tank rest on a continuous chain? Why do trucks have wider tires? Why do cutting tools have sharp edges? The answers involve thrust and pressure.

Thrust is the net force acting in a particular direction. Pressure is the force per unit area.

Consider these situations:

Situation 1: Pushing a drawing pin into a board. You apply force on the head of the pin, perpendicular to the board's surface. This force acts on a smaller area at the pin's tip.

Situation 2: Standing on loose sand. Your feet sink. Lying down, you don't sink as much. In both cases, the force is your weight, acting perpendicular to the sand's surface.

The force acting perpendicular to a surface is called thrust.

When standing on sand, your weight acts on the area of your feet. When lying down, the same force acts on the larger contact area of your body. The effects of the same force are different on different areas.

The effect of thrust depends on the area on which it acts. The thrust on unit area is called pressure:

$\text { Pressure }=\frac{\text { thrust }}{\text { area }}$

Substituting the SI units of thrust and area, the SI unit of pressure is $\text{N}/\text{m}^{2}$ or $\text{N m}^{-2}$.

The SI unit of pressure is also called pascal (Pa), named after Blaise Pascal.

Given

• Mass of the wooden block $= 5 \text{ kg}$
• Dimensions $= 40 \text{ cm} \times 20 \text{ cm} \times 10 \text{ cm}$

To Find

Pressure exerted by the wooden block on the table top with sides of dimensions (a) $20 \text{ cm} \times 10 \text{ cm}$ and (b) $40 \text{ cm} \times 20 \text{ cm}$.

Formula

$\text { Pressure }=\frac{\text { thrust }}{\text { area }}$

Solution

(a) When the block lies on its side of dimensions $20 \text{ cm} \times 10 \text{ cm}$

Thrust = Force = $m \times g = 5 \text{ kg} \times 9.8 \text{ m s}^{-2} = 49 \text{ N}$

Area = length $\times$ breadth = $20 \text{ cm} \times 10 \text{ cm} = 200 \text{ cm}^{2} = 0.02 \text{ m}^{2}$

Pressure = $\frac{49 \text{ N}}{0.02 \text{ m}^{2}} = 2450 \text{ N m}^{-2}$

Pressure exerted by the side $20 \text{ cm} \times 10 \text{ cm}$ is $2450 \text{ N m}^{-2}$

---

(b) When the block lies on its side of dimensions $40 \text{ cm} \times 20 \text{ cm}$

Thrust = Force = $m \times g = 5 \text{ kg} \times 9.8 \text{ m s}^{-2} = 49 \text{ N}$

Area = length $\times$ breadth = $40 \text{ cm} \times 20 \text{ cm} = 800 \text{ cm}^{2} = 0.08 \text{ m}^{2}$

Pressure = $\frac{49 \text{ N}}{0.08 \text{ m}^{2}} = 612.5 \text{ N m}^{-2}$

Pressure exerted by the side $40 \text{ cm} \times 20 \text{ cm}$ is $612.5 \text{ N m}^{-2}$

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Thus, the same force acting on a smaller area exerts a larger pressure, and a smaller pressure on a larger area. This is why nails have pointed tips, knives have sharp edges, and buildings have wide foundations.

Pressure in fluids

Liquids and gases are fluids. Solids exert pressure due to their weight. Fluids also have weight and exert pressure on the base and walls of their containers. Pressure exerted in a confined fluid is transmitted equally in all directions.

Buoyancy

Have you ever felt lighter while swimming? Or noticed a bucket of water feels heavier outside the well? Why doesn't a steel ship sink, while a steel sheet does? This is due to buoyancy.

When you immerse an object in a fluid, it experiences an upward force called upthrust or buoyant force. All objects experience this force when immersed in a fluid. The magnitude of buoyant force depends on the fluid's density.

Why objects float or sink WHEN PLACED ON THE SURFACE OF WATER?

An iron nail sinks in water. The gravitational force pulls it downwards, while the upthrust of water pushes it upwards. The downward force is greater, so it sinks.

A cork floats. The density of cork is less than the density of water. This means the upthrust of water on the cork is greater than the weight of the cork, so it floats.

An iron nail sinks because its density is more than the density of water. This means the upthrust of water on the nail is less than the weight of the nail.

Objects with density less than the liquid float. Objects with density greater than the liquid sink.

Archimedes' Principle

When you suspend a stone from a rubber string or spring balance and gradually lower the stone in water, the elongation of the string or the reading on the balance decreases. This is because the water exerts an upward force on the stone. This upward force exerted by water is known as the force of buoyancy.

Archimedes' principle states: When a body is immersed fully or partially in a fluid, it experiences an upward force that is equal to the weight of the fluid displaced by it.

Archimedes' principle is used in designing ships and submarines. Lactometers (used to determine milk purity) and hydrometers (used to determine liquid density) are based on this principle.