Gravitation

Mass

Mass is the measure of inertia of an object. Mass increases with increase of inertia and decreases with decrease in inertia.

Mass of an object remains the same at any place. Whether an object is on earth, on moon, in the water, in the air, anywhere, mass remains the same.

This means, the mass of an object is constant and does not change from place to place.

Weight

The earth attracts every object with a certain force towards it. This force is called gravity. The force of attraction of an object towards it depends upon the mass of the object.

The force with which an object is attracted towards earth is called weight of the object.

According to Second Law of Motion, We know that, `F=mxxa`

After replacing the acceleration (a) with acceleration due to gravity (g), we get

`F=mxxg`

This force of attraction (F) of the earth on an object is known as the weight of the object. And it is denoted by `W`.

Thus, `W=mxxg` -----------(vii)

SI unit of weight

Since weight of an object is the force with which it is attracted towards earth, thus the SI unit of weight is the same as Force (F)

That is, SI unit of Weight = N (Newton)

Weight is a force acting vertically downwards; it has magnitude and direction both. Thus, weight is a vector quantity.

Variation of Weight with mass

Since value of `g` is constant at a given place, thus at given place weight of an object is directly proportional to its mass.

This means, `W prop m`.

This is due to this reason that at a given place, we can use the weight of an object as a measure of its mass.

The mass of an object remains the same everywhere. This means on the earth and on any planet mass of an object remains the same while weight depends on its location.

Weight of an object varies with the variation of `g`.

Weight of a particular object directly proportional to the value of `g`.

Weight of an Object on the Moon

Weight of an object depends upon the acceleration due to gravity (g). And acceleration due to gravity depends upon the Mass and Radius of an object.

Since mass of the Moon is less than that of the earth. Because of this moon exerts lesser force of attraction on an object.

Let mass of an object = m

Let, weight of the object at earth = W_e

The mass of earth is = M_e

The radius of earth is = R_e

And again let, the weight of moon is = M_m

The mass of moon is = M_m

The radius of moon is = R_m

We know that by experiment and calculation that

Mass of Earth = 5.98 × 10²⁴ kg

And mass of moon = 7.36 × 10²² kg

And radius of earth = 6.37 × 10⁶ m

And radius of moon = 1.74 × 10⁶ m

Now, by Universal Law of Gravitation, we know that,

`F=G(Mxxm)/d^2`

Here, since, distance is equal to the radius, and hence `d` is replaced by `R`

Again, we know that weight on earth is the force by which earth pulls an object towards it.

Thus, weight of the given object which mass is equal to `m` on earth,

`W_e=G(M_exxm)/R_e^2`

`W_e=(5.98xx10^24kgxxm)/(6.37xx10^6m)^2`

`=>W_e = 1.474xx10^11Gxxm` -----------(viii)

And the weight of that given object which mass is equal to `m` on the moon

`W_m=G(7.36xx10^22kgxxm)/(1.74xx10^6m)^2`

`=>W_m= 1.474xx10^11\ Gxxm` ---------- (ix)

Now, ratio of weight of object at moon and weight of object at earth

`{text(weight of object on moon)}/{text(weight of object on earth)}`

`=W_m/W_e=(2.431xx10^10)/(1.474xx10^11)=0.165`

`=>W_m/W_e\ ~~1/6`

Thus,

`{text(weight of object on moon)}/{text(weight of object on earth)} = 1/6`

This means,

Weight of an object on the moon = 1/6 of the weight of the object on the earth.

Example Problem (1) If weight of an object on earth is equal to 24 N, the what will be weight of that object on the surface of moon?

Solution:

Given, weight of the given object on earth (W_e) = 24 N

Then weight of that object on moon (W_m) = ?

We know that,

Weight of an object on moon = 1/6 × weight of the object on earth.

`=1/6xx24\ N`

= 4 N

Thus, weight of the given object on the surface of moon = 4 N Answer

Example Problem (2) Weight of an object on moon is equal to 10 N, then find the weight of the object on the surface of earth.

Solution:

Given, weight of the object on moon = 10 N

Then, weight of the object on earth = ?

Weight of an object on earth = 6 × weight of the object on moon

= 6 × 10 N = 60 N

Thus, weight of the given object on earth = 60 N Answer

Example problem (3) Find the weight of an object on earth of which mass is equal to 50 kg. [value of g on earth = 9.8 m s^–2]

Solution:

Given, mass (m) = 50 kg

And, value of g on earth = 9.8 m s^–2

Weight (W) on earth = ?

We know that, W = m × g

= 50 kg 7#215; 9.8 m s^–2

= 490 kg m s^–2

= 490 N

Thus, weight of the given object on earth = 490 N Answer

Thrust And Pressure

Definition of Thrust

The force acting on an object perpendicular to the surface is called thrust.

Or the net force in a particular direction is called thrust.

SI Unit of Thrust

Since thrust is a force, hence SI unit of Thrust is Newton (N).

Definition of Pressure

The force per unit area is called pressure. Or the thrust on unit area is called pressure.

Thus, `{text(Pressure)}={text(thrust)}/{text(area)}`

SI unit of Pressure

SI unit of pressure is obtained by substituting SI units of thrust and area.

SI unit of thrust is Newton (N) and SI unit of area = m²

Thus, SI unit of pressure = N m^–2

In the honour of scientist Blaise Pascal, the SI unit of pressure is called pascal. Pascal is denoted as `Pa`

Example:

When one stand on loose sand. His feet go deep into the sand but when same person lie down on the sand, his body does not go that deep in the sand. While in both of the cases, the force exerted on the sand is the weight of his body.

In the case of standing, the whole body weight exerts pressure on sand through legs only while on lying position same weight exerts pressure on sand through larger area, i.e. through body.

The area of legs is very smaller than the area of body. This is the cause in the standing position total pressure goes downward via small area, thus pressure is greater while standing. And hence feet go deep in the loose stand.

Pressure is indirectly proportion to the area.

This means that pressure increases with decrease in area and pressure decreases with increase in area.

Why camel can run in desert easily?

Camel has broader paws. Because of broader paws the body weight of camel is distributed to larger area which decreases the pressure on sand. Less pressure on sand prevents camel paws going deep into sand. This facilitates camels run in sand easily.

Nail has one end sharp and pointed while other end is broader, why?

Pointed end of nails are meant to pierce in the wood or wall, while broader end is meant to hammer.

Because of sharp and pointed nail, in the course of hammering, great pressure exerts on sharp end of nail because of smaller pointed area. And since pressure increases with decrease in area, this ease the penetration of nail into wall or wood.

This is the cause that nails have sharp and pointed tips.

Sample Problem (4) A brick has dimension of 30 cm 15 cm and 10 cm. The mass of this brick is of 5 kg. This brick can be put on ground in three arrangements i.e.

(a) with side of its dimension 15 cm × 10 cm on ground

(b) with side of its dimension 30 cm × 10 cm on ground

(c) with side of its dimension 30 cm × 15 cm on ground

Find thrust and pressure exerted on ground in each of the three arrangements. [Value of g = 9.8 m s^–2]

Solution:

Given, dimension of brick = 30 cm × 15 cm× 10 cm

And mass of brick (m) = 5 kg

Then, thrust = ?

And pressure in given three arrangements = ?

Calculation of Thrust

We know that, Thrust = F = m 7#215; g

⇒ F = 5 kg 7#215; 9.8 m s^–2

⇒ F = 49 N

Since mass of the brick is constant, thus thrust will be same in all the conditions i.e. in all the arrangements of keeping brick on the ground.

(a) Calculation of Pressure with side of its dimension 15 cm × 10 cm on ground

Here, given, side = 15 cm × 10 cm

`15\ cm = 15/100\ m` = 0.15 m

And, `10\ cm = 10/100\ m` = 0.1 m

Thus, area of side = 0.15 m × 0.1 m

= 0.015 m²

Now, we know that, Pressure `={text(thrust)}/{text(area)}`

`= (49\ N)/(0.015\ m^2)`

= 3266.66 N m²

(b) Calculation of Pressure with side of its dimension 30 cm × 10 cm on ground

Here, given, side = 30 cm × 10 cm

`30\ cm = 30/100\ m` = 0.3 m

And, `10\ cm = 10/100\ m` = 0.1 m

Thus, area of side = 0.3 m × 0.1 m

= 0.03 m²

Now, we know that, Pressure `={text(thrust)}/{text(area)}`

`= (49\ N)/(0.03\ m^2)`

= 1633.33 N m²

(c) Calculation of Pressure with side of its dimension 30 cm × 15 cm on ground

Here, given, side = 30 cm × 15 cm

`30\ cm = 30/100\ m` = 0.3 m

And, `15\ cm = 15/100\ m` = 0.15 m

Thus, area of side = 0.3 m × 0.15 m

= 0.045 m²

Now, we know that, Pressure `={text(thrust)}/{text(area)}`

`= (49\ N)/(0.045\ m^2)`

= 1088.88 N m²

Thus, Thrust in all cases = 49 N

And, pressure (a) 3266.66 N m²

(b) = 1633.33 N m²

(c) 1088.88 N m²

Pressure in Fluids

Substances that flow are called Fluids. Since, gases and liquids flow or have tendency to flow, thus gases and liquids are called fluids. This means all gases and liquids are fluids.

A solid exerts pressure on a surface due to its weights. In similar way, fluids have weight. Fluids exert pressure on the base and walls of the container in which they are kept or enclosed.

Pressure exerted in any confined mass of fluid is transmitted equally in all directions.

Buoyancy

The upward force exerted by water or other liquid on the object which is submerged into it is known as BUOYANCY or BUOYANT FORCE. Buoyancy is known as upthrust also.

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.

Example:

When an empty bottle with closed mouth is immersed in water downwards, water pushes the bottle in upward direction. This push of water is due to buoyant force exerted by water on bottle.

When the bottle is immersed, the upward force, i.e. buoyant force exerted by the water on the bottle is greater than its weight. Therefore it rises up when released.

To keep the bottle completely immersed the upward force on the bottle due to water must be balanced. This is achieved by an externally applied force acting downwards. This force must be at least equal to the difference between the upward force and the weight of the bottle.

An empty bottle floats in water and does not sink. This happens because of buoyancy of water exerted on bottle.

Why an object does float or Sink when placed on the surface of water?

Earth pulls an object towards it because of this gravitation of earth an object stays on earth.

When and object is kept on the surface of water, earth pulls the object towards it but water pushes the object upwards because of buoyancy.

If the gravitational force applied on the object because of mass of object is greater than that of the upward thrust of water, object sinks in water, or if upward thrust is greater than that of inward force, i.e. pull of earth, then object floats in water.

This means that if downward force acting upon an object is greater than the upthrust of liquid, then object will sink. And if downward force acting upon an object is lesser than the upthrust of liquid, then object will float.

And if the density of an object is greater than the density of liquid, then object will sink in the liquid. And if the density of the liquid is greater than the density of the object, then object will float in the liquid.

Why a nail made of iron sinks in water where a cork bigger than nail does float?

The density of an iron nail is more than the density of water. This means that the

Archimedes' Principle

Archimedes' Principle states that "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 was a Greek scientist who discovered this principle after noticing that water in bathtub overflowed when he stepped into it. This is known as Archimedes' Principle after his name.

Application of Archimedes' Principle

Archimedes' Principle is used in the designing of ships and submarines.

Lactometer; which is used to determine the purity of a sample of milk works on Archimedes' Principle.

Hydrometers; which are used to determine the density of liquids work on Archimedes' Principle.

Density

Density of a substance is defined as the mass per unit volume.

This means mass of a substance in it is termed as density.

Density is denoted by Greek Letter `rho` (Rho) or `D`

Thus, Density (`rho`) `={text(mass)(m)}/{text(volume)(V)}`

The SI unit of Density

Since, density is the mass per volume thus, SI unit of density can be obtained by substituting SI unit of mass and volume.

The SI unit of mass is kilogram (kg). And the SI unit of volume is cubic meter (m³).

Thus, SI unit of density is equal to kg m^–3

The density of a given substance, under specific conditions, remains the same. Thus, density of a substance is one of the characteristic properties.

Example:

Density of gold = 19300 kg m^–3

Density of water = 1000 kg m^–3

Since, density is a characteristic properties of a substance, thus density helps in determining the purity of a substance.

Relative Density

Expressing of density of a substance generally compare to the density of water is called Relative Density.

Thus, the Relative Density of a substance is the ratio of its density to that of water.

Or, Relative density `={text(Density of a substance)}/{text(Density of water)}`

Since, relative density of a substance is the ratio of two densities, thus Relative Density has no unit.