Motion | 1st year– Class Notes

 Motion

Definition

If an object continuously changes its position with respect to its surrounding, then it is said to be in state of motion.

Rectilinear Motion
The motion along a straight line is called rectilinear motion.

Velocity

Velocity may be defined as the change of displacement of a body with respect time.
Velocity = change of displacement / time
Velocity is a vector quantity and its unit in S.I system is meter per second (m/sec).

Average Velocity
Average velocity of a body is defined as the ratio of the displacement in a certain direction to the time taken for this displacement.
Suppose a body is moving along the path AC as shown in figure. At time t1, suppose the body is at P and its position w.r.t origin O is given by vector r2.

Diagram Coming Soon

Thus, displacement of the body = r2 – r1 = Δr
Time taken for this displacement – t2 – t1 = Δt
Therefore, average velocity of the body is given by
Vav = Δr / Δt

Instantaneous Velocity
It is defined as the velocity of a body at a certain instant.
V(ins) = 1im Δr / Δt
Where Δt → 0 is read as “Δt tends to zero”, which means that the time is very small.

Velocity From Distance – Time Graph
We can determine the velocity of a body by distance – time graph such that the time is taken on x-axis and distance on y-axis.

Acceleration

Acceleration of a body may be defined as the time rate of change velocity. If the velocity of a body is changing then it is said to posses acceleration.
Acceleration = change of velocity / time
If the velocity of a body is increasing, then its acceleration will be positive and if the velocity of a body is decreasing, then its acceleration will be negative. Negative acceleration is also called retardation.
Acceleration is a vector quantity and its unit in S.I system is meter per second per second. (m/sec2 OR m.sec-2)

Average Acceleration
Average acceleration is defined as the ratio of the change in velocity of a body and the time interval during which the velocity has changed.
Suppose that at any time t1 a body is at A having velocity V1. At a later time t2, it is at point B having velocity V2. Thus,
Change in Velocity = V2 – V1 = Δ V
Time during which velocity has changed = t2 – t1 = Δ t

Instantaneous Acceleration
It is defined as the acceleration of a body at a certain instant
a(ins) = lim Δ V / Δ t
where Δt → 0 is read as “Δt tends to zero”, which means that the time is very small.

Acceleration from Velocity – Time Graph
We can determine the acceleration of a body by velocity – time graph such that the time is taken on x-axis and velocity on y-axis.

Equations of Uniformly Accelerated Rectilinear Motion
There are three basic equations of motion. The equations give relations between
Vi = the initial velocity of the body moving along a straight line.
Vf = the final velocity of the body after a certain time.
t = the time taken for the change of velocity
a = uniform acceleration in the direction of initial velocity.
S = distance covered by the body.
Equations are
1. Vf = Vi + a t
2. S = V i t + 1/2 a t2
3. 2 a S = V f2 – V i 2

Motion Under Gravity
The force of attraction exerted by the earth on a body is called gravity or pull of earth. The acceleration due to gravity is produced in a freely falling body by the force of gravity. Equations for motion under gravity are
1. Vf = Vi + g t
2. S = V i t + 1/2 g t2
3. 2 g S = Vr2 – Vi2
where g = 9.8 m / s2 in S.I system and is called acceleration due to gravity.

Law of Motion

Isaac Newton studied motion of bodies and formulated three famous laws of motion in his famous book “Mathematical Principles of Natural Philosophy” in 1687. These laws are called Newton’s Laws of Motion.

Newton’s First Law of Motion

Statement
A body in state of rest will remain at rest and a body in state of motion continues to move with uniform velocity unless acted upon by an unbalanced force.

Explanation
This law consists of two parts. According to first part a body at rest will remain at rest will remain at rest unless some external unbalanced force acts on it. It is obvious from our daily life experience. We observe that a book lying on a table will remain there unless somebody moves it by applying certain force. According to the second part of this law a body in state of uniform motion continuous to do so unless it is acted upon by some unbalanced force.
This part of the law seems to be false from our daily life experience. We observe that when a ball is rolled in a floor, after covering certain distance, it stops. Newton gave reason for this stoppage that force of gravity friction of the floor and air resistance are responsible of this stoppage which are, of course, external forces. If these forces are not present, the bodies, one set into motion, will continue to move for ever.

Qualitative Definition of Net Force
The first law of motion gives the qualitative definition of the net force. (Force is an agent which changes or tends to change the state of rest or of uniform motion of a body).

First Law as Law of Inertia
Newton’s first law of motion is also called the Law of inertia. Inertia is the property of matter by virtue of which is preserves its state of rest or of uniform motion. Inertia of a body directly related to its mass.

Newton’s Second Law of Motion

Statement
If a certain unbalanced force acts upon a body, it produces acceleration in its own direction. The magnitude of acceleration is directly proportional to the magnitude of the force and inversely proportional to the mass of the body.

Mathematical Form
According to this law
f ∞ a
F = m a → Equation of second law
Where ‘F’ is the unbalanced force acting on the body of mass ‘m’ and produces an acceleration ‘a’ in it.
From equation
1 N = 1 kg x 1 m/sec2
Hence one newton is that unbalanced force which produces an acceleration of 1 m/sec2 in a body of mass 1 kg.

Vector Form
Equation of Newton’s second law can be written in vector form as
F = m a
Where F is the vector sum of all the forces acting on the body.

Newton’s Third Law of Motion

Statement
To every action there is always an equal and opposite reaction.

Explanation
For example, if a body A exerts force on body B (F(A) on B) in the opposite direction. This force is called reaction. Then according to third law of motion.

Examples
1. When a gun is fired, the bullet flies out in forward direction. As a reaction of this action, the gun reacts in backward direction.
2. A boatman, when he wants to put his boat in water pushes the bank with his oar, The reaction of the bank pushes the boat in forward direction.
3. While walking on the ground, as an action, we push the ground in the backward direction. As a reaction ground pushes us in the forward direction.
4. In flying a kite, the string is given a downward jerk and is then released. Thereupon the reaction of the air pushes the kite upward and makes it rise higher.

Tension in a String
Consider a body of weight W supported by a person with the help of a string. A force is experienced by the hand as well as by the body. This force is known as Tension. At B the hand experiences a downward force. So the direction of force at point B is downward. But at point A direction of the force is upward.
These forces at point A and B are tensions. Its magnitude in both cases is same but the direction is opposite. At point A,
Tension = T = W = mg

Momentum of a Body

The momentum of a body is the quantity of motion in it. It depends on two things
1. The mass of the object moving (m),
2. The velocity with which it is moving (V).
Momentum is the product of mass and velocity. It is denoted by P.
P = m V
Momentum is a vector quantity an its direction is the same as that of the velocity.

Unit of Momentum
Momentum = mass x velocity
= kg x m/s
= kg x m/s x s/s
= kg x m/s2 x s
since kg. m/s2 is newton (N)
momentum = N-s
Hence the S.I unit of momentum is N-s.

Unbalanced or Net Force is equal to the Rate of Change of Momentum
i.e., F = (mVf = mVi) / t

Proof
Consider a body of mass ‘m’ moving with a velocity Vl. A net force F acts on it for a time ‘t’. Its velocity then becomes Vf.
Therefore
Initial momentum of the body = m Vi
Final momentum of the body = m Vf
Time interval = t
Unbalanced force = F
Therefore
Rate of change of momentum = (m Vf – m Vi) / t ………………….. (1)
But
(Vf – Vi) / t = a
Therefore,
Rate of change of momentum = m a = F ………………… (2)
Substituting the value of rate of change of momentum from equation (2) in equation (1), we get
F = (m Vf – m Vi) / t ……………………….. Proved

Law of Conservation of Momentum

Isolated System
When a number of bodies are such that they exert force upon one another and no external agency exerts a force on them, then they are said to constitute and isolated system.

Statement of the Law

The total momentum of an isolated system of bodies remains constant.

OR

If there is no external force applied to a system, then the total momentum of that system remains constant.

Elastic Collision
An elastic collision is that in which the momentum of the system as well as the kinetic energy of the system before and after collision, remains constant. Thus for an elastic collision.
If P momentum and K.E is kinetic energy.
P(before collision) = P(after collision)
K.E(before collision) = K.E(after collision)

Inelastic Collision
An inelastic collision is that in which the momentum of the system before and after the collision remains constant but the kinetic energy before and after the collision changes.
Thus for an inelastic collision
P(before collision) = P(after collision)

Elastic Collision in one Dimension
Consider two smooth non rotating spheres moving along the line joining their centres with velocities U1 and U2. U1 is greater than U2, therefore the spheres of mass m1 makes elastic collision with the sphere of mass m2. After collision, suppose their velocities become V1 and V2 but their direction of motion is along same line as before.

Friction

When two bodies are in contact, one upon the other and a force is applied to the upper body to make it move over the surface of the lower body, an opposing force is set up in the plane of the contract which resists the motion. This force is the force of friction or simply friction.

The force of friction always acts parallel to the surface of contact and opposite to the direction of motion.

Definition
When one body is at rest in contact with another, the friction is called Static Friction.

When one body is just on the point of sliding over the other, the friction is called Limiting Friction.

When one body is actually sliding over the other, the friction is called Dynamic Friction.

Coefficient of Friction (μ)
The ratio of limiting friction ‘F’ to the normal reaction ‘R’ acting between two surfaces in contact is called the coefficient of friction (μ).
μ = F / R
Or
F = μ R

Fluid Friction
Stoke found that bodies moving through fluids (liquids and gases) experiences a retarding force fluid friction or viscous drag. If the moving bodies are spheres then fluid friction F is given by
F = 6 π η r v
Where η is the coefficient of viscosity,
Where r is the radius of the sphere,
Where v is velocity pf the sphere.

Terminal Velocity
When the fluid friction is equal to the downward force acting on the sphere, the sphere attains a uniform velocity. This velocity is called Terminal velocity.

The Inclined Plane

A plane which makes certain angle θ with the horizontal is called an inclined plane.

Diagram Coming Soon

Consider a block of mass ‘m’ placed on an inclined plane making certain angle θ with the horizontal. The forces acting on the block are
1. W, weight of the block acting vertically downward.
2. R, reaction of the plane acting perpendicular to the plane
3. f, force of friction which opposes the motion of the block which is moving downward.

Diagram Coming Soon

Now we take x-axis along the plane and y-axis perpendicular to the plane. We resolve W into its rectangular components.
Therefore,
Component of W along x-axis = W sin θ
And
Component of W along y-axis = W cos θ

1. If the Block is at Rest
According to the first condition of equilibrium
Σ Fx = 0
Therefore,
f – W sin θ = 0
Or
f = W sin θ
Also,
Σ Fy = 0
Therefore,
R – W cos θ = 0
Or
R = W cos θ

2. If the Block Slides Down the Inclined Plane with an Acceleration
Therefore,
W sin θ > f
Net force = F = W sin θ – f
Since F = m a and W = m g
Therefore,
m a = m g sin θ – f

3. When force of Friction is Negligible
Then f ≈ 0
Therefore,
equation (3) => m a = m g sin ≈ – 0
=> m a = m g sin ≈
or a = g sin ≈ …………. (4)

Particular Cases
Case A : If the Smooth Plane is Horizontal Then 0 = 0º
Therefore,
Equation (4) => a = g sin 0º
=> a = g x 0
=> a = 0

Case B : If the Smooth Plane is Vertical Then θ = 90º
Therefore,
Equation (4) => a = g sin 90º
=> a = g x 1
=> a = g
This is the case of a freely falling body.

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Motion in two Dimension | 1st year– Class Notes

Motion in two Dimension

Projectile Motion

A body moving horizontally as well as vertically under the action of gravity simultaneously is called a projectile. The motion of projectile is called projectile motion. The path followed by a projectile is called its trajectory.

Examples of projectile motion are
1. Kicked or thrown balls
2. Jumping animals
3. A bomb released from a bomber plane
4. A shell of a gun.

Analysis of Projectile Motion

Let us consider a body of mass m, projected an angle θ with the horizontal with a velocity V0. We made the following three assumptions.

1. The value of g remains constant throughout the motion.
2. The effect of air resistance is negligible.
3. The rotation of earth does not affect the motion.

Horizontal Motion
Acceleration : ax = 0
Velocity : Vx = Vox
Displacement : X = Vox t

Vertical Motion
Acceleration : ay = – g
Velocity : Vy = Voy – gt
Displacement : Y = Voy t – 1/2 gt2

Initial Horizontal Velocity
Vox = Vo cos θ …………………. (1)

Initial Vertical Velocity
Voy = Vo sin θ …………………. (2)

Net force W is acting on the body in downward vertical direction, therefore, vertical velocity continuously changes due to the acceleration g produced by

the weight W.

There is no net force acting on the projectile in horizontal direction, therefore, its horizontal velocity remains constant throughout the motion.

X – Component of Velocity at Time t (Vx)
Vx = Vox = Vo cos θ ……………….. (3)

Y – Component of Velocity at Time t (Vy)
Data for vertical motion
Vi = Voy = Vo sin θ
a = ay = – g
t = t
Vf = Vy = ?
Using Vf = Vi + at
Vy = Vo sin θ – gt ……………….. (4)

Range of the Projectile (R)
The total distance covered by the projectile in horizontal direction (X-axis) is called is range
Let T be the time of flight of the projectile.
Therefore,
R = Vox x T ………….. {since S = Vt}
T = 2 (time taken by the projectile to reach the highest point)
T = 2 Vo sin θ / g
Vox = Vo cos θ
Therefore,
R = Vo cos θ x 2 Vo sin θ / g
R = Vo2 (2 sin θ cos θ) / g
R = Vo2 sin 2 θ / g ……………… { since 2 sin θ cos θ = sin2 θ}
Thus the range of the projectile depends on
(a) The square of the initial velocity
(b) Sine of twice the projection angle θ.

The Maximum Range
For a given value of Vo, range will be maximum when sin2 θ in R = Vo2 sin2 θ / g has maximum value. Since
0 ≤ sin2 θ ≤ 1
Hence maximum value of sin2 θ is 1.
Sin2 θ = 1
2θ = sin(-1) (1)
2θ = 90º
θ = 45º
Therefore,
R(max) = Vo2 / g ; at θ = 45º
Hence the projectile must be launched at an angle of 45º with the horizontal to attain maximum range.

Projectile Trajectory
The path followed by a projectile is referred as its trajectory.
We known that
S = Vit + 1/2 at2
For vertical motion
S = Y
a = – g
Vi = Voy = Vo sin θ
Therefore,
Y = Vo sinθ t – 1/2 g t2 ………………….. (1)
Also
X = Vox t
X = Vo cosθ t ………… { since Vox = Vo cosθ}
t = X / Vo cos θ

(1) => Y = Vo sinθ (X / Vo cos θ) – 1/2 g (X / Vo cos θ)2
Y = X tan θ – gX2 / 2Vo2 cos2 θ
For a given value of Vo and θ, the quantities tanθ, cosθ, and g are constant, therefore, put
a = tan θ
b = g / Vo2 cos2θ
Therefore
Y = a X – 1/2 b X2
Which shows that trajectory is parabola.

Uniform Circular Motion

If an object moves along a circular path with uniform speed then its motion is said to be uniform circular motion.

Recitilinear Motion
Displacement → R
Velocity → V
Acceleration → a

Circular Motion
Angular Displacement → θ
Angular Velocity → ω
Angular Acceleration → α

Angular Displacement
The angle through which a body moves, while moving along a circular path is called its angular displacement.
The angular displacement is measured in degrees, revolutions and most commonly in radian.

Diagram Coming Soon

s = arc length
r = radius of the circular path
θ = amgular displacement

It is obvious,
s ∞ θ
s = r θ
θ = s / r = arc length / radius

Radian
It is the angle subtended at the centre of a circle by an arc equal in length to its radius.
Therefore,
When s = r
θ = 1 radian = 57.3º

Angular Velocity
When a body is moving along a circular path, then the angle traversed by it in a unit time is called its angular velocity.

Diagram Coming Soon

Suppose a particle P is moving anticlockwise in a circle of radius r, then its angular displacement at P(t1) is θ1 at time t1 and at P(t2) is θ2 at time t2.
Average angular velocity = change in angular displacement / time interval
Change in angular displacement = θ2 – θ1 = Δθ
Time interval = t2 – t1 = Δt
Therefore,
ω = Δθ / Δt
Angular velocity is usually measured in rad/sec.
Angular velocity is a vector quantity. Its direction can be determined by using right hand rule according to which if the axis of rotation is grasped in right hand with fingers curled in the direction of rotation then the thumb indicates the direction of angular velocity.

Angular Acceleration
It is defined as the rate of change of angular velocity with respect to time.
Thus, if ω1 and ω2 be the initial and final angular velocity of a rotating body, then average angular acceleration “αav” is defined as
αav = (ω2 – ω1) / (t2 – t1) = Δω / Δt
The units of angular acceleration are degrees/sec2, and radian/sec2.
Instantaneous angular acceleration at any instant for a rotating body is given by
Angular acceleration is a vector quantity. When ω is increasing, α has same direction as ω. When ω is decreasing, α has direction opposite to ω.

Relation Between Linear Velocity And Angular Velocity
Consider a particle P in an object in X-Y plane rotating along a circular path of radius r about an axis through O, perpendicular to the plane of figure as shown here (z-axis).

If the particle P rotates through an angle Δθ in time Δt,
Then according to the definition of angular displacement.
Δθ = Δs / r
Dividing both sides by Δt,
Δθ / Δt = (Δs / Δt) (1/r)
=> Δs / Δt = r Δθ / Δt
For a very small interval of time
Δt → 0

Alternate Method
We know that for linear motion
S = v t ………….. (1)
And for angular motion
S = r θ …………….. (2)
Comparing (1) & (2), we get
V t = r θ
v = r θ/t
V = r ω ……………………… {since θ/t = ω}

Relation Between Linear Acceleration And Angular Acceleration
Suppose an object rotating about a fixed axis, changes its angular velocity by Δω in Δt. Then the change in tangential velocity, ΔVt, at the end of this interval is
ΔVt = r Δω
Dividing both sides by Δt, we get
ΔVt / Δt = r Δω / Δt
If the time interval is very small i.e., Δt → 0 then

Alternate Method
Linear acceleration of a body is given by
a = (Vr – Vi) / t
But Vr = r ω r and Vi = r ω i
Therefore,
a = (r ω r – r ω i) / t
=> a = r (ωr- ωi) / t
a = r α ……………………………… {since (ωr = ωi) / t = ω}

Time Period
When an object is rotating in a circular path, the time taken by it to complete one revolution or cycle is called its time period, (T).
We know that
ω = Δθ / Δt OR Δt = Δθ / ω
For one complete rotation
Δθ = 2 π
Δt = T
Therefore,
T = 2 π / ω
If ω = 2πf …………………… {since f = frequency of revolution}
Therefore,
T = 2π / 2πf
=> T = 1 / f

Tangential Velocity
When a body is moving along a circle or circular path, the velocity of the body along the tangent of the circle is called its tangential velocity.
Vt = r ω
Tangential velocity is not same for every point on the circular path.

Centripetal Acceleration
A body moving along a circular path changes its direction at every instant. Due to this change, the velocity of the body ‘V’ is changing at every instant. Thus body has an acceleration which is called its centripetal acceleration. It is denoted by a(c) or a1 and always directed towards the centre of the circle. The magnitude of the centripetal acceleration a(c) is given as follows
a(c) = V2 / r, ……………………… r = radius of the circular path

Prove That a(c) = V2 / r

Proof
Consider a body moving along a circular path of radius of r with a constant speed V. Suppose the body moves from a point P to a point Q in a small time Δt. Let the velocity of the body at P is V1 and at Q is V2. Let the angular displacement made in this time be ΔO .
Since V1 and V2 are perpendicular to the radial lines at P and Q, therefore, the angle between V1 and V2 is also Δ0, Triangles OPQ and ABC are similar.
Therefore,
|ΔV| / |V1| = Δs / r
Since the body is moving with constant speed
Therefore,
|V1| = |V2| = V
Therefore,
ΔV / V = Vs / r
ΔV = (V / r) Δs
Dividing both sides by Δt
Therefore,
ΔV / Δt = (V/r) (V/r) (Δs / Δt)
taking limit Δt → 0.

Proof That a(c) = 4π2r / T2

Proof
We know that
a(c) = V2 / r
But V = r ω
Therefore,
a(c) = r2 ω2 / r
a(c) = r ω2 …………………. (1)
But ω = Δθ / Δt
For one complete rotation Δθ = 2π, Δt = T (Time Period)
Therefore,
ω = 2π / T
(1) => a(c) = r (2π / T)2
a(c) = 4 π2 r / T2 ……………… Proved

Tangential Acceleration
The acceleration possessed by a body moving along a circular path due to its changing speed during its motion is called tangential acceleration. Its direction is along the tangent of the circular path. It is denoted by a(t). If the speed is uniform (unchanging) the body do not passes tangential acceleration.

Total Or Resultant Acceleration
The resultant of centripetal acceleration a(c) and tangential acceleration a(t) is called total or resultant acceleration denoted by a.

Centripetal Force
If a body is moving along a circular path with a constant speed, a force must be acting upon it. Direction of the force is along the radius towards the centre. This force is called the centripetal force by F(c).
F(c) = m a(c)
F(c) = m v(2) / r ………………… {since a(c) = v2 / r}
F(c) = mr2 ω2 r ………………….. {since v = r ω}
F(c) = mrω2

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Gravitation | 1st year– Class Notes

Gravitation

The property of all objects in the universe which carry mass, by virtue of which they attract one another, is called Gravitation.

Centripetal Acceleration of the Moon

Newton, after determining the centripetal acceleration of the moon, formulated the law of universal gravitation.
Suppose that the moon is orbiting around the earth in a circular orbit.
If V = velocity of the moon in its orbit,
Rm = distance between the centres of earth and moon,
T = time taken by the moon to complete one revolution around the earth.
For determining the centripetal acceleration of the moon,. Newton applied Huygen’s formula which is
a(c) = v2 / r
For moon, am = v2 / Rm ………………… (1)
But v = s/t = circumference / time period = 2πRm/T
Therefore,
v2 = 4π2Rm2 / T2
Therefore,
=> a(m) = (4π2Rm2/T2) x (1/Rm)
a(m) = 4π2Rm / T2
Put Rm = 3.84 x 10(8) m
T = 2.36 x 10(6) sec
Therefore,
a(m) = 2.72 x 10(-3) m/s2

Comparison Between ‘am’ AND ‘g’
Newton compared the centripetal acceleration of the moon ‘am’ with the gravitational acceleration ‘g’.
i.e., am / g = 1 / (60)2 …………….. (1)
If Re = radius of the earth, he found that
Re2 / Rm2 – 1 / (60)2 ……………………. (2)
Comparing (1) and (2),
am / g = Re2 / Rm2 ………………………………. (3)
From equation (3), Newton concluded that at any point gravitational acceleration is inversely to the square of the distance of that point from the centre of the earth. It is true of all bodies in the universe. This conclusion provided the basis for the Newton’ Law of Universal Gravitation.

Newton’s Law of Universal Gravitation

Consider tow bodies A and B having masses mA and mB respectively.

Diagram Coming Soon

Let,
F(AB) = Force on A by B
F(BA) = Force on B by A
r(AB) = displacement from A to B
r(BA) = displacement from B to A
r(AB) = unit vector in the direction of r(AB).
r(BA) = unit vector in the direction of r(BA).

From a(m) / g = Re2 / Rm2, we have
F(AB) ∞ 1 / r(BA)2 ……………………. (1)
Also,
F(AB) ∞ m(A) …………………………. (2)
F(BA) ∞ m(B)
According to the Newton’s third law of motion
F(AB) = F(BA) ……………….. (for magnitudes)
Therefore,
F(AB) ∞ m(B) ………………………….. (3)
Combining (1), (2) and (3), we get
F(AB) ∞ m(A)m(B) / r(BA)2
F(AB) = G m(A)m(B) / r(BA)2 ……………………… (G = 6.67 x 10(-11) N – m2 / kg2)

Vector Form
F(AB) = – (G m(A)m(B) / r(BA)2) r(BA)
F(BA) = – (G m(B)m(A) / r(AB)22) r(AB)
Negative sign indicates that gravitational force is attractive.

Statement of the Law
“Every body in the universe attracts every other body with a force which is directly proportional to the products of their masses and inversely proportional to the square of the distance between their centres.”

Mass and Average Density of Earth

Let,
M = Mass of an object placed near the surface of earth
M(e) = Mass of earth
R(e) = Radius of earth
G = Acceleration due to gravity
According to the Newton’s Law of Universal Gravitation.
F = G M Me / Re2 ……………………….. (1)
But the force with which earth attracts a body towards its centre is called weight of that body.

Diagram Coming Soon

Therefore,
F = W = Mg
(1) => M g = G M Me / Re2
g = G Me / Re2
Me = g Re2 / G ……………………………. (2)
Put
g = 9.8 m/sec2,
Re = 6.38 x 10(6) m,
G = 6.67 x 10(-11) N-m2/kg2, in equation (2)
(1) => Me = 5.98 x 10(24) kg. ………………. (In S.I system)
Me = 5.98 x 10(27) gm ……………………………… (In C.G.S system)
Me = 6.6 x 10(21) tons
For determining the average density of earth (ρ),
Let Ve be the volume of the earth.
We know that
Density = mass / volume
Therefore,
ρ = Me / Ve ……………………… [Ve = volume of earth]
ρ = Me / (4/3 Π Re3) ………….. [since Ve = 4/3 Π Re3]
ρ = 3 Me / 4 Π Re3
Put,
Me = 5.98 x 10(24) kg
& Re = 6.38 x 10(6) m
Therefore,
ρ = 5.52 x 10(3) kg / m3

Mass of Sun

Let earth is orbiting round the sun in a circular orbit with velocity V.
Me = Mass of earth
Ms = Mass of the sun
R = Distance between the centres of the sun and the earth
T = Period of revolution of earth around sun
G = Gravitational constant
According to the Law of Universal Gravitation
F = G Ms Me / R2 ……………………………… (1)
This force ‘F’ provides the earth the necessary centripetal force
F = Me V2 / R …………………………………….. (2)
(1) & (2) => Me V2 / R = G Ms Me / R2
=> Ms = V2 R / G …………………………………. (3)
V = s / Π = 2Π R / T
=> V2 = 4Π2 R2 / T2
Therefore,
(3) => Ms = (4Π R2 / T2) x (R / G)
Ms = 4Π2 R3 / GT2 …………………………………. (4)
Substituting the value of
R = 1.49 x 10(11),
G = 6.67 x 10(-11) N-m2 / kg2,
T = 365.3 x 24 x 60 x 60 seconds, in equation (4)
We get
Ms = 1.99 x 10(30) kg

Variation of ‘g’ with Altitude
Suppose earth is perfectly spherical in shape with uniform density ρ. We know that at the surface of earth
g = G Me / Re2
where
G = Gravitational constant
Me = Mass of earth
Re = Radius of earth
At a height ‘h’ above the surface of earth, gravitational acceleration is
g = G Me / (Re + h)2
Dividing (1) by (2)
g / g = [G Me / Re2] x [(Re + h)2 / G Me)
g / g = (Re + h)2 / Re2
g / g = [Re + h) / Re]2
g / g = [1 + h/Re]2
g / g = [1 + h/Re]-2
We expand R.H.S using Binomial Formula,
(1 + x)n = 1 + nx + n(n-1) x2 / 1.2 + n(n + 1)(n-2)x3 / 1.2.3 + …
If h / Re < 1, then we can neglect higher powers of h / Re.

Therefore

g / g = 1 – 2 h / Re

g = g (1 – 2h / Re) …………………………… (3)

Equation (3) gives the value of acceleration due to gravity at a height ‘h’ above the surface of earth.

From (3), we can conclude that as the value of ‘h’ increases, the value of ‘g’ decreases.

Variation of ‘g’ with Depth

Suppose earth is perfectly spherical in shape with uniform density ρ.

Let

Re = Radius of earth

Me = Mass of earth

d = Depth (between P and Q)

Me = Mass of earth at a depth ‘d’

At the surface of earth,

g = G Me / Re2 ……………………………….. (1)

At a depth ‘d’, acceleration due to gravity is

g = G Me / (Re – d)2 ……………………… (2)

Me = ρ x Ve = ρ x (4/3) π Re3 = 4/3 π Re3 ρ

Me = ρ x Ve = ρ x (4/3) π (Re – d)3 = 4/3 π (Re – d)3 ρ

Ve = Volume of earth

Substitute the value of Me in (1),

(1) => g = (G / Re2) x (4/3) π Re3 ρ
g = 4/3 π Re ρ G …………………………… (3)
Substitute the value of Me in (2)
g = [G / Re - d)2] x (4/3) π (Re – d)3 ρ
g = 4/3 π (Re – d) ρ G
Dividing (4) by (3)
g / g = [4/3 π (Re - d) ρ G] / [4/3 π Re ρ G]
g / g = (Re – d) / Re
g / g = 1 – d/Re
g / g = g (1 – d / Re) ……………………… (5)
Equation (5) gives the value of acceleration due to gravity at a depth ‘d’ below the surface of earth
From (5), we can conclude that as the value of ‘d’ increases, value of ‘g’ decreases.
At the centre of the earth.
d = Re => Re / d = 1
Therefore,
(5) => g = g (1-1)
g = 0
Thus at the centre of the earth, the value of gravitational acceleration is zero.

Weightlessness in Satellites

An apparent loss of weight experienced by a body in a spacecraft in orbit is called weightlessness.
To discuss weightlessness in artificial satellites, let us take the example of an elevator having a block of mass ;m; suspended by a spring balance attached to the coiling of the elevator. The tension in the thread indicates the weight of the block.
Consider following cases.

1. When Elevator is at Rest
T = m g

2. When Elevator is Ascending with an Acceleration ‘a’
In this case
T > m g
Therefore, Net force = T – mg
m a = T – m g
T = m g + m a
In this case of the block appears “heavier”.

3. When Elevator is Descending with an Acceleration ‘a’
In this case
m g > T
Therefore
Net force = m g – T
m a = m g – T
T = m g – m a
In this case, the body appears lighter

4. When the Elevator is Falling Freely Under the Action of Gravity
If the cable supporting the elevator breaks, the elevator will fall down with an acceleration equal to ‘g’
From (3)
T = m g – m a
But a = g
Therefore
T = m g – m g
T = 0
In this case, spring balance will read zero. This is the state of “weightlessness”.
In this case gravitation force still acts on the block due to the reason that elevator block, spring balance and string all have same acceleration when they fall freely, the weight of the block appears zero.

Artificial Gravity

In artificial satellites, artificial gravity can be created by spinning the space craft about its own axis.
Now we calculate frequency of revolution (v) of a space craft of length 2R to produce artificial gravity in it. Its time period be ‘T’ and velocity is V.

Diagram Coming Soon

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