Physics..1st Year..Chapter 2
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
No comments:
Post a Comment