## 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