Gravitation
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Newton’s Law of Gravitation
Consider two bodies of masses m1 and m2 separated by distance d from their centres. The force of attraction between them is directly proportional to the product of their masses and inversely proportional to the square of their separation.
F ∝ m1m2 … (i)
F ∝ 1/d2 … (ii)
Combining (i) and (ii):
F ∝ m1m2/d2
F = Gm1m2/d2 … (iii)
Acceleration Due to Gravity
Let a body of mass m lie on the surface of the Earth of mass M and radius R.
According to Newton’s law of gravitation:
F = GMm/R2 … (i)
If g is the acceleration due to gravity, then by Newton’s second law:
F = mg … (ii)
From (i) and (ii):
mg = GMm/R2
g = GM/R2 … (iii)
The relation shows that acceleration due to gravity is independent of the mass of the falling body.
Variation of Acceleration Due to Gravity
1. Due to the Shape of the Earth
Since g = GM/R2, therefore g ∝ 1/R2.
The polar radius Rp is smaller than the equatorial radius Re. Hence acceleration due to gravity at the pole is greater than at the equator.
gp > ge, when Re > Rp.
2. Due to Height (Altitude)
At the Earth’s surface:
At height h above the Earth’s surface:
Dividing (ii) by (i):
g′/g = R2/(R+h)2
g′/g = 1/(1 + h/R)2
g′/g = (1 + h/R)−2
Using binomial expansion and neglecting higher powers:
g′/g = 1 − 2h/R
g′ = g(1 − 2h/R) … (iii)
Thus acceleration due to gravity decreases as height increases.
3. Due to Depth
Let M be the mass of the Earth, R its radius and ρ its density.
At the surface:
g = GM/R2
Since M = (4/3)πR3ρ,
g = (4/3)πGRρ … (i)
At depth x, the effective radius is (R − x):
g′ = (4/3)πG(R − x)ρ … (ii)
Dividing (ii) by (i):
g′/g = (R − x)/R = 1 − x/R
g′ = g(1 − x/R) … (iii)
At the centre, x = R:
Thus acceleration due to gravity is zero at the centre of the Earth.
Graphical Representation of Variation of g
Gravitational Field
Gravitational Field Intensity
Consider a body of mass m at distance r from the centre of the Earth of mass M.
F = GMm/r2
E = F/m
E = GM/r2
On the surface of the Earth:
Hence, on the Earth’s surface, acceleration due to gravity is equal in magnitude to gravitational field intensity.
Gravitational Potential
For a unit mass at distance x:
F = GM/x2
For a small displacement dx toward the Earth:
dW = F dx = (GM/x2) dx
Total work in bringing unit mass from infinity to distance r:
W = ∫∞r GM x−2 dx
W = GM [−1/x]∞r
W = −GM(1/r − 1/∞)
V = −GM/r
Gravitational Potential Energy
F = GMm/x2
dW = F dx = (GMm/x2) dx
W = ∫∞r GMm x−2 dx
W = −GMm(1/r − 1/∞)
U = −GMm/r
Escape Velocity
Initial kinetic energy = ½mv2.
Work required to take the body from radius R to infinity:
W = ∫R∞ GMm/x2 dx = GMm/R
From the work-energy theorem:
GMm/R = ½mv2
v = √(2GM/R)
Since g = GM/R2,
v = √(2gR)
Escape Velocity of the Earth
Using the values written in the source, g = 9.8 m s−2 and R = 6400 km = 6.4 × 106 m:
Satellite and Orbital Velocity
Gravitational force:
F = GMm/r2 … (i)
Centripetal force:
Fc = mvo2/r … (ii)
Since gravitational force supplies the centripetal force:
GMm/r2 = mvo2/r
vo = √(GM/r) … (iii)
Using GM = gR2:
vo = R√(g/r) … (iv)
Time Period of a Satellite
T = 2πr/vo … (i)
Using vo = R√(g/r):
T = 2πr / [R√(g/r)]
T = (2πr/R)√(r/g)
Height of a Satellite
If h is the height above the Earth’s surface, then r = R + h.
T = [2π(R+h)/R] √((R+h)/g)
T2 = 4π2(R+h)3 / (R2g)
(R+h)3 = T2R2g / 4π2
R+h = [T2R2g / 4π2]1/3
h = [T2R2g / 4π2]1/3 − R
Energy of a Satellite
Consider a satellite of mass m revolving around the Earth of mass M in a circular orbit of radius r.
GMm/r2 = mv2/r
mv2 = GMm/r
K.E. = ½mv2 = GMm/(2r) … (ii)
P.E. = −GMm/r … (iii)
Total energy:
E = K.E. + P.E.
E = GMm/(2r) − GMm/r
E = −GMm/(2r)
The negative sign is retained from the source and indicates that the satellite remains gravitationally bound to the Earth.
Geostationary Satellite
The orbit of a geostationary satellite is called a parking orbit.
Solved Numericals from the Scanned Notes
Numerical 1 – Satellite 20 km Above the Earth
Question: Calculate the period of revolution of a satellite revolving at a distance of 20 km above the surface of the Earth. The source uses R = 6400 km and g = 10 m s−2.
h = 20 km
R = 6400 km = 6.4 × 106 m
r = R + h = 6420 km = 6.42 × 106 m
T = 2πr / [R√(g/r)]
Using the numerical substitution shown in the scan:
T = 5048 s
Numerical 2 – Remote Sensing Satellite at 250 km
Question: A remote sensing satellite revolves in a circular orbit at a height of 250 km above the Earth’s surface. Find its orbital speed and period of revolution.
R = 6400 km = 6.4 × 106 m
h = 250 km
r = R + h = 6.65 × 106 m
vo = R√(g/r)
vo = 7848.18 m s−1
T = 2πr/vo
T = 5321 s
Numerical 3 – Artificial Satellite with 2.5 h Period
Question: An artificial satellite revolves around the Earth in 2.5 hours in a circular orbit. Find the height of the satellite above the Earth, taking Earth’s radius as 6370 km.
T = 2.5 h = 9000 s
R = 6370 km = 6.37 × 106 m
Using the relation developed for satellite height, the scan obtains:
h = 3,040,544.941 m ≈ 3040 km
Numerical 4 – Value of g from Motion of the Moon
Question: Obtain the value of g from the motion of the Moon, taking its period around the Earth as 27 days 8 hours and radius of orbit as 60.1 times the radius of the Earth.
T = 27 days 8 hours = 2,361,600 s
r = 60.1R
Using the satellite time-period relation and the substitutions shown in the source:
g = 9.83 m s−2
Numerical 5 – Distance of the Moon from Earth
Question: The period of the Moon revolving under the gravitational force of the Earth is 27.3 days. Find the distance of the Moon from the centre of the Earth. The source uses M = 5.97 × 1024 kg.
T = 27.3 days = 2,357,720 s
The source first evaluates surface gravity from G, M and R, then uses the satellite period relation.
r = 3.83 × 108 m
Numerical 6 – Earth Satellite Moving at 6.2 km s−1
Question: An Earth satellite moves in a circular orbit with speed 6.2 km s−1. Find its time of one revolution and centripetal acceleration.
vo = 6.2 km s−1 = 6200 m s−1
R = 6400 km = 6.4 × 106 m
The source first obtains the orbital height from vo = R√(g/(R+h)).
Using the obtained orbital radius:
T = 2πr/vo
T ≈ 3 h
α = v2/r
α ≈ 3.6 m s−2
Numerical 7 – Satellite Orbit Radius 7880 km
Question: Find the period of revolution of a satellite orbiting the Earth in a circular path whose radius is 7880 km, corresponding in the source to about 1500 km above the Earth.
R = 7.88 × 106 m
h = 1500 km
r = R + h = 9.38 × 106 m
vo = r√(g/r)
vo = 8136.259625 m s−1
T = 2πr/vo
T ≈ 7239.98 s ≈ 2 h
Source Page 17 – Incomplete / Corrupted Scan
The supplied PDF page 17 is visibly corrupted across most of the page. Only the beginning of a question is readable: it starts with a man being able to jump “1.5 m on earth” and asks for an approximate height he might be able to reach elsewhere. The remainder of the question and solution are not visible reliably in the supplied source, so they have not been reconstructed or guessed.
Numerical 8 – Point of Zero Gravitational Force between Earth and Moon
Question: Calculate the point on the line joining the centres of Earth and Moon where there is no gravitational force. The source gives Me = 6 × 1024 kg, Mm = 7.4 × 1022 kg and d = 3.8 × 108 m.
Let the point be at distance x from Earth and (d − x) from the Moon.
GMe/x2 = GMm/(d−x)2
6 × 1024/x2 = 7.4 × 1022/(3.8 × 108−x)2
x = 3.75 × 108 m
Hence the point is 3.75 × 108 m from the centre of the Earth.
Numerical 9 – Energy to Raise a 1000 kg Satellite to 600 km
Question: Taking the Earth as a uniform sphere of radius 6400 km, calculate the total energy needed to raise a satellite of mass 1000 kg to a height of 600 km above the ground and set it into circular orbit.
R = 6.4 × 106 m
h = 600 km
r = R + h = 7 × 106 m
Total energy = increase in potential energy + orbital kinetic energy.
Using the source substitution:
E = 3.46 × 1010 J
Numerical 10 – Energy to Raise a 2000 kg Satellite to 800 km
Question: Taking the Earth as a uniform sphere of radius 6400 km and g = 10 m s−2, calculate the total energy needed to raise a satellite of mass 2000 kg to a height of 800 km and put it into circular orbit.
R = 6.4 × 106 m
m = 2000 kg
h = 800 km
r = R + h = 7.2 × 106 m
E = 7.07 × 1010 J
Numerical 11 – Additional Potential Energy for a 200 kg Satellite
Question: A 200 kg satellite is lifted to an orbit whose radius is written in the source as 2.2 × 104 km. Given Earth’s radius 6.37 × 106 m and mass 5.98 × 1024 kg, find the additional potential energy required.
m = 200 kg
r = 2.2 × 107 m
R = 6.37 × 106 m
M = 5.98 × 1024 kg
Additional P.E. = −GMm/r − (−GMm/R)
Additional P.E. = GMm(1/R − 1/r)
Additional P.E. = 8.89 × 109 J
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