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GRAVITATION
KEPLER’S LAWS OF PLANETARY MOTION
1. Law of orbits: All planets move in elliptical orbits with the sun situated at one of its foci
of the ellipse.
2. Law of areas: The line that joins any planet to the sun sweeps equal areas in equal
intervals of time. Or areal velocity (areal velocity = area swept by the radius vector in
unit time) of planets around the sun is a constant,
𝑑𝐴
𝑑𝑡
= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡.
(figure 8.2)
Note: Planets appear to move slower when they are farther from the sun and faster
when they are closer than sun.
3. Law of periods: The square of time period(T) of revolution of a planet is directly
proportional to the cube of the semi major axis(r) of the ellipse traced out by the planet.
𝑇 2
𝑟 3
2
2
𝑇 2 ∝ 𝑟 3 Note : 𝑇12 = 𝑟13
Q1. Page 201, question 8.3, NCRT TEXT
Q2. Page 202, question 8.14, NCERT TEXT
UNIVERSAL LAW OF GRAVITATION
Every body in the universe attracts every other body with a force which is directly
proportional to the product of their masses and inversely proportional to the square of the
distance between them
|𝐅| = G
𝑚1 𝑚2
𝑟2
𝑚 𝑚
In vector form, ⃗⃗⃗⃗⃗⃗⃗
F12 = G 𝑟1 2 2 (−𝑟̂), where −𝑟̂ indicate the direction of force which is
always attractive
Properties of gravitational force
1. Gravitational force between two masses is always attractive.
2. Gravitational force is a central force( a force which act along the line joining the centers
of two bodies)
3. Gravitational force is a conservative force
4. Gravitational force between two masses is independent of the nature of the medium
between the masses.
5. It obeys Newton’s third law of motion.
6. It obeys inverse square law.
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Intensity of gravitational field (gravitational field strength) at a point is defined as the
gravitational force per unit mass. Intensity of gravitational field of a mass ‘M’ at a distance
𝐹
‘r’ from it = 𝑚 =
𝐺𝑀
𝑅2
𝐺𝑀𝑚
𝑚 𝑟2
=
𝐺𝑀
𝑟2
. At the surface of earth(r=R) intensity of gravitational field =
= 𝑔, 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑑𝑢𝑒 𝑡𝑜 𝑔𝑟𝑎𝑣𝑖𝑡𝑦
Note:
1.The force of attraction between a hollow spherical shell of uniform density and a point
mass situated outside is same as the entire mass of the shell is concentrated at the center
of the shell.
2. The force of attraction due to a hollow spherical shell of uniform density on a point mass
situated inside is zero.
3. Weight is the gravitational force with which earth attract an object towards its center
W = mg. Even for same mass weight changes with change in value of g. Mass is the amount
of matter in a body. It is always constant for a body.
Expression for acceleration due to gravity.
ME is the mass of earth. m is the mass of an object placed at a distance r from center of the
earth.RE is the radius of earth.
According to Newton’s law of gravitation force on m due to ME , F = G
According to second law of motion F =ma or F = mg
From (1) and (2) mg = G
𝑀𝐸 𝑚
𝑟2
=G
or g
𝑀𝐸
𝑟2
Near the surface of the earth, r ≈RE ,
we have mass = volume x density ME =
𝟏
𝟒𝝅
.
𝟑
𝑟2
.
(1)
(2)
RE + h = r,
g=𝐆
𝟒𝝅
𝑀𝐸 𝑚
𝑴𝑬
𝑹𝟐𝑬
𝑹𝟑𝑬 𝝆,
𝟒
g = 𝐆 𝑹𝟐 ( 𝟑 𝑹𝟑𝑬 𝝆) = 𝟑 𝝅𝑮𝑹𝑬 𝝆
𝑬
Note 1. Acceleration due to gravity is independent of mass of falling object.
Q1. Two shot put balls A and B of masses 5kg and 10kg are releases from same heights
simultaneously. Which will reach the ground first? Why?
Binomial theorem (1 + 𝑥)𝑛 = 1 + 𝑛𝑥 +
𝑛(𝑛−1) 2
𝑥
(1)(2)
+ … ..
+ 𝑥𝑛 .
For small values of x, (1 + 𝑥)𝑛 = 1 + 𝑛𝑥
Variation g with altitude ( height from the surface of the ground) (add diagram)
At the surface of the earth, mgs = G
𝑀𝐸 𝑚
2
𝑅𝐸
or
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gs =
𝐺 𝑀𝐸
2
𝑅𝐸
Similarly at a height ‘h’ above the ground
𝐠𝐡
𝐠𝐬
𝑔ℎ
𝑔𝑠
=
=
𝐑𝟐𝐄
(𝐑 𝐄 + 𝐡)𝟐
(1 +
𝐺 𝑀𝐸
(𝑅𝐸 + ℎ)2
(This formula is applicable to all heights)
2
𝑅𝐸
2
𝑅𝐸
gh =
𝑔ℎ
ℎ 2
)
𝑅𝐸
𝑔𝑠
=
1
(1 +
𝑔ℎ
ℎ
𝑅𝐸
)2
𝑔𝑠
= (1 +
ℎ
𝑅𝐸
)−2
Applying binomial theorem and neglecting higher power of 𝑥,
𝑔ℎ
𝑔𝑠
. gh = gs (1 −
2ℎ
𝑅𝐸
2ℎ
= (1 −
𝑅𝐸
2ℎ
) or mgh = mgs (1 −
𝑅𝐸
) , for small heights (h<<RE)
2ℎ
) or Wh = Ws (1 −
𝑅𝐸
) . This is applicable to
small heights only.
Variation g with depth (distance from the surface of the earth to down ) (add diagram)
At the surface of the earth, mgs = G
𝑀𝐸 𝑚
or
2
𝑅𝐸
𝐺𝜌𝐸
gs =
2
𝑅𝐸
Similarly at a depth‘d’ below the ground
gd
gs
=
4𝜋𝐺𝜌𝐸 (𝑅𝐸 −𝑑)
3
gd = gs (1 −
𝑑
𝑅𝐸
x
3
𝑔
𝑅𝐸 4𝜋𝐺𝜌𝐸
, 𝑔𝑑 =
𝑠
) or mgd = mgs (1 −
gs =
4
x 3 𝜋𝑅𝐸3 =
4𝜋𝐺𝜌𝐸
gd =
3
(𝑅𝐸 −𝑑)
𝑔𝑑
𝑅𝐸
𝑔𝑠
𝑑
𝑅𝐸
𝐺 𝑀𝐸
2
𝑅𝐸
4𝜋𝐺𝜌𝐸
3
=
𝐺 𝑉𝐸 𝜌𝐸
2
𝑅𝐸
4
𝑉𝐸 = 3 𝜋𝑅𝐸3
𝑅𝐸
(𝑅𝐸 − 𝑑)
= (1 −
) or Wd = Ws (1 −
𝑑
𝑅𝐸
𝑑
𝑅𝐸
)
)
Q: What is the value of g at the center of the earth?
Q: A person has mass of 100 kg on the surface of the earth. Calculate the mass and weight of
person (i) at the center of the earth (ii) at a height equal to half the radius of the earth (iii) at
depth equal to half the radius of earth.
Acceleration due to gravity is not a constant. Its value decreases with height (as we go up from
the surface of earth) and depth (as we go deep in to earth from the surface of earth.)
Acceleration due to gravity is maximum at the surface of earth (at the poles)
Q: Draw a graph showing variation of acceleration due to gravity with distance from the surface
of earth.
g
RE
distance from center of earth
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Gravitational potential energy of a particle of mass ‘m’ at a distance ‘r’ from the center of
another mass ‘M’ is defined as the work done in bringing ‘m’ from infinity to ‘r’.
Consider a mass ‘m’ at a distance ‘x’ from center of another mass ‘M’
Gravitational force on ‘m’ due to ‘M’, F =
𝐺𝑀𝑚
𝑥2
work done to move ‘m’ through a small distance ‘dx’ , dw = F dx =
𝑟 𝐺𝑀𝑚
Total work done in moving ‘m’ from infinity to ‘r’, w = ∫∞
𝑥2
𝐺𝑀𝑚
𝑥2
dx
dx
𝑟 1
W = 𝐺𝑀𝑚 ∫∞ 𝑥 2 dx
𝑟
W= 𝐺𝑀𝑚 ∫∞ 𝑥 −2 dx
dx
W= 𝐺𝑀𝑚[−𝑥 −1 ] ∞𝑟
1
r
1
W = −𝐺𝑀𝑚[𝑟 − ∞]
m
M
1
W = −𝐺𝑀𝑚[𝑟 − 0]
W=
−𝑮𝑴𝒎
𝒓
Q: Derive expression for work done to move an object from surface of earth to a height ‘h’
above it. What will be the expression if h <<R
Gravitational Potential energy of a system of particles is the work done to assemble all the
articles of the system from infinity to their respective locations against the gravitational force.
Potential energy of two particle system in the absence of external field.
U12 = W12 =
−𝐺 𝑚1 𝑚2
𝑟12
Potential energy of three particle system in the absence of external field.
U123 = W12 + W13 + W23 = −
𝐺 𝑚1 𝑚2
𝑟12
+ −
𝐺 𝑚1 𝑚 3
𝑟13
+
−𝐺 𝑚2 𝑚3
𝑟23
Note: work done to dissociate a system= -( work done to assemble the system)
Q: Write expressions for potential energy of four particle system and five particle system
Gravitational potential of a mass ‘M’ at a distance r from its center is defined as the total work
done to bring unit mass (m =1kg) from infinity to that point. It is the gravitational potential
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energy per unit mass.
Gravitational potential =
𝑔𝑟𝑎𝑣𝑖𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦
𝑚𝑎𝑠𝑠
Gravitational potential at a distance r from a mass M,
V(r) =
−𝑮𝑴
𝒓
Q: Draw a graph to show variation of gravitational potential with distance from center of a mass
Q: Four identical masses ‘m’ each are placed at the four corners of a square of side ‘a’ each.
Calculate the (i) gravitational potential energy of the system (ii) gravitational potential at the
center of the system (iii) work done to dissociate the system. (iv) total gravitational force on
any one mass of the system (v) total gravitational force on unit mass placed at the center of the
system.
Escape speed (Ve) from a planet is defined as the minimum speed with which an object must be
launched from the planet such that it just escape from the gravitational field of the planet.
Let M be the mass of the planet of radius R. let m be the mass of the object to be launched so
that it just escape the gravitational field. Let Ve be the escape speed.
1
Total energy of the mass at the surface of earth = K.E. + P.E. = 2 𝑚 𝑣𝑒2 +
−𝑮𝑴𝒎
𝑹
As the object move up speed and kinetic energy decreases and just becomes zero at infinity.
Potential energy increases and becomes zero at infinity. Total energy at infinity becomes zero
according to law of conservation of energy,
Total energy at the surface of earth = total energy at infinity
1
2
𝑚 𝑣𝑒2 −
1
2
𝑮𝑴𝒎
𝑚 𝑣𝑒2 =
𝑣𝑒2 =
𝑮𝑴𝒎
𝑹
𝟐𝑮𝑴
𝑹
𝑣𝑒 = √
At the surface of earth 𝑔 =
𝐺𝑀
𝑅2
=0
𝑹
𝟐𝑮𝑴
𝑹
or gR2 = GM
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𝑣𝑒 = √
𝟐𝒈𝑹𝟐
𝑹
𝑣𝑒 = √2gR
From the surface of earth Ve = √2x9.8x 6.4x106 = 11.2 Km/s.
The escape speed from a planet is independent of the mass of the object projected. Escape
speed is same for an ant and elephant from same planet. The force needed to impart this speed
for ant and elephant is different.
Q: Calculate the escape speed from the surface of moon. Radius of the moon is 1600km and
acceleration due to gravity on the moon = 1/6 of that on earth. Hence explain why moon has
very rarer atmosphere (even no atmosphere)
The escape speed from the surface of moon is 2.3Km/s, which is about five times smaller
compared to that from the surface of earth. Thermal speed of gas molecules on the surface of
moon is greater than the escape speed from the surface of moon and the gas molecules escape
out of the gravitational pull of the moon.
Note : if initial velocity (Vi) is greater than escape speed (Ve) K.E. at infinity ≠ 0
1
1
𝑚 𝑣𝑓2 = 2 𝑚 𝑣𝑖2 +
2
If initial velocity is less than escape speed, =
−𝑮𝑴𝒎
𝑹
−𝑮𝑴𝒎
(𝑹+𝒉)
1
= 2 𝑚 𝑣𝑖2 +
−𝑮𝑴𝒎
𝑹
1. Escape speed from the surface of earth is 11.2km/s. A rocket is projected with thrice this
speed from the surface of earth. Calculate its speed far away from earth, neglecting the
presence of sun and other planets.
2. A rocket is projected with a speed of 5km/s from the surface of the earth. how far from
the surface of the earth does the rocket go before returning to earth. (M earth = , Rearth =
and G =
)
Satellites are objects which revolve around a planet.
Orbital speed of a satellite (Vo) is the speed with which a satellite in a particular orbit around
the planet. Let m be the mass of satellite revolving around a planet of mass M at a distance r
from the center of the planet. Vo = orbital speed, R + h = r , R = radius of the planet and h =
height of the satellite from the surface of the earth.
Assuming the orbit to be circular, centripetal force for circular motion is provided by the
gravitational force between the planet and the satellite.
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𝑚
𝑉𝑜2
𝑟
=
𝑉𝑜2 =
𝐺𝑀𝑚
𝑟2
𝐺𝑀
𝑟
𝑮𝑴
Vo = √
𝒓
.
The orbital speed of satellite is independent of the mass of the satellite.
If the orbit is very close to surface of earth, r≈ 𝑅 (h<<R)
𝐺𝑀
Vo = √
𝑅
Near to the surface of earth have 𝑔 =
𝑔𝑅 2
Vo = √
𝑅
𝐺𝑀
𝑅2
or gR2 = GM
Vo = √𝑔𝑅 Vo = √9.8 x6.4 x 106 = 8.3 Km/s
1. Calculate orbital speed of a satellite revolving at a height equal to half the radius of the
earth.
Relation between orbital speed and escape speed near the surface of the earth (h<<R)
Orbital speed ,Vo = √𝑔𝑅 and escape speed 𝑣𝑒 = √2gR
𝑣𝑒 = √2 Vo
Time period of satellite (T) is the time taken by the satellite to complete one revolution around
the planet.
Let m be the mass of satellite revolving around a planet of mass M at a distance r from the
center of the planet. Vo = orbital speed, R + h = r , R = radius of the planet and h = height of the
satellite from the surface of the earth.
Assuming the orbit to be circular, time period, T =
𝑇=
𝑇=
2𝜋𝑟
𝑉𝑜
2𝜋𝑟
𝑮𝑴
𝒓
√
𝒓
T = 2𝜋𝑟 √𝑮𝑴
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𝑐𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑜𝑓 𝑜𝑟𝑏𝑖𝑡
𝑜𝑟𝑏𝑖𝑡𝑎𝑙 𝑠𝑝𝑒𝑒𝑑
𝒓𝟑
T = 2𝜋 √𝑮𝑴 .
The time period of satellite is independent of the mass of the satellite.
𝑹𝟑
If the orbit is very close to the orbit r≈ 𝑅 (h<<R), T = 2𝜋√𝑮𝑴
𝐺𝑀
Near to the surface of earth have 𝑔 =
𝑅2
or gR2 = GM
𝑹𝟑
T = 2𝜋√𝒈𝑹𝟐
𝟔.𝟒 x 106
𝑹
T = 2𝜋 √𝒈 T = 2𝜋 √
𝟗.𝟖
= 1.5 hours
1. Calculate time period of a satellite revolving at a height equal to half the radius of the
earth.
Total energy of a satellite
Total energy of a satellite is the sum of potential and kinetic energy of the satellite
Let m be the mass of satellite revolving around a planet of mass M at a distance r from the
center of the planet. Vo = orbital speed, R + h = r , R = radius of the planet and h = height of the
satellite from the surface of the earth.
1
𝑮𝑴
K.E. = 2 𝑚𝑉𝑜2 using the value of orbital sped, Vo = √
K.E. =
P.E. =
𝑮𝑴𝒎
𝟐𝒓
−𝑮𝑴𝒎
𝒓
Total energy of the satellite, E= K.E. + P.E.
E=
E=
E=
𝐺𝑀𝑚
2𝑟
+
−𝐺𝑀𝑚
𝑟
𝐺𝑀𝑚 1
𝑟
𝐺𝑀𝑚
𝑟
( − 1)
2
1
(− )
2
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𝒓
E=
−𝑮𝑴𝒎
𝟐𝒓
Total energy of a satellite is negative. It means that satellite is bound to the nucleus. To free the
satellite from the gravitational field of planet, energy equal to total energy must be supplied to
the satellite in the positive sense.
1. (a)A satellite of 400kg mass is revolving at a height equal to 1600km from the surface of
earth. Calculate the (i) total energy, (ii) kinetic energy and (iii) potential energy of the
satellite. (b) How much energy is needed to double the radius of the orbit? (c) how
much energy is needed to rocket the satellite out of gravitational field of earth ( neglect
the presence of sun and other planets)
Note: 1. Energy needed to change the orbit of a satellite,
∆E = Efinal − Einitial =
−GMm −GMm
− 2r
2rf
i
𝒐𝒓
∆𝑬 =
𝑮𝑴𝒎 𝟏
𝟏
(
−
)
𝟐
𝒓𝒊
𝒓𝒇
2. Energy needed to rocket a satellite out of gravitational field of earth (planet)
Gravitational field is zero when 𝑟𝑓 = ∞
1
𝑟𝑓
= 0 , ∆𝑬 =
𝑮𝑴𝒎 𝟏
𝟐
𝒓𝒊
=
−(𝐭𝐨𝐭𝐚𝐥 𝐞𝐧𝐞𝐫𝐠𝐲 𝐨𝐟 𝐬𝐚𝐭𝐞𝐥𝐥𝐢𝐭𝐞)
Relation between kinetic energy, potential energy and total energy
K.E. =
𝐺𝑀𝑚
2𝑟
P.E. =
−𝐺𝑀𝑚
𝑟
and E =
−𝐺𝑀𝑚
2𝑟
1. Kinetic energy = −(total energy)
2. Potential energy = 2 x total energy and
3. Potential energy = −2(kinetic energy)
Satellites are classified in to natural and artificial satellites. Natural satellites are natural bodies
revolving around a planet. Moon is the natural satellite of earth. Artificial satellites are
manmade satellites revolving around the planet. Aryabhatta, INSAT series of satellites are
examples for artificial satellites.
Uses of artificial satellites
1. Television broad cast
2. Communication purposes
3. Weather fore casting
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4. Remote sensing
5. Meteorology and
6. Environmental studies
Polar satellites and geostationary satellites
Polar satellites go around the poles of earth in north south direction. These satellites orbit the
earth at low altitudes (nearly 500 to 800Km from the earth surface.) .Time period of revolution
these satellites are nearly 100 minutes. The whole earth can be scanned strip by strip in one
day. Polar satellites are used for remote sensing, meteorology, and environmental studies
Satellites which revolve in circular orbits around the earth in the equatorial plane along the
direction of rotation of earth (from west to east) with time period 24hours are called as
geostationary satellites. The height of a geostationary satellite from the surface of earth is
nearly 35800Km.
Q: Calculate the height of a geostationary satellite from the surface of earth
Weightlessness: Weight of an object is the force with which earth attracts it towards the center
of earth. Direction of gravity gives the sense of vertical downward direction for us . We
experience (feel) our own weight when we stand on a surface, since the surface exerts a force
opposite to our weight to keep us at rest. When an object is in free fall there is no force against
gravity to keep us at rest, and we will not experience our weight. This state is known as
weightlessness. In a satellite around the earth everything has acceleration towards the center
of the earth which is exactly equal to acceleration due to gravity at that height. This situation is
same as free fall. Thus in an artificial satellite people inside experience no gravity and hence
there is no sense of vertical or horizontal direction for them.
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