It is always amazing to see the interaction of planets, Sun, Stars, and other celestial objects in space which leads to astronomical events. In this chapter we will learn certain laws of physics which explains gravitation between celestial objects, free fall of body, mass and weight of the objects.

1. GRAVITATION
Gravitation, Universal Law of Gravitation, Kepler Laws of Planetary
Motion, Free Fall, Acceleration due to Gravity, Mass and Weight.
Topics to be covered
Efforts By - Yuvraj Ghaly
TGT Science Faculty
Bhavan Vidyalaya,
Panchkula

2. ● Indian Astrologer and Mathematician, Aryabhatta studied motions of planets. He
stated Earth revolves about its own axis and moves in a circular orbit around the sun.
● Ptolemy ( 100-170 BC) explains geocentric nature of universe. Earth is the centre of
universe and sun, planets and other celestial objects revolves around it.
● Copernicus (1473-1543) disproves the geocentric nature but come up with heliocentric
nature of universe in which Sun is the centre and Earth with other planets revolves
around it.
● Galileo verified the copernicus idea about universe later after his persecution.
● Tycho Brahe (1546-1601) gave accurate predictions of stars and planets location.
● Johannes Kepler gave 3 laws of planetary motion called Kepler’s law of planetary
motion.
● Newton gave law of gravitation based on 3rd Kepler law, popularly called Universal
Law of Gravitation.
GRAVITATION

3. ● Newton stated that there is force that causes the
acceleration and keeps the body along the circular
path, is acting towards the centre. This force is
called CENTRIPETAL FORCE.
GRAVITATION
●The motion of celestial bodies such as moon, earth,
planets etc. and attraction of moon towards earth and
earth towards sun is an interesting subject of study
since long time.
●All the bodies in the universe, there is a force exists between them. This force of
attraction between objects is called GRAVITATIONAL FORCE.

4. ● The gravitational force is a central force that is It acts along the line joining the centers of
two bodies.
●It is a long range force. The gravitational force is effective even at large distances.
●It is a conservative force. This means that the work done by the gravitational force in
displacing a body from one point to another is only dependent on the initial and final
positions of the body and is independent of the path followed.
●Unlike electrostatic and magnetic forces, the gravitational force is always attractive.
Characteristics of Gravitational Force

5. “Every objects in the universe attracts every other object with a force
(a) proportional to the product of their masses.
F 𝞪 M1M2
(a) inversely proportional to the square of the distance between them.
F 𝞪 1/d2
F 𝞪 M1M2
--------
d2
F = G M1M2
--------
d2
Where G is the constant called Universal Gravitational Constant
G = 6.7 x 10-11 Nm2/kg2 (measured experimentally by Henry Cavendish)
UNIVERSAL LAW OF GRAVITATION

6. ● Determine the force of gravitational attraction
between the earth (m = 5.98 x 1024 kg) and a 70-
kg physics student if the student is standing at
sea level, a distance of 6.38 x 106 m from earth's
center.
m1 = 5.98 x 1024 kg
m2 = 70 kg
d = 6.38 x 106 m
F = G M1M2
--------
d2
F = 6.7 x 10-11 x 5.98 x 1024 x 70
----------------------------
(6.38 x 106 )2
F= 686 N
NUMERICAL PRACTICE
● Determine the force of gravitational attraction
between the earth (m = 5.98 x 1024 kg) and a
70-kg physics student if the student is in an
airplane at 40000 feet above earth's surface.
This would place the student a distance of
6.39 x 106 m from earth's center.
m1 = 5.98 x 1024 kg
m2 = 70 kg
d = 6.39 x 106 m
F = G M1M2
--------
d2
F = 6.7 x 10-11 x 5.98 x 1024 x 70
----------------------------
(6.39 x 106 )2
F = 684 N

7. IMPORTANCE OF UNIVERSAL LAW OF GRAVITATION

8. Kepler's three laws emerged from the analysis of data carefully collected over a span of
several years by his Danish predecessor and teacher, Tycho Brahe. Kepler's three laws of
planetary motion can be briefly described as follows:
● The paths of the planets about the sun are elliptical in shape, with the center of the
sun being located at one focus. (The Law of Ellipses)
● An imaginary line drawn from the center of the sun to the center of the planet will
sweep out equal areas in equal intervals of time. (The Law of Equal Areas)
● The ratio of the squares of the periods of any two planets is equal to the ratio of the
cubes of their average distances from the sun. (The Law of Harmonies)
Kepler’s Law of Planetary Motion

9. CAVENDISH EXPERIMENT FOR ‘G’ VALUE
APPARATUS SET UP:
● A light, rigid rod about 2-feet long.
● Two small lead spheres were attached to the ends of the rod
● The rod was suspended by a thin wire.
PRINCIPLE: When the rod becomes twisted, the tension of the wire begins to exert
a force that is proportional to the angle of rotation of the rod. The more twist of the
wire, the more the system pushes backwards to restore itself towards the original
position.
PROCEDURE:
● Cavendish then brought two large lead spheres near the smaller spheres
attached to the rod. Since all masses attract, the large spheres exerted a
gravitational force upon the smaller spheres and twisted the rod a measurable
amount. Once the torsional force balanced the gravitational force, the rod and
spheres came to rest and Cavendish was able to determine the gravitational
force of attraction between the masses.
● By measuring m1, m2, d and Fgrav, the value of G could be determined.
Cavendish's measurements resulted in an experimentally determined value of
6.75 x 10-11 N m2/kg2. Today, the currently accepted value is 6.67259 x 10-11 N
m2/kg2.
https://www.physicsclassroom.com/Physics-Interactives/Circular-and-Satellite-Motion/Gravitational-Fields

10. ● A free falling object is an object that is
falling under the sole influence of gravity.
Any object that is being acted upon only by
the force of gravity is said to be in a state of
free fall.
● Aristotle once said ,” “The heavier a body
the faster it falls”. It was disproved by
Galileo.
Free Fall

11. Feather Coin Experiment
Robert Boyle experimented by creating vacuum in a long glass tube and then allowed the feather
and the coins to fall simultaneously. He showed that both of them reached the bottom at the same
time.
Conclusion- acceleration produced in freely falling bodies is the same for all bodies irrespective of
their masses.

12. Acceleration due to Gravity
● The force of gravitation acting on the bodies called force of gravity due to earth.
● Whenever an object falls towards the earth, an acceleration is involved. This acceleration is
called acceleration due to gravity (g).
Consider a body on the surface of Earth of mass, m. Let M be the mass
of earth and R be the distance between the body and the centre of the
earth. Let F be the force acting on the body.
According to the law of gravitation,
F = G M m
---------- (1)
R2
According to the second law of motion,
F = m x a (2)
m x a = G M m
----------
R2
a = G M
----
-----
R2
g = G M
----
------ as a = g
acceleration due to gravity is a vector quantity. Its
value is 9.8 m/s2 or 10 m/s2

13. Calculation of Acceleration Due to Gravity
g = G M
--------
--
R2
g = 6.7 x 10-11
x 6 x 1024
------------
-----------
(6.4 x 106)2
g = 6.7 x 6 x
10 -11+24
------------
-----------
6.4 x
6.4 x 1012
g = 40.2 x 10
/40.96
g = 0.981 x 101
g = 9.81 m/s2
acceleration due to gravity is 9.83 m/s2 at the poles and 9.78 m/s2 at the equator.

14. Variation of ‘g’ due to height
g’ =
- 2h) x g
----
R
Variation of ‘g’ due to depth
g’ =
(1- d) x g
----
R

15. Escape Velocity and Orbital Velocity
● Maximum vertical velocity required to take satellite just outside the earth’s gravitational field
is called escape velocity.
● Horizontal velocity imparted to the satellite so that it orbits around the earth in a stable
circular orbit with constant velocity is called orbital velocity.
Escape Velocity of a certain height = √2 orbital velocity

16. Practice Time
Ques: The orbital velocity of certain height is 28800 km per hour. How much is the escape velocity
off the same the height?
[40729 km/h]
Escape Velocity of a certain height = √2 orbital velocity
V = √2 x 28800 = 40723 km/h
Ques: What is the radius of the moon if its gravity is 1/6th of the earth. The mass of the moon is 7.35
x 1022 kg.
[1.73 x 105 m]
g = GM/R2
9.8 X ⅙ = 6.6 X 10-11 X 7.35 X 1022/ R2
1.63 = 4.245 X 1011 /R2
R2 = 30.156 X 1011
R = √30.156 X 1011
R = 1.73 X 1011 m

17. Equations of Motion for freely falling objects

18. Practice Time
Ques: An object is thrown vertically upwards and rises to a height of 18 m. Calculate the velocity
with which the object was thrown upwards? Take g=9.8 m/s2?
[14 m/s]
(0)2 = u2 + 2 x (-9.8) x 18
-u2 = - 352.8
u = 18.78 m/s
Ques: What is the time period of the stone to reach the maximum height of 250 m from the ground?
Take g = 10 m/s2
h = ut + ½ gt2
250 = 0 + ½ x 10 x t2
250/5 = t2
√50 = t
7.07 s = t
V2-U2 = 2gh
{o}2 - u2 = 2gh
u2 = 2 x 10 x 250
u2 = 5000
u = √5000
u = 70.7 m/s
v = u + gt
0 = 70.7 + 10 x t
70.7/10 = t
7.07 = t

19. Mass and Weight
S.NO MASS WEIGHT
1 Matter contained in a body Force by which Earth pulls objects towards its
centre.
2 Scalar Quantity Vector Quantity
3 Constant and cannot be zero Variable and can be zero
4 S.I. unit is kilogram (kg) S.I. unit is Newton (N)
Practice Time
Q: A man weight is 1200N on the Earth. What is is mass? (take g=10m/s2). If he were to taken to moon, his weight
would be 200N. What is his mass? what is the acceleration due to gravity on the moon?
F = 1200 N
WEIGHT = FORCE =mg
1200 = m x
10
1200/10 = m
120 kg = m
F = 200 N
WEIGHT = FORCE =mg
200 = 120 x
g
200/120 = g
1.67 m/s2 =

20. Practice Time
Q: A person of mass 70 kg weighs himself on a weighing machine first on Earth and then on a planet where
acceleration due to gravity is ½ that of the earth. Calculate the reading of the weight on the weighing machine.
Take g = 9.8 m/s2?
Weight of person on Earth
WEIGHT = FORCE =mg
W = 70 x 9.8
W = 686 N
Weight of person on the planet
WEIGHT = ½ Weight on Earth
= ½ x 686
= 343 N
Q: How far should a man go away from the centre of the Earth such that his weight is ¼th the earth? Radius of
earth is 6400 km.
weight on surface = force of gravity of earth = g = GM/R2
Weight = GM/R2
Weight on certain height = W/4
W/4 = GM/R’2
W/4 = GM/R’2
------ -------------
W GM/R2
W/4 = 1/R’2
------
-------------
W
1/R2
1 = R2
------
-----
4
R’2
R’2 = 4 x (6400)2
R’ = √4 x 6400
x6400
R’ = 2 x 6400
R’ = 12800 km