Dhanamanjuri university
Department of mathematics
A Seminar report on Gravitational Attraction
Presented by Group 9
Mechanics
Under the guidance of Dr. L. Shambu
Singh
Associate professor
Department of Mathematics
Dhanamanjuri University
Brief Summary of Classical Mechanics
Classical mechanics is the mathematical science that studies the displacement of
bodies under the action of forces. Newton solved the greatest scientific problem of his
time by applying his Universal Law of Gravitation to determine the motion of planets.
Alongside this development and refinement of the concept of force and its application to
the description of motion, the concept of energy slowly emerged. Limitation on the validity
of Newtonian mechanics appeared at the microscopic length scale. It became
increasingly clear that classical mechanics did not adequately explain a wide range of
newly discovered phenomena at the atomic and sub-atomic length scales. By the early
part of the twentieth century, quantum mechanics provided a mathematical description
of microscopic phenomena in complete agreement with our empirical knowledge.
Abstract
This presentation portrays a brief summary of classical mechanics
and its limitations. The concept of Universal law of Gravitation from
Newtonian mechanics is discussed, its relevance to the attraction of a thin
uniform rod from an external point and hence the conclusion derived from
its special case.
Introduction
The law of gravitation/ the law of attraction between any two
material bodies was discovered by Newton. It states that that every particle of
matter attracts/repels with a force which varies directly as product of the two
masses and inversely varies as the square of the distance in between them.
Thus, for two material of masses m1 and m2 having a distance 'r' in between
them , the attraction is the force given by
ɣ
Where ɣ is the universal constant which is the force in between two unit
masses placed at a unit distance apart.
Attraction of a thin uniform rod about an
external point
Let AB be the thin uniform rod of cross section “k” and density
“ρ”. Consider a small length MM’=
at a distance “x” from the
foot N of the point P, about which attraction is required. Join
PA, PM, PM’ & PB
Let PN= p, ∠APN=α, ∠M’PN=θ, ∠BPN=β.
Now, attraction of
(
about P is
)
--------------------------(1)
The component of the attraction given by (1) along PN
taking as x-axis is
(
)
And, along PB’ taken as y-axis is
(
)
where PM’=p sec
NM= , NM’= +
MM’=
, PN= p
Let (X,Y) be the components of the total attraction of AB on P
Then, X=∫
-----------------------------------(2)
.
Y=∫
-----------------------------------(3)
Equation (2) becomes
X=∫
 X=
x= NM= p tan
dx=p sec2 d
∫
=
∴X=
[sin ]
[sin -sin ] -----------------------------(4)
Equation (3) becomes Y= ∫
 Y=
∫
 Y=
[cos −cos ] --------------------------(5)
F be the resultant attraction then,
F2=X2+Y2
=(
)2 [(sin
=(
)2
=2(
-sin )2+(cosα-cosβ)2]
[sin2β+sin2α-2sinαsinβ+cos2α+cos2β-2cosαcosβ] = (
)2 [1-cos(β-α)]
 F=
2(
)[1-cos(β-α)] -----------------------(6)
 F=
2(
) 2sin(
)
[ ∵cos2x = 1 - 2sin2x]
)2
[2-2(cosαcosβ+sinαsinβ)]
sin(
 F=
If
) 2sin(
2(
 F=
[ ∵cos2x = 1 - 2sin2x]
)
be the angle made by F with PN
Then, tan
=
=

)
=
2 = + ,
 − = −
(
=
)
(
)
(
)
i.e tan
=tan (
)
--------------------------------------(7)
>
--------------------------------------(8)
Equation (8) shows that the direction of the resultant attraction of the rod AB acts along
the bisector of ∠APB, which is suspended by the rod AB at P.
Corollary
When the rod AB is of infinite length both the ends A and B extends infinity on either
side of AB through α
∴X =
 F = X∞
,
→-
=0
,β→
------------------------------------(9)
--------------------------------------------(10)
Equation (10) shows that the resultant attraction of the rod at P varies inversely on the
perpendicular distance of P from AB.
References
1
H Goldstein, Classical Mechanics,
Wesley Publication Massachusetts.
2
https://en.wikipedia.org/wiki/History_of_
classical_mechanics
Thanks for sparing your valuable time with
us, we appreciate it!!!
Thank you