Page 201 - Class-11-Physics-Part-1_Neat

GRAVITATION 187

This clearly shows that the force due to The gravitational force is attractive, i.e., the
earth’s gravity decreases with distance. If one force F is along – r. The force on point mass m
assumes that the gravitational force due to the 1
due to m is of course – F by Newton’s third law.
earth decreases in proportion to the inverse 2
Thus, the gravitational force F on the body 1
square of the distance from the centre of the 12
due to 2 and F on the body 2 due to 1 are related
21
earth, we will have a αR − 2 ; g α R − 2 and we get
m m E as F = – F .
21
12
Before we can apply Eq. (8.5) to objects under
g R m 2
= 2 3600 (8.4) consideration, we have to be careful since the
a R
m E law refers to point masses whereas we deal with
in agreement with a value of g 9.8 m s and extended objects which have finite size. If we have
-2
the value of a from Eq. (8.3). These observations a collection of point masses, the force on any
m
led Newton to propose the following Universal Law one of them is the vector sum of the gravitational
of Gravitation : forces exerted by the other point masses as
Every body in the universe attracts every other shown in Fig 8.4.
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.
The quotation is essentially from Newton’s
famous treatise called ‘Mathematical Principles
of Natural Philosophy’ (Principia for short).
Stated Mathematically, Newton’s gravitation
law reads : The force F on a point mass m due
2
to another point mass m has the magnitude
1
m m
| | =F G 1 2 2 (8.5)
r
Equation (8.5) can be expressed in vector form as
m m m m
F = G 1 2 2 ( ) – G=r 1 2 2 r
–
r r Fig. 8.4 Gravitational force on point mass m is the
1
vector sum of the gravitational forces exerted
m m by m , m and m .
2
4
3
= –G 1 2 r
3
r
The total force on m is
where G is the universal gravitational constant, 1
Gm m 1 Gm m 1 Gm m 1
3
2
4
r is the unit vector from m to m and r = r – r 1 F 1 = r 2 r 21 + r 2 r 31 + r 2 r 41
2
2
1
as shown in Fig. 8.3. 21 31 41
t Example 8.2 Three equal masses of m kg
each are fixed at the vertices of an
equilateral triangle ABC.
(a) What is the force acting on a mass 2m
placed at the centroid G of the triangle?
(b) What is the force if the mass at the
O vertex A is doubled ?
Take AG = BG = CG = 1 m (see Fig. 8.5)
Answer (a) The angle between GC and the
positive x-axis is 30° and so is the angle between
Fig. 8.3 Gravitational force on m due to m is along GB and the negative x-axis. The individual forces
1 2
r where the vector r is (r – r ). in vector notation are
2 1

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