Universal law of Gravitation (Hindi) Video Lecture - Class 9

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Sir Isaac Newton formulated a very
important law on gravity which is called
the inverse law of gravitation now
according to this law
every object attracts other object with
a force which is directly proportional
to the product of their masses and
inversely proportional to the square of
the distance between them this law is
also called inverse square law so
according to the use of inverse and
square in this expression the name is
justified now let's see there are two
objects a and B of masses M 1 and M 2
individually such that the distance
between them is small ah now as it was
stated that between every objects there
exist the force of attraction and this
force of attraction is towards the
center of two objects now according to
the gravitation law the first point came
is that the force is directly
proportional to product of their masses
that is F is directly proportional to M
1 multiplied by M 2 the second point is
that force is inversely proportional to
the square of the distance between them
that is f is inversely proportional to R
square all combining the above two
expressions we will get that F is
directly proportional to M 1 into M 2
upon R square now let's remove the
proportionality sign and for that we
require a constant and the constant is
represented by capital G which is also
called a universal gravitation constant
or it is a number that gives us the
strength of gravity now let's do cross
multiplication on the left hand side of
the equation
it turns to f x r square and RHS turn to
G which is a gravitation constant
multiplied by m1 into m2 now on
rearranging the equation said that G
becomes the subject of the formula so G
gives you F R Square upon m1 into m2 now
let's see the SI unit of G which is a
gravitation constant so G is given by
the unit for force the unit is Newton
for R which is a distance the unit is
meters should become meter square upon 4
am going into m2 which are the masses of
the objects so it is represented in kg
as m1 and m2 both are equal so it
becomes kg square so we will get the
unit of G which is Newton meter square
per kg square now let's see the value of
this constant the value of G is given by
6 point 6 7 3 into 10 raised to the
power minus 11 and the unit we have
already calculated that is Newton meter
square per kg square so this is the
mathematical expression of universal law
of gravitation
after that the Newton concluded that all
objects in the universe attract each
other so now let's see the different
value of gravitational force for
different mass or chicks in the first
case let us consider the two objects a
and B with masses M 1 and M 2 the mass M
1 and M 2 are 1 kg each and the distance
between them is 1 meter which is the
value of AA now the formula for the
gravitational force is given by F is
equals to G m1 into m2 upon R square
where G has the particular value that is
6 point 6 into 10 raised to the power
minus 11 multiplied by M 1 and M 2
both have a mass 1kg so it is multiplied
by 1 upon R square R is given to us that
is 1 meter so it becomes 1 square that
is the gravitational force is equals to
the gravitational constant that is 6
point 6 7 into 10 raised to the power
minus 11 Newton so we can see from the
above expression that if the masses are
small enough then the force of gravity
between them is also very less in the
second case again there are two objects
one is the ball of mass 1 kg and the
other is the earth and we all know that
the mass of the earth is 6 into 10
raised to the power 24 kg now let's
again use the universal law of
gravitation where the gravitational
force F is equals to G m1 m2 upon R
square now G is known to us that is 6
point 6 7 into 10 this to power minus 11
Newton meter square per kg square the
radius of the earth is given to us which
is 6 point 4 into 10 raised to the power
3 kilometer all these value that is the
radius of the earth and mass of the
earth is a constant value and it is
calculated and thus the value is already
given to us now if the radius of the
earth which is in kilometer and we have
to convert it into meter so we have to
multiply 6 point 4 into 10 raised to the
power 3 by 10 raised to the power 3 so
it will become 6 point 4 into 10 raised
to the power 6 meter so now let's put
all these values in the formula all
putting the value and rearranging it and
then numerically just solve the value
you will get F is equal to 9 point 8
Newton which is very large force applied
by the earth on the ball of mass 1 kg
now let's draw the conclusion the first
is that for larger masses large
a tional force exists the second is for
smaller masses lifts gravitational force
exists now let's summarize what we have
studied till now the first is that the
universal law of gravitation is given by
F is equals to G m1 into m2 upon R
square the second is that the unit of
gravitation is Newton meter square per
kg square the third is that the
numerical value of gravitation constant
is equals to 6 point 6 into 10 raise to
power minus 11 Newton meter square per
kg square and the fourth is when there
are objects of larger masses then
gravitational force of attraction
between them is also very large and the
fifth is that the gravitational force
exist between every objects on this
earth but they cannot attract each other
due to very small force of attraction