In his law of
gravitation, newton could not yet to determine the value of universal
gravitation constant (G). but, the determination of it, first done by Henry Cavendish
(1731 – 1810), a british physicist and chemist in 1798, after newton died.
The device used by
Henry Cavendish to determine the value of universal gravitation constant was a
torsion balance called Cavendish’s Torsion. This torsion consists of a light (
light source), a 6 – feet rigid rod which at both ends attached to a small
metal sphere with mass m and about 2 inches in diameter. The rigid rod was
suspended at its meddle using a quartz wire ( smooth wire). Two large lead
spheres which are identical with mass M and diameter approximately 8 inches are
placed near the small spheres on the rigid rod. Study the following figure :
Among the two spheres
act gravitational force (attraction) causing the rigid rod to twist and quartz
wire to rotate. The magnitude of the angle of rod torsion is detected it from
the displacement of light on the scale. After
the system os calibrated so that the magnitude of the force needed to produce
one torsion found,the attractive force between two spheres can be calculated
direcly from observational data of the angle of wire torsion.
From the experiment,
the values $m_{1}$, $m_{2}$, r, and F are found,then the value of G can be
calculated. From this calculation, Cavendish obtained the value of G = 6,754 x
$10^{-11}$ $nm^{2}/kg^{2}$. Today, the accepted value of G is 6,67259 x
$10^{-11}$ $nm^{2}/kg^{2}$. This very small value of G will have effect on object
with large mass.
Example#
:
Determine the
gravitational force between the earth ( m = 5,98 x $10^{24}$ kg) and a 60 kg
student who stands on the earth surface 6,38 x $10^{6}$ m from the earth
center.
Answer :
Because
G = 6,67 x $10^{-11}$ n$m^{2}$/$kg^{2}$
$m_{1}$ = 5,98 x $10^{24}$ kg
$m_{2}$ = 60 kg
r = 6,38 x
$10^{6}$ m
then, the gravitational
force acting between the earth and the student is :
$F_{g}=G.\frac{m_{1}.m_{2}}{r^{2}}$
$F_{g}=(6,67\times10^{-11})\frac{(5,98\times10^{24}).(60)}{(6,38\times10^{6})^{2}}$
$F_{g}$
= 587,9 N
Thus, the gravitational
force is 587,9 Newton.
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