Answer:
Let AB be the
imaginary tunnel dug across a diameter of the Earth and O be its centre
as shown in. Let P be the position at any instant of the body dropped
from one end of the tunnel, where \[OP=x\]. The Earth can be considered to be
made of two parts. Its outer shell-I does not exert any force on the particle.
The gravitational force on the particle is only due to the sphere-II of radius \[x\].
If \[\rho \] is the density of
the Earth, then mass of the sphere-II \[=\frac{4\pi }{3}{{x}^{3}}\rho \].
According
to Newton?s law of gravitation, the force of attraction acting on the body of
mass m and situated at P
is given
\[F=G\frac{[(4\pi
/3){{x}^{3}}\rho ]m}{{{x}^{2}}}=(4\pi /3)\,G\rho xm\]
This
force acts towards O.
Acceleration
produced in the body i.e.,
\[a=\frac{F}{m}=\frac{(4\pi
/3)G\rho xm}{m}\]
or \[a=(4\pi
/3)G\rho x\]
or \[a\propto
x\] [as \[(4\pi /3)G\rho \] is a constant]
Thus,
the acceleration is directly proportional to \[x\], the displacement of the
body from the fixed point O. More so, this acceleration is directed
towards the fixed point. Hence, the body executes SHM along the tunnel AB.
Note:
The
time period of the body is given by
\[T=2\pi
\sqrt{\frac{\text{displacement (}x\text{) }}{\text{acceleration (}a\text{)}}}\]
or \[T=2\pi
\sqrt{\frac{x}{(4\pi /3)G\rho x}}=2\pi \sqrt{\frac{3}{4\pi G\rho }}\]
or \[T=\sqrt{\frac{3\pi
}{G\rho }}\] ? (i)
As \[G=6.67\times
{{10}^{-11}}\,N{{m}^{2}}/k{{g}^{2}}\]
and \[\rho
=5.52\times {{10}^{3}}\,kg/{{m}^{3}},\]
\[T=\sqrt{\frac{3\times
3.14}{(6.67\times {{10}^{-11}}){{(5.52\times {{10}^{3}})}^{{}}}}s}\]
\[=5060s=84\,\min
\,20s\]
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