Answer:
Earth
is revolving around the sun. Therefore, it possesses centripetal acceleration.
Therefore, it should be treated as non inertial frame of reference. However,
value of centripetal acceleration of earth,
\[a=\frac{{{\upsilon
}^{2}}}{R}={{\omega }^{2}}R={{\left( \frac{2\pi }{T} \right)}^{2}}R=\frac{4{{\pi
}^{2}}R}{{{T}^{2}}}\]
Putting R = 1\[A.U.\,=1.496\,\times
{{10}^{11}}m\], and
\[T=365\frac{1}{4}days=\frac{1461}{4}\times
24\times 60\times 60\sec =3.16\times {{10}^{7}}\sec .\], we get
\[a=4\times
\frac{22}{7}\times \frac{22}{7}\times \frac{1.496\times {{10}^{11}}}{{{\left(
3.16\times {{10}^{7}} \right)}^{2}}}=0.006m/{{s}^{2}}\],
which is negligibly
small compared to acceleration due to gravity (\[\text{g}=\text{9}\text{.8
m}/{{\text{s}}^{\text{2}}}\]). Hence for terrestrial experiments, earth can be
considered as an inertial frame.
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