$(a)$ Earth can be thought of as a sphere of radius $6400\, km$. Any object (or a person) is performing circular motion around the axis of the earth due to the earth rotation (period $1$ day). What is acceleration of object on the surface of the earth (at equator) towards its centre ? What is it at latitude $(\theta )$ ? How does these accelerations compare with $g=9.8\,m/s^2$ ?

$(b)$ Earth also moves in circular orbit around the sun once every year with an orbital radius of $1.5 \times 10^{11} \,m$. What is the acceleration of the earth (or any object on the surface of the earth) towards the centre of the sun ? How does this acceleration compare with $g=9.8\,m/s^2$ ?

(a) Radius of the earth $=\mathrm{R}=6400 \mathrm{~km}=6.4 \times 10^{6} \mathrm{~m}$

Time period $=\mathrm{T}=1 \mathrm{day}=24 \times 60 \times 60 \mathrm{~s}$

$=86400 \mathrm{~s}$

Centripetal acceleration,

$a_{c} =\omega^{2} \mathrm{R}=\mathrm{R}\left(\frac{2 \pi}{\mathrm{T}}\right)^{2} \quad\left(\because \omega=\frac{2 \pi}{\mathrm{T}}\right)$

$=\frac{4 \pi^{2} R _{e}}{ T ^{2}}$

$=\frac{4 \times\left(\frac{22}{7}\right)^{2} \times 6.4 \times 10^{6}}{(24 \times 60 \times 60)^{2}}$

$=\frac{4 \times 484 \times 64 \times 10^{6}}{49 \times(24 \times 3600)^{2}}=0.034 \mathrm{~m} / \mathrm{s}^{2}$

At equator, latitude $\theta=0^{\circ}$

$\therefore \frac{a_{c}}{g}=\frac{0.034}{9.8}=\frac{1}{288}$

$a_{c} \ll g$

$(b)$ Orbital radius of the earth around the sun $\mathrm{R}^{\prime}=1.5 \times 10^{11} \mathrm{~m}$ Time period

$\mathrm{T}^{\prime} =1 \text { year }=365 \text { day }$

$=365 \times 24 \times 60 \times 60 \mathrm{~s}=3.15 \times 10^{7} \mathrm{~s}$

Centripetal acceleration

$a_{c}=\mathrm{R}^{\prime} \omega^{2}=\mathrm{R}^{\prime}\left(\frac{2 \pi}{\mathrm{T}^{\prime}}\right)^{2}$

$=\frac{4 \pi^{2} \mathrm{R}}{\mathrm{T}^{\prime 2}}$

$= \frac{4 \times\left(\frac{22}{7}\right)^{2} \times 1.5 \times 10^{11}}{\left(3.15 \times 10^{7}\right)^{2}}=5.97 \times 10^{-3} \mathrm{~m} / \mathrm{s}^{2}$

$\therefore \frac{a_{c}^{\prime}}{g}=\frac{5.97 \times 10^{-3}}{9.8}=\frac{1}{1642} a_{c} \ll g$