$(a)$ 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's rotation (period $1 \, \text{day}$). What is the acceleration of an object on the surface of the Earth (at the equator) towards its centre? What is it at latitude $\theta$? How do these accelerations compare with $g = 9.8 \, m/s^2$?
$(b)$ The Earth also moves in a 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 $g = 9.8 \, m/s^2$?

(N/A) Radius of the Earth $R = 6400 \, km = 6.4 \times 10^6 \, m$.
Time period $T = 1 \, \text{day} = 86400 \, s$.
Centripetal acceleration $a_c = \omega^2 R = R \left( \frac{2\pi}{T} \right)^2 = \frac{4\pi^2 R}{T^2}$.
$a_c = \frac{4 \times (3.14)^2 \times 6.4 \times 10^6}{(86400)^2} \approx 0.034 \, m/s^2$.
At latitude $\theta$, the radius of the circular path is $R \cos \theta$, so $a_c(\theta) = \omega^2 R \cos \theta = 0.034 \cos \theta \, m/s^2$.
Comparing with $g$: $\frac{a_c}{g} = \frac{0.034}{9.8} \approx \frac{1}{288}$, which is much less than $g$.
$(b)$ Orbital radius $R' = 1.5 \times 10^{11} \, m$.
Time period $T' = 1 \, \text{year} = 365 \times 24 \times 3600 \approx 3.15 \times 10^7 \, s$.
Centripetal acceleration $a_c' = \frac{4\pi^2 R'}{T'^2} = \frac{4 \times (3.14)^2 \times 1.5 \times 10^{11}}{(3.15 \times 10^7)^2} \approx 5.97 \times 10^{-3} \, m/s^2$.
Comparing with $g$: $\frac{a_c'}{g} = \frac{5.97 \times 10^{-3}}{9.8} \approx \frac{1}{1642}$, which is much less than $g$.