Ncert Exercises & Solutions

4.25. a) Earth can be thought of as a sphere of radius 6400 km. Any object is performing circular motion around the axis of earth due to earth’s rotation. What is acceleration of object on the surface of the earth towards its centre? What is it at latitude θ? How does these accelerations compare with g = 9.8 m/s²?

b) Earth also moves in circular orbit around sun once every year with an orbital radius of 1.5 × 10¹¹m. What is the acceleration of earth towards the centre of the sun? How does this acceleration compare with g = 9.8 m/s²?

(a)

Radius of the earth (R) = 6400 km 64 x $10^{6}$ m
Time period (T) = 1 day =24 x 60 x 60 s = 86400s

$\begin{array}{l} Centripetal acceleration (a_{c}) = ω^{2} R = R (\frac{2 π}{T})^{2} = \frac{4 π^{2} R}{T} \\ = \frac{4 \times (22 / 7)^{2} \times 6 .4 \times 10^{6}}{(24 \times 60 \times 60)^{2}} \\ = \frac{4 \times 484 \times 64 \times 10^{6}}{49 \times (24 \times 3600)^{2}} \\ = 0 .034 m / s^{2} \end{array}$

$\begin{array}{lr} At equator, & latitude θ = 0^{\circ} \\ ∴ \frac{a_{c}}{g} & = \frac{0 .034}{9 .8} = \frac{1}{288} \end{array}$

(b)

Orbital radius of the earth around the sun (R) = 15 x $10^{11}$ m
Time period = 1 yr = 365 day
= 365 x 24x 60 x 60s = 315 x $10^{7}$ s

$\begin{array}{l} Centripetal acceleration (q_{c}) = {Rω}^{2} = \frac{4 π^{2} R}{T^{2}} \\ = \frac{4 \times (22 / 7)^{2} \times 1 .5 \times 10^{11}}{(3 .15 \times 10^{7})^{2}} \\ = 5 .97 \times 10^{- 3} m / s^{2} \\ ∴ \frac{a_{c}}{g} = \frac{5 .97 \times 10^{- 3}}{9 .8} = \frac{1}{1642} \end{array}$

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