Consider the path of the earth around the sun as circular one of radius 1 Au. Find the linear speed of the earth around the sun. $(1 A u = 1.5 \times 10^{8} k m) .$

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Solution

$v = ω R = \frac{2 π}{365 d a y s} \times 1.5 \times 10^{8}$

Where $ω = \frac{θ}{T}$
and for one revolution, $T = 365 d a y s$

$⟹ v = \frac{2 π \times 1.5 \times 10^{8}}{365 \times 24 \times 3600}$

$⟹ v = 29.89 m / s = 107.59 k m / h$

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