A huge circular arc of length 4 . 4ly subtends an angle 4s at the...

$ℓ = 4 . 4ly = 4 . 4 \times 9 . 46 \times {10}^{15}$

length of Arc. $= ℓ = R\theta$

$4 . 4 \times 9 . 46 \times {10}^{15} = R\theta$

$\theta = 4S = 4 \times 4 . 843 \times {10}^{- 6} = 1 . 94 \times {10}^{- 5}$ rad

$4 . 4 \times 9 . 46 \times {10}^{15} = R \times 1 . 94 \times {10}^{- 5}$

$R = 2 . 1455 \times {10}^{21}$ meter

Speed $= 8Au = 8 \times 1 . 5 \times {10}^{11}{ms}^{- 1} = 12 \times {10}^{11}{ms}^{- 1}$

$4$ revolution means distance $= 4 \times 2\piR$ metre

time $= \dfrac{\text{distance}}{\text{speed}} = \dfrac{4 \times 2\piR}{12 \times {10}^{11}};$ time $= \dfrac{8 \times 3 . 14 \times 2 . 1455 \times {10}^{21}}{12 \times {10}^{11}} ⇒ 4 . 5 \times {10}^{10}sec$