Answer
Given,
Time period of star,T = 3.37 x 10⁷ s
Radius of circular orbit,R = 1.04 x 10¹¹ m
a) Angular speed of the planet
[tex]\omega = \dfrac{2\pi}{T}=\dfrac{2\pi}{3.37\times 10^{7}}[/tex]
[tex]\omega = 1.864\times 10^{-7}\ rad/s[/tex]
b) tangential speed
[tex]v = r \omega = 1.04\times 10^{11}\times 1.864 \times 10^{-7}[/tex]
v = 1.94 x 10⁴ m/s
c) centripetal acceleration magnitude
[tex]a = \dfrac{v^2}{r}= \dfrac{(1.94\times 10^4)^2}{1.04\times 10^{11}}[/tex]
a = 3.62 x 10⁻³ m/s²
