Explanation:
It is given that,
Radius of circular orbit, [tex]r=3.99\times 10^{11}\ m[/tex]
Time taken, [tex]t=2.27\times 10^{7}\ m[/tex]
(a) Angular speed of the planet is given by :
[tex]\omega=\dfrac{2\pi}{t}[/tex]
[tex]\omega=\dfrac{2\pi}{2.27\times 10^{7}\ m}[/tex]
[tex]\omega=2.76\times 10^{-7}\ rad/s[/tex]
(b) The tangential speed of the planet is given by :
[tex]v=r\times \omega[/tex]
[tex]v=3.99\times 10^{11}\times 2.76\times 10^{-7}[/tex]
v = 110124 m/s
(c) The centripetal acceleration of the planet,
[tex]a=\dfrac{v^2}{r}[/tex]
[tex]a=\dfrac{(110124)^2}{3.99\times 10^{11}}[/tex]
[tex]a=0.0303\ m/s^2[/tex]
Hence, this is the required solution.
