Answer:
82780.42123 m/s
14.45 days
Step-by-step explanation:
m = Mass of the planet
M = Mass of the star =
![0.85* 1.989* 10^(30)\ kg=1.69065* 10^(30)\ kg](https://img.qammunity.org/2020/formulas/physics/high-school/vlvk9vm8a4gmby9oa8ibvonks04afe1vtq.png)
r = Radius of orbit of planet =
![0.11* 149.6* 10^(9)\ m=16.456* 10^(9)\ m](https://img.qammunity.org/2020/formulas/physics/high-school/k1sqnikpe800kt7asid2w1uqjn3xm9pfvq.png)
v = Orbital speed
The kinetic and potential energy balance of the planet and star system is given by
![(GMm)/(r^2)=(mv^2)/(r)\\\Rightarrow v=\sqrt{(Gm)/(r)}\\\Rightarrow v=\sqrt{(6.67* 10^(-11)* 1.69065* 10^(30))/(16.456* 10^(9))}\\\Rightarrow v=82780.42123\ m/s](https://img.qammunity.org/2020/formulas/physics/high-school/jo3v9ojxnxr2z0kexy5j9zgtx8d31peojo.png)
The orbital speed is 82780.42123 m/s
The orbital period is given by
![t=(2\pi r)/(v)\\\Rightarrow t=(2\pi * 16.456* 10^(9))/(82780.42123)\\\Rightarrow t=1249040.48419\ seconds](https://img.qammunity.org/2020/formulas/physics/high-school/colk8hc8jpunsv3a5vste495zp14y4f21e.png)
Converting to days
![(1249040.48419)/(24* 60* 60)=14.45\ days](https://img.qammunity.org/2020/formulas/physics/high-school/fxigc34rter246mso2mqa9p7xirkk5lltb.png)
The orbital period is 14.45 days