Answer:
82780.42123 m/s
14.45 days
Explanation:
m = Mass of the planet
M = Mass of the star = 
r = Radius of orbit of planet = 
v = Orbital speed
The kinetic and potential energy balance is given by

The orbital speed of the star is 82780.42123 m/s
The orbital period is given by
The orbital period is 14.45 days
If you repeat the experiment, you will want to do it the same way to see if the results change.
HOPE THIS HELPS!!!
we assume the acceleration is constant. we choose the initial and final points 1.40s apart, bracketing the slowing-down process. then we have a straightforward problem about a particle under constant acceleration. the initial velocity is v xi =632mi/h=632mi/h( 1mi 1609m )( 3600s 1h )=282m/s (a) taking v xf =v xi +a x t with v xf =0 a x = t v xf −v xf = 1.40s 0−282m/s =−202m/s 2 this has a magnitude of approximately 20g (b) similarly x f −x i = 2 1 (v xi +v xf )t= 2 1 (282m/s+0)(1.40s)=198m
Vertical Free Fall and Constant Horizontal Motion