an earth satellite moves in a circular orbit with an orbital speed of 6200 m/s draw the fbd for the satellite. find the time for one revolution of the satellite. find the radial acceleration of the satelllite in its orbit

The time for one revolution is 7.4 ×[tex]10^{4}[/tex]s and the radial acceleration is 6.55 × [tex]10^{-4}[/tex]m/s².M = mass of the earthv= orbital speedv = [tex]\sqrt{\frac{GM}{r} }[/tex] or r = GM/ v²Time(t) = 2π [tex]r^{3/2}[/tex] / GM = 2πGM / v³t = 2π × 6.67 ×[tex]10^{-11}[/tex] × 6 ×[tex]10^{24}[/tex] / ( 6200)³  = 0.007388 × [tex]10^{-11}[/tex] × [tex]10 ^{18}[/tex]  = 0.007388 ×[tex]10^{7}[/tex]  = 7.4 × [tex]10^{4}[/tex]sNow we have to find the radial acceleration of the satellite in its orbit. Radial acceleration (r) = v² / r                                      = v² / (GM /v²)                                      = [tex]v^{4}[/tex]/ GM                                      = [tex]6200^{4}[/tex] / 6.67 ×[tex]10^{-11}[/tex] ×6×[tex]10^{24}[/tex]                                      = 6.55 × [tex]10^{-4}[/tex] m/s²Therefore the time is 7.4 ×[tex]10^{4}[/tex]s and radial acceleration is 6.55 ×[tex]10^{-4}[/tex]m/s².To know more about the earth satellite refer to the link given below:https://brainly.com/question/18496962#SPJ4