A satellite is revolving round the earth in a circularorbit with a velocity of 8 km/s . At a height where acceleration due to gravity is 8m//s^2 . How high is the satellite from the earth ? (Take R=6000 km)

A satellite is revolving round the earth in a circularorbit with a vel

1.	A satellite is revolving round the earth in a circularorbit with a velocity of 8 km/s . At a height where acceleration due to gravity is 8m//s^2 . How high is the satellite from the earth ? (Take R=6000 km)
Answer» Solution :As centripetal ACCELERATION equals to acceleration DUE to gravity at that HEIGHT , then `therefore a=V^2/r=g_h rArr V^2/r = 8 = (64xx10^6)/(R+h)=8` `rArr R+h=8xx10^6` `h=8xx10^6 - 6xx10^6 =2xx10^6` m = 2000 km