A satellite is revolving round the earth in a circular orbit with a velocity of 8km/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 circular orbit with a ve

1.	A satellite is revolving round the earth in a circular orbit with a velocity of 8km/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 the height, then `therefore a = (V^(2))/(r) = g_(H)rArr (V^(2))/(r) = 8 = (64 xx 10^(6))/(R+h) = 8` `rArr R + h = 8 xx 10^(6)` `h = 8 xx 10^(6) - 6 xx 10^(6) = 2 xx 10^(6)m` = 2000 km