A satellite is revolving round in a circular orbit with a speed of 8 km/`s^(-1)` at a height where the value of accleration due to gravity is 8 m/`s^(-2)`. How high is the satellite from the Earh's surface?

Explanation

Orbital velocity of satellite, `(V_(@))=` 8km/s.=`8xx10^3` m/s.
Acceleration due to gravity in the orbit =g=8 m/`s^2`.
Orbital velocity, `V=sqrt(gR)`
where R is radius of the orbit and g is acceletation due to gravity in the orbit.
`:. R=V^2`/g=`(8xx8xx10^6)/(8)=8xx10^6m =8000` km.
Height of satellite =800 - radius of earth :
Radius of earth =6000 km
Height above earth =8000-6000= 2000 km