An artificial satellite is revolving around the earth in a circular orbit. Its velocity is half the value of the escape velocity from the earth.
(i) What is the height of the satellite from the earth's surface ?

Explanation

Orbital speed of the satellite
`u=1/2xx"escape velocity" =1/2 xxsqrt(2gR)=sqrt((gR)/2)`
Again, if it is at a height h above the earth.s surface,
`u=Rsqrt(g/(R+h))=sqrt((gR^2)/(R+h))`
`therefore sqrt((gR)/2)=sqrt((gR^2)/(R+h)) or, (gR)/2 =(gR^2)/(R+h) or, R+h =2R`
`therefore h=R=6.4xx10^6m=6400 km`.