An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of the escape velocity from the earth. The height (h) of the satellite above the earth's surface is (Take radius of earth as R_ e )

Step-by-step solution

The escape velocity from earth is given by v_ e = 2 ~g R_ e (i) The orbital velocity of a satellite revolving around earth is given by v₀= G M_ e (R_ e +h ) where, M_ e = mass of earth, R_ e = radius of earth, h= height of satellite from surface of earth. By the relation G M_ e = g R_ e ² So, v₀= g R_ e ² (R_ e +h ) Dividing equation (i) by (ii), we get v_ e v₀ = 2 (R_ e +h ) (R_ e ) Given, v₀= v_ e 2 2 v_ e v_ e = 2 (R_ e +h ) R_ e Squaring on both side, we get 4= 2 (R_ e +h ) R_ e or R_ e +h=2 R_ e i.e., h=R_ e