An artificial satellite is revolving around theearth in a circular orbit. Its velocity is half the value of the escape velocity from the earth. (ii) If its revolution around the earth is stopped and the satellite is allowed to fall freely towards the earth, what will be the velocity with which it will strikethe earth's surface ? (Radius of the earth =6.4 xx10^6 m, g=9.8 m*s^(-2))

An artificial satellite is revolving around theearth in a circular orb

1.	An artificial satellite is revolving around theearth in a circular orbit. Its velocity is half the value of the escape velocity from the earth. (ii) If its revolution around the earth is stopped and the satellite is allowed to fall freely towards the earth, what will be the velocity with which it will strikethe earth's surface ? (Radius of the earth =6.4 xx10^6 m, g=9.8 m*s^(-2))
Answer» Solution :When thesatellite stops revolving and falls FREELY TOWARDS the earth, its initial velocity of fall =0 . Hence , its initial kinetic energy =0 . Since at this stage the distance of the SATELLITE from the centre of the earth is r=R+h=2R, its initial potential energy `=(GMm)/(2R)` (M=mass of the earth ,m=mass of the satellite). Hence, the total initial energy of the satellite `=0-(GMm)/(2R)=-(GMm)/(2R)`. Let the velocity of the satellite be v, just before touching the earth.s surface . So its kinetic energy `=1/2 mv^2`. At the same TIME , its distance from hte centre of the earth is r=R and its potential energy `=-(GMm)/R` Hence , total energy of the satellite `=1/2 mv^2-(GMm)/R` From the law of conservation of energy , `-(GMm)/(2R)=1/2mv^2-(GMm)/R` or, `1/2 mv^2=(GMm)/R-(GMm)/(2R)=(GMm)/(2R)` or,` v^2=(GM)/R=(gR^2)/R=gR` `or, v=sqrt(gR)=sqrt(9.8xx6.4xx10^6)` `=7.9xx10^3ms^(-1)=7.9 kms^(-1)`.